1

Chapter 1

Electronic Drives

Course Outline

The following areas will be covered during our course of study:

• Power Electronic Devices:- Characteristics, drive requirements and device protection.

• Converters: - DC-DC; DC-AC; AC-DC and Control Techniques

• Power and distortion factor; Harmonics and interference

• Applications of AC &DC motors:- motor control: variable speed drives, regenerative breaking, slip energy recovery, four

quadrant operation

1.1 Power Semiconductors Converters

The technique for controlling the electric machines has changed in significant ways. For example series DC motor are used to propel

subway cars. The speed of these cars has been controlled by use of resistances in series with the dc motors as shown in figure below. In

recent years there has been use of Choppers, which can convert a fixed voltage into a variable dc voltage hence controlling the speed of

the car.

Using the appropriate converters can also control other electric machines. The following being the various types of converters that are

frequently used to control electric machines:

AC Voltage Controller (AC to AC): - an ac voltage controller converts a fixed voltage ac to a variable voltage ac. It can be used to

control the speed of an induction motor (voltage control method) and for smooth induction motor starting.

Controlled Rectifier (AC to DC): - A controlled rectifier converts a fixed voltage ac to a variable voltage dc. It is used primarily to

control the speed of dc motors such as those used in rolling mills.

Chopper (DC to DC): - A chopper converts a fixed voltage dc to a variable DC. It is used primarily to control the speed of DC motors.

Inverter (DC to DC): - An inverter converts a fixed voltage dc to a fixed (or variable) voltage ac with variable frequency. It can be used

to control the speed of ac motors.

Cycloconverter (AC to AC): - A cycloconverter converts a fixed voltage and fixed frequency ac to a variable voltage and variable

(lower) frequency ac. It can be used to control the speed of ac motors.

High –power semiconductor devices are used in these converters to function as on-off switches. The characteristics of these devices will

be looked at first.

1.2 Power Semiconductor Devices

The power semiconductor devices generally used in converters can be grouped as follows:

• Thyristors (SCR)

• Power transistors

• Diode rectifiers

These devices are operated in the switching mode so that losses are reduced and conversion efficiency is improved.

1.2.1Thyristor (SCR)

2

The Thyristor has a four layer p-n-p-n structure with three terminals, anode (A), cathode (K), and gate (G) as shown in figure 2

below. The anode and cathode are connected to the main power circuit. The gate terminal carries a low level gate current in the

direction from the gate to cathode. The Thyristor operates in two stable states: ON or OFF.

Volt Ampere Characteristics

With zero gate current, if a forward voltage is applied across the device (anode positive with respect to cathode) junctions 1 and 3 are

forward biased while j2 is reverse biased and therefore the anode current is a small leakage current. If the anode to cathode forward

voltage reaches the critical limit called the forward breakover voltage (VBO) the device switches into high conduction. The device is

said to be latched IL . Latching current is defined as the minimum anode current that a Thyristor must attain during turn on process to

maintain conduction when the gate is removed or current required to maintain regeneration. If gate currents are introduced the VBO

is reduced. For a sufficiently high gate current the entire forward blocking region is removed and the device behaves as a diode.

When the device is conducting the gate current can be removed and the device remains in the on state.

If the anode current falls below the critical limit called the holding current Ih, the device returns to its forward blocking state. See the

I/V characteristic

Switching Characteristics

If a Thyristor is forward biased and a gate pulse is applied, the Thyristor switches on. However, once the Thyristor starts conducting

an appreciable forward current (regeneration taking place), the gate has no control on the device. The Thyristor will turn off if the

anode current becomes zero called natural commutation or is forced to become zero called forced commutation.

It is necessary to keep the device reverse biased for a finite period of time before a forward anode voltage can be applied otherwise if

applied at a less time it would conduct. This period is known as the turnoff time, toff of the Thyristor. The turnoff time of the

Thyristor is defined, as the minimum time interval between the instant the anode current becomes zero and the instant the device is

capable of blocking the forward voltage.

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See fig 3 below for the switching characteristics of a Thyristor

The Thyristor is switched on

by a gate pulse ig and when the gate pulse is applied at instant t1, anode current IA builds up and the voltage across the device VAK

falls. When the device is fully turned on the voltage across it is quite small (typically 1 to 2.5V, the higher voltage drop for higher

current devices) and for all practical purposes the device behaves as a short circuit. The device switches on very quickly, the turn-on

time ton being typically 1 to 3microseconds. Typically the gate pulse is in the range of 10 to 50microsec and its amplitude in the

range 20 to 200mA. If the current through the thyristor is to be switched off at a desired instant t2, it is momentarily reverse biased

by making the cathode positive with respect to the anode. For this forced commutation a commutation circuit is required as shown

later. In most commutation circuits a precharged capacitor is momentarily connected across the conducting thyristor to reverse bias

it. If the device is reverse biased, its current falls, becomes zero at t3,,then reverses and becomes zero at t4. at instant t5, the device is

capable of blocking a forward voltage. The time interval from t3 to t5 is known as the turnoff time of the thyristor. If a forward

voltage appears at instant t6 the time interval t3 to t6 is known as the circuit turn-off time, tq. In practical applications, the turnoff

time tq provided to the SCR by the circuit, must be greater than the device turn-off time toff by a suitable safety margin; otherwise

the device will turn on at an undesired instant, a process known as commutation failure.

Thyristors with large turnoff time 50-100microsec are called slow switching or phase control type thyristors and those having small

turnoff time (10-50microsec) are called fast switching or inverter type thyristors.

Note that during thyristor turn on, if the voltage is high, current is low and vice versa. Therefore the turn on switching loss is low.

During thyristor turnoff also, if the reverse current is small, the turnoff switching loss is low. The low switching loss in a thyristor is

a significant advantage, particularly for high frequency applications

1.2.2 TURN ON METHODS

With anode positive with respect to cathode a Thyristor can be turned on any one of the following techniques

a) Forward voltage triggering

b) Gate triggering

c) dv/dt triggering

d) Temperature triggering

e) Light triggering

Points a) and b) have already been dealt with

c) dv/dt triggering: - if the rate of rise of anode –cathode voltage is high the charging current of the capacitive junctions j1,j2, and j3

may be sufficient to turn on the thyristors. However a high value of charging current may damage the Thyristor and the device must

be protected against high dv/dt.

d) Temperature triggering: - During forward blocking most of the applied voltage appears across the reverse biased junction J2. This

voltage associated with leakage currents may raise the temperature of this junction. With increase in temperature, leakage current

through J2 further increases. This cumulative process may turn on the scr at some high temperature

e)Light triggering: - A niche is made on the p-layer as shown below. When the niche is irradiated by ,free charge carriers (holes and

electrons ) are generated just like when the gate signal is applied between the gate and the cathode

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1.2.3 Gate Characteristic

As an approximation, gate to cathode junction can be taken as a PN diode and its V/I characteristic will approach that of a diode see

figure below

This information can be used to sketch the gate characteristic and triggering region of a Thyristor.

Gate characteristic of Thyristor

For a particular type of SCR Vg and Ig characteristic has a spread between two curves, Upper and Lower limit. Ox and Oy refer to

minimum gate current and voltage to trigger the Thyristor. Vgm and Igm are maximum permissible gate voltage and gate current. Oa

non-triggering gate voltage. The preferred gate drive area for an SCR is bcdefghb.

All spurious or noise signals should be less than the voltage oa . The design of a firing circuit can be carried out with the help of

figures below.

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Considering fig a Es =Vg + IgRs. A resistance R1 is connected across gate –cathode terminals so as to provide an easy path to the

flow of leakage current between SCR terminals.

If Igmn and Vgmn are the minimum gate current and gate voltage to turn on SCR then from fig b , current through R1 is Vgmn/R1.

Es =(Igmn + Vgmn/R1)Rs+Vgmn =IRs +Vgmn

Using gate circuit of higher magnitude can reduce the SCR turn on time. Gate drive requirements can be obtained from the graph

below:

AD is the load line

Curve 3 is for a thyristor whose Vg/Ig give the operating point S and Vg=PS and Ig =OP. To minimize the turn on time and

unreliable turn on, the operating points may vary between S1 and S2 and must be close to the Pgav curve as possible. The gradient of

the load line AD =OA/OD will give the required gate source resistance Rs.

The minimum value of gate source series resistance is obtained by drawing a line AC tangent to Pgav.

With pulse triggering, greater amount of gate power dissipation can be allowed. This should however be less than peak

instantaneous gate power dissipation Pgm.

Pulse gating

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fT

PPPTfPP

T

TP gav

gmgavgmgav

gm ≥≥≥1

. where f=1/T1 =frequency of firing and T= pulse width. In the limiting case

Pgav/fT =Pgm .

A duty cycle is the ratio of ON time to periodic time fTT

T==

1

δ .

Example

Given the gate cathode characteristic of an SCR to be having a straight-line slope 130. If the trigger source voltage is 15V and the

allowable gate power dissipation is 0.5watts. Compute the gate source resistance.

Solution

Vg Ig =0.5W …………..(1) and Vg / Ig =130Vg =130Ig…………..(2) Substituting 2 into 1 gives:

VmAxIVmAII gggg 06.862130130625.0130 2 ===∴==Now for the gate circuit Es =IgRs +Vg. 15 =62mA*Rs + 8.06 Rs =112ΩΩΩΩ

Example 2

The trigger circuit of a Thyristor has a source voltage of 15V and the load line has a slope of –120V/A. The minimum gate

current to turn on the SCR is 25mA.

Determine :

a) Source resistance required in the gate circuit,

b) The trigger voltage and trigger current for an average gate power dissipation of 0-.4W

Solution

a) The slope of the load line gives Rs =120ΩΩΩΩb) VgIg =0.4W and for the gate circuit Es =RsIg +Vg 15=120Ig +0.4/Ig Ig =38.6mA or 86.4mA Vg =10.37V or

4.63V. Choose Vg =4.627V and Is =86.4mA.

1.2.4 Parallel and Series Operation of Thyristors

To accommodate high load currents a parallel connection of thyristors can be used. If the simple connection is used fig (a) then

differences in the individual thyristors will result in an unequal current sharing of current between them. This sharing can be evened

out by careful selection of matched devices , by the use of series resistance as in fig (b) or by including current sharing reactors

fig(c).

Fig b is shown below:

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Example

Two thyristors are connected as in fig a above and have forward characteristics when conducting of the form:

Thyristor 1 v= 0.96 +2.52 *10-4

i1

Thyristor 2 v= 0.92 +2.40 *10-4

i2

i) What will be the current distribution between the two thyristors when the total current is 1400A.

ii) What value of resistance added in series with each thyristo will result in the thyristors being within 5% of each other for

the same total current?

Solution

i) V= 0.96 +2.52 *10-4

i1=0.92 +2.40 *10-4

i2 and i1 + i2 = 1400

Hence 0.96 +2.52 *10-4

i1=0.92 +2.40 *10-4

(1400-i1)

Giving i1 =601.6A and i2 =798.4A

iii) Maximum current =1400A hence Maximum current difference =5%*1400=70A . We noted from above i2 is greater

than i1 hence

i2-i1 =70A and i1+12 =1400A

2i2 =1470 i2=735A and i1=665A

0.96 +2.52 *10-4

i1+Ri1 =0.92 +2.40 *10-4

i2 +Ri2

R=0.445mΩΩΩΩ

Transient or Dynamic Current Sharing

If current through T1 rises Ldi/dt across L1 increases and a corresponding voltage of opposite polarity is induced across inductor L2.

The result is a low impedance path through T2 and the current is shifted to T2. The reverse is true

At turn on the thyristors gate circuits must all be driven from the same source to force a simultaneous turn on of all devices. To

prevent any individual thyristors from turning off if its current falls below its holding current level , a continuous gate signal is

normally used to ensure immediate re-firing.

Where high voltage levels are encountered thyristors are connected in series to share the voltage. If the connection shown below is

used then the differences between devices can result in an unequal voltage sharing between them.

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R1 C and R2 C are for dynamic situation. R is used to try and equate the voltages sitting on the thyristors and the determination of R

is based on the following diagram and assumptions.

Assumptions:-

1) there are n thyristors

2) The value of R is a maximum to limit the current

3) Consider the case that T1 has a negligible leakage current and others have maximum leakage current Ileak. This is the

worst case and it indicates that the voltage V2 sits across all thyristors except T1 .

4) The source voltage Vs is the maximum that can be applied when R is used.

5) The value of V1 across T1 is the peak voltage rating of the thyristors .That is the maximum voltage T1 will support i.e

T1 will support a larger voltage than the remaining (n-1) thyristors which share voltage equally.

From Kirchhoff’s voltage law Vs =V1+V2(n-1)…………1 and from Kirchhoff’s current law Is1 =Is2

+Ileak………………….2

Also from Ohm’s law Is1 =V1/R…….3 and Is2 =V2/R ………..4.

Eliminating V2, Is1 and Is2 we have from 1 and 4 Vs =V1 +V2(n-1) but V2 =Is2 R Vs =V1 +Is2R (n-1).

From 2 Is2=Is1-Ileak Vs =V1 + (n-1)(Is1-Ileak)R

R =(nV1-Vs)/(n-1)Ileak

1.2.5 Thyristor Types

Depending on the physical construction and turn on and turn off behaviour thyristors can be classified broadly into different

categories:

1.2.5.1 A TRIAC (Bi-directional Triode Thyristor)

A TRIAC can be considered as an integration of two SCRs in inverse parallel. It can conduct in both directions and is normally used

in ac- phase control.

TRIAC

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When terminal A1 is positive

with respect to terminal A2 and the device is fired by a positive gate current +ig, it turns on. Also when terminal A2 is positive with

respect to terminal A1 and the device is fired by a negative gate current –ig, the device turns on.

A TRIAC is frequently used in many low power applications such as juice mixers, blenders, vacuum cleaners, etc. There is also what

is called a DIAC similar to a TRIAC but does not have a gate input.

1.2.5.2 A GTO (Gate Turn Off) Thyristor

A single pulse of positive gate can turn ON this thyristor, but in addition a pulse of negative gate can turn it off current. A gate

current therefore controls both the On state and Off State operation of the device. See fig below for the symbol and switching

characteristics:

GTO Thyristor

The turn on process is the same as that of a Thyristor .The turn off process is somehow different. When a negative voltage is applied

across the gate and cathode terminals, the gate and cathode terminals , the gate current ig rises. When the gate current reaches

its maximum value Igr, the anode current begins to fall and the voltage across the device Vak, begins to rise. The fall time of Ia

is abrupt typically less than 1µsec. Thereafter the anode current changes slowly, and this portion of the anode current is known as

the tail current

The ratio( Ia/Igr) of the anode current Ia (prior to turn off ) to the maximum negative gate current Igr required to turn off a Thyristor

is low typically between 3 and 5. For example , a 2500V, 100A GTO typically requires a peak negative current of 250A to turn it off.

GTOs that can operate up to 4500V, 2000 A, 5-10µsec are available. These are becoming increasingly popular in power control

equipment and GTOs will replace thyristors where forced commutation is necessary as in choppers and inverters.

1.2.6 Power Transistors

A transistor is a three layer p-n-p or n-p-n semiconductor device having two junctions. This type of a transistor is known as a bipolar

junction transistor (BJT). The structure and the symbol of an n-p-n transistor are shown below.

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Transistor

The collector and the emitter terminals are connected to the main power circuit, and the base terminal is connected to control signal.

If the base current is zero, the transistor is in the off state and behaves as an open switch. On the other hand if the base is driven hard

, that is if the base current Ib is sufficient to drive the transistor into saturation, then the transistor behaves as a closed switch. The

transistor is current driven device. The base current determines whether it is in the off or On state. To keep the device in the on state

there must be sufficient base current. Transistors with high voltage and current ratings are known as the power transistors. The

current gain Ic/Ib of power transistor can be as low as 10, although it is higher than that of a GTO. Higher current gain can be

obtained by use of a Darlington Pair.

Power transistors switch on and off much faster than thyristors. They may switch on in less than 1µsec and turn off in less than

2µsec.. Hence the power transistors can be used in situations where the frequency is as high as 100kHz. These devices however fail

under certain high voltage and high current conditions. They should be operated within specified limits known as the safe operating

areas (SOA). The SOA is partitioned into four regions defined by the following limits:

• Peak current limit (ab)

• Power dissipation limit (bc)

• Secondary breakdown limit (cd)

• Peak voltage limit (de)

Safe Operating Area (SOA)

If high voltage and high current occur simultaneously during turn off, a hot spot is formed and the device fails by thermal runaway, a

phenomenon known as second breakdown.

Figure below shows the effects of the snubber circuit on the turnoff characteristics of a power transistor. A chopper circuit with an

inductive load is considered.

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If no snubber circuit is used and the base current is removed to turn off the transistor, the voltage across the device Vce first rises, and

when it reaches the dc supply voltage Vd, the collector current falls. The power dissipation P during the turnoff interval is shown in the

figure above by the dashed line. Note that the peaks of Vce and Ic occur simultaneously, and this may lead to secondary breakdown

failure.

If the snubber circuit is used and base current is removed to turn off the transistor , the collector current is diverted to the capacitor.

The collector current therefore, decreases as the collector emitter voltage increases, avoiding the simultaneous occurrence of peak

voltage and peak current.

Power transistors of ratings as high as 1000V,500A are available

Linear approximations of Switching intervals for a purely Resistive load

The collector emitter voltage during switching on falls from Vs to almost zero and the expression for the voltage is given by:

s

on

once V

t

ttV *

−= where t is the instantaneous time

the current rises and the expression is given by

on

mc

t

tIi

*=

Similarly during turn off the collector current and the collector emitter voltage is given by:

off

offm

c

on

sce

t

ttIi

t

tVV

)(&

* −==

With the collector emitter voltage and the collector current expressions above the power losses during turn on and during turn

off can be determined.

Peak loss occur at dp/dt =0. Power loss during switching on is given by:

ttt

tIV

t

IV

dt

dp

t

t

t

tIVP on

on

ms

on

mson

onon

mson 22

)(22

2

=−=−=

That is ton is the time at which instantaneous power would be maximum. The switch on and off loss for a resistive load is given by:

12

=−==−=6

)1(*;6

*)1(0

offms

offoff

msoff

tonms

on

m

on

son

tIVdt

t

t

t

tIVW

tIVdt

t

tI

t

tVW

on

The average power loss due to switching is obtained by multiplying energy loss by the switching frequency i.e power =energy/time

=energy *frequency.

Example

A power transistor has its switching waveforms as shown below. If the average power loss in the transistor is limited to 300W,

find the switching frequency at which the transistor can be operated?

Energy loss during turn on and turn off is given by the following illustration:

sec1603.075

10**200*)

60

10*1(100

sec1067.0)40

10*1(*200*

50

10**100*

66

66

−=−=

−=−==

wattdttt

Woff

wattdttt

vdticWton

o

ton

oon

Therefore energy loss in one cycle =0.1603+0.1067 =0.267watt-sec =E

But the power loss in the transistor is limited to 300W

Hence frequency =f =300/0.267 =123.6Hz.

DIODE

A diode is a two-layer p-n semiconductor device. The structure of a diode and its symbol are in figure 7 below

From V/I characteristic it can be noted that if a reverse voltage is applied across the diode, it behaves essentially as an open circuit. If a

forward voltage is applied, it starts conducting and behaves essentially as a closed switch. Following the end of a forward conduction in a

diode, a reverse current flows for a short time . The device does not attain its full blocking capability until the reverse current ceases. The

time interval during which reverse current flows is called reverse recovery time. This reverse recovery characteristic is of great

significance in high frequency applications.

1.2.7 Protection

The reliable operation of a converter would require ensuring that at all times the circuit conditions do not exceed the ratings of the power

devices, by providing protection against overvoltage; overcurrent and overheating. In practice the power devices are protected from:

• Thermal runaway by heat sinks

• High dv/dt and di/dt by snubbers

• Reverse recovery transients

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• Supply and load side transients

• Faulty conditions by fuses

1.2.7.1 Cooling and heat sinks

Due to on state and switching losses, heat is generated within the power device. This heat must be transferred from the device to a cooling

medium to maintain the operating junction temperature within a specified range.

Although the heat transfer can be accompanied by conduction, convection, radiation or natural or forced air, convection cooling is

commonly used in industrial applications.

Cooling methods

6) Air coolers:- Air surrounding the device is heated through radiation and once heated natural convection takes place.

7) Forced Cooling:- This includes forced convection. A fan is used to push air past a heated surface. This causes the

thermal resistance to decrease. The thermal resistance decreases with air velocity. However above a certain velocity the

reduction in thermal resistance is not significant

8) Heat Sinks: - A wide variety of extruded aluminium heat sinks are commercially available and use cooling fins to

increase the heat transfer capability. The contact is between the device and heat sink is extremely important to minimise

the thermal resistance between the case and sink. The surfaces should therefore be smooth and free of dust/dirt,

corrosion and surface oxides. Silicon grease is normally applied to improve the heat transfer capability and to minimise

the formation of oxides and corrosion.

9) Heat pies:-These are pipes filled with low vapour pressure liquid. The liquid once heated it is changed to vapour and

goes to the fins and the vaporised liquid lose its heat through the fins and quickly changes back to liquid and returns to

the heat source. This is heated again and the cycle continues.

10) Liquids:- in high power applications the devices are more effectively cooled by liquids , normally water or oil. Water-

cooling is very efficient and approximately three times more effective than oil cooling. However it is necessary to use

distilled water to minimize corrosion and antifreeze to avoid freezing. Oil is flammable but eliminates problems of

freezing and corrosion.

Thermal resistance (Rθθθθ).

Heat flow by conduction is very much alike the conduction of electrical charge i.e current R

VVI

12 −= . Similarly power is the rate of

flow of heat energy and is proportional to the difference in temperature across the region through which the heat is conducted i.e:

θθ R

TT

R

TP 12 −

=∆

=

Thermal Model

Tj is the junction temperature

Rθjc:- thermal resistance from junction to case in oc/w.

Rθcs:- thermal resistance from case to sink in oc/w.

Rθsa:- thermal resistance from sink to ambient in oc/w.

Ts:- is the heat sink temperature

Ta:- is the ambient temperature.

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From the expression

θθ R

TT

R

TP 12 −

=∆

= it can be noted that power is constant hence:

ja

as

cs

sc

jc

cj

R

TT

R

TT

R

TT

R

TT

R

TP

θθθθθ

−=

−=

−=

−=

∆= 12

But sacsjcjja

ja

aj

ath RRRRR

TTIVP θθθθ

θ

++=== −&* it is from this equation and data sheets for heat sinks and thyristors that the

type and size of heat sink is derived.

Example

The collector to base junction of a certain transistor dissipates 2W. The thermal resistance from the junction to the case is 8 oc/w and the

thermal resistance from the case to air is 20 oc/w. The free air temperature is 25

oc. Determine:

a) the junction temperature

b) the case temperature

Solution

a) cT

T

R

TT

R

Tj

j

ja

aj081

208

252 =

+

−=

−=

∆=

θθ

b) Similarly P=2 Tc =65 oc.

