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Chapter - 1 Power Electronics Control

Application of Power Electronics

Power electronics combine power electronics and control. Control deals with the steady state and dynamic characteristics of the clone loop system. Power deals with the static and rotating power equipment. for the generation, transmission and distribution of electronic power. Electronic deals with the solid state device and circuits for signal processing to meet the desire control objective. Power electronics may be defined as the application electronics for the control and conversion of electric power, the inter-relationship of Potential Energy with power, electronics and control is shown in figure above. Some applications of power electronics are as follows:

1. Aerospace : Example – aircraft power system 2. Commercial : Example – advertising, heating 3. Industrial : Example – cement mill, welding 4. Residential : Example – cooking, lighting 5. Telecommunication : Example – Battery chargers 6. Transportation : Example – street car, trolley buses 7. Utility System : Example – high voltage direct current (HVDC), transmission system

History of Power Electronics 1948 → Bell Lab (Silicon Transistor) 1956 → PNPN traggiration


-By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 2

The history of power electronics begin with the mercury arc rectifier in 1900. Then metal tank rectifier, grid control, vacuum tube rectifier, ignitron, pkahotron & Hydraten were introduced gradually. These devices were applied for power control until 1950. The first electronics revolution begin in 1948 with the invention of silicon transistor at a bell telephone library by Bardeen, Brattain, Schookely. The next break through in 1956 was also f aboratory, the invention of PNPN triggering transistor which was defined a thyrastor or rom Bell Silicon Control Rectifier (SCR). The second electronics revolution begin in 1956 with the development of commercial thyrastor by the general electric company. That was the beginning of a new era of power electronics. Since then many different types of power semi-conductor devices and conversion techniques have been introudced. Power Semi-conductor Devices: Power Semi-conductor Devices can be broadly divided into 5 types. They are:

1. Power diode 2. Thyristor 3. Power BJT 4. Power MOSKET 5. Insulated Gate Bi-polar Transistor (IGBT)

or Static Induction Transistor (SIT)

Thyristor can be divided into 8 groups. They are: 1. Forced Commutated Thyristor 2. Line Commutated Thyristor 3. Gate Turn off Thyristor (GTO) 4. Reverse Conducting Thyristor (RCT) 5. Static Induction Thyristor (SITH) 6. Gate Assisted Turn off Thyristor (GATT) 7. Light Activated Silicon Control Rectifier (LASCR) 8. MOS Control Thyristor

Power diode are of two types:

1. High Speed Power Diode 2. Schotty Power Diode

Control Characteristics of Power Devices



d) MOSFET/IGBT Switch

R VVS

+

-

V

t

Thyristoror SCR

Vg

Vg

Vs

ON OFF T1t1

GTO

+

-

VoVS

+

-

A.Vg

SITH

.K

R

K

MCT

Fig : GTO/SITH/MCT Switch

V0

+

-t1

T

t

t

T

t1

Vo

VB

VS

+

-

VB

Fig : Transistor Switch



1) Uncontrolled turn on & off : e.g. diode 2) Controlled turn on and uncontrolled turn off : e.g. SCR 3) Controlled turn on and off : e.g. GTO 4) Continuous Gate Signal Requirement : e.g. Transistor Types of Power Electronics Circuit:

1. Diode Rectifiers 2. AC-DC Converter (Controlled Rectifier) 3. AC-AC Convert (AC Voltage Controllers) 4. DC-AC Converter (Inverter) 5. DC-DC Converter (DC Choppers) 6. Static Switches

1) Diode Rectifier This rectifier converts AC voltage into fixed DC voltage.

vs

+G

IGBT

C

E

G

SVgs

D+

-

+

- Tt

vgs

vs

t1tT

t1

1

V0

AC Supply

D2

D1

vs

vs=vmSinwt load resistance

R

+-+

-

a) Circuit Diagram



2) AC-DC Converter (Controlled Rectifier) By this rectifier, average output voltage can be controlled by varying the conduction time of thyristor or firing angle α . 3. AC-AC Converter

• Such types of converter are used to obtain a variable ac voltage from fixed ac voltage. • Output Voltage is controlled by varying the conduction time or firing angle α .

Vm Vs=Vm Sinwt

VmV0

b) Voltage waveform

wt

wt

Vm

π 2π

π 2π

AC Supply

T2

T1

vs

vs=SinwtLead result

Fig : Circuit Diagram

Vm Vs=Vm Sinwt

VmV0

wt

wt

Vm

π 2π

2ππ

d

d

Fig : Voltage Waveform



4. DC-AC Converter (Inverter) 5. DC-DC Converter 6. Static Switch Since the power devices can be operated as static switches, the supply to there switches could be either DC or AC and the switches are called as AC Static Switches or DC Switches. Design of Power Electronic Equipment The design of power electronics can be divided into four parts: 1. Design of power circuit 2. Protection of power devices 3. Determination of control strategy 4. Design of logic and gating circuit

Peripheral Effects The operations of power converter are based mainly on the switching of the power save conductor devices and as a result, the converter introduces current and voltage harmonics into the supply system and on the output of converter. This can cause problem of distortion of output voltage, harmonic generation into the supply system and interference with the communication signaling circuit. It is normally necessary to introduce filters on the input and output of the converter system to reduce a harmonic level to an acceptable magnitude. Figure shows the block diagram of generalized power converter. Power converter can cause radio frequency interference due to electromagnetic radiation and the getting circuit may generate erroneous signal and this interference can be avoided by grounded shielding.



Chapter - 2 Semi-conductor Diodes

• A power diode is two terminal P-N Junction diode. • When anode potential is higher than that of cathode, the diode is said to be forward biased

and the diode conducts. • When cathode junction is higher than that of anode, the diode is said to be reversed. Under

reverse condition, a small reverse current (also known as leakage amount) is the range of micro or milli ampere flows and this leakage amount of current increases slowly in magnitude with reverse voltage until the avalanche or zenor voltage is reached.

v-i characteristics can be expressed by Shockley diode equation as given below:

( )1−= Tp nVVsD eII

where, A dioderough Current th=DI

2 to1 from varies valuesefactor whoideality or efficient -coemission asknown constant Empericalcurrent Leakage

cathode w.r.t.positive anode with voltageDiode

===

nIV

s

D

For Ge diode : n=1 For Si diode : n=2

qKT Voltage Thermal ==TV

where, K = Boltzmann constant = 1.3806×10-23 J q = Electron charge = 1.6022×10-19 C T = Absolute temperature in Kelvin At a junction temperature of 25°c, we have

mVq

KTVT 8.25106022.1

298103806.119

23

≈×

××== −

−

The diode characteristics can be divided into three regions. They are: 1. Forward Biased Region, 0>DV

2. Reverse Biased Region, 0<DV

0

ID

I

VVD

I

V

Reversel eakagecurrent

A K

V

D

a) Practical b) IdealFig : V-I characteristics of diode



3. Breakdown Region, zkD VV −<

Consider a small diode voltage, mVVnVV TD 8.25&1,1.0 ===

( ) ( )( )( ) [ ]

errorwithIIneglectingI

eIeII

s

ss

snVV

sDRD

%1.223.48123.48

11 0288.011.0

=−=

−=−= ×

For VD>0.1 V, ID>>Is ( ) errorwitheIeII TDTD nVV

snVV

sD %211 ≈−=

Reverse Biased Region, 0<DV

For ,1.0 VVD −< the exponential term becomes negligible as compared to unity.

( ) snVV

sD IeII TD −≈−= − 1

In Breakdown Region, the reverse voltage is high usually greater than 1000 V, the magnitude of the reverse voltage exceeds the specified voltage known as breakdown voltage BrV . The reverse voltage

increases rapidly with small change in reverse voltage beyond BrV . Reverse Recovery Characteristics The current in a forward leased junction diode is due to the net effect of majority and minority carriers. Once a diode is in forward conductor mode and then it’s forward current is reduced to zero, the diode continuity to conduct due to minority carrier which remain store in P-N Junction and bulk semi-conductor material. The minority carrier requires a certain time to recombine with opposite charges and to be neutralized. This time is called reverse recovery time of the diode. Figure shows two reverse recovery characteristics of a junction diode.

( )

materialtor semiconducbulk in storage charge todue regiondepletion in storage charge todue

1

==

−+=

b

a

barr

tt

ttt

The ratio ba tt / is known as Softness Factor, denoted by SF.

( )2−=dtditI aRR

tta

trr

t1IRR

IRR0.25

IF

t

trr

t6IRR

IF

a) Soft Recovery b) Abrupt Recovery

Fig : Reverse Recovery Characteristics



Reverse recovery time (TRR) may be defined as the time interval between the instant the current passes through zero during the change over from forward conduction to reverse blocking conduction and the movement, the reverse current has decayed to 25% of it’s peak reverse value RRI . Reverse Recovery Charge ( )RRθ It is the amount of charge carrier that flow across the diode in the reverse direction due to charge over from forward conduction to reverse blocking condition.

( )3.

21

.21.

21

−=

+≅

rrRR

RRbRRaRR

tI

ItItθ

( )42−=

RR

RRRR t

I θ

From (2) and (4), we have:

( )52−=

rr

RRa tdt

dit θ

If bt is negligible as compared to arra ttt ≅,

( )

( )7.2

62

22

−≈

−≈

≈

dtdiI

dtdit

dtdit

RRRR

RRrr

RRrr

θ

θ

θ

Types of Power Diode Depending upon the recovery characteristics and manufacturing technique, the power diode can be classified into three categories. They are:

1. General or Standard Purpose Diode 2. Fast Recovery Diode 3. Schottky Diode

General Purpose Diode These diodes have relatively high reverse recovery time of the order of about 25 µ s. Their current ratings vary from 1A to several thousand amperes and range of voltage rating is from 50V to 5KV. Application of power diode of this type include battery charging, UPS, etc. Fast Recovery Diode The diodes with low reverse recovery time of about 5 sµ or less are classified as fast recovery diode. These diode convert current rating from less than 1A to 100A with voltage rating from 50V to 3KV. They are used in DC to DC, DC to AC converter circuit.



Schottky Diode It uses metal to semi-conductor junction for rectification purpose instead of P-N junction. They have very fast recovery time and low forward voltage drop. The maximum available voltage is generally limited to 100V. The current rating of schottky diode varied from 1A to 300A and they are used in high frequency instrumentation system. Effects of forward & reverse recovery time If the switch is turned on at t=0 and remains on long enough a steady state current of

RVI so /= would flow through the load and the free wheeling diode mD will be reverse bias.

If the switch is turned off at t=t1 diode Dm would circulate through Dm. Again, the switch is turned on at t=t2. The rate of rise of forward current of diode D1 and the rate of fall of forward current of diode Dm would be very high tending to be infinity.

According to the equation, dtdiI RRRR θ2= , the peak reverse current of diode would be very

high and diodes D1 and Dm may be damaged. This problem is normally overcome by connecting di/dt limiting inductor Ls as shown in the figure.

Sw D1

DmL

R

Vs

I1I2

+

-I2

Ls

a) Circuit Diagram

t

t

t1 t2

t1 t2

I0

I0

I0

I1

I2

I3

VsI0

R IP=Is/R

Due to reverse recovery of Dmb) Wave Form



The practical diodes require a certain turn on time before the entire area of junction becomes conductive and dtdi / must be kept low to meet the turn on time limit. This time is some time known as forward recovery time (tfr). The rate of rise of current through the diode D1 which should be same

as the rate of fall of current through diode Dm is L

Vdtdi s= . If trr is the recovery time of Dm, the peak

reverse current of Dm is

s

srr

rrRR

LV

t

dtditI

=

=

and, the peak current through the inductor sL would be s

srrorrop L

VtIIII +=+=

When the inductor current becomes pI diode mD turn off suddenly. Due to a highly inductive

lead, the current cannot change suddenly from Po ItoI . The excess energy stored in sL would induce

high reverse voltage across mD and this may damage diode mD .

The excess energy stored as a result of reverse recovery time is

[ ]

( )[ ]( )[ ]22

22

22

212121

orros

oRRos

opsR

ItIL

IIIL

IILw

−+=

−+=

−=

This excess energy can be transfer from inductor ss CL capacitor a to which is connected across

diode mD .

2

2

221

c

Rs

Rcs

VwC

wVC

=

=

where, CV is the allowable reverse voltage of the diode. A resistor Rs is connected in series with Cs

to damp out any transient oscillation.

Sw D1

DmL

R

Vs

I1I2

+

-I2

Ls

a) Circuit Diagram

t

t

t1 t2

t1 t2

I0

I0

I0

I1

I1

I2

tm

Ip

b) Waveform



Series Connected Diode Let us consider two diode connected in series. In practical v-i characteristics for the same type of diode differ due to tolerances in their production process. In forward bias condition, both diodes conduct the same amount of current and the forward voltage drop of each diode would be all most equal. However in reverse blocking condition each diode has to carry the same leakage current and as a result the blocking voltage will be differ significantly. A simple solution to this problem is to force equal voltage sharing by connecting a resistor across a diode.

( )( )iiIII

iIII

ssR

sRs

−=+

−+=

11

2

From (i) and (ii),

2211 sRsR IIII +=+

then ,R If 21

212

2

1

1

RR

IR

VI

RV

sD

sD

==

+=+

( )iiiIR

VI

RV

sD

sD −+=+

2

2

1

1

Parallel Connected Diode In high power application diodes are connected in parallel to increase the current carrying capability to meet the desire current requirement. The current sharing of diodes would in accord with their respective forward voltage drop. Uniform current sharing can be achieved by connecting current sharing resistor. Q. Two diodes are connected in series as shown in the figure to share a total voltage of KVVD 5= ,

the reverse leakage current of the two diodes are mAImAI ss 35 and 3021

== .

-

+

+

VD1

VD2

Is

i=-Is

IR1

IR2

Is1

Is2

i

V-VD1=-VD2

-Is1-Is2

-

+

-

+-

+

VD1

VD2

I=-Is

D1

D2

Is

a) Circuit Diagram

VVD1 VD2

Is

b) v-i chracteristics

i



a. Find the diode voltages if the voltage sharing resistances are equal. Ω=== kRRR 10021 .

b. Find the voltage sharing resistances 21 & RR if the diode voltages are equal.

Solution:

Ω===

==

==

kRRR

VV

mAImAI

DD

ss

100

??

3530

21

21

21

21

2

2

1

1

2211

RV

IR

VI

IIII

DR

DR

RsRs

==

+=+

( )

5000

500

21

21

21

2

2

1

121

=+

=−∴

−+=

+=+

DD

DD

DDD

Ds

Ds

VV

VV

iiVVVR

VI

RV

I

VV

VV

D

D

2250&

2750

2

1

=

=∴

Q. The reverse recovery time of a diode is strr µ5= and the rate of fall of diode current

./80 sAdtdi µ= If the softness factor, SF = 0.5, then determine storage charge.