2) di/dt protection

If the current in a thyristor rises at too high a rate, that is high di/dt, the device can be destroyed. Some inductance must be present or

inserted in series with the thyristor so that di/dt is below a safe limit specified by the manufacturer. Typically di/dt limit values for SCRs

are 20 500A/microsec

3) dv/dt Protection

If the entire anode to cathode forward voltage Va appears across J2, the charge on the junction is Q, then the charging current is given by:

dt

dVaCj

dt

dCjVa

dt

dVaCjdtCjVad

dt

dqi =+=== /)( Since Cj is a constant. If dVa/dt is high, the charging current will be more.

This charging current will play the role of gate current and turns on the SCR even when the gate signal is zero. As a result dv/dt must be

kept low. Typical values are 20-500V/microsec.

Power devices are protected from excessive di/dt and dv/dt by adding an inductor and snubber circuits see fig below:

A snubber circuit consists of a series combination of resistance R and capacitor C in parallel with thyristor. Before the thyristor is fired, C

charges to full supply voltage Vs. Once on the capacitor discharges through the thyristor hence the need for an R to limit the discharge of

C.

Example

Figure below shows a thyristor controlling the power to the load of resistance RL. The supply voltage is 240V dc and the specified limits

for di/dt and dv/dt for the SCR are 50A/µsec and 300V/µsec respectively. Determine the values of the inductance and the snubber circuit

parameters R and C.

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Solution:

When the switch is closed the capacitor behaves like a short circuit and the SCR is in forward blocking state and offers a high resistance.

RLR

L

RLR

VsIwhereeIi

dt

LdiiRLRVs

t

+=

+=−−−=++=

−ττ &1).......1()(

Now di/dt from 1 gives:

ττ

τ

tt

eL

VsIe

dt

di−−

==1

* The value of di/dt is a maximum when t=0 that is at point of switching on.

H

dtdi

VsL

L

Vs

dt

diµ8.450/10*240 6 ==== −

.

The voltage across the SCR is given by

Ω===−== 6*///* RL

VsRdtdvLVsdtdibutdt

diR

dtdvaiRVa

The voltage across the capacitor is given by :

610*)300(6240

)(*)1( ≥≥=−=

−

C

dtdVcR

VsCe

CR

Vs

dt

dVceVsVc CR

t

CRt

1.2.7.2 Overvoltage Protection

Switching transients usually appear across the converter and these cause overvoltages. This effect of overvoltages is usually minimised

by use of snubber circuits, non linear resistors called voltage clamping devices (VARISTORS); Metal oxide Varistors –MOVs as well as

selenium diodes.

The VARISTORS, snubber circuits, and selenium diodes are connected in parallel with the converter.

The voltage clamping devices have falling resistance characteristic with increasing voltage . When a voltage surge appears, the voltage-

clamping device operates in the low resistance region and produces a virtual short circuit across the power device limiting the amount of

the transient voltage. After the surge energy is dissipated in the non linear resistor , the operation of the voltage clamping device returns

to its high resistance region. Under normal operating conditions these devices draw very small current.

1.2.7.3 Overcurrent Protection

a) The power converters may develop short circuits or faults and the resultant fault currents must leared quickly. Fast acting fuses

are normally used to protect the semiconductor devices. As the faulty current increases the fuse opens and clears the fault current

in milliseconds

The operation of the fast acting fuse is illustrated below

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The current limiting fuse consists of one or more fine silver ribbons having very short fusing times. If we assume that the a fault occurs at

t=0, without the fuse the fault current would rise up to A and up to or beyond D.

A properly selected fuse melts at A. For a brief moment or interval after A, the current continues to rise depending upon the circuit

parameters and fuse design. This current reaches the peak let through current B, after B the arc resistance increases and fault current

decreases. At point C arcing stops and the fault current is cleared. The total arcing time is tc =tm + ta.

The current time curves of the devices and fuses may be used for the coordination of a fuse for a device see figures below:

The fault is normally cleared in 10msec. The fuse must be rated to carry full load current plus marginal overload current for an indefinite

period, but the peak let through current of fuse must be less than the surge current of the SCR.

If R is the resistance of the fault circuit and I is the instantaneous fault current between the instant of fault occurring and the instant of arc

extinction, the energy fed to the circuit is = dtRiWe 2.

If R remains constant , the value of i2t is proportional to the energy fed to the circuit. The i

2t value is termed the let through energy and is

responsible for melting the fuse. The fuse manufacturers specify the i2t of characteristic of a fuse. In selecting the fuse it is necessary to

estimate the fault current and then to satisfy the following requirements:

These fuses must take account of :

a) the need to permit the continuous passage of the steady- state load current.

b) Permitted overload conditions including transients.

c) Prospective fault conditions

d) The I2t rating of the device –the fuse must clear the fault before the I

2t limit is reached

e) The fuse must be able to withstand the voltage after the arc extinction.

f) The peak arc voltage must be less than the peak voltage rating of the device

Crowbar Protection Circuit

17

A current sensing resistor R1 detects the value of the converter current. If it exceeds preset value , gate circuit provides the signal to the

craw bar SCR and turns it on. The input terminals are then short circuited by the craw bar SCR and it shunts away the converter over

current. Thus causing F1 to breakdown and interrupting the fault current. Also R2 and R3 can be used for overvoltage protection. If Vfb

exceeds the set voltages, the SCR is also switched on eliminating the overvoltage.

Chapter 2

2.1 Diode Rectifiers

A rectifier is a circuit that converts ac signal to DC. Diodes are taken as ideal i.e Voltage across a diode =0 .

2.1.1 Single phase half wave Rectifiers

Performance Parameters

As noted above the output voltage is discontinuous and contains harmonics. The power processing quality of a rectifier requires the

determination of harmonic contents of the input current, the output voltage. The average output load voltage is = Vdc and load current is

Idc.

Therefore the output dc power is Pdc =Vdc*Idc. The rms value of voltage and current is Vrms and Irms. Therefore output ac power Pac

=Vrms*Irms.

The efficiency or the rectification ratio of a rectifier is given by ηηηη =Pdc/Pac

The output voltage voltage can be considered to be composed of two components:-

1) the dc value

2) the ac component or ripple.

The effective rms value of the ac component of output voltage is determined from:

22222

dcrmsacdcacrms VVVVVV −=+=

The form factor, which is a measure of the shape of output voltage, is: FF =Vrms/Vdc.

The ripple factor, which is a measure of the ripple content is defined as

RF =Vac/Vdc 12

22

−=−

= FFVdc

VVRF

dcrms

Transformer utilization factor is defined as TUF =Pdc/Vs*Is where Vs and Is are the rms voltage and rms current of the

transformer secondary, respectively (VA of the transformer secondary)

18

Let us consider the waveforms of figure below where Vs is the instantaneous, sinusoidal input voltage, is is the instantaneous input

current and is1 is its fundamental component. See figure below

If φ is the angle between the fundamental component of the input current and voltage it is called the displacement angle. The

displacement factor is defined as φCosDF = .

The harmonic factor ( a measure of the distortions of a waveform) is defined as

2

1

2

1

221

2

1

2

1

2

1

−=

−=

s

s

s

ss

I

I

I

IIHF where Is1 is the fundamental component of the input current Is. Both Is1 and Is are expressed in

rms. The input power factor PF is defined as

PF =Vs Is1Cosφ/ Vs Is =Is1 Cos φ/Is.

Crest Factor (CF) is a measure of the peak input current Ipeak as compared with its r.m.s value Is.

CF =Ipeak/Is This helps in specifying the peak current ratings of devices and components..

Note :

If the input current is is purely sinusoidal Is1 =Is and the power factor PF equals the displacement factor DF. The displacement angle

becomes the impedance angle θ= tan-(ωL/R ) for an RL load. 2. Displacement factor is often known as displacement power factor (DPF)

Example :-

The rectifier above has a purely resistive load of R. determine :

a) the efficiency

b) The FF

c) The RF d) The TUF

e) The PIV of diode

f) The CF of the input current

Solution

First we determine the average output voltage Vdc:

( )R

VR

VI

VtCosT

VdtTSinV

TV mdc

dcmm

T

mdc

318.01

2*

1 2

0===−

−== π

ωω

ω

and

R

V

R

VI

VdttSinV

T

IV mrms

rmsm

T

mrms

5.0

2*

21

2

0

22 ===

= ω

a) efficiency =Pdc/Pac =Vdc*Idc/Vrms*Irms =40.5%

b) FF =Vrms/Vdc =0.5Vm/0.318Vm =157%

c) RF =121%

d) The transformer secondary rms voltage is :

R

V

R

VIV

VdttSinV

T

IV mrms

rmsmm

T

mrms

5.0707.0

2*

21

21

1

0

22 ====

= ω

Note the rms current comes in only when the diode is conducting.

The Voltage Ampere of the transformer is VA =Vs*Is =0.707Vm *0.5Vm/R

Therefore TUF =Pdc/VsIs =0.3182/0.707*0.5 =0.286

d) The PIV =Vm

19

e) Ipeak =Vm/R and Is =0.5Vm/R therefore the CF =Ipeak/Is =2

Half wave rectifier with RL load

Due to inductive load , the conduction period of diode D1 will extend beyond 180 deg until the current becomes zero at ωt =π +α:

The average output voltage is given by:

( )R

VICos

VtdtSinVV dc

dcm

mdc =++−== +

112

*2

1 1

0

αππ

ωωπ

απ

It can be noted that the average voltage and current can be increased by making α =0, which is possible by adding a freewheeling diode

Dm.

If the output of the circuit is connected to the battery the rectifier becomes a battery charger seefig below:

For Vs >E diode D1 conducts ,from α to β. The angle α can be determined from VmSinα =E

α =Sin-1

(E/Vm) and diode D1 is turned off when Vs< E at β.=π -α. . And the charging current is :

io =(Vs-E)/R =(Vm Sinω t –E)/R for α <ω t< β.

Example

The battery voltage in figure above is E=12V and its capacity is 100Wh. The average charging current should be Idc =5A.The primary

input voltage is Vp 120V, 60Hz and the transformer has a turns ratio of n=2 calculate:-

a) Conduction angle of the diode

b) The current limiting resistance R

c) The power rating of R

d) The charging time in hours (ho)

e) The rectifier efficiency

f) The PIV of the diode

Solution

E=12V;Vp =120V; Vs =Vp/2 =120/2 =60V and Vm =60* 20.5= 84.85V

a) α =Sin-1

(12/84.85) =8.13 deg and β.=π -α. =180-8.13 =171.87 therefore the Conduction angle = β. -α. =171.87-8.13 =163.74

deg.

b) The average charging current Idc is −

=β

αω

ωπ

tdR

EtSinVIdc m *

2

1 remember

β.=π -α. R =4.26Ω

c) The rms battery current is : tdR

EtSinVI m

rms ωω

π

β

α−

=2

22 )(

2

1I=8.2A The power rating of R is Pr =8.2*8.2*4.26

=286.4W

20

d) The power delivered Pdc to the battery is Pdc =E*Idc =12*5 =60W. Now the battery capacity =100Wh =Power * Time;

charging time ho =100Wh/Powerho =100Wh/60W =1.67hrs.

e) The rectifier efficiency = %3.174.28660

60=

+=

+=

Rdc

dc

PP

Pη

f) The peak inverse voltage of the diode =PIV =Vm +E =84.85 +12 =96.85V

Single-phase full wave Rectifiers

A full wave rectifier circuit with a center tapped transformer circuit is shown in fig above. The average output voltage is:

VV

tdtSinVT

Vdc mT

m 637.022 2

0=== πω

Instead of using center tapped transformer four diodes can be used as shown below:

The peak inverse voltage of a

diode is Vm. This is a bridge rectifier and it is commonly used in industrial applications.

Three phase Bridge Rectifiers

A three-phase bridge rectifier is commonly used in high power applications see the circuit diagram below:

21

This can operate with/out a transformer and gives six pulses/ ripples on the output voltage. Each diode conducts 120 deg. The

conduction sequence for diodes is D1-D2; D3-D2; D3-D4;D4-D5; D5-D6 and D1-D6. The pair of diodes having the highest of

amount of instantaneous line-to-line voltage will conduct.

If Vm is the peak value of the phase voltage then the instantaneous voltages are given by:: ;tSinVVan m ω=

)120;( −= tSinVVbn m ω );240( −= tSinVVcn m ω

CONTROLLED RECTIFIERS

A controlled rectifier converts ac power to dc power. Controlling the instants at which the semiconductor devices switch can control the

output voltage and power.

Thus controlled rectifiers can be used to control the speed of a DC motor. Some controlled rectifiers can convert DC to ac and that is

Inversion.

Thyristor Rectifier, Resistive load

A simple thyristor rectifier circuit consisting of a single thyristor and a resistance load is shown in below.

Fig Thyristor rectifier with resistive load circuit and waveforms

The thyristor is forward biased during the intervals 0<ωt <π, 2π<ωt<3π, etc. A gate pulse is applied at an angle α. This angle is

known as the firing angle of the thyristor. The thyristor current becomes zero at ωt =π,3π, etc, and the thyristor conducts from α to

π,2π +α, to 3π, etc. During the interval when the thyristor conducts, known as conduction interval, the load voltage is the same as the supply voltage, Vo =V and the average voltage is determined below:

We noted for a single phase diode rectifier circuit consisting of a resistive load, is:

tSinVv p ω2= Then the average value of the load is given by πωωπ

πp

po

VtdtSinVv

22

2

1

0

==

22

Thyristor Rectifier, Resistive Load

The thyristor is forward biased during the interval 0-180 and 360-540 degrees.

The thyristor is fired at an angle α and conduct for a conduction angle β. The average value of the load is :

)1(2

);1(2

)(2

10 απαπωω

π

π

α

CosR

VICos

VttdSinVVo

Lo

mm +=+==

The load voltage will vary with the variation of αααα and becomes maximum at αααα=0 and zero at αααα=ππππ.

Vdc max=Vm /ππππ and occurs at αααα=0, Idc max = Vm /ππππR = Im /ππππ.

The rms load voltage for a given firing angle αααα is:

212

1

22 )2

2sin(

1

2)(

2

1

+−=

=

ααπ

πωω

π

π

α

mmrms

VtdtSinVV

And the rms load current is given by the expression:

[ ]2

;)2

2(12

max

21

mrms

L

m

L

rmsrms

VVSin

R

V

RV

I =+−== ααππ

The average power P in the load for any trigger angle αααα is given by P =Vdc*Idc.

Example1

A 100Ω load is connected to a peak supply of 300V through a controlled half wave thyristor rectifier. The load average power is to be

varied from 25W to 80W. What is the angular firing control required? Neglect forward drop of the thyristor.

Solution

Power =Vdc*Idc=Vo*Io

And )1(2

);1(2

)(2

1απαπωω

π

π

α

CosR

VICos

VttdSinVVo

L

mo

mm +=+==

Power =P =22

2

22

2

2

)1(5.22)1(100*4

300)1(

4αα

πα

πCosCosCos

R

V

L

m +=+=+

When P=80W then the firing angle can be determined by the following relationship:

8857.0556.31)1(556.3)1(5.2280 22 ==++=+= αααα CosCosCosCos Therefore αααα =27.660 .At a

power of 25W the firing angle is 87,30 .

Thyristor half Wave Rectifier, Reactive Load

23

Most practical Loads have resistance and inductance L, e.g. the armature of a dc motor load has resistance (R) and inductance (L). When

a thyristor is fired at αααα the inductance in the load forces the current to lag the voltage and decays to zero at ββββ. The waveform of the

load current and voltage is shown below:

From αααα to ππππ the load current is driven by the supply source and from ππππ to ββββ the load voltage is negative and the current is maintained

by the voltage induced in the inductance

Half wave Controlled Rectifier With RL load and a free Wheeling Diode

The diode D bypasses the load current during the intervals of negative transformer voltage and this prevents the output voltage from

assuming a negative value.

• The diode prevents the voltage across the load from reversing during the negative half cycle of the supply voltage.

• The load average voltage would be determined only by the positive half cycle wave of the transformer secondary emf from αααα to

ππππ .

The average voltage is )1(2

)1(22

2α

πα

πωω

π

π

α

CosR

VICos

VttdSin

VV

l

m

dc

m

dc +=+==

Thyristor Full Converter

Half wave rectification contains a significant amount of ripples. Therefore this is not suitable for speed control of dc motors.

A full converter circuit shown in fig below is used for the speed control of DC motors.

The load current io has the same waveform as the load voltage vo..

The average load voltage is given by:

)1();1()(2

2απαπωω

π

π

α

CosR

VICos

VttdSinVVo

L

mo

mm +=+==

The rms voltage is given by the following expression:

24

212

1

22 )2

2sin(

2

1)(

1

+−=

=

ααπ

πωω

π

π

αmmrms VtdtSinVV

and the rms current is

[ ] 21

)2

2(2

1 ααππSin

R

V

RV

IL

m

L

rmsrms +−==

DC Motor Load / RL load

When terminal A1 is more positive as compared to A2 T1 and T2 are fired simultaneously and current flows through T1 load and T2. In

the reverse half cycle T3 and T4 are fired and the reverse voltage applied across them commutates T1 and T2. Due to inductive load, T1

and T2 will conduct beyond ππππ ,even though the input voltage is already negative see the waveforms above.

Depending on the size of the inductance the conduction can be discontinuous or continuous as shown above. If XL>>R then the load

current will be continuous.

Discontinuous load conduction

The average voltage is given by )(2

2βα

πωω

π

β

α

CosCosV

ttdSinV

V mm

dc −== The conduction angle of the thyristors is given by θθθθ =ββββ-αααα, hence ββββ=θθθθ +αααα

Now

)2

(2

22

*2

2θ

αθβαβα

βα +=−+

−=− SinSinSinSinCosCos

Therefore the average voltage for discontinuous load conductions is

l

dc

dc

m

dcR

VISinSin

VV =+= )

2(

2

2 θα

θπ

Continuous load Conduction

If we assume that sufficient inductance is present in the dc armature circuit to ensure that the motor current is continuous (present all the

time).

If conduction is continuous the limits are from αααα to (ππππ+αααα) and the average voltage is given by:

l

m

l

dc

dc

mm

dcR

CosV

R

VICos

VttdSin

VV

πα

απ

ωωπ

απ

α

22

2

2====

+

The rms voltage is determined from the following relationship:

21

22

=

+απ

α

ωωπ

ttdSinV

V mrms

Depending on the value of αααα the average output voltage could be either positive or negative and it provides two-quadrant operation.

25

For α=450 it can be noted that T1 and T2 conduct the motor current during the interval α< ωt < (π+α) and connects the motor to the

supply v=Vs . At (π+α), T3 and T4 are fired. The supply voltage appears immediately across thyristorsT1 and T2 as reverse bias voltage

and turns them off. This is called Line Commutation. The motor current , ia which was flowing from the supply through T1 and T2 is

transferred to T3 and T4. T3 and T4 conduct the motor current during the interval (π+α)<ωt < (2π+α) and connects the motor to the

supply (vo =-V)

If conduction is continuous the limits are from αααα to (ππππ+αααα) and the average voltage is given by:

l

m

l

dc

dc

mm

dcR

CosV

R

VICos

VttdSin

VV

πα

απ

ωωπ

απ

α

22

2

2====

+

The inductance (La) does not sustain any average voltage. Therefore Vdc =Vo =IoRa +Ea Where Io is the average motor current; Ea is the armature back emf, which is constant if the speed and field current are constant

The variation of the motor terminal voltage Vo as a function of the angle α based on the equation Vdc. For firing angles in the range 0o<

α <900 , the average output is positive. Since the current can flow in one direction in the load circuit because of the thyristors , the power

VoIo is positive ;power is flowing from the supply to the dc machine , and the dc machine operates as a motor. For firing angles 90o< α

<1800 ,the output voltage is negative and therefore the power VoIo is negative; that is power flow is from the machine to the ac supply.

This is known as inversion Operation of the converter, and this mode of operation is used for the regenerative braking of the motor. Note

that for inversion operation, the polarity of the motor back emf Ea must be negative. It can be reversed by reversing the field current if

so that the dc machine behaves as dc generator.

Converter output characteristics for continuous load current

26

Two full converters can be connected back to back as shown below . This arrangement is known as the dual–Converter connection. If one

converter is used, it causes motor current to flow in one direction. If the other converter is used, the motor current reverses and so does

the speed

Dual Converter

Example .

A full wave fully controlled single phase bridge rectifier feeds a load resistance R and inductance L from an AC source of Vrms voltage

Vs.

Assuming the operation is continuous:

a) show that the mean or average voltage value of the DC output voltage Vo is given by αCosVV so 90.0= .

If R=4Ω,L500mH, αααα=300 ,Vs =240V and the supply frequency is 50Hz.

b) Calculate:

i) the average and the rms currents output

ii) the thyristor average and rms currents and

iii) the power factor of the ac input to the rectifier

Solution

a) the mean value of the dc output for continuous operation is

ααπ

απ

ωωπ

απ

α

CosVCosV

CosV

ttdSinV

V s

smm

dc 90.0*222

2

2====

+

b) (i) Xl=2πfL=2*π*50*0.5H =157Ω This implies that the load current is going to be continuous. Therefore

AiVCosCosVV dcsdc 8.464

1.1871.18730*240*90.090.0 ===== α .

The current is almost constant since the ripple is almost negligible because of high circuit inductance. Thus this current is both the

average and the rms value i.e io =irms =46.78A

(ii) The thyristor average current is Ai

tdiI otha 4.232

78.46)(*

22

1 0 ==−+== +

ααππ

ωπ

απ

α

the rms thyristor current is:

Ai

tdiI oothrms 1.33)(

22

12

122

1

2 =

−+=

=

+

ααππ

ωπ

απ

σ

iii) The output power to the load is :

27

WRiP oo 48.87534*78.46* 22 === This is also the input power to the converter. The VA drawn from the supply is given by

: 8.02.11227

48.87532.11227240*78.46 ==== PFWVI ss

Thyristor Semiconverter

A thyristor semiconverter that can be used for the speed control of a dc motor consists of thyristors and diodes. See fig below:

Thyristor Semiconverter

Thyristors T1 and T2 are fired at αααα and (2ππππ+αααα), respectively. See the waveforms. The motor is connected to the input supply for the

period αααα<ωωωωt<ππππ through T1 and D2, and the load voltage is the same as the input voltage. Beyond ππππ, Vo tends to reverse as the

input voltage changes polarity. As Vo tends to reverse , the diode Dfw (known as the free wheeling diode ) becomes forward biased

and starts to conduct. The motor current which was flowing from the supply through T1, is transferred to Dfw, (i.e T1 commutates).

The output terminals are shorted through the freewheeling diode during the interval (ππππ<ωωωωt<ππππ+αααα) making Vo =0. At ωωωωt=ππππ+αααα , T2 is

fired and it take over the motor current ia from Dfw. The load current now flows through T2 and D1.At ωωωωt=2ππππ, Dfw, becomes

forward biased again and takes over the current from T2, and the process continues. See the waveforms. Note that if Dfw is not used,

freewheeling action will take place through T1 and D2 during the interval (2ππππ<ωωωωt<2ππππ+αααα).

The average value of the output voltage is:

)1(*2

)()**2(1

απ

ωωπ

CosV

tdtSinVVo

+=

=

Note; Vo is always positive and therefore Vo*Io is always positive that is power flow is from the ac supply to the dc load.

Semiconverters do not invert power. However they are cheaper than full converters.

Waveforms

Example

28

A single –phase full converter is used to control the speed of a 5hp, 110V, 1200rpm, separately excited dc motor. The converter is

connected to a single phase 120V,60Hz supply. The armature resistance is Ra =0.4Ω and armature circuit inductance is La=5mH.

The motor voltage constant Kφ =0.09V/rpm.

1. Rectifier(or motoring) operation. The dc machine operates as a motor runs at 1000rpm, and carries an armature current of 30A.

Assume that the motor current is ripple free.

a) Determine the firing angle αb) Determine the power to the motor

c) Determine the supply power factor

2. Inverter Operation (Regenerating Action). Reversing the field excitation reverses the polarity of the motor back emf Ea.

a) Determine the firing angle to keep the motor at 30A when speed is 100rpm

b) Determine the power fed back to the supply at 1000rpm.

Solution

a) Ea =0.09*1000 =90V; Vo =Ea +IoRa =90+30 =102V, 02.19

120*22102 == αα

πCos

b) P =Io2*Ra + EaIo =VoIo =102*30 =3060W

c)The supply current has a square waveform with amplitude 30A (=Io). The rms supply current is I=30A .

The supply Volt-Amperes are 120*30A =3600VA

If the losses in the converter are neglected, the power from the supply is the same as the power to the motor Ps =3060W

Thus the supply power factor is PF =Ps/S=3060/3600 =0.85

2. (a) At the time of polarity reversal the back emf is Ea =90V and from Vo =Ea +IoRa =-90+30*0.4 =-90+12 =-78V

Now 02.13678*

120*22=−== αα

πVCosVo

b) Power from the dc machine Pdc =90*30 =2700W

Power lost in Ra =302 *0.4 =360W

Power fed back to the ac supply =Ps =2700-360 =2340W

Three Phase Circuits

For higher power applications, several kilowatts three phase circuits are preferable. The magnitude of harmonic voltage is lower in

three phase circuits than in single-phase circuits. This is due to increasing number of pulses.

Half Wave Controlled Rectifier With Resistive load

29

)120(**2)240(**2

)120(**2

**2

00

0

+=−=

−=

=

tSinVtSinVVcn

tSinVVbn

tSinVVan

ωω

ω

ω

If thyristor T1 is fired at ωt =300 ,T2 is fired at ωt =1500 , and T3 ωt =2700 that is at the crossing points of the phase voltages. Firing

at these instants will also result in maximum output voltage. The reference for the firing angle α is therefore the crossing point of the

phase voltages. The firing of the thyristors can be delayed from these crossing points. In other words the firing angle is measured

from the crossing points of the phase voltages. Recall that in a single-phase converter, the firing angle was measured from the zero

crossing of the input supply voltage.

For the above waveform thyristor T1 is fired at ωt =300 + α, and the output voltage is Vo=Van. The output voltage io =Vo/R, and

becomes zero at ωt =π. Thyristor T1 turns off at this instant through natural commutation. Thyristor T2 is fired atωt =1500 + α,

making Vo= Vbn. T2 turns off at ωt =3000 .T3 is fired at

ωt =2700 + α. Making Vo =Vcn.

Example The load in the fig above now consists of a resistance and a very large inductance. The inductance is so large that the output current

io can be assumed to be continuous and ripple- free. For α=600 . a) Draw the waveforms of vo and io.

b) Determine the average value of the output voltage if phase voltage Vp =120V

30

Solution

b) The average voltage Vdc is given by the following relationship:

VCosVVVV

CosV

ttdSinVttdSinVV

dcmp

m

mmdc

2.70)60(*2

120*2*33120*2120

*2

33*

2

3*

2

3

0

12030

30

65

6

====∴

=== ++

+

+

+

π

απ

ωωπ

ωωπ

α

α

απ

απ

Thyristor Full Converter

This consists of six thyristor switches as shown in figure below and this is the most commonly used controlled rectifier circuit in

industrial applications up to 120kW.

Thyristors T1 ,T3 and T5 are

fired during the positive half cycle of the voltages of the phases to which they are connected, and thyristors T2,T4,T6 are fired during

the negative cycle of the phase voltages. The references for the firing angles are the crossing points of the phase voltages. See

waveforms below:

31

The times at which the

thyristors fire are marked in the figure above for α=300.

Assuming that the output current io ( i.e the dc motor current ) is continuous and ripple free as shown above.

• At ωt= 300 +α =π/6+ α T1 turns ON. Prior to this T6 was ON.∴ during interval (π/6+ α< ) ω t < (π/6+ α+π/3), T1 and T6 conduct the output current and the motor terminals are connected to phase A and B making the output voltage vo =Van

–Vbn. The output voltage vo is the distance between the envelopes of the phase voltages Van and Vbn as shown by the

arrows in the waveforms above.Note Van is positive and Vbn is negative hence Van –Vbn becomes Van +Vab. Under

phase A there is T1 and T4; while under B there is T3 and T6 and under C there is T5 and T2. this can be noted also on

each compltete cycle for each phase

• At ω t = (π/6+ α+π/3), T2 is fired and immediately voltage Vcb appears T6, which reverse-biases it and turns it off (line commutation. The current from T6 is transferred to S2. The motor terminals are connected to phase A through T1 and

phase C through T2, making Vo =Vac. This process repeats after every 600 whenever a thyristor is fired. Note that the

thyristors are numbered in a sequence in which they are fired.

• Each thyristor conducts for 1200 in a cycle. The thyristor current iT1 is shown in the graph. The line current such as iA, is a

quasi square wave having a pulse width of 1200 as shown above.

• The conduction pattern or firing sequence for the above circuit is

• Remember for a three phase star connected system Iline =Iphase and Vline =√3*Vphase and Ineutral =IR +IY +IB.

• Remember for a three phase Delta connected system Iline =√3*Iphase and Vline =Vphase.

The average value of the output voltage is: T1T6; T1T2; T2T3; T3T4; T4T5; T5T6; T6T1etc.

The average value of the output voltage is: απ

απ

ωπ

παπ

απ

CosV

CosVtdVbnVanV mso

3363)(*)(

2

6 36

6

==−= ++

+

32

The average output voltage varies with the firing angle α, and this variation is shown in fig below:

For α varying in the range 0 < α < 900 , Vo is positive and power flow is from the ac supply to the DC motor. For 900 < α <1800 ,Vo is negative and the converter operates in the inversion mode. The power can be transferred from the dc motor to the

ac supply a process known as regeneration.

Thyristor Semiconverter

The 3-phase semiconverter consists of three thyristors and three diodes as shown in fig below:

Voltages and current waveforms are shown in fig below for a firing angle of α =900 . The instants of firing a thyristor and the duration of conduction of the diodes are shown in the figure below showing the waveforms.

33

• We will assume that the output current is continuous and ripple free. At ωt =π/6+ α, T1 turns ON and T1 and D3 conduct

the output current io, making Vo=Vac. At ωt= 2100 ,Vo is zero, and from this instant onwards Vo tends to be negative. The diode

D1 will become forward –biased (and start conducting). The output current io will freewheel through T1 and D1, making Vo =0.

• When T2 is triggered on, the output current io will conduct through T2 and and D1 making Vo =Vba.

• The process repeats every 1200 whenever a thyristor is fired.

• Note that the line current iA starts at ωt =π/6+ α and terminates at ωt= 2100.

The average value of the output voltage is:

)1(2

33*)120(

2

3

2

3 06

6

6

6

απ

ωωωπ

ωπ

ππ

απ

ππ

απ

CosV

tdtSinVtSinVtdVacV mmmo +=+−==

+

+

+

+

Note that the output voltage

cannot reverse. Hence this converter does not operate in the inversion mode.

Dual Coverters

Two full converters can be connected back to back to form a dual converter just as in the single-phase situation and both the voltage Vo

and current Io can reverse in a dual converter.

Example

A 3phase full converter is used to control the speed of a 100hp, 600V,1800rpm, separately excited motor. The converter is operated from

a star connected 3phase, 480V, 60Hz supply. The motor parameters are Ra =0.1Ω, La =5mH ,Kφ =0.3V/rpm (Ea=Kφn). The rated

armature current is 130A

1. The rectifier (or motoring) operation:- The machine operates as a motor, draws rated current, and runs at 1500rpm. Assume that

the motor current is ripple free

a) Determine the firing angle.

b) Determine the supply power factor

2. Inverter operation :- The dc machine is operated in the regenerative braking mode. At 1000rpm and rated motor current

a) Determine the firing angle.

b) Determine the power fed back to the supply and the supply power factor

34

Solution

a) Phase voltage is VRIEVVEVV aoaoap 4631.0*1304504501500*3.02773

480=+=+=∴==== but

we know :

04.44*3*277*23

46346333

==== ααπ

απ

CosVCosV

V mo

b) Since the ripple in the motor current is neglected, the supply current iA is a square wave of magnitude 130A and width 1200 .

The rms value of the supply current is Atdii orms 1.106130*3

2

3

2*130*

11 21

2

21

32

0

2 ==

=

=

ππ

ωπ

π

The supply voltage amperes are S =3Via =3*277*106.1 =88169.1VA

Assuming no losses in the converter, the power from the supply Ps is the same as the power input to the motor. Hence

Ps =Vo*Io =463*130 =60190W

Therefore the supply power factor is PF =Ps/S =60190/88169.1 =0.68

2. (a) Ea =0.3*1000 = 300V

For inversion the polarity of Ea is reversed

Vo =Ea +IoRa

=-300 +130*0.1

= -287V

03.116*3*277*23

28728733

==−−== ααπ

απ

CosVCosV

V mo

b) Power from the DC machine (operating as a generator): Pdc =300*130 =39000W

Power lost in Ra: Pr =1302 * 0.1 =1690W

Power to the source: Ps =39000 –1690 =37310 W

Supply Volt amperes S = 88169.1V and Supply power factor =PF =37310/88169.1 =0.423

Assignment 2

Question 1

A three phase half wave converter in fig below is operated from a three phase Y connected 415V, 50Hz supply and the load

resistance is R =10ΩΩΩΩ. If it is required to obtain an average output voltage of 50% of the maximum possible output voltage.

Calculate:

a) the delay angle αααα, [4]

b) the rms and average output currents [4]

c) the average and rms thyristor currents [4]

d) the rectification efficiency [4]

e) the input power factor [4]

Question 2

A three phase full wave converter is operated from a three phase Y connected 415V, 50Hz supply and the load is highly inductive

with resistance R =10ΩΩΩΩ. If this converter is required to obtain an average output voltage of 50% of the maximum possible output

voltage. Sketch the circuit diagram. [2]

Calculate:

35

a) the delay angle αααα, [3]

b) the rms and average output currents [4]

c) the average and rms thyristor currents [4]

d) the rectification efficiency [4]

e) the input power factor [3]

Question 3

A full wave fully controlled single-phase bridge rectifier feeds a load resistance and inductance L from an AC source of rms

voltage Vs. A freewheeling diode is fitted across the load.

a) Show that the output voltage is given by 2

)1(9.0 αCosVV s

o

+= [4]

Given that R =4ΩΩΩΩ, L=500mH, αααα=30 ,Vs =240V and the supply frequency to be 50Hz, Calculate:

b) The average and rms output currents [4]

c) The thyristor average and rms currents [4]

d) The diode average and rms currents [4]

e) The power factor of the ac input to the rectifier [4]

AC Voltage Controllers

AC voltage controllers

These convert a fixed –voltage ac supply into a variable voltage ac supply. They can be used to control the speed of induction motors.

Single phase AC Voltage Controllers

The waveforms are as shown below:

36

T1 is fired at α and T2 is fired at π+α. If T1 is fire the current builds up at α and decays to zero at β. When T2 turns on at π+α a negative

current flows in the load.

During conduction interval say for T1:

tVSindt

diLiRV oo ω*2* =+=

The load current is:

τφωt

transientesteadystato AetSinZ

Viii

−+−=+= )(

2……..1

where :

RL

RLLRZ ==+= − τωφω ;tan;)( 122

At ωt =α, io =0 then:

αωφα)(

)(2

LR

eSinZ

VA −−= ……..2

From equations 1 and 2

])()([2 ))(( t

LR

o eSintSinZ

Vi

ωαωφαφω−

−−−= ……….3

Letting α=φ from 3:

)(2

φω −= tSinZ

Vio ………………4

From equation 4, the load current io is sinusoidal, which indicates that if the firing angle is the same as the impedance angle i.e., α=φ, the load current becomes purely sinusoidal.

Each thyristor conducts for 1800 and the full supply appears across the load. Fig below shows the waveforms of load current for two

different firing angles.

For α > φ, io is non-sinusoidal, and for α=φ, io is sinusoidal. Even for α<φ will be sinusoidal.

To determine β, when the current io falls to zero and T1 is turned off, can be found from the condition io(ωt=β)=0 and this gives a

relation :

37

))(()()(

βαωφαφβ−

−=− LR

eSinSin

The angle β, which is also known as the extinction angle, can be determined from this equation by way of an iterative method of solution.

Once β is known then the conduction angle of T1 δ can then be determined from δδδδ=ββββ-αααα.

The rms output voltage is :

21

21

22

02

2

2

21)(*2

2

2

−+−=

=

βααβ

πωω

π

β

α

SinSinVtdtSinVV Where V is the rms supply voltage.

The rms thyristor current is obtained from:

21

2

0 )(2

1

=

β

α

ωπ

tdiI rth and the rms output current can then be found by combining the rms current of each thyristor as

( ) rthrthrth IIII *221

22

0 =+= and the average thyristor current can be determined from

=

β

α

ωπ

)(2

10 tdiI a

Example

A single-phase full wave controller supplies an RL load. The input rms voltage is V =120V, 60Hz. The load is such that L=6.5mH and

R=2.5Ω. The delay angle of thyristors are equal:α1=α2=π/2. Determine

a) the conduction angle of thyristor T1,δ;

b) the rms output voltage Vo,

c) the rms thyristor current Irth;

d) the rms output current Io;

e) the average current of a thyristor Ia; and

f) the input power factor.

Solution:

R=2.5Ω, L=6.5mH, f=60Hz, ω=2π*60 =377rad/s, V=120V, α=π/2. and

φ=tan-1

(ωL/R)=44.430.

a) an iteration solution yields 220.350 hence the conduction angle =δ =220.350-900=130.430

b) Vo =68.09V

c) Integrating from α toβ gives Irth =15.07A

d) Io =21.3A

e) Ia=8.23A

f) Output power =21.32 2.5=1134.2W and the input VA is VA =120*21.3=2556W Therefore PF =1134.2/2556 =0.444lag

Three Phase AC voltage Controllers

For higher power loads such as the induction motors driving fans, or pumps, three phase controllers are used. Figure below shows the two

types of three phase ac voltage controllers.

In one circuit, the thyristors switches are in the lines and the load can be connected either in Star or Delta. While in the other circuit the

thyristors are connected in series with the phase loads to form a delta connection

38

The firing sequence for the Y connected will be T1;T2;T3;T4;T5;T6 or see the waveforms below:

The operation of the delta-connected controller can be studied on a per phase basis because each phase is connected across a known

supply voltage.

Example:

A three phase ac voltage controller is used to start and control the speed of a 3 phase 100Hp ,460V, four pole induction motor driving a

centrifugal pump. At full –load output the power factor of the motor is 0.85 and the efficiency is 80%. The motor current is sinusoidal.

The controller and the motor are connected in delta as shown above.

a) determine the rms current rating of the thyristors

b) determine the peak voltage rating of the thyristor

c) determine the control range of the firing angle α

Solution

a) output power is given SPFPo **η=S=100*0.746/0.8*0.85 =109.71kVA

We know VA=3Vph*Iph Iph =109.71kVA/3*460 =79.5A (remember Vph=VL for a delta connected load and phL II 3= )

∴thyristor rms current =79.5/(2)1/2

= 56.22A

b) Peak voltage across the thyristor =(2)1/2

*460=650.4V

c) the control range of the firing angle is determined from the fact that :

39

01 8.3185.0 === −CosCosPF φφ the control range is 31.8

0< α <1800.

Thyristor Commutation

A thyristor needs a commutation circuit whenever a dc source supplies power to be modulated. For example when a thyristor is

being used on inverters and Choppers once ON it will remain being on until a positive voltage is applied on its cathode that is when

it can be turned off and this is termed forced commutation.

There is natural commutation which is to do with the switching off of the thyristor by virtue of current falling below its holding

current and this happens naturally as in the case of ac supply. See fig below:

At 90 the thyristor switches off naturally and awaits another firing signal for the next positive pulse

a) Parallel Capacitor turn off Circuit

Vs and load Rl comprise the main circuit. T1 modulates the power in the load. C,R and T2 are assumed to be off . If T1 is switched

on, so that there is load current i at the same time the capacitor charges up to +Vs on plate Y through R, C and T1. Plate X of

capacitor is virtually at ground potential because T1 is in the zero impedance state. This sets the stage for commutation.

When it is desired to interrupt the load current, T2 is turned ON. This puts the capacitor C in parallel with T1 and the reverse

voltage sitting across C reverse biases T1. If T1 is reverse biased long enough, it will turn off. Capacitor C discharges and charges,

this time becoming positive at X through Rl, C and T2. At some particular time T1 is turned ON again, now the voltage across C

40

reverse biases T2 to turn it off. The capacitor C becomes positively charged at plate Y so that the cycle can be repeated. By

controlling the times that T1 and T2 are allowed to conduct the average power in the load can be adjusted. To be able to do this the

value of the capacitance must be determined.

During Turn Off:

CRt

L

LeR

Vsti

−

=2

)( current through the circuit. 2Vs comes from the supply and the capacitor.

The voltage across the thyristor T1 is : )21(1

CRt

sLsTLeViRVV

−

−=−=The time t1 for the voltage VT1 to rise to zero is :

)21(1

CRt

sLsTLeViRVV

−

−=−= = 0 t1 =0.69R L C and this time must be larger than the turn off time of the thyristor T1.

Turn off time of the device is :

L

offLoff R

tCCRt

69.069.0 ≥≤

If T1 and T2 have the same turn off times and if R≥ RL then the above value of C will allow satisfactory commutation.

If the capacitor is to charge fully after each switching action, it will take at least 5τ (5RL C) after T2 has been triggered for this to be

accomplished.

)(511)(5)55(

LLL RRT

CRRRCCRT +=+=+≥

Example

Consider a case of parallel capacitance turn off as shown above .The load resistance RL =5Ω and the supply voltage is Vs =120V.

Calculate the minimum value of the capacitor C if the manufacturer’s specified turn off time for the thyristor is 15µs. What is a

suitable value of the resistor R, if the thyristor T1 is triggered on every millisecond?

Solution

C =15µs /0.69Rl =4.3µF a reasonable value of C could be 5µF.

The minimum time of cycle of operation is T =5(RL C +RC).

The value of R provides the only adjustments of this time since other parameters are fixed.

Now 5(RL C +RC)=1*10-3R =41.5Ω

The value of R can be less than 41.5 and this would result in the capacitor charging up faster while T1 is conducting.

b). Auxiliary Resonant Commutation Circuit

This circuit is popular and is used in many inverter and chopper circuits see the circuit diagram first:

T2 is triggered first to order to charge up the capacitor C in the polarity shown. As soon as the capacitor C is charged T2 is

commutated owing to lack of current. When T1 is triggered current flows in two paths, the load current flows in the in Rl and the

commutating current through C, T1, L and D. The charge on the capacitor is reversed and is held with the hold off diode D. At any

desired time T2 may be triggered which then places C across T1 thus turning off T1. The waveforms are shown below.

41

At to T1 is assumed to be conducting

At t4 T1 is switched on and the capacitor will begin to discharge via T1, diode D and the inductance L. During the first period of this

oscillatory discharge, until time t5 a reverse voltage is placed across the commutating thyristor T2. After one half cycle of the oscillation, at time t6, the current through the diode attempts to reverse and the diode ceases to conduct and the capacitor is charged

in the opposite direction of the figure ready for the next commutation cycle. If say we want to commutate the T1 then T2 is fired and

this places Vc across T1 commutating T1.

The frequency of Oscillation for this circuit is determined by the values of capacitance and the inductor according to the

relationship:

)(1

LC=ω

c) Resonant turn off

There is a series resonant and parallel resonant commutation circuit

Series Resonant Circuit

If an LC network is included as part of the load circuit, the current reversing properties of a tuned LC will force the device to commutate.

When T1 is triggered, a current flows through the LC circuit charging the capacitor towards the supply voltage.

42

After some time , the magnitude of the current reverses and tries to flow through T1 in opposite direction owing to the resonating

effect of Land C. For proper commutation the LCR circuit must be under-damped to allow current oscillation which will turn off the

thyristor at first current zero. The capacitor will then discharge through the load . The on time of the thyristor is determined by the

frequency of the oscillatory circuit (LC) )(2)(2

12)(

1 LCTLC

ffLC

ππ

πω ====

The current and voltage waveforms are shown below:

Parallel Resonant Commutation Circuit

When the circuit is switched on and in the absence of gate trigger pulse, the capacitor C is charged up in the polarity shown.

When T1 is triggered , current flows through load resistance Rl and a pulse current through the resonating LC circuit. The capacitor

is discharged from the initial polarity and charges it in the reverse direction. The resonant circuit current then reverses and tries to

flow through T1 in opposite direction to the load current and during this process T1 is turned off.

The thyristor ON period is again a function of the frequency of the oscillation of the LC circuit, while the off period must be sufficient

to allow the capacitor to be adequately charged.

Commutation is accomplished if the resonant current ic reaches a value greater than the load current il for a time greater than toff

turn off time of the thyristor.

The circuit equation for the resonant circuit is:

43

==+=+=+ )(*)(0)(]1[0)(*101 2

2

2

tSinL

CVtisIC

LStiCdt

diLidt

Cdt

diL s ω The capacitor

voltage at any given time is given my the following relationship:

tCosVV sc ω=By design the peak discharge current is:

LCVs And this should considerably in excess of the load current. Hence:

CLR

L

C

RLCVii

R

Vl

l

sc

L

s =====2max

1

When the thyristor turns off, capacitor has negative charge. The capacitor then charges to a positive value through the load R, L and

C forming a damped resonant circuit.

Example :

A parallel resonant turn off thyristor chopper consisting of an LC circuit in parallel with an SCR supplies an inductive load current

of 20A from a DC source of 400V. Choose suitable values for L and C for turn off so that the trigger pulses have a minimum period

of 0.1msec.