(a) θRR (b) peak reverse current Solution:

( )

( )iitI

idtditI

rrRRRR

aRR

−=

−=

.21θ

5.0==a

b

tt

SF

( )itt ab −= 5.0

( )iisttt barr −=+= µ5

sta µ33.3=



Chapter - 3 Diodes Circuit & Rectifier

Diodes with RC Load When switch 1s is closed at t=0

( )01=++=

+=

∫ tVidtc

R

VVV

ci

cRs

With initial condition, ( ) 00 ==tVc

( )ssIsRI

sVs +≠ )(

s

VsI

csRor s=⎟

⎠⎞

⎜⎝⎛ + )(1,

( )s

VsI

csRcsor s=⎟

⎠⎞

⎜⎝⎛ +1,

( ))1(

,RcsRc

VscsIor+

=

( ) ( )RcsRVssIor

/1,

+=

Taking Inverse Laplace, Rcts e

RV

ti /)( −=

Capacitor Voltage

dte

RV

c

idtc

tV

Rctt R

t

c

−∫

∫

=

=

0

0

1

1)(

[ ][ ]Tt

s

Rcts

tRcts

eV

eV

Rce

RcV

/

/

/

1

1

/1

−

−

−

−=

−=

⎥⎦

⎤⎢⎣

⎡−

=

T = Rc = Time constant of Rc load. The rate of change of capacitor voltage is

( ) RctsRc

sc e

RcV

eRc

Vdt

tdV t //11 −− =⎟⎠⎞

⎜⎝⎛−−=

D1R

Vs

+

-

C

VR1

Vc

+

-+

-

S1



& the initial rate of change of capacitor voltage (at t=0) is

( )

RcV

dttdV s

t

c ==0

Diode Circuit with RL load When switch S is closed at t=0

dtdiLRi

VVV LRs

+=

+=z

( ) ( ) ( )[ ]0) issILsRIs

Vs −+=

With initial condition i(0) = 0

)()( sLsIsRIs

Vs +=

( )

( ) ( )LsRsV

sI

sV

ILsR

s

ss

+=

=+

( )LRsLsVs

/+=

RL

LRssLV

sI s⎥⎦⎤

⎢⎣⎡

+−=

/11)(

τt

VS/R

0.36 VS/R

i

τ=Rc

τ

0.632VsVs

Vc

t

D1R

Vs

Ls

L

VR

+

-

+

-VL

a) Circuit Diagram



⎥⎦⎤

⎢⎣⎡

+−=

LRssRV

sI s

/11)(

By inversion,

t

LR

stLR

s eL

Ve

RV

ti−

−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−= 1)(

L

Vdtdi s

t

==0

Voltage across inductor,

Tt

s

tLR

sL eVeVdtdiLV

−−===

where, ==RLT time constant RL load

Q. A diode circuit is shown in figure with R = 44 Ω and C = 0.1 .Fµ The capacitor has an initial

voltage .220VVo = If switch 1s is closed at t=0, determine:

a) Peaked diode current b) Energy dissipated in a resistor R c) The capacitor voltage at t=2 sµ Solution: at t=0, s is closed

( )

( ) oc

C

t

i

VtV

tVidtc

R

−==

==++ ∫0

0010

Taking laplace transform,

( ) ( )0=−+

sV

CssIsRI o

τt

VS

0.36 VS

VL

Ob) waveformτ

0.63VsVs/R

t

i

R

C

VRi

V0

+

-V

C



( )

( )

( )⎟⎠⎞

⎜⎝⎛ +

=

=⎟⎠⎞

⎜⎝⎛ +

=⎟⎠⎞

⎜⎝⎛ +

RCsRC

CVsI

sV

sICs

RCss

VsI

CsR

o

o

o

1

1

1

( )⎟⎠⎞

⎜⎝⎛ +

=

RCsR

VsI o

1

By inversion,

( ) RCto eR

Vti /−=

a) Peak diode current = AR

Vo 544

220==

b) Energy dissipated in resistor ( ) mJCVo 42.2220101.021

21 262 =×××== −

c) Voltage across capacitor, ( ) ( )010

=+= ∫ tVidtc

tV c

t

C

o

RCto

t

oRCto

V

RC

eRCV

VdteR

Vc

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=

−=

−

−∫

1

1

/

0

/

( )

RCto

oRCt

o

eV

VeV/

/1−

−

−=

−−=

At st µ2= , ( ) VestVc 64.139220)2(96 101.04/102 −=−==

−− ×××−µ

Diode with LC Load At t=0, switch s is closed.

)0(10∫ =++=t

cs tVidtcdt

diLV

O

Vc(t)

-Vo

tO

V0/R

i(t)



( ) 00 ==tVc

Using Laplace Transform:

( )[ ]

( )

( )⎟⎠⎞

⎜⎝⎛ +

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +

=⎟⎠⎞

⎜⎝⎛ +

=

+−=

LCsLC

CVsI

sV

sIcs

Lcs

sV

sIcs

Ls

iCs

sIissILs

V

s

s

s

s

1)

1

)(10)0(

)(0)(

2

2

( ) ( ) LCwhere

sLV

sI s 1,22 =+

= ωω

By inversion,

t

LCV

tL

Vti

s

s

ω

ωω

sin.

sin.1.)(

=

=

( )LCVIwheretIti spp == ωsin

The rate of current is

( )2cos.

cos.1.cos.

−=

==

tL

Vdtdi

tLCL

CVtIdtdi

s

sp

ω

ωωω

Initial rate of rise of current,

( )30

−== L

Vdtdi s

t

Voltage across capacitor

t

s

t

p

t

ct

LCV

cdttI

cidt

ctV

000

cos.1sin11)( ⎥⎦⎤

⎢⎣⎡−=== ∫∫ ω

ωω

( )tLCVs ω

ωcos11. −=

( )4)cos1( −−= tVs ω

LCπ== 1tAt t , the diode current i falls to zero and the capacitor is changed to 2Vs.

D

L

Vs

S

+

-+

-

VL

Vc



Q. A diode circuit with Lc load is shown in figure with the capacitor having initial voltage Vo=220V, capacitance C=20 Fµ and inductance .80 HL µ= If switch s1 is closed at t=0, determine: a) Peak current through the diode b) Conduction time of the diode c) Steady state capacitor voltage Solution: At t=0, s1 is closed.

( )

( ) oc

t

c

VtV

tVidtCdt

diL

−==

==++ ∫0

0010

Taking laplace transform,

( ) ( )[ ] ( )

( )s

VsI

CsLs

sV

CssIisSIL

o

o

=⎟⎠⎞

⎜⎝⎛ +

=−+−

1

00

ALCVti o 110)( ==

Conduction time, sLCt µπ 66.1251 ==

( ) ( )∫ =+=t

cc tVidtC

tV0

01

o

t

o VtdtLCV

C−= ∫ ωsin1

0

o

to Vt

LC

CV

−⎥⎦⎤

⎢⎣⎡−=

0

cosω

ω

( ) oo Vt

LCV

−−= ωω

cos11.

( ) tVtV oc ωcos−=

Steady-state Capacitor Voltage

tO

i(t)

tO

Vc(t)t1= 2ct1/2

Vs2Vs

VL

S1

Vc

+

-+

-

iL

V0



( ) VVVLCLC

VtVttV ooooc 220cos.1coscos1 ==−=−=−== ππω

Q. In the diode `/WHLC is shown in the figure. The capacitor is initially charged to voltage cV

with upper plate +ve. Switch s is closed at t=0, device expressions for the current through and voltage across C. Find the conduction time of the diode, peak current through the diode and final steady state voltage across C. In case .30,100,100,400 µµ ==== CHLVVVV os Determine also the voltage

across the diode after stage conduction. Solution: At t=0, switch s is closed,

( )

( ) oc

c

t

s

VtV

tVidtCdt

diLV

==

=++= ∫0

010

Using Laplace Transform,

( )[ ] ( )s

VCs

sIsissILs

V os ++−= )(

( )s

VVsI

csLs os −

=⎟⎠⎞

⎜⎝⎛ +

1

( )s

VVsI

csLcs os −

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +12

( ) ( )

⎟⎠⎞

⎜⎝⎛ +

−=

LCsLC

VVcsI os

12

( )LC

wheresL

VVsIor os 11, 22 =

+⎟⎠⎞

⎜⎝⎛ −

= ωω

tO

t

t1V0

V0

-V0

t1/2 t1

L

Vs

Ls

C V0

+

-+

-Vc

+

-

VL

i



tO

i(t)

t

t1/2

Vs

Vc(t)t1

VsV°

2Vs-V°

-Vs+V°t

V1-V° C/2

VR

Vs

S1

V°

+

-+

-C

+

-

L VL

Vc

+

-

R

+ -

By inversion

( ) tLCVVt

LVV

ti osos ωω

ωsin.sin1.)( −=

−=

Voltage across capacitor,

( ) ( )010

=+= ∫ tVidtC

tV c

t

c

( ) o

t

os VtLCVV

C+−= ∫ ωsin.1

0

o

tos Vt

LC

CVV

+⎥⎦⎤

⎢⎣⎡−

−=

0

cos.ω

ω

( )( ) oos VtVV +−−= ωcos1

a) sLCt µπ 77.541 ==

b) ( ) ALCVVI osp 32.164=−=

c) Steady state voltage across capacitor = VVV os 7002 =−

d) ( ) VVVVVVVVVVVVV osossosscLso 300220 −=+−=+−=−−−=−−=

Diode Circuit with RLC Load At t=0, switch s1 is closed.

( )010

=+++= ∫ tVidtCdt

diLRV c

t

is

With initial condition ( ) oVti == 0

By differentiation,

0

0

2

2

2

2

=++

=++

LCi

dtdi

LR

dtid

ci

dtdiR

dtidL

Changing equation in s domain,

02 =++LCis

LRs

LCL

RL

Rs 122

2

2,1 −⎟⎠⎞

⎜⎝⎛±−=

We define, damping factor L

R2

=α

resonance frequency, LCo1

=ω

222,1 os ωαα −±−=

1st case



If οωα = , the roots are equal 21 ss = and the circuit is called critically damp and the solution will be

of the form ( ) tsetAAti 1

21)( +=

If oωα > , the roots are real and the circuit is called over damped. The solution takes the form: tsts eAeAti 21

21)( +=

If oωα < , the roots are complex and the circuit is called under damped.

The roots are rjs ωα ±−=2,1

where, 221 αωω −=r

( ) ( )tAtAt rreti ωωα sincos 21 +−= Q. The 2nd order RLC circuit shown in the figure has the source voltage ,220VVs = Inductance

,12 += mL Capacitance FC µ05.0= and Resistance Ω= 160R . The initial value of the capacitor

voltage is 0=oV . If switch s1 is closed at t=0, determine:

a) Expression for the current i(t) b) Conduction time of diode Solution:

sec/101

sec/000,401022

1602

5

3

radLC

radL

R

o ==

=××

== −

ω

α

1. oωα < it is an under-damped circuit . The solution is

( ) ( ) ( )itAtAeti t −+= − ωωα sincos 21

sec/652,9122 rador =−= αωω

At t=0 i(t=0)=0 From (i), i(0)=A1=0 ( ) tAeti t ωα sin. 2

−=

( )teteAdtdi

rt

rrt ωαωω αα sincos.2

−− −=

rt

Adtdi ω2

0

==

Initial rate of rise of current is limited only by inductor L.

30 102

220−

= ×==

LV

dtdi s

t

L

VA s

r =ω2

2.12 ==L

VA

r

s

ω



D1

R

Vs

S1

L

Dm

if

i+

-

i2

t1 t

t2t1

if

I1

i2

i

Waveform

( ) ( )teti t 652,91sin2.1 000,40−=

b) Conduction time of diode, str

µπωπ 27.34

652,911 ===

Freewheeling Diodes In the circuit steady state current after switch s1 is close is

equal to R

Vs . If switch s1 is now open the energy stored

2

21

⎟⎠⎞

⎜⎝⎛

RV

L s in inductance L will appear in the form of arc at

the opening contact of switch s1. In order to avoid such occurrence a diode called freewheeling diode is connected across RL as shown in the figure. The circuit operation can be divided into two mode. The circuit operation has two modes. Mode 1 begin when the switch is closed at t=0. Mode 2 begin when the switch is then opened. Mode-1 Diode Current

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

tLR

s eR

Vi 11

When the switch is open at t=t1, current at that time becomes

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−==

1111

tLR

s eR

Vtti

Mode-2

220 Ri

dtdiL +=

( ) ( )[ ] ( ) 00 222 =+− sRIissIL

( ) 12 0 IR

Vi s ==

( ) ( )R

VLsIRL s

s .2 =+

( ) ( )LRsRLRLV

sI s

/2 +=

( ) tLReIti == 12

Single Phase Half-wave Rectifier: Rectifier is a circuit that converts an AC signal into unidirectional signal.

i2 R

L



During +ve half cycle of i/p voltage diode D1 conduct and the i/p voltage appears across the lead. During –ve half cycle of input voltage, the diode is in blocking condition and the o/p voltage is zero. Performance Parameter A rectifier is a power processor that should give a DC output voltage with minimum amount of harmonic contains. There are different types of rectifier circuit. The performances of rectifier are normally evaluated in terms of following parameters. Average value of o/p (lead) voltage Vdc Average value of o/p current Idc Output DC Power Pdc = Vdc × Idc RMS value of o/p voltage Vrms RMS value of o/p current Irms Output AC Power Pac = Vrms . Irms

Efficiency (rectification ratio) ac

dc

PP

η =

Output Voltage can be considered as being composed of two components (1) DC value (2) AC component or ripple.

Effective AC component of output voltage is 22dcac VVV

rms−=

From factor, measure shape of output voltage is dc

rms

VV

FF =

Ripple factor measures ripple content is 11 2222

−=−⎟⎟⎠

⎞⎜⎜⎝

⎛=

−== FF

VV

VVV

VV

Rdc

rms

dc

dcrms

dc

acF

Transformer Utilization Factor is A

dc

ss

dc

VP

IVP

TUF ==

where, Vs & Is are rms voltage and rms current of transformer secondary respectively. Let us consider the waveform

+

-

Vs=Vm sinwt R VL

Is

AC, Supply Vp

/2 2

2wt

wt

wt22

/2

/2

o

o

o

o

Vm

V2

Vm

IsVm/R

VD

-Vm



Here, Vs = sinusoidal input voltage is = instantaneous input current is1 = fundamental component of instantaneous input current If φ is the angle between fundamental component of input current and voltage, then the angle φ is known as displacement angle. Displacement Factor is defined as φφ cos i.e.cos =DF . The harmonic factor of input current is

defined as 21

221

2

22

111

1

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−

⎥⎥⎦

⎤

⎢⎢⎣

⎡=⎟

⎟⎠

⎞⎜⎜⎝

⎛ −=

s

s

s

ss

II

III

HF

Both 1

and ss II are expressed in rms. Input Power Factor is defined as:

VAP

II

IVIV

PF ac

s

s

ss

ss ⇒== φφ

coscos

11

Crest Factor, CF is to specify the peak current ratings of devices and components CF of input current is defined as:

( )

s

s

IpeakI

CF =

Q. The half wave rectifier has a purely resistive lead of R. Determine: a. Efficiency b. Foam Factor c. Ripple Factor d. Transformer Utilization Factor e. Peak Inverse Voltage of Diode D1 f. Crest Factor of i/p current Solution:

dttV

TV

PP

T

Ldc

ac

dc

)(10∫=

=η

dttVT

T

m∫=2

0sin1 ω t

2

0

cosT

m tT

V⎥⎦⎤

⎢⎣⎡−=

ωω

[ ]2/cos1 TTVm ω

ω−=

Ip

o

-Ip

VsIs

Is1

Input Current

FundamentalComponent

Input Voltage

wt



( )ππ

cos12

−= mV

mm V

V318.0==

π

R

VR

VI mdc

dc318.0

==

( ) 21

0

21⎥⎦⎤

⎢⎣⎡= ∫

T

Lrms dttVT

V

21

2

0

2 sin1⎥⎦⎤

⎢⎣⎡= ∫

T

m tdtVT

ω

21

2

0

2

22cos1

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

= ∫Tm dtt

TV ω

21

2

0

2

22sin

2 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎥⎦⎤

⎢⎣⎡ −=

Tm ttT

Vω

ω

21

2

2sin

22 ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −=

ωωTT

TVm

21

2

2.

2 ⎥⎦

⎤⎢⎣

⎡=

TT

Vm

mm

rms VV

V 5.02

==

RV

RV

I mrmsrms

5.0==

( )

RV

IVP mrmsrmsac

25.0==

dcdcdc IVP =

( )

RVm

2

=31.0

( )

( )%5.40

5.0318.0

2

2

===m

m

ac

dc

VRV

PP

η

b) 57.1318.0

5.0===

m

m

dc

rms

VV

VV

FF

c) 21.112 =−= FFRF

d) ( )2

sin1 21

2

0

mT

msV

dttVT

V =⎥⎦⎤

⎢⎣⎡= ∫ ω



current load5.0

==RV

I ms

( )

286.05.0.

.2.318.0 2

=×

==mm

m

ss

dc

VVRRV

IVP

TUF

e) PIV of diode = Vm

f) RV

IR

VpeakI m

sm

s5.0

)( =⇒=

2)(

==s

s

IpeakI

CF

Single Phase Half wave Rectifier with RL load

Current oI continuous to flow even after source voltage sV has become –ve, this is because of

presence of inductance L in the lead circuit. Voltage RIV oR = has the same wave shape as that of

oI . Conduction period of 1D will extend beyound 180° until the current becomes zero at σπω +=t .

Average Output Voltage, ( )[ ]σππ

ωωπ

σπ+−== ∫

+cos1

2sin

21

0

mmdc

VtdtVV

Average Output Current, R

VI dc

dc =

Vs=Vm sinwt R

Dm

is

AC, Supply Vp

D1

VR

VL

io+ -

VSVS=VS-VR

VR=i°R

2+6is=i

°

v°

VD

D1

6



+

-

R

Vp

D1

iL+

-

Vs E

The average output voltage and current can be increased by making 0=σ , which is possible by adding freewheeling diode mD . The effect of this diode is to prevent a negative voltage appearing

across the load and as a result the magnetic stored energy is increased. At ωπ

== 1tt , the current

from diode 1D is transferred to mD and this process is called commutation of diode.

Half-wave Rectifier as battery charges If the output is connected to a battery, the rectifier can be used as a battery charger. For EVs > , the

diode 1D conducts

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

=

−

m

m

m

VE

VE

EV

1sin

sin

sin

α

α

α

⎥⎦

⎤⎢⎣

⎡−=

<<−=

−=

απββωαω tfor

REtV

REV

I msL

sin Current, Charging

Q. The battery shown in the figure is VE 12= and its capacity is 100 watt hour. The average charging current should be AI dc 5= . The primary input voltage HzVVp 60,120= and transfer has a

turn ratio of 1:2=N . Then calculate a) The conduction angle δ of the diode b) Current turning resistance R c) Power ratings RP of R

d) Charging time oh in hours

e) Rectifier efficiency η f) Peak inverse voltage Solution: VE 12=

V

nV

V

VV

ps

p

602

120

120

===

=

VVVV sm 85.842 ==

+

- E

R

V1

D1

i2+

-

Vs

a) Circuit Diagram

wt

wtβ π 2π

Vm+E

β

α

α

Vs-EVm-E

2ππ 3π

Vm-ER

i

Vs=Vmsinωt



87.171180

13.8sin 1

=−°=

°=⎟⎟⎠

⎞⎜⎜⎝

⎛= −

αβ

αmV

E

a) Conduction Angle, °=−= 74.163αβδ

b) Average Charging Current, tdR

EtVI m

dc ωω

πβ

α∫ ⎟⎠⎞

⎜⎝⎛ −

=sin

21

[ ]

( ) ( )( )αααπαππ

ααββπ

ωωπ

βα

EVEVR

EVEVR

VR

mm

mm

m

++−+−−=

++−−=

−=

coscos2

1

.cos.cos2

1

tE-tcos2

1

( )πααπ

EEVR

I mdc −+= 2cos22

1

Ω= 26.4R

c) RMS Battery Current, tdR

EtVI m

rms ωω

πβ

α

2

∫ ⎟⎠⎞

⎜⎝⎛ −

=sin

212

( )

4.67

cos42sin2

222

1 22

2

2

=⎥⎥⎦

⎤

⎢⎢⎣

⎡−+−⎟⎟

⎠

⎞⎜⎜⎝

⎛+= αααπ

πEV

VE

VR m

mm

AI rms 2.8=

Power Rating of R, WRIP rmsR 4.28626.44.67.2 =×==

d) Power deliver to the battery, WEIP dcdc 60512 =×==

100=dco Ph

hrsho 667.160

100==

e) Rectifier efficiency, powerinputtotal

batterythetodeliverPower=η

%32.174.28660

60=

+=

+=

Rdc

dc

PPP

f) PIV diode VEVm 85.961285.84 =+=+=

Single Phase Full Wave Rectifier



+

- EVp

VD1

i2+

-

+ -

D2

VD2

+ -

-+ Vs i2

D12

2wt

wt2o

o

o

Vm

V2

-2Vm

VD

Vm

-Vm

VD1=0

VD2

VD2=0

VD1

Vs



wt2

o

Vm

VL

Single Phase Full Wave Bridge Rectifier During positive half cycle of input voltage, the power is supplied to the load through diode

21 DandD during negative half cycle the diode 43 DandD conduct.

Q. A single phase full wave center tapped transformer has a purely resistive load of R. Determine: a) Efficiency b) Wave-form Factor c) Ripple Factor d) TUF e) PIVof diode D f) CF of input current. Solution: Average Output Voltage:

tdVV Ldc ωπ

π

∫=0

1

mm V

V6366.0

2==

π

Average Load Current:

( )26366.0

.

6366.0

RV

IVP

RV

RV

I

mdcdcdc

mdcdc

==

==

RMS value of output voltage:

21

0

21⎥⎦⎤

⎢⎣⎡= ∫

πω

πtdVV Lrms

21

0

2

21

0

2

2cos1

sin1

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

=

⎥⎦⎤

⎢⎣⎡=

∫

∫ 2

tdtV

tdtV

m

m

ωωπ

ωπ

π

π

21

0

2

22sin

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎥⎦⎤

⎢⎣⎡ −=

πωωπ

ttVm

Vp

+

-

D1

D2 D4

D3

R

+

-V1

2

2wt

2o

o

Vm

-Vm

Vs

Vm

-VmVL

Vs

wt



mm

m

VV

V

707.02

.2

21

2

==

⎥⎦

⎤⎢⎣

⎡= π

π

RMS Value of Load Current:

R

VR

VI mrms

rms707.0

==

( )2707.0

RV

IVP mrmsrmsac ==

a) ( )( )

%81707.0.

.6366.0 2

=== 2m

m

ac

dc

VRRV

PP

η

b) 11.16366.0707.0

===m

m

dc

rms

VV

VV

FF

c) 482.012 =−= FFRF

2) m

sV

Vd =

2

.2.2

2

rmss

rmss

II

RRI

I

=

=

RV

RV

RV

m

mm

5.022..2

=

==

( )%32.57

5.0226366.0

2

2

=××××

××==

mm

m

ss

dc

VVRRV

IVP

TUF

e) PIV of diode = mV2

f) CF = ( )

2.

2==

m

m

rms

s

VRRV

IpeakI

Q. A single phase full wave bridge rectifier has a purely resistive load of with the supply voltage 220V and supply frequency 50Hz. Then calculate:

a) Efficiency b) Ripple Factor c) TUF d) PIV of diode e) CF Solution:

VVVR m 31122025 =×=Ω=

mrmsmdc VVVV 707.06366.0 ==

RV

IR

VI m

rmsm

dc707.06366.0

==



a) ( )

( )%81

707.0..6366.0 2

=== 2m

m

ac

dc

VRRV

PP

η

b) 11.1==dc

rms

VV

FF

482.012 =−= FFRF

c) ( ) %81707.0.

6366.0===

mm

m

ss

dc

VVRV

IVP

TUF

d) mVPIV =

e) ( )

2.

.2.===

m

m

s

s

VRRV

IpeakI

CF

Single Phase Full Wave Rectifier with RL Load: With resistive load, the load current is identical in shape to the output voltage. In practice, most loads are inductive to a certain extent and the load current depends on the value of load resistance R and load inductance L. This is shown in the figure.

Fig: Full-wave Bridge Rectifier with R-L Load If sms VtVV 2sin == ω is the input voltage, the load current Li can be found from:

tVERidtdi

L sLL ωsin2=++

which has the solution of the form,

( )REeAt

Vi

tLR

sL −+−=

⎟⎠⎞

⎜⎝⎛ −

1sin22

θω

where, ( ) 21

22⎥⎦⎤

⎢⎣⎡ += LRz ω

⎟⎠⎞

⎜⎝⎛= −

RLωθ 1tan

Case 1: Continuous Load Current At 1, Iit L == πω

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−+= ω

πθ .

11sin2 L

Rs ez

VREIA

D1

D2 D4

D3 R

L

-

-

+

E

iL

a) Circuit Diagram

2

2

iL

IL

iLMax

iLmin

/2

b) Wave Form



( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−++−= θθω sin

22

sin22

1ss

LV

REIt

VI

At ( ) 10,0 Itit L === ωω

RE

e

eVI

LR

LR

s −

−

+=

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

ωπ

ωπ

θ

1

1sin22

1

After substitution, we get

( ) 0;0sin.

1

2sin22

≥≤≤−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+−=⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ −

L

tLR

LR

sL it

REeR

e

tV

i πωθθωωπ

Case 2: Discontinuous Load Current Load Current Flows only during the period βωα ≤≤ t

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

mVE1sinα

At ( ) 0; == tit L ωαω

( )⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−= ω

π

θα LR

s ezV

REA sin

21

( ) ( ) 0sin2

sin2

=−⎥⎥⎦

⎤

⎢⎢⎣

⎡−−+−=

⎟⎠⎞

⎜⎝⎛ −

REe

zV

REt

zV

i LR

ssL

ωβα

θαθω

β can be determined from this transidental equation 0sin =− − xex by an iterative method of solution. rms diode current

21

2

21

⎥⎦⎤

⎢⎣⎡= ∫

β

αω

πtdiI Lr

Average diode current

( )∫=β

αω

πtdiI Ld 2

1

wto

Vm

V

E

iL

α β

π

Wave Form



Multiphase Star Rectifier: In single phase full wave rectifier, the output contains the harmonics and the frequency of fundamental component is 2 times the source frequency ( )f2 . In case of multi phase rectifier, the fundamental frequency of the harmonics is also increased and is 9 times the source frequency ( )f9 . This rectifier is also known as star rectifier. For simplicity, let us assume three phase star rectifier.

( )

( ) ⎟⎠⎞

⎜⎝⎛ +=°+=

⎟⎠⎞

⎜⎝⎛ −=°−=

=

32sin120sin

32sin120sin

sin

πωω

πωω

ω

tVtVV

tVtVV

tVV

mmb

mmy

mr

v

a

12

3

4

N

Va

R

+

-

VL

iL

D1

D2

D3

D4

Circuit Diagram



At °= 30tω

↓=

−=

↑=

2

2

mb

my

mr

VV

VV

VV

So, rV is most positive phase voltage and 1D conducts from 30° to 150°. At °= 150tω , yV is most

positive phase voltage and 2D conducts from 150° to 270°. At °= 270tω

↑=

↓=

−=

2

2m

b

my

mr

VV

VV

VV

So, bV is most positive phase voltage and 3D conducts from 270° to 30° next cycle. Hence output

consists of 3 phase voltages .,, byR VVV Each lasting for a period of 120°.

Q. A φ3 star rectifier has a purely resistive load with ΩR . Determine a) efficiency b) foam factor c) Ripple factor d) TUF e) PIV of each voltage f) Peak current through a diode. If the rectifier deliver

.30AI dc = At an output of VVdc 140= .

Solution: Average Output Voltage

∫⎟⎠⎞

⎜⎝⎛

= 65

6

sin

321 π

π ωωπ

ttdVV mdc

65

6

cos

321

π

πωω

π ⎥⎦

⎤⎢⎣

⎡=tVm

mV827.0=

Y R B

-120 -60 30 60 120 180 240150 270 300 360

At °= 150tω

mb

my

mr

VV

VV

VV

−=

↑=

↓=

2

2



RV

RV

I mdcdc

827.0==

( )21

65

6

2sin

321

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

= ∫π

π ωωπ

tdtVV mrms

mV84068.0=

RV

I mrms

84068.0=

( )R

VIVP m

dcdcdc

2827.0=×=

( )284068.0R

VIVP m

rmsrmsac =×=

a) ( )( )

%77.9684068.08272.0

2

2

===ac

dc

PP

η

b) 0165.1827.0

84068.0===

m

m

dc

rms

VV

VV

F

c) ( ) 1824.010165.11 22 =−=−= FFRF

d) 2m

sV

V =

RIRI rrmss =23

R

VII mrms

s4855.0

3==

( )

( ) 6643.04855.03

2827.03

2

=××

×==

mm

m

ss

dc

VVV

IVP

TUF

e) mVPIV 3=

3φ bridge rectifier: This is a full wave rectifier. The pair of diodes which are connected between that pair of supply lines having the highest amount of instantaneous line to line voltage will conduct. Phase Voltages:

D1

D4 D6

D3

R +

-

RVL

D2

D5

RNr

VB

Vy

iL

Fig : 3φ Bridge Rectifier



( )°−==

120sinsin

tVVtVV

my

mr

ωω

( )°+= 120sin tVV mb ω ( )°−−=−= 120sinsin tVtVVVV mmyrry ωω

⎟⎠⎞

⎜⎝⎛ °−+

⎟⎠⎞

⎜⎝⎛ °+−

=2

120cos.2

120sin2. ttttVmωωωω

( )°−= 60cos23.2 tVm ω

( )( )tVm ω+−°= 3090cos3

( )°+= 30sin3 tVm ω

( )°−= 90sin3 tVV myb ω

( )°−= 200sin3 tVV mby ω

Average Output Voltage, tdVV rydc ωπ

π

π∫

⎟⎠⎞

⎜⎝⎛

=2

63

1

tdtVm ωπωπ

π

π ⎟⎠⎞

⎜⎝⎛ += ∫ 6

sin3

3

1 2

6

mV6539.1=

21

2

6

22 t6

sin

3

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ += ∫

π

π ωπωπ

dtVV mrms

mV6554.1= Q 1. A 3φ bridge rectifier has a purely resistive load of R. Determine a) efficiency b) foam factor c) Ripple factor d) TUF e) PIV of each diode f) Peak current through a diode. The rectifier deliver

AI dc 60= at an output voltage of VVdc 7.280= and the source frequency is Hz60 . Solution:

mrmsmdc VVVV 6554.1,6539.1 ==

RV

IR

VI rms

rmsdc

dc == ,

( )26539.1R

VIVP m

dcdcac ==

30 60 90 120 150 180 210 240 270 300 330 360

wt

-30

VDV6yVry Vrb Vyb Vyr Vbr Vby



R

L

- E

( )26554.1R

VIVP m

rmsrmsac ==

a) ( )( )