Solution:

Peak current due to oscillatory circuit is Ipeak =L

CVs ≥20A

4001

201 ==

LC

LC

Minimum period (highest chopper frequency) corresponds to half (1/2) periodic time of the oscillatory current that is:

mHLFCLCLC 636.0;10*5.110*013.110*1.0 693 ≥=== −−−π

Remember capacitor voltage leads current by 90 deg see the current response graph below:

The maximum time ton that the thyristor conducts is from the time it is turned on to the time that the capacitor current reaches its

maximum value in the reverse direction. In terms of the period T this time is 3/4T hence in limit this is :

LCton π4

3= .

DC –DC CONVERTER

(CHOPPER)

A chopper directly converts a fixed –voltage dc supply to a variable –voltage dc supply. The chopper can be used to control the speed of a

DC motor.

44

Applications

-Switched –mode power supply (SMPS), DC motor Control, Battery chargers

Step Down Chopper (Buck Converter)

A schematic diagram of a step down chopper with a motor load is shown in fig below.

The chopper can be a conventional thyristor (SCR), a GTO thyristor, a power transistor, or a MOSFET.

When the chopper is turned ON say at t=0, the supply is connected to the supply and Vo =V. The load current io builds up. When the

chopper/switch is turned off at t=ton, the load current freewheels through Dfw and Vo =0. At t=T the switch is turned on again and the cycle repeats. The waveforms of the load voltage and load current are shown in fig above. It

is assumed that the current is continuous, while the voltage is chopped.

The average value of the output voltage is VVT

tonV *0 δ==

Where ton is the on time of the chopper

T is the chopping period

δ=the duty ratio of the chopper

The output voltage can be controlled in the range 0< Vo < V .

If the switch is a GTO thyristor, a positive gate pulse will turn it on and a negative gate pulse will turn it off. If the switch is a transistor,

the base current will control the on and off period of the switch. If the switch is an SCR, a commutation circuit is required to turn it off.

There are many forms of commutation circuits that can be used to force commutate a thyristor.

The power supplied to the motor is 00 ** iViVP oo δ== .

If we assume a loss less dc-dc converter Pi= 00 ** iViVP oo δ==

45

The average value of the input current is 00 *** iIiVIV ss δδ ==

The equivalent input resistance of the dc-dc converter drive seen by the source is: δ1*

0i

V

I

VR

s

eq == . By varying the duty cycle, the

power flow to the motor and the speed can be controlled.

Example

A dc converter as shown above from a 600V dc source powers a dc separately excited motor. The armature resistance is Ra =0.05Ω. The

back emf constant of the motor is kv = 1.527V/Arads-1

.The average armature current is io =250A. The field current is if =2.5A. The

armature current is continuous and has negligible ripple. If the duty cycle of the dc-dc converter is 60% determine:

a) the input power from the source,

b) the equivalent input resistance of the dc-dc converter drive,

c) the motor speed , and

d) the developed torque

Solution

a) the input power from the source is 00 ** iViVP oo δ== =0.6*600*250 =90kW

b) the input resistance of the dc-dc chopper = δ1*

0i

V

I

VR

s

eq == =600/(250*0.6)=4Ω

c) the motor speed is determined from ω** fvg IkE = but aaaao RiVEERiV 000 * −=+= but

VVT

tonV *0 δ== =0.6*600 =360V Ea =360 –250*0.05 =347.5V 347.5

= rpmrpsIkE fva 27.86903.915.2*527.1

5.347** ==== ωω

d) The developed torque is afv IIkT **= =1.527*250*2.5 =954.4NM

Step Up Chopper (Boost Converter)

A change in chopper configuration as shown in fig below provides higher load voltages.

When the chopper is ON, the inductor is connected to the supply V and the energy from the supply is stored in it. When the chopper is

off, the inductor current is forced to flow through the diode and the load.

The induced voltage Vl across the inductor is negative. The inductor voltage adds to the source voltage to force the inductor current into

the load. Thus the stored energy in the inductor is released to the load.

If the ripple in the source current is neglected, then during the time the chopper is ON (ton) the energy input to the inductor from the

source is:

onVItEi =

During the time the chopper is off (toff), the energy released by the inductor to the load is: offo ItVVEo )( −=For a loss less system in the stead state, the two energies will be the same hence:

δδ −=

−=

−=

+=−=

1***)(

VV

TT

TV

tT

TV

t

ttVItVVVIt

onoff

offon

ooffoon

Thus for a variation of δ in the range 0< δ <1, the voltage Vo varies in the range 0< Vo < ∞.

Two Quadrant Chopper

46

A combination of step up and step down configurations can form a two-quadrant chopper. See figure below:

If chopper C1 and diode D1 are operated the system operates as a step down chopper and the dc machine operates as a motor. The output

voltage is either V (when C1 is on) or zero (when C1 is off and D1 conducts).

• The average output voltage is positive and the output current io flows as shown in the circuit. The chopper therefore operates in

the first quadrant,

• If , however the chopper C2 and the diode D2 are operated , the system operates as a step up chopper with Ea as a source and the

DC machine operates in a regenerative braking mode.

• The output is either zero (when C2 is ON) or V (when C2 is Off and D2 conducting).

• The average output voltage is positive, but the output current now flows in the negative direction. The chopper then operates in

the fourth quadrant.

Note

The armature of the separately excited motor is rotating due to inertia of the motor and the load; and in case of transportation system, the

kinetic energy of the vehicle or train would rotate the armature shaft. Then if the transistor or chopper is switched on, the armature

current rises due to the short-circuiting of the motor terminals. If the dc –dc converter (C2) is turned off, diode D2 would be turned on

and the energy stored in the armature circuit inductance would be transferred to the supply provided the supply is receptive. A single

chopper can be used for both powering and regenerative breaking. In regenerative breaking C1 and D1 are rearranged from powering

mode to regenerative braking. That is D1 and C1 are interchanged.

Example

The two-quadrant chopper shown in fig above is used to control the speed of the dc motor and also for regenerative braking of the motor.

The motor constant is Kφ =0.1V/rpm (Ea=Kφn). The chopping frequency is fc =250Hz and the motor armature resistance is Ra =0,2Ω. The inductance La is sufficiently large and the motor current io can be assumed to be ripple –free. The supply voltage is 120V.

a) Chopper C1 and diode D1 are operated to control the speed of the motor. At n=400rpm and io=100A (ripple –free),

i) Draw waveforms of vo, io and is

ii) Determine the turn on time ton of the chopper

iii) Determine the power developed by the motor, power absorbed by Ra, and power from the source b) In the two-quadrant chopper C2 and diode D2 are operated for regenerative breaking of the motor. At n=350rpm and io = -100A

(ripple free),

i) Draw waveforms of vo, io and is

ii) Determine the turn on time ton of the chopper

iii) Determine the power developed by the motor, power absorbed by Ra, and power to the source

Solution

a) ( i)

47

ii) aaa RIEV +=0 =0.1*400 +100*0.2 =60V

2120**60

Tt

T

tV

T

ton

onon === but T =1/250 =0.004seconds =4ms hence ton =2msec

iii) Pmotor =Ea*Io =0.1*400*100 =4000W

PR =(io)2*Ra =100

2*0.2 =2000W

Ps =V*(is)avg =120*100*2/4 =6000W

b) the waveforms are shown below

i)

ii) aaa RIEV +=0 =0.1*350 +(-100*0.2) =15V

sec5.34*8

7120**15 mt

T

tTV

T

tTon

onon ==−

=−

=

iii) Pmotor =Ea*Io =0.1*350*(-100) = -3500W

PR =(io)2*Ra =100

2*0.2 =2000W

Ps =V*(is) avg =120*(-100*1/8) =-1500W

The average voltage across the chopper is given for the following circuit used in regenerative brake control:

48

)1()(1

δδδ

−=−== VTTT

VdtV

TV s

T

T

ch

If Ia is the armature current, the regenerated power is )1( δ−== VIVIPg acha

The voltage generated by the motor acting as a generator is:

aaamchfvg IRVIRVIkE +−=+== )1( δωWhere kv is the machine constant and ω is the machine speed in rads/sec. Ra is the total armature circuit resistance.

Therefore the equivalent load resistance of the motor acting as a generator is:

a

aa

g

eq RI

V

I

ER +

−==

)1( δ

By varying the duty cycle δ, the equivalent load resistance seen by the motor can be varied from Ra to (V/Ia +Ra) and the regenerative

power can be controlled.

The conditions for the permissible potentials and polarity of two voltages are:

,0 minmin

fv

aa

aafvaaaaIk

IRIRIkEVRIE ===≤−≤ ωω and ω>or=ωmin . The maximum braking speed of a motor can

be determined from

maxmaxmax ;0 ωωωω ≤+=−=≤−≤fv

aa

fv

aafvaaaIk

IR

Ik

VIRIkVVRIE .

The regenerative braking would be effective only if the motor speed is between these two speed limits (eg ωmin <ω<ω max). At any speed

less than ω min an alternate braking arrangement would be required. Which could be Rheostatic brake control or combined Regenerative

and Rheostatic brake control

Example

A dc-dc converter is used in regenerative braking of a dc series motor. The DC supply is 600V. The armature resistance is Ra =0.02Ω and

the field resistance is Rf= 0.03Ω. The back emf constant is kv=15.27mv/A rads-1

.The average armature current is maintained constant at

Ia =250A.The armature current is continuous and has negligible ripple. If the duty cycle of the dc- dc converter is 60%, determine:

a) the average voltage across the dc-dc converter Vch

b) the power regenerated to the dc supply Pg

c) The equivalent load resistance of the motor acting as a generator Req

d) The minimum permissible braking speed ωmin ,

e) The maximum permissible braking speed ω max.

f) The motor speed

Solution

a) )1( δ−= VVch = (1-0.6)*600 = 240V

b) Pg =Ia*Vch =250*240 =60kW

c) Req =V/Is = Ea/Ia =(Vch + IaRm) where Rm =0.02+0.03 =0.05Ω Req =(240+250*0.05)/250 =1.01Ω.

d) ,0 minmin

fv

aa

aafvaaaaIk

IRIRIkEVRIE ===≤−≤ ωω ωmin =0.05*250/(15.27mV*250)=3.274rad/s

=31.26rpm

49

e) maxmaxmax ;0 ωωωω ≤+=−=≤−≤fv

aa

fv

aafvaaaIk

IR

Ik

VIRIkVVRIE ω max.=600/(0.01527*250)

+0.05/0.01527 =160.445rad/s =15232.14rpm

f) aaamchfvg IRVIRVIkE +−=+== )1( δω Eg =Vch +IaRa =240 +250*0.05 =252.5V ω

=252.5/(15.27mV*250)=66.14rad/s =631.6rpm

Inverters

These are static circuits that convert power from a dc source to ac power at a specified output voltage and frequency. Inverters are used in

many industrial applications.

For example:

a) Variable – speed ac motor drives

b) Induction heating

c) Aircraft power supplies

d) Uninterruptible power supplies (UPS) for computers.

There are two types of inverters namely:

i) Voltage source inverters (VSI)

ii) Current Source Inverter (CSI)

Voltage Source Inverter (VSI)

In the VSI the input is a dc voltage supply, e.g., battery, fuel cell, soar cell and other dc sources such as output of controlled rectifier.

Both Single phase and three phase voltage sources are used in industry.

Figures illustrating Inverter Configurations

The switching devices can be a conventional thyristor(with its commutation circuit), a GTO thyristor, or a power transistor.

Single Phase VSI- Principle of operation

50

The inverter consists of two choppers. When Q1 is turned on for a time T/2, the instantaneous voltage across the load is V/2.

If only Q2 is turned on for a time T/2, -V/2 appears across the load.

The logic circuit should be designed such that Q1 and Q2 are not turned on at the same time. Third and fourth graphs are for resistive and

inductive loads.

Note that prior to turning on a switch, the other one must be turned off . Otherwise both switches will conduct and short circuit the DC

supply. If the load is inductive, that is a lagging power factor, the output current io also lags the output voltage Vo as shown in the fig

above.

The load current can not change immediately with the output voltage. If Q1 is switched off at T/2, the load current will continue to flow

through D2, load and the lower half of the dc source until the current falls to zero.

Similarly when Q2 is turned off at t=T, the load current flows, through D1, load and upper half of the DC source. When the diodes D1

and D2, conduct, energy is fed back to the dc source and these diodes are known as feedback diodes.

The rms output voltage is :

RVdt

RV

TPVdt

V

TV

TT

o

orms 44

2

24

2 22/

0

2

21

2

0

2

===

=

E.g

A center tapped source inverter as shown above modulates power from a 200V source to a purely resistive load whose value is R=2Ω.If

the thyristors have a duty cycle of 0.4=m. Sketch VT1 ,VT2 , IL , IT1 , and determine:

a) the average power absorbed by the load,

b) the voltage and current ratings of the switches

Solution

a) Pav =16kW

b) Current rating =100A

51

Waveforms

Single Phase Bridge Inverter

Q1 ,Q2 ,Q3, Q4, are four choppers. Where Q1 and Q2 are turned on simultaneously, the input voltage Vs appears across the load while

Q3 and Q4, the voltage across the load is reversed and is –Vs.

If m=duty cycle= Ton/T then the rms voltage is :

Vrms =Vs√(2m)

The average power absorbed by the load is P =2mV2s/R

Example

The dc source voltage is Vs =600V and the load has a resistance of R=20Ω if the inverter is to operate at 500Hz with an rms load voltage

of 500V. Find:

a) the average power absorbed by the load

b) the average source current

c) the average current in each thyristor

d) the thyristor on time each period

Solution:

a) 12.5kW

b) 20.83A =Isav

c) Ithav =20.83/2=10.42A

d) Ton =0.694x 10-3

s

1

Chapter 1

Electronic Drives

Course Outline

The following areas will be covered during our course of study:

• Power Electronic Devices:- Characteristics, drive requirements and device protection.

• Converters: - DC-DC; DC-AC; AC-DC and Control Techniques

• Power and distortion factor; Harmonics and interference

• Applications of AC &DC motors:- motor control: variable speed drives, regenerative

breaking, slip energy recovery, four quadrant operation

1.1 Power Semiconductors Converters

The technique for controlling the electric machines has changed in significant ways. For example series

DC motor are used to propel subway cars. The speed of these cars has been controlled by use of

resistances in series with the dc motors as shown in figure below. In recent years there has been use of

Choppers, which can convert a fixed voltage into a variable dc voltage hence controlling the speed of

the car.

Using the appropriate converters can also control other electric machines. The following being the

various types of converters that are frequently used to control electric machines:

AC Voltage Controller (AC to AC): - an ac voltage controller converts a fixed voltage ac to a

variable voltage ac. It can be used to control the speed of an induction motor (voltage control method)

and for smooth induction motor starting.

Controlled Rectifier (AC to DC): - A controlled rectifier converts a fixed voltage ac to a variable

voltage dc. It is used primarily to control the speed of dc motors such as those used in rolling mills.

Chopper (DC to DC): - A chopper converts a fixed voltage dc to a variable DC. It is used primarily to

control the speed of DC motors.

Inverter (DC to AC): - An inverter converts a fixed voltage dc to a fixed (or variable) voltage ac with

variable frequency. It can be used to control the speed of ac motors.

Cycloconverter (AC to AC): - A cycloconverter converts a fixed voltage and fixed frequency ac to a

variable voltage and variable (lower) frequency ac. It can be used to control the speed of ac motors.

High –power semiconductor devices are used in these converters to function as on-off switches. The

characteristics of these devices will be looked at first.

2

1.2 Power Semiconductor Devices The power semiconductor devices generally used in converters can be grouped as follows:

• Thyristors (SCR)

• Power transistors

• Diode rectifiers

These devices are operated in the switching mode so that losses are reduced and

conversion efficiency is improved.

1.2.1Thyristor (SCR)

The Thyristor has a four layer p-n-p-n structure with three terminals, anode (A),

cathode (K), and gate (G) as shown in figure 2 below. The anode and cathode are

connected to the main power circuit. The gate terminal carries a low level gate

current in the direction from the gate to cathode. The Thyristor operates in two

stable states: ON or OFF.

Volt Ampere Characteristics With zero gate current, if a forward voltage is applied across the device (anode positive with

respect to cathode) junctions 1 and 3 are forward biased while j2 is reverse biased and therefore

the anode current is a small leakage current. If the anode to cathode forward voltage reaches the

critical limit called the forward breakover voltage (VBO) the device switches into high conduction.

The device is said to be latched IL . Latching current is defined as the minimum anode current that

a Thyristor must attain during turn on process to maintain conduction when the gate is removed or

current required to maintain regeneration. If gate currents are introduced the VBO is reduced. For

a sufficiently high gate current the entire forward blocking region is removed and the device

behaves as a diode. When the device is conducting the gate current can be removed and the device

remains in the on state.

If the anode current falls below the critical limit called the holding current Ih, the device returns to

its forward blocking state. See the I/V characteristic

3

Switching Characteristics If a Thyristor is forward biased and a gate pulse is applied, the Thyristor switches on. However,

once the Thyristor starts conducting an appreciable forward current (regeneration taking place), the

gate has no control on the device. The Thyristor will turn off if the anode current becomes zero

called natural commutation or is forced to become zero called forced commutation.

It is necessary to keep the device reverse biased for a finite period of time before a forward anode

voltage can be applied otherwise if applied at a less time it would conduct. This period is known as

the turnoff time, toff of the Thyristor. The turnoff time of the Thyristor is defined, as the minimum

time interval between the instant the anode current becomes zero and the instant the device is

capable of blocking the forward voltage.

See fig 3 below for the switching characteristics of a Thyristor

The Thyristor is switched on by a gate pulse ig and when the gate pulse is applied at instant t1,

anode current IA builds up and the voltage across the device VAK falls. When the device is fully

turned on the voltage across it is quite small (typically 1 to 2.5V, the higher voltage drop for higher

current devices) and for all practical purposes the device behaves as a short circuit. The device

switches on very quickly, the turn-on time ton being typically 1 to 3microseconds. Typically the

gate pulse is in the range of 10 to 50microsec and its amplitude in the range 20 to 200mA. If the

current through the thyristor is to be switched off at a desired instant t2, it is momentarily reverse

biased by making the cathode positive with respect to the anode. For this forced commutation a

commutation circuit is required as shown later. In most commutation circuits a pre-charged

capacitor is momentarily connected across the conducting thyristor to reverse bias it. If the device

is reverse biased, its current falls, becomes zero at t3, then reverses and becomes zero at t4. At

instant t5, the device is capable of blocking a forward voltage. The time interval from t3 to t5 is

known as the turnoff time of the thyristor. If a forward voltage appears at instant t6 the time

4

interval t3 to t6 is known as the circuit turn-off time, tq. In practical applications, the turnoff time

tq provided to the SCR by the circuit, must be greater than the device turn-off time toff by a

suitable safety margin; otherwise the device will turn on at an undesired instant, a process known

as commutation failure.

Thyristors with large turnoff time 50-100microsec are called slow switching or phase control type

thyristors and those having small turnoff time (10-50microsec) are called fast switching or inverter

type thyristors.

Note that during thyristor turn on, if the voltage is high, current is low and vice versa. Therefore

the turn on switching loss is low. During thyristor turnoff also, if the reverse current is small, the

turnoff switching loss is low. The low switching loss in a thyristor is a significant advantage,

particularly for high frequency applications

1.2.2 TURN ON METHODS

With anode positive with respect to cathode a Thyristor can be turned on any one

of the following techniques a) Forward voltage triggering

b) Gate triggering

c) dv/dt triggering

d) Temperature triggering

e) Light triggering

Points a) and b) have already been dealt with

c) dv/dt triggering: - if the rate of rise of anode –cathode voltage is high the charging current of

the capacitive junctions j1,j2, and j3 may be sufficient to turn on the thyristors. However a high

value of charging current may damage the Thyristor and the device must be protected against high

dv/dt.

d) Temperature triggering: - During forward blocking most of the applied voltage appears across

the reverse biased junction J2. This voltage associated with leakage currents may raise the

temperature of this junction. With increase in temperature, leakage current through J2 further

increases. This cumulative process may turn on the scr at some high temperature

e) Light triggering: - A niche is made on the p-layer as shown below. When the niche is irradiated

by light, free charge carriers (holes and electrons) are generated just like when the gate signal is

applied between the gate and the cathode

1.2.3 Gate Characteristic

As an approximation, gate to cathode junction can be taken as a PN diode and its

V/I characteristic will approach that of a diode see figure below

5

This information can be used to sketch the gate characteristic and triggering region of a Thyristor.

Gate characteristic of Thyristor

For a particular type of SCR Vg and Ig characteristic has a spread between two curves, Upper and

Lower limit. Ox and Oy refer to minimum gate current and voltage to trigger the Thyristor. Vgm

and Igm are maximum permissible gate voltage and gate current. Oa non-triggering gate voltage.

The preferred gate drive area for an SCR is bcdefghb.

All spurious or noise signals should be less than the voltage oa . The design of a firing circuit can

be carried out with the help of figures below.

Considering fig a Es =Vg + IgRs. A resistance R1 is connected across gate –cathode terminals so

as to provide an easy path to the flow of leakage current between SCR terminals.

6

If Igmn and Vgmn are the minimum gate current and gate voltage to turn on SCR then from fig b

, current through R1 is Vgmn/R1.

Es =(Igmn + Vgmn/R1)Rs+Vgmn =IRs +Vgmn

Using gate circuit of higher magnitude can reduce the SCR turn on time. Gate drive requirements

can be obtained from the graph below:

AD is the load line

Curve 3 is for a thyristor whose Vg/Ig give the operating point S and Vg=PS and Ig =OP. To

minimize the turn on time and unreliable turn on, the operating points may vary between S1 and

S2 and must be close to the Pgav curve as possible. The gradient of the load line AD =OA/OD will

give the required gate source resistance Rs.

The minimum value of gate source series resistance is obtained by drawing a line AC tangent to

Pgav.

With pulse triggering, greater amount of gate power dissipation can be allowed. This should

however be less than peak instantaneous gate power dissipation Pgm.

Pulse gating

fT

PPPTfPP

T

TP gav

gmgavgmgav

gm ≥≥≥1

. where f=1/T1 =frequency of firing and T=

pulse width. In the limiting case Pgav/fT =Pgm .

A duty cycle is the ratio of ON time to periodic time fTT

T==

1

δ .

Example

Given the gate cathode characteristic of an SCR to be having a straight-line

slope 130. If the trigger source voltage is 15V and the allowable gate power

dissipation is 0.5watts. Compute the gate source resistance.

7

Solution

Vg Ig =0.5W …………..(1) and Vg / Ig =130Vg =130Ig…………..(2) Substituting 2 into 1

gives:

VmAxIVmAII gggg 06.862130130625.0130 2 ===∴==Now for the gate circuit Es =IgRs +Vg. 15 =62mA*Rs + 8.06 Rs =112ΩΩΩΩ

Example 2The trigger circuit of a Thyristor has a source voltage of 15V and the load line has a slope of –

120V/A. The minimum gate current to turn on the SCR is 25mA.