%8.99.6554.1.6539.1

2

2

===RVRV

PP

m

m

ac

dcφ

b) 009.16539.16554.1

===m

m

dc

rms

VV

VV

FF

c) 042.012 =−= FFRF

d) 2

V voltage,secondary er transformof voltagerms smV

=

current diode of valuerms=aI RIRI rmsa

223 =

3rms

aI

I =

RIRI as22 2=

rmsas III322 ==

e) ss

dc

IVPTUF

3=

( )mm

m

VVRRV

6554.123.326539.1 2

××××

××=

f) VVPIV m 9.29372.16933 =×==

g) ARV

CurrentdiodePeak m 9.6267.4

72.16933=

×==

RV

I dcdc =

Ω== 67.4dc

dc

IV

R

φ3 bridge rectifier with RL Load:

The output voltage becomes

32

3sin2 πωπω ≤≤×= tfortVV ryry

When voltagerms line toline =ryV , the lead current oi can be found from,

form theofsolution thehaswhich sin2 tVERdtdiL ryio

ω=++

( ) ( )1sin2

1 −−+−=⎟⎠⎞

⎜⎝⎛−

REeAt

zV

itR

ryo

ωθω

Where, load impedance, ( )21

222 LRz ω+=

And load impedance angle, ⎟⎠⎞

⎜⎝⎛= −

RLωθ 1tan

A1 in the equation can be determined from the condition at oo Iit == ,3πω



⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−+=

ωπ

θπ 31 3

sin2 c

Rry

o ezV

REIA

After substitution,

( ) ( )23

sin2

sin2 3 −−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−++−=

⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛

REe

zV

REIt

zV

it

cR

ryo

ryo

π

θπθω

ooo Ititi =⎟⎠⎞

⎜⎝⎛ ==⎟

⎠⎞

⎜⎝⎛ =

332 πωπω

( ) 03

1

3sin

32sin2

3

3

≥−−

−

⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −

=⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ −

o

LR

LR

ry

o IForRE

e

eVI

ωπ

ωπ

θπθπ

After substitution,

( )0&

32

31

3sin

32sinsin

2

3

≥≤≤−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −+−

=⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛−

ot

LR

mo itFor

RE

e

t

zV

i πωπθπθπθω

ωπ

Rectifier Circuit Design: The design of rectifier involves determining the ratings of diodes like PIV, peak current, etc. The output of rectifier contains harmonics and filters can be used to smooth out DC output voltage of the rectifier and these are known as DC Filter. Due to rectification action, the input current of the rectifier contains harmonics also. An AC Filter is used to filter out some of the harmonics from the supply system. Q. A φ3 bridge rectifier supplies a highly inductive load such that the average load current Aldc 60=

and ripple contain is negligible. Determine the ratings of the diodes if the line to neutral voltage of star connected supply 120V at 60Hz. Solution:

Average diode current AtdtdI dc 206021

21 6

5

6

65

6

=××== ∫∫π

π

π

π ωπ

ωπ

RMS diode current = AtdI dc 64.3421 2

1

65

6

2 =⎥⎦

⎤⎢⎣

⎡∫

π

π ωπ

PIV = VVm 294120233 =××=×

Output Voltage with LC Filter: If dcV is less than mV current Li will begin to flow at αω =t which is given by:

Figure



dcm VV =αsin which gives ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

m

dc

VV1sinα

The output current Li is given by

dcmL

c VtVdtdiL −= ωsin

which can be solved for Li

( ) ( ) ( ) αωαωω

ωαω

ωωω

α≥−−−=−= ∫ tfort

LV

tL

VtdVtV

LCi

c

dc

c

mt

dcmL coscossin1



Chapter – 4 Thyristor Characteristics

A thyrisor is 4 layered semi-conductor device of PNPN structure with 3 PN Junction. It has three terminals. When anode voltage is made positive with respect to cathode, the junction

2 1 JandJ are forward bias, the junction 2J is reverse bias and only a small leakage current flows from anode to cathode. The thyristor is then said to be in forward blocking or off state condition and leakage current is known as off state current dI . If anode to cathode voltage akV is increased to

sufficiently large value, the reverse bias junction 2J will break. This is known as avalancy

breakdown and the corresponding voltage is called forward break down voltage ( )BDV . Since the

other junction oJandJ1 already forward bias there will be free movement of carriers across all the

junctions resulting in a large forward anode current. The device will then be in a conducting state or on state. The anode current must be more than a value known as latching current LI . The latching

current LI is the minimum anode current required to maintain the thyristor in on state immediately after a thyristor has been turned on and the gate signal has been removed. Once a thyristor conducts, it behaves liek a conducting diode and there is no control over the device. However, if the anode current is induced below a level known as holding current hI , a

depletion region is developed around the junction 2J due to reduce no. of carriers and thyristor will be in blocking state.

G

R

A p

n

p

n

J1

J2

J3Ggate

Gcathod

Acathod

Fig : Thyristor Symbol and Thyristor Junction

Forwardvoltage drop

gate troggered

Forwardbreak downvoltage

Forward leavage current

VAkVBR

Reverseleavagecurrent

IL

IH

Fig : v-i characteristics



When the cathode voltage is positive with respect to anode, the junction 2J is forward biased

but the junction 31 JandJ are reverse biased. The thyristor will be in reverse blocking state an a

reverse leakage current would flow through the device. In practice, the forward voltage is maintain below BOV and the thyristor is turn only by applying a positive voltage between its gate and cathode.

Two transistor model of thyristor: A thyristor can be consider as two complementary transistor one PNP transistor 1θ and the

other NPN transistor 2θ as shown in the figure. The collector current is

CBOEc III += α

where, junction baseor collect ofamount leakage=CBOI

gaincurrent basecommon =α

For transistor θ1, ( )iIII CBOAc −+=

11 1α

where, 11 for gain current θα =

junction basecollector ofcurrent leakage1

=CBOI

For transistor θ2,

( )iiIII CBOkc −+=22 1α

where, 22 for gain current θα =

2for gain current leakage2

θ=CBOI

Combining equ(i) and (ii), we have:

21 ccA III +=

( )321 21 −+++= CBOkCBOA IIII αα

But, ( )4−+= GAk III

n

p

n

pJ2

P

P

GIG

θ1

θ2

AIT

I2

I3

Ik

K

IGG IB2

IG1

θ1 α1

IA=ITIB1=IC2

α2θ2

Ik

K

A

a) Basic Structure b) Equivalent Circuit

Fig : Two transistor model of Thyristor



Now,

21 221 CBOGACBOAA IIIIII ++++= ααα

( ) ( )51 21

2 21 −+−

++=

ααα CBOCBOG

A

IIII

( )[ ] GCBOCBOA IIII 221 211 ααα ++=+−

If the gate current GI is suddenly increased, this will immediately increase anode current AI , which would

further increase 21 αα and . The increase in the value of 21 αα and would furhter inrease AI I. Therefore,

there is a regenerative of positive feed back effect. If 21 αα + tends to infinity, the denominator of equation

(5) approaches zero, resulting in a large value of anode current AI and the thyristor will turn on with a small

gate current. Thryristor turn on:

1. Thermal 2. Light 3. High Voltage

4. dtdv

5. Gate Current A thyristor a turn on by increasing the anode current. This can be done in one of the following ways: 1. Thermal : If the temperature of thyristor is high, there will be increse in the number of electro-hole pair, which would increase the leakage current and cause 21 αα and to increase due to regenerative action.

21 αα + may tend to be unity and the thyristor may be turn on. This type of turn on may cause thermal

runaway and is normally avoided. 2. Light : If light is allowed to strike the junction of thyrsitor, the electron hole pair will increase and the thyristor may be turn on. 3. High Voltage : If forward anode to cathode voltage is greater than the forward breakdown voltage BOV

sufficient, leakage current will flow to initiate regenerative turn on. This type of turn on may be destructive and should be avoided.

4. dtdv

( ) ( )dt

diV

dtdV

CVCdtda

dtdi j

ji

jjjjj2

2

2

22222+===

If the rate of rise of anode to cathode voltage is high, the charging current of capacitive junction may be sufficient enough to turn the thyristor, a high value of charging current may damage the thyristor and device

must be protected against high .dtdv

5. Gate Current: If a thyristor is forward bias, the injection of gate current by applying positive voltage between the gate and cathode terminals would turn on the thyristor. As the gate current is increased, the forward blocking voltage is decreased as shown in the figure.



:Protectiondtdi

A thyristor requires a minimum time to spread the current conduction uniformly through the junction. If the rate of rise of anode current is very fast compared to the spreading velocity of turn on process, a localize “hot spot” heating will occur due to high current density and the device may fail as a result of excessive temperature.

The practical devices must be protected against high dtdi . Under steady state operation mD

conduct when the thyristor 1T is off. If 1T is free when mD is still conducting dtdi can be very high

and limited only by the stray inductance of the circuit. In practice dtdi is limited by adding a series

inductor sL as shown in the figure.

ss LVdtdi

=

Protectiondtdv :

VAk

ILIA

O V3 V2 V1 VB0

IG=O

IG1 > IG2 > IG3

V1 > V2 >V3

iT

IG1IG2

Fig : Effect of Gate Current on Forward Blocking Voltage

iT1 Im

loadR2

C2

Dm

+

-

Ls

CsVs

+

-

CRs

A

k

T1

S1

0.632VsVs

t=τ

VAk

O t



If switch 1s is close at ,0=t a step voltage will be applied across thyristor 1T and dtdv may be

enough to turn on the device. dtdv can be limited by connecting capacitor sC . When 1T is turn on, the

discharge current of capacitor is limited by resistor sR .

i

CRV

dtdvor

Vdtdv

ss

s

s

9632.0

,

632.0

−=

=τ

where, ssCR=τ

current discharge where, == TDTD

ss I

IV

R

It is possible to use more resistor for dtdv and discharging current such that:

21 RR

VI s

RD +=

Types of Thyristor: Depending upon the physical construction, turn on and turn off behaviour, thyristor can be broadly classified into following categories:

1. Phase Control Thyristor (SCR) 2. Fast Switching Thyristor (SCR) 3. Gate Turn off Thyristor (GTO) 4. Bidirectional Triode Thyristor (TRIAC) 5. Reverse Conducting Thyristor (RCT) 6. Static Induction Thyristor (SITH) 7. Light Activated Silicon-Controlled Rectifier (LASCR) 8. FET Controlled Thyristor (FET-CTH) 9. MOS Controlled Thyristor (MCT)

R1

Vs

+

-

R2

T1

Cs

Ds



Phase Control Thyristor: This type of thyristor generally operate at the line frequency and is turn off by natural communication. The turn off time is of the order 500-100 sµ . This is most suited for low speed switching application and is also known as converter thyristor. The modern thyristor use an amplifying gate where auxiliary thyristor AT is gated by a gate

signal and then the amplified output of the AT is applied as a gate signal to the main thyristor mT .

Fast Switching Thyristor: These are used in high speed switching application with forced communication (example chopper, inverter). They may have fast turn off time. Generally in a range 5-15 sµ depending on the voltage range. Gate Turn-off Thyristor: A GTO line an SCR can be turn on by applying a positive gate signal. However it can be turn off by negative gate signal. It eliminates the requirement of equipment. Bi-directional Triod Thyristor: A TRIAC can conduct in both directions and is normally used in AC phase control. It can be considered as two SCR connected in anti-parallel with a common gate connection as shown in the figure. FET Controlled Thyristor:

A

TmTAIG

Fig : Amplifying Gate ThyristorK

MT1

MT2

MT2

TA

MT2TRIAC Sumbol Equivalent of TRIAC

M1T1

R

G



A FET controlled thyristor combines a MOSFET and thyristor in parallel as shown in the figure. If a sufficient voltage is applied to the gate of MOSFET typically 3r, a triggering current for the thyristor generated internally. MOS Controlled Thyristor: It is basically a thyristor with two MOFET built into the get structure. One MOSFET is used for turning on the MCT and other for turning off the device with the application of negative voltage pulse, on FET gets turn on and off FET is off. With on FET on, current begins to flow from anode A through ON FET and then as the base current and emitter current of NPN transistor and then to the cathode K. This turn on NPN transistor as a result collector current begins to flow in NPN transistor as OFF FET is off. This collector current of NPN Transistor acts as the base current of PNP transistor. Subsequently, PNP transistor is also turn on, MCT turn on. For turning off, OFF FET is energise by positive voltage pulse at the gate. With the application of positive voltage off FET is turn on and ON FET is turn off. After off FET is turn on emitter base terminal of PNP transistor are short circuited by off FET. So anode current begins to flow through off FET and therefore base current of PNP transistor begins to decrease. Further, the collector current of PNP transistor that forms a base current of NPN transistor also become to decrease. As a result MCT turn off. Thyristor Turn Off:

a) Equivalent Circuit

on FET(p-channel)

npn

(off-FET)G

(n-channel)

pnp

G

A

k

off

on

channelPvechannelNve

−→−−→+

O

i

α

2ππ

Vm

VmRL

O

wt

wt

v

i

VAK+ -

R1V

Fig: Line Commutated Thyristor Circuit



A thyristor which is on the ON state can be turned off by reducing the forward current to a level below the holding current .HI Series Operation of Thyristor: For high voltage application, two or more transistors can be connected in series to provide the voltage rating. However, due to production spread the characteristics of thyristor of the same type are not identical. Series Operation of Thyristor: Consider n thyristor connected in series as shown in the figure:

Is

O

V1 V2

T1

T2

ON-state

V

off-state

Fig: Off State Characteristics of Two Thyristor



rmsbmn IIIIII −=−= 21 & where, I = total string current Voltage across SCR1 = RIVbm 1= Voltage across (n+1) SCR = ( ) RIn 21− Now, ( ) RInRIVs 21 1−+=

( )( )RIInV bmxbm −−+= 1 ( ) ( )[ ]RIIInV bmnbmxbm −−−+= 11 ( ) ( ) bbm IRnRInV ∆−−−+= 11 1

where, bmnbmxb III −=∆ As, bmVRI =1

( ) bbmns InnVV ∆×−−= 1 Parallel Operation of Thyristor: When thyristor are connected in parallel, the load current is not shared equally due to the difference in their characteristics. If a thyristor carries more current then that of others, it’s power dissipation increases there by increasing the junction temperature and decreasing the internal resistance. This inturn will increase it’s current sharing and may damage the thyristor. This thermal runaway may be avoided by having a common heat sink so that all unit operate at same temperature. A small resistance may be connected in series with each thyristor to force equal current sharing but there will be considerable power loss in series resistance. Thyristor Firing Circuit: In thyristor converter, different potential existed various terminals. The power circuit is subjected to a high voltage usually greater than 100V and gate circuit is held at low voltage typically

VGm

R

String voltage

String current

( ) b

sbm

InVnV

R∆−−

=1

I2

I1IT

R1

R2

T2

T1

Fig : Current Sharing Thyristor



12V to 30V and isolation circuit is required between individual thyristor and it’s gate pulse generating circuit. The isolation can be accomplish by either pulse trf or opto couples. An optocoupler could be a phototransistor or photoSCR as shown in the figure. A short pulse to the input of ILED 1D turns on the photo SCR 1T and power thyristor LT is trigger.

A simple isolation arrangement with pulse transformer is shown in the figure. When a pulse of adequate voltage is applied to the base of switching transistor θ1, the transistor saturate and DC voltage ccV appears across the transformer primary, inducing a pulse voltage on the transformer

secondarily which is applied between the thyristor gate and cathode terminal. When pulse is removed from base of the transfer turn off and voltage of opposite polarity is induced across the primary and the freewheeling diode mD conducts.

Uni-junction Transistor (UJT)

T1

kG

A

ILEDR1

V1 D1

Photoseroptocoupler

+

-

+Vc0

R

gate voltage

V1

N2N1

C1

G

D1θ1

R1

IE

EB2

RB2R

B1

RB1

+

-

VB1

Vs

+

-

C

Fig : Circuit

Ip Iv

IE

Peak PointVp

VE(Set)

Vv

SaturationRegion

Valley Point

Negativeresistance region

VE

IE0(VA)

Fig : Characteristics

Vv

Vp

VE

T 2T

T 2T

VB1

Fig : Waveform



UJT is commonly used for generating triggering signals for SCR. A basic UJT triggering circuit is shown in the figure. UJT has three terminals called emitters E, B1 and B2. When DC supply voltage sV is applied, the capacitor C is charged through resistance R. Since

the emitter circuit of UJT is in the open state, the time constant of charging circuit is RCT =1 . When

the emitter voltage EV which is same as the capacitor voltage cV reaches the peak voltage pV , the

UJT turns on and the capacitor C will discharge through RB1 at a rate determine by time constant XCRB12 =τ . When the emitter voltage EV decays to the valley point VV , the emitter ceases to

conduct, the UJT turns off and the charging cycle is repeated. The waveform of emitter and triggering voltage are shown in the figure C. The period of oscillation is approximately given by:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

≈=η1

1ln1 RCf

T

where η is intrinsic stand off ratio. 82.051.0 to=η UJT will turn on if pps VRIV >−

( )iI

VVR

p

ps −−

<

At valley point, VEVE VVII == , for turning off.

VVs VRIV <−

( )iiI

VVR

V

Vs −−

>

The width of triggering pulse is CRBtg .1=

( )iiiV

RBs

−=η

4

210

- (iv)

Q. Design triggering circuit of figure between the parameter of UJT are 51.0,30 == ηVVs V, .10,5.3,10 mAIVVAI Vvp === µ . The frequency of oscillation is Hzf 60= and width of triggering

pulse is stg µ50= . Solution:

msf

T 67.166011

===

VVV sp 8.155.03051.05.0 =+×=+= η Let FC µ5.0=

( )VVVV Dsp 5.0=+≈ η



Ω=−

=−

< mI

VVR

p

ps 42.110

8.1530

Ω=−

=−

> KI

VVR

V

Vs 65.210

5.330

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=η1

1lnRCT

⎟⎠⎞

⎜⎝⎛

−×=

5.011ln5.0

601 R

Ω= KR 7.46 which falls within limiting value we have CRt Bg .

1=

Ω== 1005.0

501BR

Ω== 654104

2s

B VR

η

Programmable Unijunction Transistors (PUT): PUT is a small thyristor shown in figure (a). PUT can be used as relaxation oscillator as shown in figure (b).The gate voltage GV is maintained from the supply by the resistor divided by 21 RandR and determine the peak voltage pV in case of UJT pV is fixed for a device by DC supply voltage. But

pV of PUT can be varied by varying resistor dividor 21 RandR . If anode AV is less than ate voltage

GV , the device will remain in its off state. If AV exceeds gaet voltage GV by one diode forward voltage dV , the peak is reached and the device turns on.

sp VRR

RV .21

2

+=

Intrinsic Ratio

21

2

s

p

VV

RRR+

==η

Period of Oscillation

⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎟⎠

⎞⎜⎜⎝

⎛

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=1

21lnln1

1lnRR

RCVV

VRCRCT

ps

s

η

The gate current GI at valley point is given by

( )G

sG R

VI η−= 1

anode

Gate

Cathod

Rv R1

R2

+

-+

-

+

-VA

anod

+V3

VG

Rk VRk

b) Circuit

a) Symbol



where, 21

21

RRRRRG +

=

η

GRR =1

η−

=11

GRR

Q. Design triggering circuit of figure (i) the parameter of PUT are mAIVV Gs 1,30 == frequency of oscillation ,60Hzf = pulse width is stg µ50= . Peak triggering voltage is .10VVRK = Solution:

msf

T 67.166011

===

VVV Rkp 10== Let, FC µ5.0= CRt kg .=

Ω== 1005.0

50KR

31

3010

===s

p

VV

η

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=η1

1lnRCT

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−×=

311

1ln5.0601 R

Ω= KR 2.82

( )2

1RV

I sG η−=

Ω=⎟⎠⎞

⎜⎝⎛ −= KR 20

130

3112



Chapter - 5 Thyristor Commutation Technique

Introduction: A thyristor is normally switched on by applying a pulse of gate signal. When a thyristor is in conduction mode it’s voltage drop is small. Once the thyristor is turn on, and the output requirement are satisfied, it is normally necessary to turn it off. Commutation is the process of turning off a thyristor and it normally causes transfer of current flow to other parts of the circuit. A commutation circuit normally uses additional component to accomplish the turn off. There are many techniques to commutate a thyristor. However, these can be broadly classified into two types:

1. Natural Commutation 2. Forced Commutation

Natural Commutation: If the source voltage is AC, the thyristor current goes through a natural zero and a reverse voltage appears across the thyristor. The device is then automatically turn off due to the natural behaviour of source voltage. This is known as natural commutation of line commutation . Forced Commutation: In some thyristor circuit, the input voltage is DC and the forward current of thyristor is forced to zero by an additional circuitry called commutation circuit to turn off the thyristor. This technique is called as forced commutation. The forced commutation of thyristor can be achieved by seven ways and can be classified as:

1) Self Commutation 2) Impulse Commutation 3) Resonant Pulse Commutation 4) Complementary Commutation 5) External Pulse Commutation 6) Load Side Commutation 7) Line Side Commutation

Vs= Vmsinwt

+

-

a) Circuit Diagram

Vo

+

-

io

R

wt

V0

i0=

V0R

2π

b) Wave-Form

Fig : Thyristor with natural commutation



Self Commutation: In this type of commutation, a thyristor is turn off due to the natural characteristics of the circuit. When the thyristor 1T is switch on, the capacitor charging current is given by:

( )01=++=+= ∫ tVidT

cdtdiLVVV ccLs

With initial conditions ( ) ( )0;00 === ttVc

The charging current is

( ) tLCVti ms ωsin=

And the capacitor voltage as : ( ) ( )tVtV msc ωcos1−=

where LCm1

=ω

At time LCtt o π== charging current becomes zero and thyristor 1t is switch off itself, ot is commutation time. This method of turning off a thyristor is called self commutation and the thyristor

1t is said to be self commutated. The figure shows a typical circuit where the capacitor is an initial voltage of oV− when the thyristor 1t is fire the current that will flow through a circuit is given by:

Vs

+

-V2 L

CVc+-

+

-

i

T1

a) Circuit Diagram

b) Waveform

w2t2π

πt0

C/2Vs

OO

O

i(E)

LCt

t

t

o

mo

om

π

ωπ

πω

=

=

=



( ) 001==++ ∫ tVidt

CdtdiL c

With ( ) ( ) 00, ==== tiVtV oc

( ) tLCVti mo ωsin=

Capacitor Voltage, ( ) tVtV moc ωcos−=

After time LCtt o π== , the current becomes zero and the capacitor voltage is reverse to ,oV rt is called reversing time. Q. A thyristor circuit is shown in the figure. If thyristor 1T is switched on at t=0, determine the conduction time of thyristor 1T and the capacitor voltage after 1T is turned off. The circuit parameters are .200&50,10 VVFCHL s === µµ The inductor carries an initial current of

AI m 250= . Solution:

( )01=++= ∫ tVidt

CdtdiLV cs

With initial condition ( ) ( ) socm VVtVIti ===== 0&0 By solving above equation, ( ) tIti mm ωcos=

where LCm1

=ω

( ) ( ) smmcc VtCLItVidt

CtV +==+= ∫ ωsin01

-L

CVc+-

i

T1

a) Circuit

wmtπ

π

i(t)

O

Oπ/L

V0

-V0

CV0

+

-

iIm

Vs

T1



Conduction time of thyristor 1T = sLCm µπωπ 12.35

22==

Capacitor Voltage after 1T is turned off sm VCLI +=

Impulse Commutation: It is assumed that the capacitor is initially charge to voltage of oV− . Let us assume that thyristor 1T is initially conducting and carrying a load current of mI . When the auxillary thyristor 2T is fire, thyristor 1T is reverse bias by the capacitor voltage and 1T is turned off. The current through 1T and cease to flow and the capacitor would carry the load current .The capacitor will discharge from oV− to 0 and then charge to the DC input voltage sV . When the capaitor current falls to zero and thyristor

2T turns off. The charge reversal of capacitor from ( )so VV = to oV− is then done by firing thyristor 3T . Thyristor 3T is self commutated. Time required for capacitor to discharge from oV− to zero is called circuit turn off time ( )offt .

CtI

dtIC

V offmt

mooff == ∫0

1

Q. An impulse commutated thyristor circuit is shown in the figure. Determine the available turn off time of the circuit if sos VVFCRVV ==Ω== &5,10,210 µ .

i(t)Im

Oπ/2

wmt

Vc(t)

π/2wmt

Dm

Im

T3

T2

T1-

+Vs

Fig : Impulse Commutated Circuit

m

ooff I

CVt =



Solution:

( ) icics RtVidtC

RVV +=+=+= ∫ 01 v

With initial condition, ( ) soc VVtV −=−== 0

( ) RCt

s eRV

ti−

=2

Capacitor Voltage

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−==+=

−

∫ RCt

scc eVtVidtC

tV 2101

At offtt = ( ) 0== offc ttV

021 =⎟⎟⎠

⎞⎜⎜⎝

⎛−

−RCt

s

off

eV

sRCtoff µ347ln 2 == Line-side Commutation:

In this type of commutation, the discharging and recharging of capacitor are not accomplished through the load and the commutation circuit can be tested without connecting the load. When the thyristor 2T is fired, the capacitor is charged to sV2 and 2T is self commutated. Thyristor 3T is fired to reverse the voltage of capacitor to sV2− and 3T is also self commutated. Assuming that thyristor 1T is conducting and carries a load current of mI thyristor 2T is fired to turn off 1T . The inductor L carries the load current mI and equivalent circuit during the commutation period is shown in the figure.

( )01=++= ∫ tVidt

CdtdiLV cs

With initial condition ( ) mIti == 0 and sc VtV 2)0( −= , Capacitor current

R

i

VsT2

T1-

+V0=Vs

+

-

Im

Vs

T1

T2

T3

Lr

L

Dm

C

Vc(t)Vs+

-

Im

Vs

L+

-

i(t)+ -



( ) ( )1sin3cos −+= tLCVtIti msmm ωω

And capacitor voltage,

( ) ( )iiVtVtCLItV smsmmc −+−= ωω cos3sin

where, LCm1

=ω

The circuit turn off time is obtained from the condition ( ) 0== offc ttV of equation (2). Commutation Circuit Design and Commutation Capacitor: The design of commutation circuit require determining the value of capacitor C and inductor L. In selecting a commutation capacitor, the specification of peak, rms and average current and peak to peak voltage must be satisfied.



Chapter - 6 Controlled Rectifier

Introduction: A diode rectifier provides a fixed output voltage only. In order to obtain controlled output voltage, phase controlled thyristor are used instead of diode. The output voltage can be varied by controlling the decay or fixing angle of thyristor. Since, these rectifiers convert from AC to DC, these controlled rectifiers are also called AC to DC converter and are used extensively in industrial applications specially variable speed drive ranging from fractional h.p. (horse power) to mega watt power level. The phase controlled converter can be classified into two types depending on the input supply.

1. Single Phase Converter 2. Three Phase Converter

Each type can be sub-divided into 1. Semi Converter 2. Full Converter 3. Dual Converter

Semi Converter is the one quadrant converter and it has one polarity of voltage and current. Full Converter is a two quadrant converter and the polarity of its output voltage can be either positive or negative. However, the output current of full converter has only one polarity. Dual Converter can operate in four quadrant and both the output voltage and current can be either positive or negative. Principle of Phase Controlled Converter Operation:

Vo

Vo

Vo

Io

Io

Fig : Semi Converter Fig : Full Converter Fig : Dual Converter

Io



During positive half of input voltage the thyristor anode is positive with respect to its cathode and the thyristor is said to be forward bias. When the thyristor 1T is fired αω =tat , thyristor 1T conduct and input voltage appears across the load. When input voltage starts to be negative at πω =t the thyristor anode is negative with respect to its cathode and thyristor 1T is said to be reverse bias and it is turned off. The average output voltage can be formed from

( )απ

ωωπ

π

αcos1

2sin

21

+== ∫ mmdc

VttdVV

When, 0,

,0

==

==

dc

mdc

V

VV

παπ

α

RMS Output Voltage is given by:

( )21

21

22sin1

2sin

21

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−=⎥⎦

⎤⎢⎣⎡=

2

∫ααπ

πωω

ππ

α

mmrms

VtdtVV

Q. If the converter of above figure has a purely resistive load of R and the delay angle is 2πα = ,

determine a) Rectification Efficiency b) Foam Factor c) Ripple Factor d) TUF e) PIV of thyristor 1T Solution:

2πα =

VsVp+

-

iv

Vo

T1

a) Circuit Diagram

wt2π

wt2π

wt2π

2ππ

π

π

πα

-Vm

o

o

VT1

i0

Vm

V0

Vm

Vs

α

α

α

V0/R

b) Wave Form



( ) mm

dc VV

V 1592.0cos1 =+= απ

R

VR

VI mdc

dc1592.0

==

mm

rms VαVV 3536.0

22sin1

221

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−= απ

π

RV

RV

I mrmsrms

3536.0==

( )R

VIVP m

dcdcdc

21592.0=×=

( )R

VIVP m

rmsrmsac

23536.0. ==

a) Rectification Efficiency, ( )

( )%27.20

3536.0.1592.0

2

2

===m

m

ac

dc

VRRV

PP

η

b) Foam Factor, 21.21592.03536.0

===m

m

dc

rms

VV

VV

FF

c) 983.112 =−= FFRF

d) R

VI

VV m

sm

s3536.0

.2

==

( )

( ) 1014.03536.0

21592.0 2

=×

==mm

m

ss

dc

VRVRV

IIP

TUF

e) mVPIV =

Single Phase Semi Converter: Fig shows a single phase semi converter with highly inductive lead such that current is as same continuous and ripple free.

Vp

T1

T2

R

L

T3

T4E

Vs Vo

+

-

io

a) Circuit

wt2π

wt2π

wt2π

2ππ

π

πα

IA

o

IT1,ID2

v0

Vs

α

α

α

π+α

π+απ

IT2,ID1

IDma

i0

Ia

wt

Ia

b) Wave Form



During a positive half cycle, thyristor 1T is forward bias when 1T is fired at αω =t the load is

connected to the input supply through 21 T andT . During the period πωα ≤≤ t . During the period

from απωα +≤≤ t , the input voltage is negative and freewheeling diode mD is forward bias.

During negative half cycle of input voltage thyristor 2T is forward bias and firing of 2T at

απω +=t will reverse bias mD . The load is connected to the supply through 12 and DT . Average

output voltage is ( )απ

ωωπ

π

αcos1sin

22

+== ∫ mmdc

VttdVV

21

22sin1

2 ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−=

ααππ

mV

Single Phase Semi Converter with R-L Load: In practice a load has finite inductance. The load current depends on the values of load resistor R, load inductor L and battery voltage E as shown in the figure above. A converter operation can be divided into two mode: mode 1 and mode 2. Mode 1: This mode is valid for at ≥≤ ω0 during which the freewheeling diode mD conducts.