Determine :

a) Source resistance required in the gate circuit,

b) The trigger voltage and trigger current for an average gate power dissipation of 0-.4W

Solution a) The slope of the load line gives Rs =120ΩΩΩΩb) VgIg =0.4W and for the gate circuit Es =RsIg +Vg 15=120Ig +0.4/Ig Ig =38.6mA

or 86.4mA Vg =10.37V or 4.63V. Choose Vg =4.627V and Is =86.4mA.

1.2.4 Parallel and Series Operation of Thyristors

To accommodate high load currents a parallel connection of thyristors can be

used. If the simple connection is used fig (a) then differences in the individual

thyristors will result in an unequal current sharing of current between them. This

sharing can be evened out by careful selection of matched devices, by the use of

series resistance as in fig (b) or by including current sharing reactors fig(c).

8

Fig b is shown below:

Example

Two thyristors are connected as in fig a above and have forward characteristics when

conducting of the form:

Thyristor 1 v= 0.96 +2.52 *10-4

i1

Thyristor 2 v= 0.92 +2.40 *10-4

i2

i) What will be the current distribution between the two thyristors when the total current

is 1400A.

ii) What value of resistance added in series with each thyristor will result in the thyristors

being within 5% of each other for the same total current?

Solution

i) V= 0.96 +2.52 *10-4

i1=0.92 +2.40 *10-4

i2 and i1 + i2 = 1400

Hence 0.96 +2.52 *10-4

i1=0.92 +2.40 *10-4

(1400-i1)

Giving i1 =601.6A and i2 =798.4A

iii) Maximum current =1400A hence Maximum current difference =5%*1400=70A . We

noted from above i2 is greater than i1 hence

i2-i1 =70A and i1+12 =1400A

2i2 =1470 i2=735A and i1=665A

0.96 +2.52 *10-4

i1+Ri1 =0.92 +2.40 *10-4

i2 +Ri2

R=0.445mΩΩΩΩ

Transient or Dynamic Current Sharing

If current through T1 rises Ldi/dt across L1 increases and a corresponding voltage of opposite

polarity is induced across inductor L2. The result is a low impedance path through T2 and the

current is shifted to T2. The reverse is true

At turn on the thyristors gate circuits must all be driven from the same source to force a

simultaneous turn on of all devices. To prevent any individual thyristors from turning off if its

9

current falls below its holding current level , a continuous gate signal is normally used to ensure

immediate re-firing.

Where high voltage levels are encountered thyristors are connected in series to share the voltage.

If the connection shown below is used then the differences between devices can result in an

unequal voltage sharing between them.

R1 C and R2 C are for dynamic situation. R is used to try and equate the voltages sitting on the

thyristors and the determination of R is based on the following diagram and assumptions.

Assumptions:-

1) there are n thyristors 2) The value of R is a maximum to limit the current

3) Consider the case that T1 has a negligible leakage current and others have maximum

leakage current Ileak. This is the worst case and it indicates that the voltage V2 sits

across all thyristors except T1 .

4) The source voltage Vs is the maximum that can be applied when R is used.

5) The value of V1 across T1 is the peak voltage rating of the thyristors .That is the

maximum voltage T1 will support i.e T1 will support a larger voltage than the

remaining (n-1) thyristors which share voltage equally.

From Kirchhoff’s voltage law Vs =V1+V2(n-1)…………1 and from Kirchhoff’s current

law Is1 =Is2 +Ileak………………….2

Also from Ohm’s law Is1 =V1/R…….3 and Is2 =V2/R ………..4.

Eliminating V2, Is1 and Is2 we have from 1 and 4 Vs =V1 +V2(n-1) but V2 =Is2 RVs =V1 +Is2R (n-1).

From 2 Is2=Is1-Ileak Vs =V1 + (n-1)(Is1-Ileak)R

R =(nV1-Vs)/(n-1)Ileak

10

1.2.5 Thyristor Types

Depending on the physical construction and turn on and turn off behaviour

thyristors can be classified broadly into different categories:

1.2.5.1 A TRIAC (Bi-directional Triode Thyristor)

A TRIAC can be considered as an integration of two SCRs in inverse parallel. It can conduct in

both directions and is normally used in ac- phase control.

TRIAC

When terminal A1 is positive with respect to terminal A2 and the device is fired by a positive gate

current +ig, it turns on. Also when terminal A2 is positive with respect to terminal A1 and the

device is fired by a negative gate current –ig, the device turns on.

A TRIAC is frequently used in many low power applications such as juice mixers, blenders,

vacuum cleaners, etc. There is also what is called a DIAC similar to a TRIAC but does not have a

gate input.

1.2.5.2 A GTO (Gate Turn Off) Thyristor A single pulse of positive gate can turn ON this thyristor, but in addition a pulse of negative gate

can turn it off current. A gate current therefore controls both the On state and Off State operation

of the device. See fig below for the symbol and switching characteristics:

GTO Thyristor

The turn on process is the same as that of a Thyristor .The turn off process is somehow different.

When a negative voltage is applied across the gate and cathode terminals, the gate and cathode

terminals , the gate current ig rises. When the gate current reaches its maximum value Igr, the

anode current begins to fall and the voltage across the device Vak, begins to rise. The fall time

of Ia is abrupt typically less than 1µsec. Thereafter the anode current changes slowly, and this

portion of the anode current is known as the tail current

The ratio( Ia/Igr) of the anode current Ia (prior to turn off ) to the maximum negative gate current

Igr required to turn off a Thyristor is low typically between 3 and 5. For example , a 2500V, 100A

11

GTO typically requires a peak negative current of 250A to turn it off. GTOs that can operate up to

4500V, 2000 A, 5-10µsec are available. These are becoming increasingly popular in power control

equipment and GTOs will replace thyristors where forced commutation is necessary as in choppers

and inverters.

1.2.6 Power Transistors

A transistor is a three layer p-n-p or n-p-n semiconductor device having two junctions. This type of

a transistor is known as a bipolar junction transistor (BJT). The structure and the symbol of an n-p-

n transistor are shown below.

Transistor

The collector and the emitter terminals are connected to the main power circuit, and the base

terminal is connected to control signal. If the base current is zero, the transistor is in the off state

and behaves as an open switch. On the other hand if the base is driven hard , that is if the base

current Ib is sufficient to drive the transistor into saturation, then the transistor behaves as a closed

switch. The transistor is current driven device. The base current determines whether it is in the off

or On state. To keep the device in the on state there must be sufficient base current. Transistors

with high voltage and current ratings are known as the power transistors. The current gain Ic/Ib of

power transistor can be as low as 10, although it is higher than that of a GTO. Higher current gain

can be obtained by use of a Darlington Pair.

Power transistors switch on and off much faster than thyristors. They may switch on in less than

1µsec and turn off in less than 2µsec.. Hence the power transistors can be used in situations where

the frequency is as high as 100kHz. These devices however fail under certain high voltage and

high current conditions. They should be operated within specified limits known as the safe

operating areas (SOA). The SOA is partitioned into four regions defined by the following limits:

• Peak current limit (ab)

• Power dissipation limit (bc)

• Secondary breakdown limit (cd)

• Peak voltage limit (de)

Safe Operating Area (SOA)

12

If high voltage and high current occur simultaneously during turn off, a hot spot is formed and the

device fails by thermal runaway, a phenomenon known as second breakdown.

Figure below shows the effects of the snubber circuit on the turnoff characteristics of a power

transistor. A chopper circuit with an inductive load is considered.

If no snubber circuit is used and the base current is removed to turn off the transistor, the voltage across

the device Vce first rises, and when it reaches the dc supply voltage Vd, the collector current falls. The

power dissipation P during the turnoff interval is shown in the figure above by the dashed line. Note

that the peaks of Vce and Ic occur simultaneously, and this may lead to secondary breakdown

failure.

If the snubber circuit is used and base current is removed to turn off the transistor ,

the collector current is diverted to the capacitor. The collector current therefore,

decreases as the collector emitter voltage increases, avoiding the simultaneous

occurrence of peak voltage and peak current.

Power transistors of ratings as high as 1000V,500A are available

13

Linear approximations of Switching intervals for a purely Resistive load

The collector emitter voltage during switching on falls from Vs to almost zero and the expression for

the voltage is given by:

s

on

once V

t

ttV *

−= where t is the instantaneous time

the current rises and the expression is given by

on

mc

t

tIi

*=

Similarly during turn off the collector current and the collector emitter voltage is given by:

off

offm

c

on

sce

t

ttIi

t

tVV

)(&

* −==

With the collector emitter voltage and the collector current expressions above the power losses

during turn on and during turn off can be determined.

Peak loss occur at dp/dt =0. Power loss during switching on is given by:

ttt

tIV

t

IV

dt

dp

t

t

t

tIVP on

on

ms

on

mson

onon

mson 22

)(22

2

=−=−=

That is ton is the time at which instantaneous power would be maximum. The switch on and off loss

for a resistive load is given by:

=−==−=6

)1(*;6

*)1(0

offms

offoff

msoff

tonms

on

m

on

son

tIVdt

t

t

t

tIVW

tIVdt

t

tI

t

tVW

on

The average power loss due to switching is obtained by multiplying energy loss by the switching

frequency i.e power =energy/time =energy *frequency.

Example A power transistor has its switching waveforms as shown below. If the average power loss in the

transistor is limited to 300W, find the switching frequency at which the transistor can be

operated?

14

Energy loss during turn on and turn off is given by the following illustration:

sec1603.075

10**200*)

60

10*1(100

sec1067.0)40

10*1(*200*

50

10**100*

66

66

−=−=

−=−==

wattdttt

Woff

wattdttt

vdticWton

o

ton

oon

Therefore energy loss in one cycle =0.1603+0.1067 =0.267watt-sec =E

But the power loss in the transistor is limited to 300W

Hence frequency =f =300/0.267 =123.6Hz.

DIODE A diode is a two-layer p-n semiconductor device. The structure of a diode and its symbol are in

figure 7 below

From V/I characteristic it can be noted that if a reverse voltage is applied across the diode, it behaves

essentially as an open circuit. If a forward voltage is applied, it starts conducting and behaves

essentially as a closed switch. Following the end of a forward conduction in a diode, a reverse current

flows for a short time . The device does not attain its full blocking capability until the reverse current

ceases. The time interval during which reverse current flows is called reverse recovery time. This

reverse recovery characteristic is of great significance in high frequency applications.

1.2.7 Protection The reliable operation of a converter would require ensuring that at all times the circuit conditions do

not exceed the ratings of the power devices, by providing protection against overvoltage; overcurrent

and overheating. In practice the power devices are protected from:

• Thermal runaway by heat sinks

• High dv/dt and di/dt by snubbers

• Reverse recovery transients

• Supply and load side transients

• Faulty conditions by fuses

15

1.2.7.1 Cooling and heat sinks Due to on state and switching losses, heat is generated within the power device. This heat must be

transferred from the device to a cooling medium to maintain the operating junction temperature within

a specified range.

Although the heat transfer can be accompanied by conduction, convection, radiation or natural or

forced air, convection cooling is commonly used in industrial applications.

Cooling methods

6) Air coolers:- Air surrounding the device is heated through radiation and once heated

natural convection takes place.

7) Forced Cooling:- This includes forced convection. A fan is used to push air past a

heated surface. This causes the thermal resistance to decrease. The thermal resistance

decreases with air velocity. However above a certain velocity the reduction in

thermal resistance is not significant

8) Heat Sinks: - A wide variety of extruded aluminium heat sinks are commercially

available and use cooling fins to increase the heat transfer capability. The contact is

between the device and heat sink is extremely important to minimise the thermal

resistance between the case and sink. The surfaces should therefore be smooth and

free of dust/dirt, corrosion and surface oxides. Silicon grease is normally applied to

improve the heat transfer capability and to minimise the formation of oxides and

corrosion.

9) Heat pies:-These are pipes filled with low vapour pressure liquid. The liquid once

heated it is changed to vapour and goes to the fins and the vaporised liquid lose its

heat through the fins and quickly changes back to liquid and returns to the heat

source. This is heated again and the cycle continues.

10) Liquids:- in high power applications the devices are more effectively cooled by

liquids , normally water or oil. Water-cooling is very efficient and approximately

three times more effective than oil cooling. However it is necessary to use distilled

water to minimize corrosion and antifreeze to avoid freezing. Oil is flammable but

eliminates problems of freezing and corrosion.

Thermal resistance (Rθθθθ). Heat flow by conduction is very much alike the conduction of electrical charge i.e current

R

VVI

12 −= . Similarly power is the rate of flow of heat energy and is proportional to the difference

in temperature across the region through which the heat is conducted i.e:

θθ R

TT

R

TP 12 −

=∆

=

Thermal Model

16

Tj is the junction temperature

Rθjc:- thermal resistance from junction to case in oc/w.

Rθcs:- thermal resistance from case to sink in oc/w.

Rθsa:- thermal resistance from sink to ambient in oc/w.

Ts:- is the heat sink temperature

Ta:- is the ambient temperature.

From the expression

θθ R

TT

R

TP 12 −

=∆

= it can be noted that power is constant hence:

ja

as

cs

sc

jc

cj

R

TT

R

TT

R

TT

R

TT

R

TP

θθθθθ

−=

−=

−=

−=

∆= 12

But sacsjcjja

ja

aj

ath RRRRR

TTIVP θθθθ

θ

++=== −&* it is from this equation and data sheets for

heat sinks and thyristors that the type and size of heat sink is derived.

Example

The collector to base junction of a certain transistor dissipates 2W. The thermal resistance from the

junction to the case is 8 oc/w and the thermal resistance from the case to air is 20

oc/w. The free air

temperature is 25 oc. Determine:

a) the junction temperature

b) the case temperature

Solution

a) cT

T

R

TT

R

Tj

j

ja

aj081

208

252 =

+

−=

−=

∆=

θθ

b) Similarly P=2 Tc =65 oc.

2) di/dt protection

If the current in a thyristor rises at too high a rate, that is high di/dt, the device can be destroyed. Some

inductance must be present or inserted in series with the thyristor so that di/dt is below a safe limit

specified by the manufacturer. Typically di/dt limit values for SCRs are 20 500A/microsec

3) dv/dt Protection

If the entire anode to cathode forward voltage Va appears across J2, the charge on the junction is Q,

then the charging current is given by:

dt

dVaCj

dt

dCjVa

dt

dVaCjdtCjVad

dt

dqi =+=== /)( Since Cj is a constant. If dVa/dt is high,

the charging current will be more. This charging current will play the role of gate current and turns on

the SCR even when the gate signal is zero. As a result dv/dt must be kept low. Typical values are 20-

500V/microsec.

Power devices are protected from excessive di/dt and dv/dt by adding an inductor and snubber circuits

see fig below:

17

A snubber circuit consists of a series combination of resistance R and capacitor C in parallel with

thyristor. Before the thyristor is fired, C charges to full supply voltage Vs. Once on the capacitor

discharges through the thyristor hence the need for an R to limit the discharge of C.

Example

Figure below shows a thyristor controlling the power to the load of resistance RL. The supply voltage

is 240V dc and the specified limits for di/dt and dv/dt for the SCR are 50A/µsec and 300V/µsec

respectively. Determine the values of the inductance and the snubber circuit parameters R and C.

Solution:

When the switch is closed the capacitor behaves like a short circuit and the SCR is in forward blocking

state and offers a high resistance.

RLR

L

RLR

VsIwhereeIi

dt

LdiiRLRVs

t

+=

+=−−−=++=

−ττ &1).......1()(

Now di/dt from 1 gives:

ττ

τ

tt

eL

VsIe

dt

di−−

==1

* The value of di/dt is a maximum when t=0 that is at point of switching on.

H

dtdi

VsL

L

Vs

dt

diµ8.450/10*240 6 ==== −

.

The voltage across the SCR is given by

Ω===−== 6*///* RL

VsRdtdvLVsdtdibutdt

diR

dtdvaiRVa

The voltage across the capacitor is given by :

610*)300(6240

)(*)1( ≥≥=−=

−

C

dtdVcR

VsCe

CR

Vs

dt

dVceVsVc CR

t

CRt

1.2.7.2 Overvoltage Protection

Switching transients usually appear across the converter and these cause overvoltages. This effect of

overvoltages is usually minimised by use of snubber circuits, non linear resistors called voltage

clamping devices (VARISTORS); Metal oxide Varistors –MOVs as well as selenium diodes.

18

The VARISTORS, snubber circuits, and selenium diodes are connected in parallel with the converter.

The voltage clamping devices have falling resistance characteristic with increasing voltage . When a

voltage surge appears, the voltage-clamping device operates in the low resistance region and produces

a virtual short circuit across the power device limiting the amount of the transient voltage. After the

surge energy is dissipated in the non linear resistor , the operation of the voltage clamping device

returns to its high resistance region. Under normal operating conditions these devices draw very small

current.

1.2.7.3 Overcurrent Protection

a) The power converters may develop short circuits or faults and the resultant fault currents must

leared quickly. Fast acting fuses are normally used to protect the semiconductor devices. As

the faulty current increases the fuse opens and clears the fault current in milliseconds

The operation of the fast acting fuse is illustrated below

The current limiting fuse consists of one or more fine silver ribbons having very short fusing times. If

we assume that the a fault occurs at t=0, without the fuse the fault current would rise up to A and up to

or beyond D.

A properly selected fuse melts at A. For a brief moment or interval after A, the current continues to rise

depending upon the circuit parameters and fuse design. This current reaches the peak let through

current B, after B the arc resistance increases and fault current decreases. At point C arcing stops and

the fault current is cleared. The total arcing time is tc =tm + ta.

The current time curves of the devices and fuses may be used for the coordination of a fuse for a

device see figures below:

19

The fault is normally cleared in 10msec. The fuse must be rated to carry full load current plus marginal

overload current for an indefinite period, but the peak let through current of fuse must be less than the

surge current of the SCR.

If R is the resistance of the fault circuit and I is the instantaneous fault current between the instant of

fault occurring and the instant of arc extinction, the energy fed to the circuit is = dtRiWe 2.

If R remains constant , the value of i2t is proportional to the energy fed to the circuit. The i

2t value is

termed the let through energy and is responsible for melting the fuse. The fuse manufacturers specify

the i2t of characteristic of a fuse. In selecting the fuse it is necessary to estimate the fault current and

then to satisfy the following requirements:

These fuses must take account of :

a) the need to permit the continuous passage of the steady- state load current.

b) Permitted overload conditions including transients.

c) Prospective fault conditions

d) The I2t rating of the device –the fuse must clear the fault before the I

2t limit is reached

e) The fuse must be able to withstand the voltage after the arc extinction.

f) The peak arc voltage must be less than the peak voltage rating of the device

Crowbar Protection Circuit

A current sensing resistor R1 detects the value of the converter current. If it exceeds preset value , gate

circuit provides the signal to the craw bar SCR and turns it on. The input terminals are then short

circuited by the craw bar SCR and it shunts away the converter over current. Thus causing F1 to

breakdown and interrupting the fault current. Also R2 and R3 can be used for overvoltage protection. If

Vfb exceeds the set voltages, the SCR is also switched on eliminating the overvoltage.

20

Chapter 2

2.1 Diode Rectifiers

A rectifier is a circuit that converts ac signal to DC. Diodes are taken as ideal i.e Voltage across a diode

=0 .

2.1.1 Single phase half wave Rectifiers

Performance Parameters

As noted above the output voltage is discontinuous and contains harmonics. The power processing

quality of a rectifier requires the determination of harmonic contents of the input current, the output

voltage. The average output load voltage is = Vdc and load current is Idc.

Therefore the output dc power is Pdc =Vdc*Idc. The rms value of voltage and current is Vrms and

Irms. Therefore output ac power Pac =Vrms*Irms.

The efficiency or the rectification ratio of a rectifier is given by ηηηη =Pdc/Pac

The output voltage voltage can be considered to be composed of two components:-

1) the dc value

2) the ac component or ripple.

The effective rms value of the ac component of output voltage is determined from:

22222

dcrmsacdcacrms VVVVVV −=+=

The form factor, which is a measure of the shape of output voltage, is: FF =Vrms/Vdc.

The ripple factor, which is a measure of the ripple content is defined as

RF =Vac/Vdc 12

22

−=−

= FFVdc

VVRF

dcrms

Transformer utilization factor is defined as TUF =Pdc/Vs*Is where Vs and Is are the rms

voltage and rms current of the transformer secondary, respectively (VA of the transformer

secondary)

Let us consider the waveforms of figure below where Vs is the instantaneous, sinusoidal input voltage,

is is the instantaneous input current and is1 is its fundamental component. See figure below

21

If φ is the angle between the fundamental component of the input current and voltage it is called the

displacement angle. The displacement factor is defined as φCosDF = .

The harmonic factor ( a measure of the distortions of a waveform) is defined as

2

1

2

1

221

2

1

2

1

2

1

−=

−=

s

s

s

ss

I

I

I

IIHF where Is1 is the fundamental component of the input current Is.

Both Is1 and Is are expressed in rms. The input power factor PF is defined as

PF =Vs Is1Cosφ/ Vs Is =Is1 Cos φ/Is.

Crest Factor (CF) is a measure of the peak input current Ipeak as compared with its r.m.s value Is.

CF =Ipeak/Is This helps in specifying the peak current ratings of devices and components..

Note :

If the input current is is purely sinusoidal Is1 =Is and the power factor PF equals the displacement

factor DF. The displacement angle becomes the impedance angle θ= tan-(ωL/R ) for an RL load. 2. Displacement factor is often known as displacement power factor (DPF)

Example :-

The rectifier above has a purely resistive load of R. determine : a) the efficiency

b) The FF

c) The RF

d) The TUF

e) The PIV of diode

f) The CF of the input current

Solution

First we determine the average output voltage Vdc:

( )R

VR

VI

VtCosT

VdtTSinV

TV mdc

dcmm

T

mdc

318.01

2*

1 2

0===−

−== π

ωω

ω

and

R

V

R

VI

VdttSinV

T

IV mrms

rmsm

T

mrms

5.0

2*

21

2

0

22 ===

= ω

a) efficiency =Pdc/Pac =Vdc*Idc/Vrms*Irms =40.5% b) FF =Vrms/Vdc =0.5Vm/0.318Vm =157%

c) RF =121%

d) The transformer secondary rms voltage is :

R

V

R

VIV

VdttSinV

T

IV mrms

rmsmm

T

mrms

5.0707.0

2*

21

21

1

0

22 ====

= ω

Note the rms current comes in only when the diode is conducting. The Voltage Ampere of the transformer is VA =Vs*Is =0.707Vm *0.5Vm/R

Therefore TUF =Pdc/VsIs =0.3182/0.707*0.5 =0.286

d) The PIV =Vm

e) Ipeak =Vm/R and Is =0.5Vm/R therefore the CF =Ipeak/Is =2

Half wave rectifier with RL load

Due to inductive load , the conduction period of diode D1 will extend beyond 180 deg until the current

becomes zero at ωt =π +α:

22

The average output voltage is given by:

( )R

VICos

VtdtSinVV dc

dcm

mdc =++−== +

112

*2

1 1

0

αππ

ωωπ

απ

It can be noted that the average voltage and current can be increased by making α =0, which is

possible by adding a freewheeling diode Dm.

If the output of the circuit is connected to the battery the rectifier becomes a battery charger seefig

below:

For Vs >E diode D1 conducts ,from α to β. The angle α can be determined from VmSinα =E

α =Sin-1

(E/Vm) and diode D1 is turned off when Vs< E at β.=π -α. . And the charging current is :

io =(Vs-E)/R =(Vm Sinω t –E)/R for α <ω t< β.

Example

The battery voltage in figure above is E=12V and its capacity is 100Wh. The average charging current

shuld be Idc =5A.The primary input voltage is Vp 120V, 60Hz and the transformer has a turns ratio of

n=2 calculate:-

a) Conduction angle of the diode

b) The current limiting resistance R

c) The power rating of R

d) The charging time in hours (ho)

e) The rectifier efficiency

f) The PIV of the diode

Solution

E=12V;Vp =120V; Vs =Vp/2 =120/2 =60V and Vm =60* 20.5= 84.85V

a) α =Sin-1

(12/84.85) =8.13 deg and β.=π -α. =180-8.13 =171.87 therefore the Conduction

angle = β. -α. =171.87-8.13 =163.74 deg.

b) The average charging current Idc is −

=β

αω

ωπ

tdR

EtSinVIdc m *

2

1 remember

β.=π -α. R =4.26Ω

c) The rms battery current is : tdR

EtSinVI m

rms ωω

π

β

α−

=2

22 )(

2

1I=8.2A The power

rating of R is Pr =8.2*8.2*4.26 =286.4W

23

d) The power delivered Pdc to the battery is Pdc =E*Idc =12*5 =60W. Now the battery

capacity =100Wh =Power * Time; charging time ho =100Wh/Powerho =100Wh/60W

=1.67hrs.

e) The rectifier efficiency = %3.174.28660

60=

+=

+=

Rdc

dc

PP

Pη

f) The peak inverse voltage of the diode =PIV =Vm +E =84.85 +12 =96.85V

Single-phase full wave Rectifiers

A full wave rectifier circuit with a center tapped transformer circuit is shown in fig above. The average output voltage is:

VV

tdtSinVT

Vdc mT

m 637.022 2

0=== πω

Instead of using center tapped transformer four diodes can be used as shown below:

The peak inverse voltage of a diode is Vm. This is a bridge rectifier and it is commonly used in

industrial applications.

Three phase Bridge Rectifiers

A three-phase bridge rectifier is commonly used in high power applications see the circuit

diagram below:

24

This can operate with/out a transformer and gives six pulses/ ripples on the output voltage. Each diode conducts 120 deg. The conduction sequence for diodes is D1-

D2; D3-D2; D3-D4;D4-D5; D5-D6 and D1-D6. The pair of diodes having the

highest of amount of instantaneous line-to-line voltage will conduct.

If Vm is the peak value of the phase voltage then the instantaneous voltages are

given by:: ;tSinVVan m ω= )120;( −= tSinVVbn m ω );240( −= tSinVVcn m ω

CONTROLLED RECTIFIERS A controlled rectifier converts ac power to dc power. Controlling the instants at which the

semiconductor devices switch can control the output voltage and power.

Thus controlled rectifiers can be used to control the speed of a DC motor. Some controlled rectifiers

can convert DC to ac and that is Inversion.

Thyristor Rectifier, Resistive loadA simple thyristor rectifier circuit consisting of a single thyristor and a resistance load is shown in

below.

Fig Thyristor rectifier with resistive load circuit and waveforms

The thyristor is forward biased during the intervals 0<ωt <π, 2π<ωt<3π, etc. A

gate pulse is applied at an angle α. This angle is known as the firing angle of the

thyristor. The thyristor current becomes zero at ωt =π,3π, etc, and the thyristor

conducts from α to π,2π +α, to 3π, etc. During the interval when the thyristor

25

conducts, known as conduction interval, the load voltage is the same as the supply

voltage, Vo =V and the average voltage is determined below:

We noted for a single phase diode rectifier circuit consisting of a resistive load, is:

tSinVv p ω2= Then the average value of the load is given by

πωωπ

πp

po

VtdtSinVv

22

2

1

0

== Thyristor Rectifier, Resistive Load

The thyristor is forward biased during the interval 0-180 and 360-540 degrees.

The thyristor is fired at an angle α and conduct for a conduction angle β. The average value of the load

is :

)1(2

);1(2

)(2

10 απαπωω

π

π

α

CosR

VICos

VttdSinVVo

Lo

mm +=+==

The load voltage will vary with the variation of αααα and becomes maximum at αααα=0 and zero at αααα=ππππ.

Vdc max=Vm /ππππ and occurs at αααα=0, Idc max = Vm /ππππR = Im /ππππ.

The rms load voltage for a given firing angle αααα is:

212

1

22 )2

2sin(

1

2)(

2

1

+−=

=

ααπ

πωω

π

π

α

mmrms

VtdtSinVV

And the rms load current is given by the expression:

[ ]2

;)2

2(12

max

21

mrms

L

m

L

rmsrms

VVSin

R

V

RV

I =+−== ααππ

The average power P in the load for any trigger angle αααα is given by P =Vdc*Idc.

Example1

A 100Ω load is connected to a peak supply of 300V through a controlled half wave thyristor rectifier.

The load average power is to be varied from 25W to 80W. What is the angular firing control required?

Neglect forward drop of the thyristor.

26

Solution Power =Vdc*Idc=Vo*Io

And )1(2

);1(2

)(2

1απαπωω

π

π

α

CosR

VICos

VttdSinVVo

L

mo

mm +=+==

Power =P =22

2

22

2

2

)1(5.22)1(100*4

300)1(

4αα

πα

πCosCosCos

R

V

L

m +=+=+

When P=80W then the firing angle can be determined by the following relationship:

8857.0556.31)1(556.3)1(5.2280 22 ==++=+= αααα CosCosCosCos

Therefore αααα =27.660 .At a power of 25W the firing angle is 87,3

0 .

Thyristor half Wave Rectifier, Reactive Load

Most practical Loads have resistance and inductance L, e.g. the armature of a dc motor load has

resistance (R) and inductance (L). When a thyristor is fired at αααα the inductance in the load forces the

current to lag the voltage and decays to zero at ββββ. The waveform of the load current and voltage is

shown below:

From αααα to ππππ the load current is driven by the supply source and from ππππ to ββββ the load voltage is

negative and the current is maintained by the voltage induced in the inductance

Half wave Controlled Rectifier With RL load and a free Wheeling Diode

The diode D bypasses the load current during the intervals of negative transformer voltage and this

prevents the output voltage from assuming a negative value.

• The diode prevents the voltage across the load from reversing during the negative half cycle of

the supply voltage.

• The load average voltage would be determined only by the positive half cycle wave of the

transformer secondary emf from αααα to ππππ .

27

The average voltage is )1(2

)1(22

2α

πα

πωω

π

π

α

CosR

VICos

VttdSin

VV

l

m

dc

m

dc +=+==

Thyristor Full Converter

Half wave rectification contains a significant amount of ripples. Therefore this is not suitable for speed

control of dc motors.

A full converter circuit shown in fig below is used for the speed control of DC motors.

The load current io has the same waveform as the load voltage vo.. The average load voltage is given by:

)1();1()(2

2απαπωω

π

π

α

CosR

VICos

VttdSinVVo

L

mo

mm +=+==

The rms voltage is given by the following expression:

212

1

22 )2

2sin(

2

1)(

1

+−=

=

ααπ

πωω

π

π

αmmrms VtdtSinVV

and the rms current is

[ ] 21

)2

2(2

1 ααππSin

R

V

RV

IL

m

L

rmsrms +−==

DC Motor Load / RL load

When terminal A1 is more positive as compared to A2 T1 and T2 are fired simultaneously and current

flows through T1 load and T2. In the reverse half cycle T3 and T4 are fired and the reverse voltage

applied across them commutates T1 and T2. Due to inductive load, T1 and T2 will conduct beyond ππππ,even though the input voltage is already negative see the waveforms above.

Depending on the size of the inductance the conduction can be discontinuous or continuous as shown

above. If XL>>R then the load current will be continuous.

28

Discontinuous load conduction

The average voltage is given by )(2

2βα

πωω

π

β

α

CosCosV

ttdSinV

V mm

dc −== The conduction angle of the thyristors is given by θθθθ =ββββ-αααα, hence ββββ=θθθθ +αααα

Now

)2

(2

22

*2

2θ

αθβαβα

βα +=−+

−=− SinSinSinSinCosCos

Therefore the average voltage for discontinuous load conductions is

l

dc

dc

m

dcR

VISinSin

VV =+= )

2(

2

2 θα

θπ

Continuous load Conduction

If we assume that sufficient inductance is present in the dc armature circuit to ensure that the motor

current is continuous (present all the time).

If conduction is continuous the limits are from αααα to (ππππ+αααα) and the average voltage is given by:

l

m

l

dc

dc

mm

dcR

CosV

R

VICos

VttdSin

VV

πα

απ

ωωπ

απ

α

22

2

2====

+

The rms voltage is determined from the following relationship:

21

22

=

+απ

α

ωωπ

ttdSinV

V mrms

Depending on the value of αααα the average output voltage could be either positive or negative and it

provides two-quadrant operation.

For α=450 it can be noted that T1 and T2 conduct the motor current during the interval α< ωt < (π+α)

and connects the motor to the supply v=Vs . At (π+α), T3 and T4 are fired. The supply voltage appears

immediately across thyristorsT1 and T2 as reverse bias voltage and turns them off. This is called Line

Commutation. The motor current , ia which was flowing from the supply through T1 and T2 is

29

transferred to T3 and T4. T3 and T4 conduct the motor current during the interval (π+α)<ωt < (2π+α)

and connects the motor to the supply (vo =-V)

If conduction is continuous the limits are from αααα to (ππππ+αααα) and the average voltage is given by:

l

m

l

dc

dc

mm

dcR

CosV

R

VICos

VttdSin

VV

πα

απ

ωωπ

απ

α

22

2

2====

+

The inductance (La) does not sustain any average voltage. Therefore Vdc =Vo =IoRa +Ea Where Io is the average motor current; Ea is the armature back emf, which is constant if the speed and

field current are constant

The variation of the motor terminal voltage Vo as a function of the angle α based on the equation

Vdc. For firing angles in the range 0o< α <90

0 , the average output is positive. Since the current can

flow in one direction in the load circuit because of the thyristors , the power VoIo is positive ;power is

flowing from the supply to the dc machine , and the dc machine operates as a motor. For firing angles

90o< α <180

0 ,the output voltage is negative and therefore the power VoIo is negative; that is power

flow is from the machine to the ac supply. This is known as inversion Operation of the converter, and

this mode of operation is used for the regenerative braking of the motor. Note that for inversion

operation, the polarity of the motor back emf Ea must be negative. It can be reversed by reversing the field current if so that the dc machine behaves as dc generator.

Converter output characteristics for continuous load current

Two full converters can be connected back to back as shown below . This arrangement is known as the

dual–Converter connection. If one converter is used, it causes motor current to flow in one direction. If

the other converter is used, the motor current reverses and so does the speed

Dual Converter

Example .

A full wave fully controlled single phase bridge rectifier feeds a load resistance R and inductance L

from an AC source of Vrms voltage Vs.

Assuming the operation is continuous:

30

a) show that the mean or average voltage value of the DC output voltage Vo is given by

αCosVV so 90.0= .

If R=4Ω,L500mH, αααα=300 ,Vs =240V and the supply frequency is 50Hz.

b) Calculate:

i) the average and the rms currents output

ii) the thyristor average and rms currents and

iii) the power factor of the ac input to the rectifier

Solution

a) the mean value of the dc output for continuous operation is

ααπ

απ

ωωπ

απ

α

CosVCosV

CosV

ttdSinV

V s

smm

dc 90.0*222

2

2====

+

b) (i) Xl=2πfL=2*π*50*0.5H =157Ω This implies that the load current is going to be

continuous. Therefore

AiVCosCosVV dcsdc 8.464

1.1871.18730*240*90.090.0 ===== α .

The current is almost constant since the ripple is almost negligible because of high circuit

inductance. Thus this current is both the average and the rms value i.e io =irms =46.78A

(ii) The thyristor average current is

Ai

tdiI otha 4.232

78.46)(*

22

1 0 ==−+== +

ααππ

ωπ

απ

α

the rms thyristor current is:

Ai

tdiI oothrms 1.33)(

22

12

122

1

2 =

−+=

=

+

ααππ

ωπ

απ

σ

iii) The output power to the load is :

WRiP oo 48.87534*78.46* 22 === This is also the input power to the converter. The VA

drawn from the supply is given by :

8.02.11227

48.87532.11227240*78.46 ==== PFWVI ss

Thyristor Semiconverter

A thyristor semiconverter that can be used for the speed control of a dc motor consists of thyristors

and diodes. See fig below:

31

Thyristor Semiconverter

Thyristors T1 and T2 are fired at αααα and (2ππππ+αααα), respectively. See the waveforms. The motor is

connected to the input supply for the period αααα<ωωωωt<ππππ through T1 and D2, and the load voltage is

the same as the input voltage. Beyond ππππ, Vo tends to reverse as the input voltage changes

polarity. As Vo tends to reverse , the diode Dfw (known as the free wheeling diode ) becomes

forward biased and starts to conduct. The motor current which was flowing from the supply

through T1, is transferred to Dfw, (i.e T1 commutates). The output terminals are shorted through

the freewheeling diode during the interval (ππππ<ωωωωt<ππππ+αααα) making Vo =0. At ωωωωt=ππππ+αααα , T2 is fired

and it take over the motor current ia from Dfw. The load current now flows through T2 and D1.At

ωωωωt=2ππππ, Dfw, becomes forward biased again and takes over the current from T2, and the process

continues. See the waveforms. Note that if Dfw is not used, freewheeling action will take place

through T1 and D2 during the interval (2ππππ<ωωωωt<2ππππ+αααα).

The average value of the output voltage is:

)1(*2

)()**2(1

απ

ωωπ

CosV

tdtSinVVo

+=

=

Note; Vo is always positive and therefore Vo*Io is always positive that is power flow is from the

ac supply to the dc load. Semiconverters do not invert power. However they are cheaper than full

converters.

Waveforms

Example

A single –phase full converter is used to control the speed of a 5hp, 110V, 1200rpm, separately

excited dc motor. The converter is connected to a single phase 120V,60Hz supply. The armature

resistance is Ra =0.4Ω and armature circuit inductance is La=5mH. The motor voltage constant Kφ=0.09V/rpm.

1. Rectifier(or motoring) operation. The dc machine operates as a motor runs at 1000rpm, and

carries an armature current of 30A. Assume that the motor current is ripple free.

32

a) Determine the firing angle αb) Determine the power to the motor

c) Determine the supply power factor

2. Inverter Operation (Regenerating Action). Reversing the field excitation reverses the polarity of

the motor back emf Ea.

a) Determine the firing angle to keep the motor at 30A when speed is 100rpm

b) Determine the power fed back to the supply at 1000rpm.

Solution

a) Ea =0.09*1000 =90V; Vo =Ea +IoRa =90+30 =102V,

02.19120*22

102 == ααπ

Cos

b) P =Io2*Ra + EaIo =VoIo =102*30 =3060W

c)The supply current has a square waveform with amplitude 30A (=Io). The rms supply current is

I=30A .

The supply Volt-Amperes are 120*30A =3600VA

If the losses in the converter are neglected, the power from the supply is the same as the power to the

motor Ps =3060W

Thus the supply power factor is PF =Ps/S=3060/3600 =0.85

2. (a) At the time of polarity reversal the back emf is Ea =90V and from Vo =Ea +IoRa =-90+30*0.4

=-90+12 =-78V

Now 02.13678*

120*22=−== αα

πVCosVo

b) Power from the dc machine Pdc =90*30 =2700W

Power lost in Ra =302 *0.4 =360W

Power fed back to the ac supply =Ps =2700-360 =2340W

Three Phase Circuits For higher power applications, several kilowatts three phase circuits are preferable. The magnitude

of harmonic voltage is lower in three phase circuits than in single-phase circuits. This is due to

increasing number of pulses.

Half Wave Controlled Rectifier With Resistive load

33

)120(**2)240(**2

)120(**2

**2

00

0

+=−=

−=

=

tSinVtSinVVcn

tSinVVbn

tSinVVan

ωω

ω

ω

If thyristor T1 is fired at ωt =300 ,T2 is fired at ωt =1500 , and T3 ωt =2700 that is at the crossing

points of the phase voltages. Firing at these instants will also result in maximum output voltage.

The reference for the firing angle α is therefore the crossing point of the phase voltages. The firing

of the thyristors can be delayed from these crossing points. In other words the firing angle is

measured from the crossing points of the phase voltages. Recall that in a single-phase converter,

the firing angle was measured from the zero crossing of the input supply voltage.

For the above waveform thyristor T1 is fired at ωt =300 + α, and the output voltage is Vo=Van.

The output voltage io =Vo/R, and becomes zero at ωt =π. Thyristor T1 turns off at this instant

through natural commutation. Thyristor T2 is fired atωt =1500 + α, making Vo= Vbn. T2 turns off

at ωt =3000 .T3 is fired at

ωt =2700 + α. Making Vo =Vcn.

Example The load in the fig above now consists of a resistance and a very large inductance. The inductance

is so large that the output current io can be assumed to be continuous and ripple- free. For α=600

.

a) Draw the waveforms of vo and io.

b) Determine the average value of the output voltage if phase voltage Vp =120V

34

Solution

b) The average voltage Vdc is given by the following relationship:

VCosVVVV

CosV

ttdSinVttdSinVV

dcmp

m

mmdc

2.70)60(*2

120*2*33120*2120

*2

33*

2

3*

2

3

0

12030

30

65

6

====∴

=== ++

+

+

+

π

απ

ωωπ

ωωπ

α

α

απ

απ

Thyristor Full Converter

This consists of six thyristor switches as shown in figure below and this is the most commonly

used controlled rectifier circuit in industrial applications up to 120kW.

35

Thyristors T1 ,T3 and T5 are fired during the positive half cycle of the voltages of the phases to

which they are connected, and thyristors T2,T4,T6 are fired during the negative cycle of the phase

voltages. The references for the firing angles are the crossing points of the phase voltages. See

waveforms below:

The times at which the thyristors fire are marked in the figure above for α=300.

Assuming that the output current io ( i.e the dc motor current ) is continuous and ripple free as

shown above.

• At ωt= 300 +α =π/6+ α T1 turns ON. Prior to this T6 was ON.∴ during interval (π/6+

α< ) ω t < (π/6+ α+π/3), T1 and T6 conduct the output current and the motor terminals are connected to phase A and B making the output voltage vo =Van –Vbn. The output voltage vo is the distance between the envelopes of the phase voltages Van and Vbn as

shown by the arrows in the waveforms above.Note Van is positive and Vbn is negative

hence Van –Vbn becomes Van +Vab. Under phase A there is T1 and T4; while under

36

B there is T3 and T6 and under C there is T5 and T2. this can be noted also on each

compltete cycle for each phase

• At ω t = (π/6+ α+π/3), T2 is fired and immediately voltage Vcb appears T6, which

reverse-biases it and turns it off (line commutation. The current from T6 is transferred to S2. The motor terminals are connected to phase A through T1 and phase C through T2,

making Vo =Vac. This process repeats after every 600 whenever a thyristor is fired. Note

that the thyristors are numbered in a sequence in which they are fired.

• Each thyristor conducts for 1200 in a cycle. The thyristor current iT1 is shown in the

graph. The line current such as iA, is a quasi square wave having a pulse width of 1200 as

shown above.

• The conduction pattern or firing sequence for the above circuit is

• Remember for a three phase star connected system Iline =Iphase and Vline =√3*Vphase

and Ineutral =IR +IY +IB.

• Remember for a three phase Delta connected system Iline =√3*Iphase and Vline

=Vphase.

The average value of the output voltage is: T1T6; T1T2; T2T3; T3T4; T4T5; T5T6; T6T1etc.

The average value of the output voltage is:

απ

απ

ωπ

παπ

απ

CosV

CosVtdVbnVanV mso

3363)(*)(

2

6 36

6

==−= ++

+

The average output voltage varies with the firing angle α, and this variation is shown in fig

below:

For α varying in the range 0 < α < 900 , Vo is positive and power flow is from the ac supply to

the DC motor. For 900 < α < 1800 ,Vo is negative and the converter operates in the

inversion mode. The power can be transferred from the dc motor to the ac supply a process known as regeneration.

37

Thyristor Semiconverter

The 3-phase semiconverter consists of three thyristors and three diodes as shown in fig

below:

Voltages and current waveforms are shown in fig below for a firing angle of α =900 . The instants of firing a thyristor and the duration of conduction of the diodes are shown in the

figure below showing the waveforms.

• We will assume that the output current is continuous and ripple free. At ωt =π/6+ α, T1

turns ON and T1 and D3 conduct the output current io, making Vo=Vac. At ωt= 2100 ,Vo is

zero, and from this instant onwards Vo tends to be negative. The diode D1 will become

forward –biased (and start conducting). The output current io will freewheel through T1 and

D1, making Vo =0.

• When T2 is triggered on, the output current io will conduct through T2 and and D1 making Vo =Vba.

• The process repeats every 1200 whenever a thyristor is fired.

• Note that the line current iA starts at ωt =π/6+ α and terminates at ωt= 2100.

38

The average value of the output voltage is:

)1(2

33*)120(

2

3

2

3 06

6

6

6

απ

ωωωπ

ωπ

ππ

απ

ππ

απ

CosV

tdtSinVtSinVtdVacV mmmo +=+−==

+

+

+

+

Note that the output voltage cannot reverse. Hence this converter does not operate in the inversion

mode.

Dual Coverters

Two full converters can be connected back to back to form a dual converter just as in the single-phase

situation and both the voltage Vo and current Io can reverse in a dual converter.