( )10 −=++ ERidtdiL L

Initial conduction, ( )oLL Iti == 0ω

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

−−tL

RtL

R

LL eREeII

o

.. 11

for ( )20 −≥Li

At αω =t , load current becomes 011

≥⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ −

LLR

LR

L iforeREeI

o

ωα

ωα

Mode 2: This mode is valid for πωα ≤≤ t while thyristor 1T conducts if tVV oo ωsin2−= is the

input voltage the load current 2,1I during mode 2 can be found as

tVERILdiI sL ωsin222 =++ , whose solution is of the form,

( ) ( ) ( )50sin/222

.

1 −≥−+−=⎟⎠⎞

⎜⎝⎛−

L

tIR

sL IForREeAtZVI θω where,

( )[ ] 21

2 −2+= LRz ω

⎟⎠⎞

⎜⎝⎛= −

RLωθ 1tan

A1 can be determined from the initial condition at ;12 LL iit == αω

( ) ( )6sin.211 −⎥

⎦

⎤⎢⎣

⎡−−+=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

ωα

θα LR

sL eVzR

EIA

With substitution of 1A ,



( ) ( ) ( )70sin.2sin.2212

−≥⎥⎦

⎤⎢⎣

⎡−−++−−=

⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛

L

tLR

sLsL IforeVzR

EIREtV

zI ω

α

θαθω

Applying initial condition ( ) :get we;2 oLL Iti == πω

( ) ( )( )

( )801sinsin.2−≥−−

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−−=⎟⎠⎞

⎜⎝⎛ −

−⎟⎠⎞

⎜⎝⎛

oo LLR

LR

sL IforREeeV

zI ω

πω

πα

θαθπ R

ms current of thyristor, 21

22

21

⎥⎦⎤

⎢⎣⎡= ∫ tdLII

T

r ωπ α

Average current of thyristor, ∫=T

Lav tdiIα

ωπ 221

Single Phase Full Converter: For shown a single phase full wave converter with highly inductive load so that the load current is continue and ripple free. During the positive half cycle thyristor 1T and 2T are forward biased and ao ii = when these two thyristor are fired simultaneously at αω =t , the load is connected to the input supply through 1T and

2T . Due to inductive load thyristor 1T and 2T will continue to conduct beyound πω =t even those the input voltage is already negative. Single Phase Dual Converter:

Vp

T1

T2

R

L

T3

T4E

Vs

wt2ππα π+α

wt

wt

Io=Ia

Vo

Vo

T1,T2 T3,T4Vs

Vs+

-

T1

T4 T2

T3

V0V01

+

-

i0

V02

T12 T4

2

T13 T1

1

Vs+

-

Lr/2 Lr/2

a) Circuit Diagram



If two of these full converters are connected back to back as shown in the figure above both output voltage and load current flow can be reversed. The system will provide four quadrant operation and is called dual converter. The delay angles are controlled such that one converter operates as rectifier and the other as inverter. If 21 αα and are delay angles of converter 1 and converter 2 respectively,

the corresponding output voltages are 21 DCDC VandV .

2211 cos2

&cos2

απ

απ

mDC

mDC

VV

VV ==

21 DCDC VV −=

( )221 coscoscos απαα −=−=

21 απα −=

12 απα −= Since the instances output voltages of two converter are out of phase there will be instantaneous voltage difference. This will result in circulating current between the two converters. The circuiting current will not flow through the load and is normally limited by a circulating current reactor rL as shown in the figure.

( )∫ −=

t

rr

r tdL

iω

απωθ

ω 12

1

( ) tdVVL

t

oor

ωω

ω

απ∫ −+=

1212

1

wt2ππ1 π=α1

wt

Vo1

Vm

Vs

Vm sinwt

Converter 1 o/p

Vm sinwt

α1

2ππ π=α1α1

-Vm sinwt

Converter 2 o/p

Vo2

Vm sinwt



⎥⎦⎤

⎢⎣⎡ −+−= ∫∫ −−2 1

tdtVttdVL

t

m

t

mr

ωωωωω

ω

απ

ω

απ 12sinsin1

( )1coscos2

αωω

−= tLV

ir

mr

Q. A single phase dual converter is operated from 120 V, 60Hz supply and the load resistance is Ω= 10R , the circulating inductance is mHLr 40= . Delay angles are °=°= 120&60 21 αα .

Calcuate the peak circulating current and peak current of converter 1. Solution: At πω 2=t

( ) ( ) ALV

Ir

mr 25.1160cos2cos

2max =°−= π

ω

Peak Load Current = AR

Vm 97.16=

Peak Current of Converter1 A22.2825.1197.16 =+= Three Phase Half Wave Converter: The circuit functions in a manner such that only one thyristor is conducting at any given instant, the one which is connected to the phase voltage having highest instantaneous positive value. Here no any thyristor can be trigger below phase angle of 30° because it remains reverse bias by other conducting phases. The firing angle α for a particular thyristor connected in a particular phase is therefore measure from 30° w.r.t. corresponding phase voltage. 1T can conduct from °°= 15030 totω as phase

1 is most positive in the period 30° to 150°. Similarly 2T can conduct from °°= 270150 totω and 3T

from °°= 30270 totω of next cycle. With the resistive load, there are two modes of conduction:

1. Continuous Conduction Mode ( )°≤ 30α

2. Discontinuous Conduction Mode ( )°> 30α When the firing angle α is taken in between 0° to 30° from the cross over point, the load current is continuous. This is due to the fact that the maximum value of conduction angle of thyristor is 120°. If α is more than 30°, conduction angle will be less than 120° and hence output voltage and current becomes discontinuous.

3

1R

T1

T2

T3

2

a) Circuit Diagram



Average load voltage for continuous conduction mode is ∫+

+= 6

5

6

sin

321 π

α

πα

ωωπ

ttdVV mdc

απ

cos233 mV

=

Average load voltage for discontinuous conduction mode is ∫ +=

π

απ ωω

π 6

sin

321 ttdVV mdc

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++= απ

π 6cos1

23 mV

Rms Output Voltage are

conduction continuousfor 2cos8

3613

21

⎟⎟⎠

⎞⎜⎜⎝

⎛+= α

πmrms VV

conduction ousdiscontinufor 23

sin81

4243

21

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++−= απ

ππαρ

mrms VV

Design of Converter Circuits: The design of converter circuit requires determining the rating of thyristors and diodes. The thyristors and diodes are specified by average current, r.m.s. current, peak current and PIV in the case of controlled rectifier, the current rating of device depend on the delay angle α . The output of converter contains harmonics that depend on delay angle and filters must be design to remove the harmonics. Q. A φ3 half wave converter is operated from φ3 star connected 208V, 60Hz supply and the load resistance Ω= 10R . If it is required to obtain an average output of 50% maximum possible output voltage then calculate a) Delay angle α b) Rms and Average output current c) Average and rms thyristor current d) Rectification efficiency e) TUF f) Input Power Factor (IPF) Solution: 208V, 60Hz R=10 Ω

o 120

V1V2 V3

30 60 150 180 240 230 360

α=300 Continuous conduction)

wt



Phase Voltage ( ) VVs 3208

=

VVV sm 83.1692 ==

( ) ππ 289.16933

233

max×

== mdc

VV

( ) VVV dcdc 23.705.0 max =×=

AR

VI dc

dc 023.710

23.70===

απ

cos233 m

dcV

V =

απ

cos233

23.70 mVV =

( )mode conduction ousdiscontinuin workingisit so309.59 °>=α So, we have:

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++= απω

π 61

23 m

dcV

V

70.67=α

AVV mrms 74.9723

sin81

42453

21

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++−= απ

ππα

AR

VI rms

rms 474.910

74.94===

c) Average current of thyristor, AI

I dcA 34.2

3023.7

3===

Rms current of thyristor, AI

I rmsR 47.5

3==

d) %95.54474.974.94023.723.70

=××

=η

e) 25.047.5

32083

023.723.703

=××

×==

ss

dc

IVP

TUF

f) Output Power, WRIP rmso 81.8961047.9. 22 =×==

Input Power , 84.19703 == ssi IVP

Input P.F. )(455.084.197081.896 lag==



Chapter – 8 DC Choppers

DC Choppers: DC Choppers converts fixed DC input voltage to variable DC output voltage. It can step up or step down DC voltage. The main principal is to chop a constant DC voltage for a particular time interval and by changing the chopping time, output DC voltage can be controlled. Generally, power transistors are used as switch for chopper operation. Principal of Step-Down Operation: When switch ws is closed for a time 1t , the input voltage sV appears across the load. If the switch ws

remains off for a time 2t , the voltage across the load is zero. The chopper switch can be implemented by using

1. Power BJT 2. Power MOSFET 3. Power GTO 4. Forced Commutated Thyristor

The average output voltage is given by:

∫ ∫∫ ====== 1

0 110.11 t

ss

sls

o

t

oa kVtTV

VttTV

dtVT

dtVT

V

where, cycledutyTt

k == 1

periodchoppingT = frequencychoppingf =

Rms Output Voltage is

Vo

Vs

io

Is=Vs/R

t1 t2

t1 t2

T

t

t

Vs

+

-

R

sw

Vo

+

-

io-+

VH

Chopper

a) Circuit Diagram

b) Wave Form

a) Fig : Step down chopper with resistive load



s

kT

so VkdtVT

V =⎥⎦⎤

⎢⎣⎡= ∫

21

0

21

Assuming a loss less chopper, the input power to the chopper is same as the output power and is given by

R

Vkdt

RV

TidtV

TP sokT

oi

20

0

2

0

11=== ∫∫

The effective input resistance seen by the source is kR

RV

k

VIV

Rs

s

a

si ===

The duty cycle k can be vary from 0 to 1 by varying .,1 forTt Therefore, the output voltage can be

varied from 0 to sV by controlling k/ Two types of operation can be defined:

a) Constant Frequency Operation: The chopping frequency f or chopping period ''T is kept constant and on time 1t is varied and this type of control is known pulse width modulation (PWM) control. b) Variable Frequency Operation: The chopping frequency '' f is varied either ''on time 1t or off time 2t is kept constant. This is called frequency modulation. Q. A step down DC chopper has a resistive load of Ω= 10R and the input voltage VVs 220= . When

the chopper remains on its voltage and chopping frequency KHzf 1= . If the duty cycle is %50 determine (i) average output voltage VA ii) Rms output voltage Vo iii) chopper efficiency iv) effective input resistance Ri of the chopper v) on-time and off-time of the chopper. Solution: KmzfkVVRVV chs 1,5.0,2,10,220 ===Ω==

i) ( ) VVVkV chsa 109=−=

ii) ( ) 15.154=−= chso VVkV

iii) Output Power, ( ) ( )

kWR

VVkdt

RVV

fdt

RV

TP chskT chskT o

o 239811 2

0

2

0

2

=−

=−

== ∫∫

Input Power, ( )

kTR

VVVT

dtR

VVV

TidtV

TP chsskT chs

s

kT

si ..11100

−=⎟

⎠⎞

⎜⎝⎛ −

== ∫∫

( )

kWR

VVkV chss 2398=−

=

iv) %09.992398

2.2376===

i

o

PP

η

v) Ω=== 205.0

10kRRi

vi) on-time mskTt 5.015.01 =×==

off-time mstTt 5.012 =−=

Vs Vo

Vch



R

L

E

+

-

Vs

Step Down Chopper with R-L Load: The operation of the chopper can be divided into two modes. During mode1, the chopper is switch on and current flows from supply to the load. During mode2, the chopper is switch off and the load current continuous to flow through freewheeling diode. The load current for mode1 can be found

( )iEdtdiLRV is −++=

1

With initial current ( ) 11 0 Iti == gives the load current as

( ) ( ) ( )2111 −⎟⎟⎠

⎞⎜⎜⎝

⎛−

−+=

−−LR

tsLR

te

REV

eIti

This mode is valid for ( )kTtt =≤≤ 10 at the end of this mode, the load current becomes

( ) ( )3211 −=== IkTttI The load current for mode 2 can be found from

( )40 2 −++= Edtdi

LRLi

With initial condition ( ) 22 0 Iti ==

( ) ( )5122 −⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

−−LR

tLR

te

REeIti

This mode is valid for ( )( )Tktt −=≤≤ 10 2 at the end of this mode, load current becomes

( ) ( )6322 −== Itti

Vs

+

-

R

Dm

+

-

i

L

E

Vo

Vo

Vs

IV

t1 t2

kt

T

t

t

I1

i1 i2

I2

F

DiscountineousCurrent

CountineousCurrent

(1-k)T

T

Fig : Wave Form

Fig : Chopper with R-L Load



Under steady state condition ( )731 −= II

Now from (2)

( ) ( )8112 −⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−+=

−−LR

kTsLR

kTe

REV

eII

From (5)

( ) ( ) ( )91

11

213 −⎟⎟⎠

⎞⎜⎜⎝

⎛−−==

−−−−LRTk

LRTR

eREeIII

Peak to peak ripple current is 12 III −=∆ which after simplification becomes

( )

LTR

LTRk

LRT

LRkT

s

e

eeeR

VI

−

−−−−

−

−+−=∆

1

11

The condition for maximum ripple

( )

5.0

0

=

=∆

k

givesdk

Id

The maximum peak to peak ripple current at ( )5.0=k is

fLR

RV

I s

4tanhmax =∆

For &tanh,4 θθ ≈>> RFL Q. Step down chopper is feeding RL load with

.05.0,1,5.7,5,220 ====Ω== EandKKHzfmHLRVVs Calculate: a) Minimum instantaneous load current '' 1I b) Peak instantaneous load current ‘ '2I c) Maximum peak to peak load ripple Solution:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−+=

−−LR

KTsLR

KIe

REV

eII 112

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

−−−LRkT

LRTK

eREeI 1

1

1

FLV

I s

4max =∆

I2

I1

E



( )iII −+= 473.127165.0 12

( )iiII −= 21 7165.0 Solving these two equations, we get:

AIAI 63.25,37.18 21 == I

AIII 26.712 =−=∆ Q. The step down chopper has a load resistance Ω= 25.0R , input voltage VVs 550= and VE 0= .

The average load current AI A 220= in chopping frequency Hzf 250= . Use the average output voltage. Calculate the load inductance ‘L’ which would limit the maximum load ripple current to 90% of .AI Solution: AIofi a 202001.0%10 =×==∆

sec004.01==

fT

Average output voltage, asa RIkVV ==

Voltage across inductor is

( ) sssas VkkVVRIVdtdiL −=−=−= 1

( ) kTL

Vki s .1−=∆

( ) kTVkLi s .1. −=∆

5.0,0 ==∆ kdk

i

Principle of Step-up operation:

Vs

+

_

iL

+D1

CL

ChopperLoad

_a) Step-up Arrangement

I2

I1

E

t1 t2

i1i2

tT

i

1

2

3

4

5

6

0.2 0.4 0.6 0.8 1.0

Vo/Vs

K

b) Current Wave Form c) Output Voltage



When a switch is closed for time 1t , the inductor current rises and energy is stored in the inductor L. If the switch is opened for time 2t , the energy stored in the inductor is transfer to the load through diode 1D and the inductor current falls. When the chopper is turn, the voltage across the inductor is

dtdiLVV sL ==

And, this gives the peak to peak ripple current in the inductor as

LtV

i s 1.=∆

The instantaneous output voltage is

dtdiLVV so +=

2tiLVV so

∆+=

2

1.t

tVV s

s +=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

2

11tt

Vs

2

.tTVs=

1

.tT

TVs −=

Tt

Vs11

1.−

=

If a large capacitor LC is connected across the load as shown by dash line in figure, the output voltage will be continuous. The voltage across the load can be step up by varying the duty cycle k. This principle can be applied to transfer energy from one voltage source to another as shown in the figure. The inductor current for mode 1 is given by:

dtdiLVs =

And, it is expressed as

( ) 11 ItL

Vti s += where, 1I is initial current.