Example

A 3phase full converter is used to control the speed of a 100hp, 600V,1800rpm, separately excited

motor. The converter is operated from a star connected 3phase, 480V, 60Hz supply. The motor

parameters are Ra =0.1Ω, La =5mH ,Kφ =0.3V/rpm (Ea=Kφn). The rated armature current is 130A

1. The rectifier (or motoring) operation:- The machine operates as a motor, draws rated current,

and runs at 1500rpm. Assume that the motor current is ripple free

a) Determine the firing angle.

b) Determine the supply power factor

2. Inverter operation :- The dc machine is operated in the regenerative braking mode. At

1000rpm and rated motor current

a) Determine the firing angle.

b) Determine the power fed back to the supply and the supply power factor

Solution

a) Phase voltage is

VRIEVVEVV aoaoap 4631.0*1304504501500*3.02773

480=+=+=∴====

but we know :

04.44*3*277*23

46346333

==== ααπ

απ

CosVCosV

V mo

b) Since the ripple in the motor current is neglected, the supply current iA is a square wave of

magnitude 130A and width 1200 . The rms value of the supply current is

Atdii orms 1.106130*3

2

3

2*130*

11 21

2

21

32

0

2 ==

=

=

ππ

ωπ

π

The supply voltage amperes are S =3Via =3*277*106.1 =88169.1VA

Assuming no losses in the converter, the power from the supply Ps is the same as the power

input to the motor. Hence

Ps =Vo*Io =463*130 =60190W

Therefore the supply power factor is PF =Ps/S =60190/88169.1 =0.68

2. (a) Ea =0.3*1000 = 300V

For inversion the polarity of Ea is reversed

Vo =Ea +IoRa

=-300 +130*0.1

= -287V

03.116*3*277*23

28728733

==−−== ααπ

απ

CosVCosV

V mo

b) Power from the DC machine (operating as a generator): Pdc =300*130 =39000W

Power lost in Ra: Pr =1302 * 0.1 =1690W

39

Power to the source: Ps =39000 –1690 =37310 W

Supply Volt amperes S = 88169.1V and Supply power factor =PF =37310/88169.1 =0.423

Assignment 2

Question 1

A three phase half wave converter in fig below is operated from a three phase Y connected

415V, 50Hz supply and the load resistance is R =10ΩΩΩΩ. If it is required to obtain an average

output voltage of 50% of the maximum possible output voltage. Calculate:

a) the delay angle αααα, [4]

b) the rms and average output currents [4]

c) the average and rms thyristor currents [4]

d) the rectification efficiency [4]

e) the input power factor [4]

Question 2

A three phase full wave converter is operated from a three phase Y connected 415V, 50Hz

supply and the load is highly inductive with resistance R =10ΩΩΩΩ. If this converter is required to

obtain an average output voltage of 50% of the maximum possible output voltage. Sketch the

circuit diagram. [2]

Calculate:

a) the delay angle αααα, [3]

b) the rms and average output currents [4]

c) the average and rms thyristor currents [4]

d) the rectification efficiency [4]

e) the input power factor [3]

Question 3

A full wave fully controlled single-phase bridge rectifier feeds a load resistance and inductance

L from an AC source of rms voltage Vs. A freewheeling diode is fitted across the load.

a) Show that the output voltage is given by 2

)1(9.0 αCosVV s

o

+= [4]

Given that R =4ΩΩΩΩ, L=500mH, αααα=30 ,Vs =240V and the supply frequency to be 50Hz,

Calculate:

b) The average and rms output currents [4]

c) The thyristor average and rms currents [4]

d) The diode average and rms currents [4]

e) The power factor of the ac input to the rectifier [4]

40

AC Voltage Controllers

AC voltage controllers

These convert a fixed –voltage ac supply into a variable voltage ac supply. They can be used to control

the speed of induction motors.

Single phase AC Voltage Controllers

The waveforms are as shown below:

T1 is fired at α and T2 is fired at π+α. If T1 is fire the current builds up at α and decays to zero at β.

When T2 turns on at π+α a negative current flows in the load.

During conduction interval say for T1:

tVSindt

diLiRV oo ω*2* =+=

The load current is:

τφωt

transientesteadystato AetSinZ

Viii

−+−=+= )(

2……..1

where :

RL

RLLRZ ==+= − τωφω ;tan;)( 122

At ωt =α, io =0 then:

41

αωφα)(

)(2

LR

eSinZ

VA −−= ……..2

From equations 1 and 2

])()([2 ))(( t

LR

o eSintSinZ

Vi

ωαωφαφω−

−−−= ……….3

Letting α=φ from 3:

)(2

φω −= tSinZ

Vio ………………4

From equation 4, the load current io is sinusoidal, which indicates that if the firing angle is the same as

the impedance angle i.e., α=φ, the load current becomes purely sinusoidal. Each thyristor conducts for 180

0 and the full supply appears across the load. Fig below shows the

waveforms of load current for two different firing angles.

For α > φ, io is non-sinusoidal, and for α=φ, io is sinusoidal. Even for α<φ will be sinusoidal.

To determine β, when the current io falls to zero and T1 is turned off, can be found from the condition

io(ωt=β)=0 and this gives a relation :

))(()()(

βαωφαφβ−

−=− LR

eSinSin

The angle β, which is also known as the extinction angle, can be determined from this equation by way

of an iterative method of solution. Once β is known then the conduction angle of T1 δ can then be

determined from δδδδ=ββββ-αααα.

The rms output voltage is :

21

21

22

02

2

2

21)(*2

2

2

−+−=

=

βααβ

πωω

π

β

α

SinSinVtdtSinVV Where V is the

rms supply voltage.

The rms thyristor current is obtained from:

21

2

0 )(2

1

=

β

α

ωπ

tdiI rth and the rms output current can then be found by combining the rms current

of each thyristor as ( ) rthrthrth IIII *221

22

0 =+= and the average thyristor current can be

determined from

=

β

α

ωπ

)(2

10 tdiI a

Example

A single-phase full wave controller supplies an RL load. The input rms voltage is V =120V, 60Hz. The

load is such that L=6.5mH and R=2.5Ω. The delay angle of thyristors are equal:α1=α2=π/2. Determine

a) the conduction angle of thyristor T1,δ;

42

b) the rms output voltage Vo,

c) the rms thyristor current Irth;

d) the rms output current Io;

e) the average current of a thyristor Ia; and

f) the input power factor.

Solution:

R=2.5Ω, L=6.5mH, f=60Hz, ω=2π*60 =377rad/s, V=120V, α=π/2. and

φ=tan-1

(ωL/R)=44.430.

a) an iteration solution yields 220.350 hence the conduction angle =δ =220.350-

900=130.430

b) Vo =68.09V

c) Integrating from α toβ gives Irth =15.07A

d) Io =21.3A

e) Ia=8.23A

f) Output power =21.32 2.5=1134.2W and the input VA is VA =120*21.3=2556W Therefore PF

=1134.2/2556 =0.444lag

Three Phase AC voltage Controllers

For higher power loads such as the induction motors driving fans, or pumps, three phase controllers are

used. Figure below shows the two types of three phase ac voltage controllers.

In one circuit, the thyristors switches are in the lines and the load can be connected either in Star or

Delta. While in the other circuit the thyristors are connected in series with the phase loads to form a

delta connection

The firing sequence for the Y connected will be T1;T2;T3;T4;T5;T6 or see the waveforms below:

43

The operation of the delta-connected controller can be studied on a per phase basis because each phase

is connected across a known supply voltage.

Example:

A three phase ac voltage controller is used to start and control the speed of a 3 phase 100Hp ,460V,

four pole induction motor driving a centrifugal pump. At full –load output the power factor of the

motor is 0.85 and the efficiency is 80%. The motor current is sinusoidal. The controller and the motor

are connected in delta as shown above.

a) determine the rms current rating of the thyristors

b) determine the peak voltage rating of the thyristor

c) determine the control range of the firing angle α

Solution

a) output power is given SPFPo **η=S=100*0.746/0.8*0.85 =109.71kVA

We know VA=3Vph*Iph Iph =109.71kVA/3*460 =79.5A (remember Vph=VL for a delta connected

load and phL II 3= )

∴thyristor rms current =79.5/(2)1/2

= 56.22A

b) Peak voltage across the thyristor =(2)1/2

*460=650.4V

c) the control range of the firing angle is determined from the fact that :

01 8.3185.0 === −CosCosPF φφ the control range is 31.8

0< α

<1800.

Thyristor Commutation

A thyristor needs a commutation circuit whenever a dc source supplies power to

be modulated. For example when a thyristor is being used on inverters and

Choppers once ON it will remain being on until a positive voltage is applied on its

cathode that is when it can be turned off and this is termed forced commutation.

There is natural commutation which is to do with the switching off of the thyristor

by virtue of current falling below its holding current and this happens naturally as

in the case of ac supply. See fig below:

44

At 90 the thyristor switches off naturally and awaits another firing signal for the

next positive pulse

a) Parallel Capacitor turn off Circuit

Vs and load Rl comprise the main circuit. T1 modulates the power in the load.

C,R and T2 are assumed to be off . If T1 is switched on, so that there is load

current i at the same time the capacitor charges up to +Vs on plate Y through R,

C and T1. Plate X of capacitor is virtually at ground potential because T1 is in

the zero impedance state. This sets the stage for commutation.

When it is desired to interrupt the load current, T2 is turned ON. This puts the

capacitor C in parallel with T1 and the reverse voltage sitting across C reverse

biases T1. If T1 is reverse biased long enough, it will turn off. Capacitor C

discharges and charges, this time becoming positive at X through Rl, C and T2.

At some particular time T1 is turned ON again, now the voltage across C reverse

45

biases T2 to turn it off. The capacitor C becomes positively charged at plate Y so

that the cycle can be repeated. By controlling the times that T1 and T2 are

allowed to conduct the average power in the load can be adjusted. To be able to

do this the value of the capacitance must be determined.

During Turn Off:

CRt

L

LeR

Vsti

−

=2

)( current through the circuit. 2Vs comes from the supply and the

capacitor.

The voltage across the thyristor T1 is : )21(1

CRt

sLsTLeViRVV

−

−=−=The time t1 for the voltage VT1 to rise to zero is :

)21(1

CRt

sLsTLeViRVV

−

−=−= = 0 t1 =0.69R L C and this time must be larger

than the turn off time of the thyristor T1.

Turn off time of the device is :

L

offLoff R

tCCRt

69.069.0 ≥≤

If T1 and T2 have the same turn off times and if R≥ RL then the above value of C

will allow satisfactory commutation.

If the capacitor is to charge fully after each switching action, it will take at least 5τ(5RL C) after T2 has been triggered for this to be accomplished.

)(511)(5)55(

LLL RRT

CRRRCCRT +=+=+≥

Example

Consider a case of parallel capacitance turn off as shown above .The load

resistance RL =5Ω and the supply voltage is Vs =120V. Calculate the minimum

value of the capacitor C if the manufacturer’s specified turn off time for the

thyristor is 15µs. What is a suitable value of the resistor R, if the thyristor T1 is

triggered on every millisecond?

Solution

C =15µs /0.69Rl =4.3µF a reasonable value of C could be 5µF.

The minimum time of cycle of operation is T =5(RL C +RC).

The value of R provides the only adjustments of this time since other parameters

are fixed.

Now 5(RL C +RC)=1*10-3R =41.5Ω

The value of R can be less than 41.5 and this would result in the capacitor

charging up faster while T1 is conducting.

b). Auxiliary Resonant Commutation Circuit

46

This circuit is popular and is used in many inverter and chopper circuits see the

circuit diagram first:

T2 is triggered first to order to charge up the capacitor C in the polarity shown.

As soon as the capacitor C is charged T2 is commutated owing to lack of current.

When T1 is triggered current flows in two paths, the load current flows in the in

Rl and the commutating current through C, T1, L and D. The charge on the

capacitor is reversed and is held with the hold off diode D. At any desired time T2

may be triggered which then places C across T1 thus turning off T1. The

waveforms are shown below.

At to T1 is assumed to be conducting

At t4 T1 is switched on and the capacitor will begin to discharge via T1, diode D

and the inductance L. During the first period of this oscillatory discharge, until

time t5 a reverse voltage is placed across the commutating thyristor T2. After one

47

half cycle of the oscillation, at time t6, the current through the diode attempts to

reverse and the diode ceases to conduct and the capacitor is charged in the

opposite direction of the figure ready for the next commutation cycle. If say we

want to commutate the T1 then T2 is fired and this places Vc across T1

commutating T1.

The frequency of Oscillation for this circuit is determined by the values of

capacitance and the inductor according to the relationship:

)(1

LC=ω

c) Resonant turn off

There is a series resonant and parallel resonant commutation circuit

Series Resonant Circuit

If an LC network is included as part of the load circuit, the current reversing

properties of a tuned LC will force the device to commutate.

When T1 is triggered, a current flows through the LC circuit charging the

capacitor towards the supply voltage.

After some time , the magnitude of the current reverses and tries to flow through

T1 in opposite direction owing to the resonating effect of Land C. For proper

commutation the LCR circuit must be under-damped to allow current oscillation

which will turn off the thyristor at first current zero. The capacitor will then

discharge through the load . The on time of the thyristor is determined by the

frequency of the oscillatory circuit (LC)

)(2)(2

12)(

1 LCTLC

ffLC

ππ

πω ====

The current and voltage waveforms are shown below:

48

Parallel Resonant Commutation Circuit

When the circuit is switched on and in the absence of gate trigger pulse, the

capacitor C is charged up in the polarity shown.

When T1 is triggered , current flows through load resistance Rl and a pulse

current through the resonating LC circuit. The capacitor is discharged from the

initial polarity and charges it in the reverse direction. The resonant circuit

current then reverses and tries to flow through T1 in opposite direction to the load

current and during this process T1 is turned off.

The thyristor ON period is again a function of the frequency of the oscillation of

the LC circuit, while the off period must be sufficient to allow the capacitor to be

adequately charged.

49

Commutation is accomplished if the resonant current ic reaches a value greater

than the load current il for a time greater than toff turn off time of the thyristor.

The circuit equation for the resonant circuit is:

==+=+=+ )(*)(0)(]1[0)(*101 2

2

2

tSinL

CVtisIC

LStiCdt

diLidt

Cdt

diL s ω

The capacitor voltage at any given time is given my the following relationship:

tCosVV sc ω=By design the peak discharge current is:

LCVs And this should considerably in excess of the load current. Hence:

CLR

L

C

RLCVii

R

Vl

l

sc

L

s =====2max

1

When the thyristor turns off, capacitor has negative charge. The capacitor then

charges to a positive value through the load R, L and C forming a damped

resonant circuit.

Example :

A parallel resonant turn off thyristor chopper consisting of an LC circuit in

parallel with an SCR supplies an inductive load current of 20A from a DC source

of 400V. Choose suitable values for L and C for turn off so that the trigger pulses

have a minimum period of 0.1msec.

Solution:

Peak current due to oscillatory circuit is Ipeak =L

CVs ≥20A

4001

201 ==

LC

LC

Minimum period (highest chopper frequency) corresponds to half (1/2) periodic

time of the oscillatory current that is:

mHLFCLCLC 636.0;10*5.110*013.110*1.0 693 ≥=== −−−π

Remember capacitor voltage leads current by 90 deg see the current response

graph below:

50

The maximum time ton that the thyristor conducts is from the time it is turned

on to the time that the capacitor current reaches its maximum value in the

reverse direction. In terms of the period T this time is 3/4T hence in limit this is

:

LCton π4

3= .

DC –DC CONVERTER

(CHOPPER)

A chopper directly converts a fixed –voltage dc supply to a variable –voltage dc supply. The chopper

can be used to control the speed of a DC motor.

Applications

-Switched –mode power supply (SMPS), DC motor Control, Battery chargers

Step Down Chopper (Buck Converter)

A schematic diagram of a step down chopper with a motor load is shown in fig below.

51

The chopper can be a conventional thyristor (SCR), a GTO thyristor, a power transistor, or a MOSFET.

When the chopper is turned ON say at t=0, the supply is connected to the supply and Vo =V. The load

current io builds up. When the chopper/switch is turned off at t=ton, the load current freewheels

through Dfw and Vo =0. At t=T the switch is turned on again and the cycle repeats. The waveforms of the load voltage and load

current are shown in fig above. It is assumed that the current is continuous, while the voltage is

chopped.

The average value of the output voltage is VVT

tonV *0 δ==

Where ton is the on time of the chopper

T is the chopping period

δ=the duty ratio of the chopper

The output voltage can be controlled in the range 0< Vo < V .

If the switch is a GTO thyristor, a positive gate pulse will turn it on and a negative gate pulse will turn

it off. If the switch is a transistor, the base current will control the on and off period of the switch. If the

switch is an SCR, a commutation circuit is required to turn it off. There are many forms of

commutation circuits that can be used to force commutate a thyristor.

The power supplied to the motor is 00 ** iViVP oo δ== .

If we assume a loss less dc-dc converter Pi= 00 ** iViVP oo δ==

The average value of the input current is 00 *** iIiVIV ss δδ ==The equivalent input resistance of the dc-dc converter drive seen by the source is:

δ1*

0i

V

I

VR

s

eq == . By varying the duty cycle, the power flow to the motor and the speed can be

controlled.

Example

52

A dc converter as shown above from a 600V dc source powers a dc separately excited motor. The

armature resistance is Ra =0.05Ω. The back emf constant of the motor is kv = 1.527V/Arads-1

.The

average armature current is io =250A. The field current is if =2.5A. The armature current is

continuous and has negligible ripple. If the duty cycle of the dc-dc converter is 60% determine:

a) the input power from the source,

b) the equivalent input resistance of the dc-dc converter drive,

c) the motor speed , and

d) the developed torque

Solution

a) the input power from the source is 00 ** iViVP oo δ== =0.6*600*250 =90kW

b) the input resistance of the dc-dc chopper = δ1*

0i

V

I

VR

s

eq == =600/(250*0.6)=4Ω

c) the motor speed is determined from ω** fvg IkE = but

aaaao RiVEERiV 000 * −=+= but VVT

tonV *0 δ== =0.6*600 =360V Ea

=360 –250*0.05 =347.5V 347.5

= rpmrpsIkE fva 27.86903.915.2*527.1

5.347** ==== ωω

d) The developed torque is afv IIkT **= =1.527*250*2.5 =954.4NM

Step Up Chopper (Boost Converter)

A change in chopper configuration as shown in fig below provides higher load voltages.

When the chopper is ON, the inductor is connected to the supply V and the energy from the supply is

stored in it. When the chopper is off, the inductor current is forced to flow through the diode and the

load.

The induced voltage Vl across the inductor is negative. The inductor voltage adds to the source voltage

to force the inductor current into the load. Thus the stored energy in the inductor is released to the load.

If the ripple in the source current is neglected, then during the time the chopper is ON (ton) the energy

input to the inductor from the source is:

onVItEi =During the time the chopper is off (toff), the energy released by the inductor to the load is:

offo ItVVEo )( −=For a loss less system in the stead state, the two energies will be the same hence:

δδ −=

−=

−=

+=−=

1***)(

VV

TT

TV

tT

TV

t

ttVItVVVIt

onoff

offon

ooffoon

Thus for a variation of δ in the range 0< δ <1, the voltage Vo varies in the range 0< Vo < ∞.

53

Two Quadrant Chopper

A combination of step up and step down configurations can form a two-quadrant chopper. See figure

below:

If chopper C1 and diode D1 are operated the system operates as a step down chopper and the dc

machine operates as a motor. The output voltage is either V (when C1 is on) or zero (when C1 is off

and D1 conducts).

• The average output voltage is positive and the output current io flows as shown in the circuit.

The chopper therefore operates in the first quadrant,

• If , however the chopper C2 and the diode D2 are operated , the system operates as a step up

chopper with Ea as a source and the DC machine operates in a regenerative braking mode.

• The output is either zero (when C2 is ON) or V (when C2 is Off and D2 conducting).

• The average output voltage is positive, but the output current now flows in the negative

direction. The chopper then operates in the fourth quadrant.

Note

The armature of the separately excited motor is rotating due to inertia of the motor and the load; and in

case of transportation system, the kinetic energy of the vehicle or train would rotate the armature shaft.

Then if the transistor or chopper is switched on, the armature current rises due to the short-circuiting of

the motor terminals. If the dc –dc converter (C2) is turned off, diode D2 would be turned on and the

energy stored in the armature circuit inductance would be transferred to the supply provided the supply

is receptive. A single chopper can be used for both powering and regenerative breaking. In regenerative

breaking C1 and D1 are rearranged from powering mode to regenerative braking. That is D1 and C1

are interchanged.

Example

The two-quadrant chopper shown in fig above is used to control the speed of the dc motor and also for

regenerative braking of the motor. The motor constant is Kφ =0.1V/rpm (Ea=Kφn). The chopping

frequency is fc =250Hz and the motor armature resistance is Ra =0,2Ω. The inductance La is

sufficiently large and the motor current io can be assumed to be ripple –free. The supply voltage is 120V.

a) Chopper C1 and diode D1 are operated to control the speed of the motor. At n=400rpm and

io=100A (ripple –free),

i) Draw waveforms of vo, io and is

ii) Determine the turn on time ton of the chopper iii) Determine the power developed by the motor, power absorbed by Ra, and power

from the source

b) In the two-quadrant chopper C2 and diode D2 are operated for regenerative breaking of the

motor. At n=350rpm and io = -100A (ripple free),

i) Draw waveforms of vo, io and is

ii) Determine the turn on time ton of the chopper iii) Determine the power developed by the motor, power absorbed by Ra, and power

to the source

54

Solution

a) ( i)

ii) aaa RIEV +=0 =0.1*400 +100*0.2 =60V

2120**60

Tt

T

tV

T

ton

onon === but T =1/250 =0.004seconds =4ms hence ton =2msec

iii) Pmotor =Ea*Io =0.1*400*100 =4000W

PR =(io)2*Ra =100

2*0.2 =2000W

Ps =V*(is)avg =120*100*2/4 =6000W

b) the waveforms are shown below

i)

ii) aaa RIEV +=0 =0.1*350 +(-100*0.2) =15V

sec5.34*8

7120**15 mt

T

tTV

T

tTon

onon ==−

=−

=

iii) Pmotor =Ea*Io =0.1*350*(-100) = -3500W

PR =(io)2*Ra =100

2*0.2 =2000W

Ps =V*(is) avg =120*(-100*1/8) =-1500W

55

The average voltage across the chopper is given for the following circuit used in regenerative brake

control:

)1()(1

δδδ

−=−== VTTT

VdtV

TV s

T

T

ch

If Ia is the armature current, the regenerated power is )1( δ−== VIVIPg acha

The voltage generated by the motor acting as a generator is:

aaamchfvg IRVIRVIkE +−=+== )1( δωWhere kv is the machine constant and ω is the machine speed in rads/sec. Ra is the total armature circuit

resistance.

Therefore the equivalent load resistance of the motor acting as a generator is:

a

aa

g

eq RI

V

I

ER +

−==

)1( δ

By varying the duty cycle δ, the equivalent load resistance seen by the motor can be varied from Ra to

(V/Ia +Ra) and the regenerative power can be controlled.

The conditions for the permissible potentials and polarity of two voltages are:

,0 minmin

fv

aa

aafvaaaaIk

IRIRIkEVRIE ===≤−≤ ωω and ω>or=ωmin . The

maximum braking speed of a motor can be determined from

maxmaxmax ;0 ωωωω ≤+=−=≤−≤fv

aa

fv

aafvaaaIk

IR

Ik

VIRIkVVRIE .

The regenerative braking would be effective only if the motor speed is between these two speed limits

(eg ωmin <ω<ω max). At any speed less than ω min an alternate braking arrangement would be required.