During mode 1, current must rises and necessary condition

kV

V so −

=1

Vs

+i

D1

Chopper

_

L

iL +i

_

L

Mode 1

Vs

+iL

D1

_

L

+_

I2

I1t1 t2

Hi2

t

i



00 >> sVordtdi

The current for mode 2 is given by

Edtdi

LVs += 2

And is solved as

( ) 22 ItL

EVti s +⎟

⎠⎞

⎜⎝⎛ −

=

where, 2I is initial current for mode 2. For stable system, the current must fall and condition is:

EVordtdi

s << 02

Performance Parameter: The power semi-conductor devices require a minimum time to turn on and turn off. Therefore, the duty cycle k can only be controlled between a minimum value minK & maxK , thereby limiting the

minimum and maximum value of output voltage, we have:

fL

VI s

4max =∆

Load ripple current depends inversely on chopping frequency f and frequency should be as high as possible to reduce load ripple current. Chopper Classification: Depending on the direction of current and voltage flow chopper can be classified into 5 types:

1. Class A Chopper 2. Class B Chopper 3. Class C Chopper 4. Class D Chopper 5. Class E Chopper

Class A Chopper:

νi

VL

ILiL

vL

VL

o

Class A Chopper

o

VL

νL

-IL ILClass B Chopper

-IL ILClass C Chopper



Class A Chopper: The load current flows into the load. Both the load voltage and load current are positive. This is a single quadrant chopper and is to be operated as rectifier. Its example is step-down chopper. Class B Chopper: The load current flows out of the load, the load voltage is positive but the load current is negative. This is also a single quadrant chopper but operates in second quadrant and said to be operated as inviter. When switch 1s is closed, the voltage E drives current through inductor L and load voltage

LV becomes zero. The current mI which rises is described by:

ERidtdiL L ++=0

With initial current ( ) ,0 1ItiL == gives

kTt0for 111 ≤≤⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−−=

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−

LRt

LRt

LR

L eREe

RieIi

at 1tt = , ( ) 21 IkTttiL ==== When switch is turned off, a magnitude of energy stored in inductor L is returned to the supply

.sV iV diode 1D . The load current falls. The load current is described by,

ERidtdiLV Ls ++=

An initial current of 2I ;

22 01 ttforeR

EVeIi

tLR

st

LR

L ≤≤⎟⎟⎠

⎞⎜⎜⎝

⎛−

−+=

−⎟⎠⎞

⎜⎝⎛−

VL

vL

iLo +IL

-VL

+IL

VL

vL

iL

-VL

Class D Chopper Class E Chopper

R

VLS1 EVs

is iL

+ -

a) Circuit Diagram

I2

kt

t

I1

Load CurrentT

i2

VL

Vs



Class C Chopper: The load current is either positive or negative. The load voltage is always positive. This is known as two quadrant chopper. Class A and Class B chopper can be combined to form Class C chopper. S1 and D2 operate as Class A Chopper whereas S2 and D1 operates as Class B Chopper. Care must be taken to insure that two switches are not fire together; otherwise, the supply Vs will be short circuited. Class C Chopper can be operated either as a rectifier or as an invitor. Class D Chopper: Class D Chopper can also operate as rectifier or as inverter. If S1 and S4 are turn on VL, iL become positive. If S1 and S4 are turn off, load current will be positive and continue to flow for a highly inductive load. Class E Chopper: 1st Quadrant Operation: S2 and S3 are kept off, S4 is kept on and S1 is operated when S1 is on, VL and iL are positive. When S1

is switched off, current free wheels through S4 and D2. 2nd Quadrant Operation:

VLS1

RVs

S2 D2

D1L

+

-

iL

a) Circuit Diagram

S1 EVs

L

+

D2

-

D3

a) Circuit Diagram

VLS1

RVs

S2 D2

D1

L

V2

D3

D4S4

S3

a) Circuit Diagram

RectifyingVL +veiL +ve

InvertingVL +veiL -ve

VL -veiL -veRectifying

VL -veiL +veInverting

VL

iL

b) Polarities



S2 is operated while S1, S3, S4 are off when S2 on reverse current flows through R-L-E leand and D4

and inductor stores energy. With S2 off, energy stored by inductor is turned to supply through D1 and D4. 3rd Quadrant Operation: S1, S2, S3 are kept off and only S4 is operated. With S4 on the current freewheels through D2. With S4 turn off current is feed to the source through D2 and D3. DC chopper can be used as switching mode regulator to convert DC voltage normally unregulated to a regulated DC output voltage. The regulation is normally achieved by pulse width modulation (PWM) at a fix frequency and the switching device is normally BJT or MOSFET. Control voltage cV

is obtained by comparing the output voltage with its desired value. Vc can be compared with saw tooth voltage to generate the PWM control signal for DC Choppers. Therefore, there are four basic topologies of switching regulator.

1. Buck Regulator 2. Boost Regulator 3. Buck-Boost Regulator 4. Cuk Regulator

Buck Regulator: In a Buck regular, the output voltage aV is less than the output voltage sV .This is like a step

down chopper. The circuit operation can be divided into two modes. Mode 1 begins when 1θ is

switch on and mode 2 begins when transistor 2θ is switch off.

Assuming the inductor current rises linearly from 21 ItoI in time 1t

( )

( )2

1

1

1

−−∆

=

−∆

=−

as

as

IVILt

tILVV

dcchopper

ControlVr

Input+

-

Vref

+

-

AmplifierVc

VsVg Va

Ve

a) Block Diagram

VrVc

To

oKT T

V

t

b) Control Signal

Figures



sa kVV =

& inductor current falls linearly from 12 ItoI in time 2t ,

( )32

−=tdiLVa

aVILt ∆

=2

where, =−=∆ 12 III peak to peak ripple current of inductor from (1) and (2),

( )

LtV

LtVV

I aas 21 =−

=∆

1

2,tt

VVV

ora

as =−

( )kT

TkVV

a

s −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

11

kk

kVV

a

s 111 =−

+=

-(5) Assuming a lossless circuit, ,asaass IkVIVIV == average input current

( )6−= as kII

The switching period T can be expressed as,

( ) ( )( )42121 andFrom

VI

VVILtt

fT

a

L

as

+−

∆=+==

( ) ( )7−−

∆=

asa

s

VVVILV

which gives peak to peak ripple current as

( ) ( )8−

−=∆

s

asa

fLVVVV

I

( ) ( )91

, −−

=∆fL

kkVIor s

We can write inductor current Li as

DCL iii +=

If we assume Di∆ is small and negligible,

currentcapacitorAverageii CL &∆=∆

which flows into for 222

21 Ttt=+ is

4IIC

∆=



& Peak to peak ripple voltage of capacitor

( )1088

.4

1 20

−∆

=∆

=∆

=∆ ∫ fCI

CTIdtI

CV

T

C

from (8) and (10)

( ) ( )11

8 2 −−

=∆s

asac VLCf

VVVV

( ) ( )12

81

2 −−

=∆LCf

kkVV s

C

Q. The buck regulator has an input voltage of VVs 12= . The required average output voltage is

VVa 5= & peak to peak ripple voltage is 20 mv. The switching frequency is 29 kHz. If the peak to

peak ripple current of inductor is limited to 0.8A, determine (a) Duty Cycle K (b) Filter inductance L (c) Filter capacitor. Solution: kHzfmVVcVVVV as 25,20,5,12 ==∆==

a) 4167.0

125

==

=

k

kVV sa

b) ( )

s

asa

fLVVVV

I−

=∆

( )

s

asa

IfVVVV

L∆

−=

HL µ83.148=

c) fCIVc 8

∆=∆

FVf

Cc

µ2008

=∆∆Ι

=

Boost Regulator: In boost regular the output voltage is greater than the input voltage. The circuit operation can be divided into two modes. Mode 1 begins when MOSFET M1 is switch on and mode 2 begins when MOSFET is switch off. Mode-1

( )11

−∆

=tILVs

( )2, 1 −∆

=sVILtor

Mode-2

Figures



( )32

−∆

−=−tILVV as

( )4.., 2 −

∆=

sa VVLItor

( )2

1.t

LVV

LtV

I sas −==∆ - (5) ( )( )[ ]TktkTt /1, 21 −==

Assuming loss less circuit, aass IVIV =

as

ss Ik

VIV ⎟

⎠⎞

⎜⎝⎛

−=

1

( )61

−−

=k

II a

s

Now, 211 ttf

T +==

sas VVIL

VIL

−∆

+∆

=.

( )sas

a

VVVILV

−∆

=

( ) ( )7−−

=∆a

sas

fLVVVV

I

- (8)

Peak to peak ripple voltage is

dtIC

Vt

Cc ∫=∆ 1

0

1

∫= 1

0

1 t

adtIC

CtI a 1.

=

( ) ( )9−−

=fCV

VVI

a

saa

- (10)

Q. A boost regular has an input voltage VVs 5= , Average output voltage V15 , Average output load current AI a 5.0= . The switching frequency &1500,25 HLkHzf µ== CC µ220= . Determine a) Duty cycle b) ripple current of inductor iδ c) peak current of indcutor sI d) ripple voltage of filter capacitor.

kV

V sa −

=1

fLkV

I s .=∆

fCkI

V aC

.=∆



Solution:

a) k

VV s

a =

b) AfL

kVI s 89.0

.==∆

c) Ak

II a

s 5.11

=−

=

d) fC

kIV a

C =∆

AI s 5.2=

AIII s 945.1269.05.1

22 =+=∆

+=

mVfC

kIV a

c 61.80.

==∆

Buck Boost Regulator: It provides the output voltage which is less than or greater than the inut voltage. But the output voltage polarity is opposite to that of the input voltage. The circuit operation can be divided into two modes. Mode 1 begins when T1 is turn on and mode 2 when T1 is turn off. Mode 1:

( )1.

1

−∆

=t

ILVs

or, ( )2.1 −∆

=sVILt

Mode 2:

( )32

−∆

−=tILVa

( )42 −∆

−=aVILt

LtV

LtV

I as 21.−==∆

- (5) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=Tkt

kTt12

1

Assuming loss less circuit,

k

kVIVIV a

aass −=−=

1

- (6)

Figure

Figures

kkV

V sa −

−=1

kkI

I as −

=1

.



Now, ( )

as

sa

as VVVVIL

VIL

VILtt

fT

−∆=

∆−

∆=+== 21

1

- (7)

( )8.

−=∆fL

kVI s

Peak to peak ripple voltage of capacitor is

( ) ( )91.111.00

11 −−

====∆ ∫∫ fCVVVI

tIC

dtIC

dtIC

Vsa

saa

t

a

t

Cc

- (10)

Q. The buck boost regulator has an input voltage VVs 10= , duty cycle Vk 5.0= and switching frequency is 25kHz. The inductance HL µ150= and fileter capacitance FC µ220= . The average load current AI a µ25.1= . Determine a) average output voltage aV b) Peak to peak output ripple voltage cV∆ c) Peak to peak ripple current of inductor. Solution:

a) VkkV

V sa 4

1−=

−−=

b) mVfC

kIV a

c 8.56==∆

c) AfL

kVI s 8.0==∆

It provides an output voltage which is less than or greater than an input voltage but the output voltage polarity is opposite to that of input voltage. The circuit operation can be divided into two modes. Mode 1 begins when 1θ is turn on and mode 2 begins when 1θ is turn off. Mode-1

( )11

11 −∆

=t

ILVs

( )2, 111 −

∆=

sVIL

tor

( )32

111

−∆−

=−t

ILVV cs

1

112

.

cs VVLI

t−

∆−=

When 11121 LL III −=∆

( )2

11

11

1.t

LVV

LtV

I css−−

==∆

( )sa

as

VVfLVV

I−

=∆

fCkI

V aC

.=∆

Figures



( )511

−−

=k

VV s

c

Now, ( )61

221

−∆

=+tI

LVV ac

( )71

2.21 −

+∆

=ac VV

LIt

2

22

tIL

Va∆

−=

( )8. 22

2 −∆−

=aV

LIt

( )

2

2

2

12

1

LtV

LtVV

I aac −=+

=∆

( )92

−−=k

VV a

c

From (5) and (9), we get:

( )101

−−

−=K

KVV s

a

( )11.

21 −=∆

fLKV

I s

)12(2

2 −=∆fL

KVI s

( ) ( )131

11

−−

=∆fc

kIV s

c

222

2 8 fLcKV

V sc =∆

Limitation of single stage conversion: 1. The four regulator use only one transistor employing one stage conversion due to current

handling limitation of single transistor, the output power of these regulator is small, there is no isolation between input and output voltage which is highly desirable characteristics in most applications. For high power application, multi-stage conversion are used where a dc-voltage is converted to ac by inverter. The ac output is isolated by the transformer and converted to dc by rectifier.

Chopper circuit design:

1. Identify the modes of operation for cheaper circuit 2. Determine the equivalent circuit for the various mode. 3. Determine current and voltage for the modes and their waveform. 4. Evaluate value of L and C that would satisfy design limit. 5. Determine voltage and current rating requirement of all components.



Chapter 10 DC Drives

Basic Characteristics of DC motor: When a separately excited DC motor is excited by a filed current of If and armature current of Ia flows in the armature circuit, the motor develops a back emf and a torque to balance the lead torque at a particular speed. The instantaneous field current if is described as:

dt

diLiRL f

fffef +=

The instantaneous armature current can be found from ga

aaaa edtdi

LiRV ++=

The motor back emf is expressed as: fg ikVe ω=

The torque developed by motor is aftd iikT =

The developed torque must be equal to load torque,

Ld TBdtdIT ++= ωω

Where, sradspeedmotor /,=ω sradmNtconsfrictionviscous //tan −=Β

sradAVtConsVoltageKv //tan −=

tconstorqueKK Vt tan== HceinduccircuitarmatureLa ,tan=

HceinduccircuitfieldL f ,tan=

Ω= ,tan ceresiscircuitarmatureRa

Ω= ,tan ceresiscircuitfieldR f

mNtorqueLoadTL −= ,

La

If, if+

-

vf, Vf

-

va VaRa Rf

Lf

+

-

eg

w Td Tc

BFig : Equivalent Circuit of Separately Excited DC Motor



Under steady state conditions, 0=dt

didtdi fa

fff IRV =

fvg KE Ι= ω

fvaaa IKIRV ω+=

Laftd TBIIKT +== ω

Developed power is ω.dd TP =

The speed of separately excited motor can be found from,

f

fv

aaa

fv

aaa

RV

K

RIVIK

RIV

.