Which could be Rheostatic brake control or combined Regenerative and Rheostatic brake control

Example

A dc-dc converter is used in regenerative braking of a dc series motor. The DC supply is 600V. The

armature resistance is Ra =0.02Ω and the field resistance is Rf= 0.03Ω. The back emf constant is

kv=15.27mv/A rads-1

.The average armature current is maintained constant at Ia =250A.The armature

current is continuous and has negligible ripple. If the duty cycle of the dc- dc converter is 60%,

determine:

a) the average voltage across the dc-dc converter Vch

b) the power regenerated to the dc supply Pg

c) The equivalent load resistance of the motor acting as a generator Req

d) The minimum permissible braking speed ωmin ,

e) The maximum permissible braking speed ω max.

f) The motor speed

56

Solution

a) )1( δ−= VVch = (1-0.6)*600 = 240V

b) Pg =Ia*Vch =250*240 =60kW

c) Req =V/Is = Ea/Ia =(Vch + IaRm) where Rm =0.02+0.03 =0.05Ω Req

=(240+250*0.05)/250 =1.01Ω.

d) ,0 minmin

fv

aa

aafvaaaaIk

IRIRIkEVRIE ===≤−≤ ωω ωmin

=0.05*250/(15.27mV*250)=3.274rad/s =31.26rpm

e) maxmaxmax ;0 ωωωω ≤+=−=≤−≤fv

aa

fv

aafvaaaIk

IR

Ik

VIRIkVVRIE

ω max.=600/(0.01527*250) +0.05/0.01527 =160.445rad/s =15232.14rpm

f) aaamchfvg IRVIRVIkE +−=+== )1( δω Eg =Vch +IaRa =240 +250*0.05

=252.5V ω =252.5/(15.27mV*250)=66.14rad/s =631.6rpm

Inverters

These are static circuits that convert power from a dc source to ac power at a specified output voltage

and frequency. Inverters are used in many industrial applications.

For example:

a) Variable – speed ac motor drives

b) Induction heating

c) Aircraft power supplies

d) Uninterruptible power supplies (UPS) for computers.

There are two types of inverters namely:

i) Voltage source inverters (VSI)

ii) Current Source Inverter (CSI)

Voltage Source Inverter (VSI)

In the VSI the input is a dc voltage supply, e.g., battery, fuel cell, soar cell and other dc sources such as

output of controlled rectifier.

Both Single phase and three phase voltage sources are used in industry.

Figures illustrating Inverter Configurations

The switching devices can be a conventional thyristor(with its commutation circuit), a GTO thyristor,

or a power transistor.

57

Single Phase VSI- Principle of operation

The inverter consists of two choppers. When Q1 is turned on for a time T/2, the instantaneous voltage

across the load is V/2.

If only Q2 is turned on for a time T/2, -V/2 appears across the load.

The logic circuit should be designed such that Q1 and Q2 are not turned on at the same time. Third and

fourth graphs are for resistive and inductive loads.

Note that prior to turning on a switch, the other one must be turned off . Otherwise both switches will

conduct and short circuit the DC supply. If the load is inductive, that is a lagging power factor, the

output current io also lags the output voltage Vo as shown in the fig above.

The load current can not change immediately with the output voltage. If Q1 is switched off at T/2, the

load current will continue to flow through D2, load and the lower half of the dc source until the current

falls to zero.

Similarly when Q2 is turned off at t=T, the load current flows, through D1, load and upper half of the

DC source. When the diodes D1 and D2, conduct, energy is fed back to the dc source and these diodes

are known as feedback diodes.

The rms output voltage is :

RVdt

RV

TPVdt

V

TV

TT

o

orms 44

2

24

2 22/

0

2

21

2

0

2

===

=

E.g

A center tapped source inverter as shown above modulates power from a 200V source to a purely

resistive load whose value is R=2Ω.If the thyristors have a duty cycle of 0.4=m. Sketch VT1 ,VT2 , IL ,

IT1 , and determine:

a) the average power absorbed by the load,

b) the voltage and current ratings of the switches

Solution

a) Pav =16kW

b) Current rating =100A

58

Waveforms

Single Phase Bridge Inverter

Q1 ,Q2 ,Q3, Q4, are four choppers. Where Q1 and Q2 are turned on simultaneously, the input voltage

Vs appears across the load while Q3 and Q4, the voltage across the load is reversed and is –Vs.

If m=duty cycle= Ton/T then the rms voltage is :

Vrms =Vs√(2m)

The average power absorbed by the load is P =2mV2s/R

Example

The dc source voltage is Vs =600V and the load has a resistance of R=20Ω if the inverter is to operate

at 500Hz with an rms load voltage of 500V. Find:

a) the average power absorbed by the load

b) the average source current

c) the average current in each thyristor

d) the thyristor on time each period

Solution:

a) 12.5kW

b) 20.83A =Isav

c) Ithav =20.83/2=10.42A

d) Ton =0.694x 10-3

s

Solution Power =Vdc*Idc=Vo*Io

And )1(2

);1(2

)(2

1απαπωω

π

π

α

CosR

VICos

VttdSinVVo

L

mo

mm +=+==

Power =P =22

2

22

2

2

)1(5.22)1(100*4

300)1(

4αα

πα

πCosCosCos

R

V

L

m +=+=+

When P=80W then the firing angle can be determined by the following relationship:

8857.0556.31)1(556.3)1(5.2280 22 ==++=+= αααα CosCosCosCos Th

erefore αααα =27.660 .At a power of 25W the firing angle is 87,3

0 .

Thyristor half Wave Rectifier, Reactive Load

Most practical Loads have resistance and inductance L, e.g. the armature of a dc motor load has resistance

(R) and inductance (L). When a thyristor is fired at αααα the inductance in the load forces the current to lag

the voltage and decays to zero at ββββ. The waveform of the load current and voltage is shown below:

From αααα to ππππ the load current is driven by the supply source and from ππππ to ββββ the load voltage is negative

and the current is maintained by the voltage induced in the inductance

Half wave Controlled Rectifier With RL load and a free Wheeling Diode

The diode D bypasses the load current during the intervals of negative transformer voltage and this

prevents the output voltage from assuming a negative value.

• The diode prevents the voltage across the load from reversing during the negative half cycle of the

supply voltage.

• The load average voltage would be determined only by the positive half cycle wave of the

transformer secondary emf from αααα to ππππ .

The average voltage is )1(2

)1(22

2α

πα

πωω

π

π

α

CosR

VICos

VttdSin

VV

l

mdc

mdc +=+==

Thyristor Full Converter

Half wave rectification contains a significant amount of ripples. Therefore this is not suitable for speed

control of dc motors.

A full converter circuit shown in fig below is used for the speed control of DC motors.

The load current io has the same waveform as the load voltage vo..

The average load voltage is given by:

)1();1()(2

2απαπωω

π

π

α

CosR

VICos

VttdSinVVo

L

mo

mm +=+==

The rms voltage is given by the following expression:

212

1

22 )2

2sin(

2

1)(

1

+−=

=

ααπ

πωω

π

π

αmmrms VtdtSinVV

and the rms current is

[ ] 21

)2

2(2

1 ααππSin

R

V

RV

IL

m

L

rmsrms +−==

DC Motor Load / RL load

When terminal A1 is more positive as compared to A2 T1 and T2 are fired simultaneously and current flows

through T1 load and T2. In the reverse half cycle T3 and T4 are fired and the reverse voltage applied across

them commutates T1 and T2. Due to inductive load, T1 and T2 will conduct beyond ππππ ,even though the

input voltage is already negative see the waveforms above.

Depending on the size of the inductance the conduction can be discontinuous or continuous as shown

above. If XL>>R then the load current will be continuous.

Discontinuous load conduction

The average voltage is given by )(2

2βα

πωω

π

β

α

CosCosV

ttdSinV

V mm

dc −== The conduction angle of the thyristors is given by θθθθ =ββββ-αααα, hence ββββ=θθθθ +αααα

Now

)2

(2

22

*2

2θ

αθβαβα

βα +=−+

−=− SinSinSinSinCosCos

Therefore the average voltage for discontinuous load conductions is

l

dcdc

mdc

R

VISinSin

VV =+= )

2(

2

2 θα

θπ

Continuous load Conduction

If we assume that sufficient inductance is present in the dc armature circuit to ensure that the motor current

is continuous (present all the time).

If conduction is continuous the limits are from αααα to (ππππ+αααα) and the average voltage is given by:

l

m

l

dcdc

mmdc

R

CosV

R

VICos

VttdSin

VV

πα

απ

ωωπ

απ

α

22

2

2====

+

The rms voltage is determined from the following relationship:

21

22

=

+απ

α

ωωπ

ttdSinV

V mrms

Depending on the value of αααα the average output voltage could be either positive or negative and it provides

two-quadrant operation.

For α=450 it can be noted that T1 and T2 conduct the motor current during the interval α< ωt < (π+α) and

connects the motor to the supply v=Vs . At (π+α), T3 and T4 are fired. The supply voltage appears

immediately across thyristorsT1 and T2 as reverse bias voltage and turns them off. This is called Line

Commutation. The motor current , ia which was flowing from the supply through T1 and T2 is transferred

to T3 and T4. T3 and T4 conduct the motor current during the interval (π+α)<ωt < (2π+α) and connects the

motor to the supply (vo =-V)

If conduction is continuous the limits are from αααα to (ππππ+αααα) and the average voltage is given by:

l

m

l

dcdc

mmdc

R

CosV

R

VICos

VttdSin

VV

πα

απ

ωωπ

απ

α

22

2

2====

+

The inductance (La) does not sustain any average voltage. Therefore Vdc =Vo =IoRa +Ea

Where Io is the average motor current; Ea is the armature back emf, which is constant if the speed and

field current are constant

The variation of the motor terminal voltage Vo as a function of the angle α based on the equation Vdc.

For firing angles in the range 0o< α <90

0 , the average output is positive. Since the current can flow in one

direction in the load circuit because of the thyristors , the power VoIo is positive ;power is flowing from the

supply to the dc machine , and the dc machine operates as a motor. For firing angles 90o< α <180

0 ,the

output voltage is negative and therefore the power VoIo is negative; that is power flow is from the machine

to the ac supply. This is known as inversion Operation of the converter, and this mode of operation is used

for the regenerative braking of the motor. Note that for inversion operation, the polarity of the motor back emf Ea must be negative. It can be reversed by reversing the field current if so that the dc machine behaves

as dc generator.

Converter output characteristics for continuous load current

Two full converters can be connected back to back as shown below . This arrangement is known as the

dual–Converter connection. If one converter is used, it causes motor current to flow in one direction. If the

other converter is used, the motor current reverses and so does the speed

Dual Converter

Example .

A full wave fully controlled single phase bridge rectifier feeds a load resistance R and inductance L from

an AC source of Vrms voltage Vs.

Assuming the operation is continuous:

a) show that the mean or average voltage value of the DC output voltage Vo is given by

αCosVV so 90.0= .

If R=4Ω,L500mH, αααα=300 ,Vs =240V and the supply frequency is 50Hz.

b) Calculate:

i) the average and the rms currents output

ii) the thyristor average and rms currents and

iii) the power factor of the ac input to the rectifier

Solution

a) the mean value of the dc output for continuous operation is

ααπ

απ

ωωπ

απ

α

CosVCosV

CosV

ttdSinV

V s

smm

dc 90.0*222

2

2====

+

b) (i) Xl=2πfL=2*π*50*0.5H =157Ω This implies that the load current is going to be continuous.

Therefore AiVCosCosVV dcsdc 8.464

1.1871.18730*240*90.090.0 ===== α .

The current is almost constant since the ripple is almost negligible because of high circuit inductance.

Thus this current is both the average and the rms value i.e io =irms =46.78A

(ii) The thyristor average current is

Ai

tdiI otha 4.232

78.46)(*

22

1 0 ==−+== +

ααππ

ωπ

απ

α

the rms thyristor current is:

Ai

tdiI oothrms 1.33)(

22

12

122

1

2 =

−+=

=

+

ααππ

ωπ

απ

σ

iii) The output power to the load is :

WRiP oo 48.87534*78.46* 22 === This is also the input power to the converter. The VA

drawn from the supply is given by : 8.02.11227

48.87532.11227240*78.46 ==== PFWVI ss

Thyristor Semiconverter

A thyristor semiconverter that can be used for the speed control of a dc motor consists of thyristors and

diodes. See fig below:

Thyristor Semiconverter

Thyristors T1 and T2 are fired at αααα and (2ππππ+αααα), respectively. See the waveforms. The motor is

connected to the input supply for the period αααα<ωωωωt<ππππ through T1 and D2, and the load voltage is the

same as the input voltage. Beyond ππππ, Vo tends to reverse as the input voltage changes polarity. As Vo

tends to reverse , the diode Dfw (known as the free wheeling diode ) becomes forward biased and

starts to conduct. The motor current which was flowing from the supply through T1, is transferred to

Dfw, (i.e T1 commutates). The output terminals are shorted through the freewheeling diode during the

interval (ππππ<ωωωωt<ππππ+αααα) making Vo =0. At ωωωωt=ππππ+αααα , T2 is fired and it take over the motor current ia from

Dfw. The load current now flows through T2 and D1.At ωωωωt=2ππππ, Dfw, becomes forward biased again

and takes over the current from T2, and the process continues. See the waveforms. Note that if Dfw is

not used, freewheeling action will take place through T1 and D2 during the interval (2ππππ<ωωωωt<2ππππ+αααα).

The average value of the output voltage is:

)1(*2

)()**2(1

απ

ωωπ

CosV

tdtSinVVo

+=

=

Note; Vo is always positive and therefore Vo*Io is always positive that is power flow is from the ac

supply to the dc load. Semiconverters do not invert power. However they are cheaper than full

converters.

Waveforms

ExampleA single –phase full converter is used to control the speed of a 5hp, 110V, 1200rpm, separately excited

dc motor. The converter is connected to a single phase 120V,60Hz supply. The armature resistance is

Ra =0.4Ω and armature circuit inductance is La=5mH. The motor voltage constant Kφ =0.09V/rpm.

1. Rectifier(or motoring) operation. The dc machine operates as a motor runs at 1000rpm, and carries

an armature current of 30A. Assume that the motor current is ripple free.

a) Determine the firing angle αb) Determine the power to the motor

c) Determine the supply power factor

2. Inverter Operation (Regenerating Action). Reversing the field excitation reverses the polarity of the

motor back emf Ea.

a) Determine the firing angle to keep the motor at 30A when speed is 100rpm

b) Determine the power fed back to the supply at 1000rpm.

Solution

a) Ea =0.09*1000 =90V; Vo =Ea +IoRa =90+30 =102V, 02.19

120*22102 == αα

πCos

b) P =Io2*Ra + EaIo =VoIo =102*30 =3060W

c)The supply current has a square waveform with amplitude 30A (=Io). The rms supply current is I=30A .

The supply Volt-Amperes are 120*30A =3600VA

If the losses in the converter are neglected, the power from the supply is the same as the power to the motor

Ps =3060W

Thus the supply power factor is PF =Ps/S=3060/3600 =0.85

2. (a) At the time of polarity reversal the back emf is Ea =90V and from Vo =Ea +IoRa =-90+30*0.4 =-

90+12 =-78V

Now 02.13678*

120*22=−== αα

πVCosVo

b) Power from the dc machine Pdc =90*30 =2700W

Power lost in Ra =302 *0.4 =360W

Power fed back to the ac supply =Ps =2700-360 =2340W

Three Phase Circuits For higher power applications, several kilowatts three phase circuits are preferable. The magnitude of

harmonic voltage is lower in three phase circuits than in single-phase circuits. This is due to increasing

number of pulses.

Half Wave Controlled Rectifier With Resistive load

)120(**2)240(**2

)120(**2

**2

00

0

+=−=

−=

=

tSinVtSinVVcn

tSinVVbn

tSinVVan

ωω

ω

ω

If thyristor T1 is fired at ωt =300 ,T2 is fired at ωt =1500 , and T3 ωt =2700 that is at the crossing

points of the phase voltages. Firing at these instants will also result in maximum output voltage. The

reference for the firing angle α is therefore the crossing point of the phase voltages. The firing of the

thyristors can be delayed from these crossing points. In other words the firing angle is measured from

the crossing points of the phase voltages. Recall that in a single-phase converter, the firing angle was

measured from the zero crossing of the input supply voltage.

For the above waveform thyristor T1 is fired at ωt =300

+ α, and the output voltage is Vo=Van. The

output voltage io =Vo/R, and becomes zero at ωt =π. Thyristor T1 turns off at this instant through

natural commutation. Thyristor T2 is fired atωt =1500

+ α, making Vo= Vbn. T2 turns off at ωt =3000

.T3 is fired at

ωt =2700

+ α. Making Vo =Vcn.

Example

The load in the fig above now consists of a resistance and a very large inductance. The inductance is so

large that the output current io can be assumed to be continuous and ripple- free. For α=600 .

a) Draw the waveforms of vo and io.

b) Determine the average value of the output voltage if phase voltage Vp =120V

Electronic Engineering Department

Electronic Drives Assignment 1

Tee 2102

Due Date 27/02/2006

Question 1

The source voltage is Vs =200V and the resistive load is 4Ω. For this condition the

thyristor has a delay time of td =0.5µs and a rise time of 3µs at turn ON. The thyristor

leakage current is Ileak =2mA and the ON state voltage drop is V th(ON) =1.5V.

a) Sketch graphs illustrating the voltage and current response with respect to time

at turn on and show the delay time as well as the rise time. [4]

b) Determine the components of energy loss incurred in the thyristor during turn

on and, [6]

c) Compare the turn on loss with the thyristor loss during the ON state over an

equivalent interval of rise time (tri) [10]

Question 2

The maximum permissible junction temperature of a certain power transistor is

1500C. It is desired to operate the transistor with a power dissipation of 15W in an

ambient temperature of 400C. The thermal resistances are as follows Rθjc =0.5

0C/W;

Rθca =100C/W.

a) Determine whether a heat sink is required for this application [5]

b) If a heatsink is required determine the maximum thermal resistance it can

have. Assume that a mica washer having a thermal resistance of Rθw

=0.500C/W must be used between the case and the heat sink [10]

Question 3

A semiconductor dissipates 25W through its case and its heat sink to the surrounding

air . The thermal resistances are as follows, device to case 10C/W; case to heat sink

1.20C/W and heat sink to air 0.7

0C/W. In what maximum air temperature can the

device be operated if its temperature cannot exceed 1000C. [10]

Question 4

Two diodes having approximate characteristics in the forward direction:

Diode 1 110*44.288.0 4 iv −+=

Diode 2 210*32.296.0 4 iv −+=

Are connected in parallel. Find the current in each diode if the total current is :

a) 400,

b) 1200,

c) 2000, [9]

What single value of resistance connected in series with each diode will bring the

diode currents to within 8% of equal current sharing with a total current of 1200?

How will this affect the current sharing at total currents of 400A and 2000A ?[11]

Electonic drives assignment

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Assignment3

Question1

The speed of a separately excited motor is controlled by a single phase full wave

converter in figure below

A full converter also controls the field circuit and the field current is set to the

maximum possible value. The ac supply voltage to the armature and field

converters is a single phase of 440V, 60Hz. The armature resistance is Ra

=0.25ΩΩΩΩ, the field circuit resistance is Rf =175ΩΩΩΩ, and the motor voltage constant is

Kv =1.4V/Arad/s. The armature current corresponding to the load demand is Ia

=45A. The viscous friction and no load losses are negligible. The inductances of

the armature and field circuits are sufficient to make the armature and field

currents continuous and ripple free. If the delay angle of the armature converter

ααααa=600 and the armature current is I a =45A. Determine:

a) the torque developed by the motor Td [7]

b) the speed ,ωωωω [8]

c) the input power factor of the drive [5]

Question 2

The speed of a 20HP, 300V, 900r.p.m separately excited dc motor controlled by a

three phase full converter. A three phase full converter also controls the field

circuit. The ac input to the armature and field converters is 3 phase, Y

connected, 208V, and 60Hz. The armature resistance is Ra=0.25ΩΩΩΩ,the field

circuit resistance is Rf =145ΩΩΩΩ,and the motor voltage constant is Kv =1.2V/A

rad/s. The viscous friction and no load losses can be considered negligible. The

armature and field currents are continuous and ripple free.

a) if the field converter is operated at the maximum field current and the

developed torque is Td= 116NM at 900r.p.m determine the delay angle of

the armature converter ααααa. [10]

b) if the circuit converter is set for the maximum field current, thye

developed torque is Td= 116Nm and the delay angle of the armature

converter is ααααa=0, determine the speed of the motor. [5]

c) For the same load demand in b), determine the delay angle of the field

converter if the speed has been increased to 1800r.p.m [5]

Question 3

A dc-dc converter is used in regenerative braking of a dc series motor .The dc

supply voltage is 600V. The armature resistance is Ra= 0.02ΩΩΩΩ and the field

resistance is Rf =0.03ΩΩΩΩ. The back emf constant Kv =15.27mV/Arad/s. The

average armature current is maintained constant at Ia=250A. The armature

current is continuous and has negligible ripple . If the duty cycle of the dc-dc

converter is 60%, determine ;

a) the average voltage across the dc-dc converter, Vch [2]

b) the power regenerated to the dc supply Pg, [3]

c) the equivalent load resistance of the motor acting as a generator, Req,

d) the minimum permissible braking speed ωωωωmin, [4]

e) the maximum permissible braking speed ωωωωmax and [4]

f) the motor speed. [4]

Electronic Engineering Department

Electronic Drives Assignment 1

Tee 2102

Due Date 4/03/2008

Question 1

The source voltage is Vs =200V and the resistive load is 4Ω. For this condition the

thyristor has a delay time of td =0.5µs and a rise time of 3µs at turn ON. The thyristor

leakage current is Ileak =2mA and the ON state voltage drop is V th(ON) =1.5V.

a) Sketch graphs illustrating the voltage and current response with respect to time

at turn on and show the delay time as well as the rise time. [4]

b) Determine the components of energy loss incurred in the thyristor during turn

on and, [6]

c) Compare the turn on loss with the thyristor loss during the ON state over an

equivalent interval of rise time (tri) [10]

Question 2

The maximum permissible junction temperature of a certain power transistor is

1500C. It is desired to operate the transistor with a power dissipation of 15W in an

ambient temperature of 400C. The thermal resistances are as follows Rθjc =0.5

0C/W;

Rθca =100C/W.

a) Determine whether a heat sink is required for this application [5]

b) If a heatsink is required determine the maximum thermal resistance it can

have. Assume that a mica washer having a thermal resistance of Rθw

=0.500C/W must be used between the case and the heat sink [10]

Question 3

A semiconductor dissipates 25W through its case and its heat sink to the surrounding

air . The thermal resistances are as follows, device to case 10C/W; case to heat sink

1.20C/W and heat sink to air 0.7

0C/W. In what maximum air temperature can the

device be operated if its temperature cannot exceed 1000C. [10]

Question 4

Two diodes having approximate characteristics in the forward direction:

Diode 1 110*44.288.0 4 iv −+=

Diode 2 210*32.296.0 4 iv −+=

Are connected in parallel. Find the current in each diode if the total current is :

a) 400,

b) 1200,

c) 2000, [9]

What single value of resistance connected in series with each diode will bring the

diode currents to within 8% of equal current sharing with a total current of 1200?

How will this affect the current sharing at total currents of 400A and 2000A ?[11]