−=

−=ω

The motor speed can be varied by: 1. Controlling armature voltage Va known as voltage control. 2. Controlling the field current, if , known as field control. 3. Torque demand, which corresponds to rated armature voltage, rated field current and rated

armature current is known as base speed.

w = constant

applcoximatelylinear region

Eg

IfFig : Magnetization Characteristics

o

Constanttorque

Constantpower

field current

Torque Td

Speed, w

armature current

If

Iaia, if

Speed, w

Speed

Id, Pd

Fig : Characteristics of separately excited motor



For a speed less than base speed, armature current and field current are maintained constant to meet the torque demand and armature voltage Va is varied. For speed higher than base speed, armature voltage is maintained at rated value and field current is varied to control the speed. However, the power developed by motor remains constant.

fvg IKE ω=

( ) ( ) fvafagafaa IKIRREIRRV ω++=++=

faftd tIBIIKT ω==

( )fv

afaa

IKIRRV +−

=ω

The speed can be varied by controlling (1) armature voltage Va or (2) armature current Ia which is a measure of torque demand. Numericals: Q. A 15 HP, 220 V, 2200 r.p.m. separately excited d.c. motor controls a load requiring a torque of TL = 45 Nm at a speed of 1200 r.p.m. The field circuit resistance is Rf = 147 Ω and the armature circuit resistance is Ω= 25.0aR and voltage constant of motor is AVKv /7032.0= rad/s. The field voltage

La, Ra

+

-

va Va

-w

Td Tc

B

Lf, Rf

Ia = If

ia = if

Constanttorque

Constantpower

armature current

Torque Td

Speed, w

Speed Id, Pd

Power Pd

Fig : Characteristics of DC Series Motor



VV f 220= . The viscous friction and no load losses are negligible. The armature current may to

amuse continuous and ripple free; determine: a) Back emf gE

b) Required armature voltage aV

c) Rated armature current of motor. Solution: Ω=Ω= 25.0,147 af RR

sradAVKKVV fvf //7032.0220 ===

NmTT Ld 45==

srad /66.12560

12202 =×= πω

ARV

If

ff 497.1

147220

===

a) VIEE fvg 28.132497.166.1257032.0 =××== ω

b) gaaa ERIV +=

fatd IIKT =

AKIT

If

da 75.42

497.17032.045

=×

==

VVa 97.14225.13225.075.42 =+×=

c) 74600=phI

AI rated 87.50220

74615=

×=

Operating Modes:

(i) Motoring

Ra

-

La

Ia = If

Va

-

Ra, La

Ia

MEg

+

Rf, If

Series motorSeparately Excited Motor



Back emf is less than supply voltage aV . Both armature and field current are +ve, the motor develops

torque to meet the load demand. Operating Modes: ii) Regenerating Braking The motor acts a generator and develop induce emf voltage gE . gE must be greater than the supply voltage AV . The armature current is –ve. iii) Dynamic Braking It is similar to regenerate braking except the supply voltage AV is replaced by braking resistance bR .

The kinetic energy of motor is dissipated in bR .

iv) Plugging

Ra

La

Ia

Va

Eg+

-

RF

Ia

iF

Ra, La

Eg+

-

Rb

M

M

Ra

La

Ia

Eg+

-

RF

Ia

iF

Ra, La

Eg

RbRb

Series motor

+

-

Rb

LaVa

Eg+

-

RF

If

iF

M

Vf

Eg+

-

Va

M

Ia=If Rf, Lf

Seperately excited moter



v) 4 Quadrants Single Phase Motor: If the armature circuit of dc motor is connected to the output of single phase control rectifier, the armature voltage can be vary by varying the delay angle of converter aα . A

converter is also applied in field circuit to control the field current by varying delay angle . Depending on the type of single phase converter, single phase drive may be sub-divided into:

1. Single Phase Half-wave Converter Drive 2. Single Phase Semi-conductor Device 3. Single Phase Full-converter Drive 4. Single Phase Dual-converter Drive

Single Phase Half-wave Converter Drive:

Va<Eg

M

Ia

Forward BrakingForward Motering

Ra, ta

M

Ia

+

-Eg

Va>Eg

M

Ia

Va +

-+

-Eg M

Ia

Reverse Braking

Va +

-+

-Eg

Ra, La

Va < Eg

Torque

Speed

Va > EgReverse Motoring

Fig : Condition for four quadrature

La, Ra

M

Ia

+

-Eg

Va

is

id

Dm

+

-LF, Rf

Vc

if

a) circuit

Ia

isIa

wt

wt

wt

Ia

idαa π

αa π 2π



With a single phase half wave converter in the armature circuit, the average armature voltage is

( ) πααπ

≤≤+= aam

a forV

V 0cos12

Where, mV = peak voltage of ac supply

With a semi-converter in the field circuit average field current is

( ) πααπ

≤≤+= ffm

f forV

V 0cos1

Single Phase Semi-converter: With a single phase semiconductor in armature circuit, the average armature is

( ) πααπ

≤≤+= aam

a forV

V 0cos1

When, =mV peak voltage of ac supply.

With a semiconductor in filed circuit, average field voltage ( )fm

fV

V απ

cos1+= for πα ≤≤ f0 .

La, Ra

M

+

-Egid

LF, RfVs

if

ia

Vs

is

Ia

isId

wt

wt

wt

Ia

id

αa π

αa π 2π

2π

π=αa

π=αa

La, Ra

M

+

-Eg LF, Rf

Vs

if

Va

a) Circuit

Is

Va

b) Quadrature

Ia

isIa

wt

wt

-Ia

αa π

ia

π+αa

o

c) Waveform



πααπ

≤≤= aam

a forV

V 0cos2

, where, mV = peak voltage of supply

πααπ

≤≤= ffm

f forV

V 0cos2

Single Phase Dual Converter: If converter 1 operates with a delay angle of 1aα , the average armature voltage is

πααπ

≤≤= 11 0cos2

aam

a forV

V

If converter 1 operates with a delay angle of 2aα , the average armature voltage is

πααπ

≤≤= 22 0cos2

aam

a forV

V where 12 aa απα −=

With a full converter in the field circuit, the average field voltage is

πααπ

≤≤= ffm

f forV

V 0cos2

Q. The speed of the separately excited motor is a controlled by φ1 semi-converter. The field current which is also controlled by a semi-converter is to the maximum possible value. The ac supply to the armature .60,208,1 HzVφ The armature resistance Ω= 25.0aR , Ω= 147fR the motor voltage

constant sradAVKv //7032.0 −= load torque NmTL 45= at 1000 rpm. The viscous friction and

no load losses are negligible. The inductance of the armature and field circuit are sufficient enough to make the armatural field current, continuous and ripple free. Determine a) Field current fi

b) Delay angle of converter in the armature circuit aα

c) Input power factor of armature circuit converter Solution:

VVVVV sms 16.2942,208 =×==

Ω=Ω= 147,25.0 fa RR

sradAVKK tv //7032.0==

NmTT Ld 45==

srad /72.10460

10002 =×= πω

La,Ra

Ia

LF, RfVs

if

Vs

converter-1 αa1 converter-2 αa2



a) ( )fm

fV

V απ

cos1+=

When, 0=fα

VV

V mf 27.18716.29422

=×

==ππ

ARV

If

ff 274.1

14727.187

===

b) ?=aα

( )am

aV

V απ

cos1+=

fatd IIkT =

AIk

TI

ft

da 20.50

274.17032.045

=×

==

VEIRE gaag 38.10682.9323.5025.0 =+×=+=

( )aαπ

cos116.29438.106 +=

°= 2.82aα

c) The output power, WIVP aao 5.534323.5038.106 =×==

If the losses in armature converter are neglected, the power from supply is WPP oa 5.5343==

The rms input current of the armature converter is

AItdII aaasa

a

03.3722 2

121

2 =⎟⎠⎞

⎜⎝⎛ −

=⎥⎦⎤

⎢⎣⎡= ∫ π

απω

ππ

α

Input volt-amp rating WIVVI as 24.770203.37208 =×==

Input Power Factor, )(694.024.77025.5343 lagging

VIP

PF a ===

Closed Loop Control of DC drives: The speed of dc motor changes with the load torque to maintain a constant speed, the armature (and or field) voltage should be varied by continuously by varying the delay angle of ac to dc converter or duty cycle of chopper. A close loop control system has the advantages of improve accuracy fast dynamic response and reduce effects of load disturbances and system non linearity.

SpeedConverter Converter

Seed sesing

Vr Ve+

-

Power supply

Vc Vadc motor

T2

w

Fig : Block Diagram of Closed Loop Converter Field DC motor drive



If the speed of the motor decreases due to the application of additional load torque, the speed error

eV increases. The speed controller responses with increase controlled signal cV , change the delay

angle or duty cycle of converter and increase the armature voltage of the motor and increase armature voltage develops more torque to restore the motor speed to the original value. Open Loop Transfer Function: The motor speed is adjusted by setting reference voltage .rV Assuming a linear power conver of

gain 2K . The armature of the motor is ( )iVkV ra −= 2

Assuming filed current aI remains constant during any transient disturbances, the system equation

are: ( )2−= ωfvg IKe

( )3−++= ωfVa

mama IKdtdi

LiRV

( )4−= aftd iIKT

( )5−++= Ld TBdtdJT ωω

Taking Laplace transform of above equation: ( ) ( ) ( )612 −= sVKsVa

( ) ( ) ( ) ( )7)( −++= sIKssILsIRsV fVamama ω

( ) ( ) ( ) ( ) ( )8−+== sTssJsIIKsT Laftd ω

From (7), the armature current is

( ) ( ) ( )( ) ( )9−

+

−=

mm

fvaa RsL

sIKsVsI

ω

( )

( ) ( )101

)(−

+

−=

am

fva

sTRsIksV

Tω

Where, ==a

ma R

LT time constant of motor armature circuit

From (8), motor speed is

( ) ( ) ( )( ) ( )11−

+−

=BSI

sTsTsw Ld

Lm, Rm

+

-

w Td T2

B

VsConerter of

gainK2

If

Vf

+

-

Fig : Converter Fed Separately Excited dc motor



( ) ( )( ) ( )12

1−

+−

=m

Ld

sTBsTsT

Where, == BJTm / mechanical time constant of motor

Two possible disturbances are control voltage rV and load torque LT . The response due to a step

change in the reference voltage is obtained by setting LT to zero.

( )( )

( )( ) ( ) ( ) ( )

( )13./1

./22

2 −+++

=BRIKTTsTTs

BRIKKsVs

mfvmamo

mfr

r

ω

The response due to a change in load torque LT can be obtained by setting rV to zero.

( )( )

( )( )( ) ( ) ( ) ( )

( )14./1

1122

−++++

+−=

BRIkTTsITssTB

sTs

mfVmama

a

L

ω

Using Final value theorem steady-stage relationship of change in speed

( ) ( )1522 −∆+

=∆ rfvm

fv VIKBR

Ikkω

( ) ( )1612 −∆+

=∆ TIKBR

R

fVm

mω

Close Loop Transfer Function:

1B(STm+1)

Kt If

-Kv If

K2

-

+

1Rm(STa-1)

w(s)Ia(s)

Vr(s)

Td(s)

TL(s)

Fig : Open Loop Block Diagram of Separetly Excited d.c. motor

Vrs VcK2

Eg(s)

laKv If

-

Va w(s)

TL(s)

K1

Kf If

+-

-



To change open loop arrangement into closed loop system a speed senser connected to the output shaft. The closed loop step response due to a change in reference voltage can be setting rV to zero.

( )( ) ( ) ( ) ( )[ ] ( )

( )1/1

/

2122

2 −+++++

=BRIKKKIKTSTS

BRiKKsVsW

mfvfvmama

mfr

r ττ

The response due to change in load torque LI can be obtained by setting rV to zero.

( )( )

( )( )( ) ( ) ( )[ ] ( )

( )2/1

1/1

2122

−+++++

+−=

BRIKKKIKTSTSSB

sVsW

mfvfvmama

a

r τττ

Using final value theorem the steady change in speed due to step change in control voltage LI∆ can be found from (1) and (2) by substituting S=0.

( ) ( )321

22 −∆

++=∆ r

fvfvm

fV VIKKKIKBR

IKKW

( ) ( )421

2 −∆++

=∆ Lfvfvm

m TIKKKIKBR

RW

Phase Locked Loop ( PLL) Control: In PLL control system, the motor speed is converted to a digital pulse train by using speed encoder the output of encoder acts as a speed signal of frequency to the phase detector compares a reference pulse train rf with the feed back frequency of and provides a pulse with modulated output

voltage cV to a continuous dc level cV which varies the output of power converter and in turn the

motor speed. When the motor runs at same speed as the reference pulse train the low frequencies could be synchronized (or blocked) with a phase difference. The output of the phase detector would be constant voltage proportional to the phase difference and motor speed would be main trained at first value. Any disturbances contributing to the speed change would result in phase differences in the output of phase detector could respond immediately to vary the speed of motor. In such direction and magnitude as to retain the locking of reference and feedback frequencies. Micro-computer control of DC Drives:

Phase detector Low pas filter Converter k2 DC mortor

Speed encoder



A micro computer control reduces the size and cost of hardware electronics improving reliability and control performance. This control scheme is implemented in software and it flexible to change the control strategy to meet difference performance characteristics. A micro computer control system can also perform various desirable functions: ON/OFF of main power supply, start/stop of drive, speed control, current control, etc. The speed signal is fed into micro-computer using A/D converter to limit armature current of motor inner current control loop is used the line synchronizing circuit is required to synchronize the generation of firing pulses with a supply line frequency. The pulse amplifier provides the necessary isolation and produces gate pulses of required magnitude and duration.

Determinationof thyristor of be fired

Delay angle generator

Current controller

Line synchronizing ckt

TimeandLogic

Pulseamplifier

MotorCurrent

Speed controller

Speed comparator

Current comparator

SpeedSignal

AD

AD

IfRfLf

M

Speed reference (ωr)

start/stop command



Chapter 11 Properties of Devices & Circuit

Introduction: Due to the reverse recovery process of power devices and switching action in the presence of circuit inductance voltage transient occurring converter circuit, short circuit fault conditions may exist resulting in excessive current flow through the devices. In practice power devices are protected from:

1. Thermal runaway by heat sink

2. High dtdv and

dtdi by shutter circuit

3. Reverse recovery transient 4. Supply and load side transient 5. Fault condition by fuse

Cooling & heat sink: Due to on state and switching loses, heat is generated within the power devices. This heat must be transfer from the devices to a cooling medium to maintain the operating junction temperature within the specified values. Although heat transfer can be accomplished by conduction, convection and radiation, natural or forced air, convection cooling is commonly used in industrial application. Heat must be flow from the device to the case and then to the heat sink in cooling medium. If aP is

the average power loss in the device, the electrical analog of a device which is mounted on a heat sink is shown in the figure: The junction temperature of a device is given by: ( )SACSJCAj RRRPT ++=

Where, case ojunction t from resistance thermal=JCR

=CSR thermal resistance from case to sink c/w

=SAR thermal resistance from sink to ambient c/w

mperatureambient te =AT In high power application, the devices are more effectively cooled by liquids, normally oil or water. Snubber Circuits:

Ts

RSA

TA

PA

T1Ric Tc

Rcs



An RC-snubber is normally connected across the semiconductor device to limit dtdv within the

maximum allowable rating. The snubber could be polarized or unpolarized. The forward polarized

snubber is suitable when thyristor is connected with anti-parallel diode; R limit the forward dtdv and

1R limit the discharge current of capacitor when the device is turn on.

A reverse polarized snubber limits reverse dtdv where 1R limits do change current of capacitor.

When a pair of thyristor is connected in inverse parallel, the snubber must be effective in either direction. Design of snubber circuit:

( )dtdiLiRRV Lss ++= which gives

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−Tt

eIi 1

Where, LsLs

s

RRLT

RRV

I+

=+

= &

Now,

Tt

sTt

Ls

Ls

sTt

eL

Ve

RRR

RRV

TIe

dtdi −−−

=+

+== ..1.

R1

C

R

D

T1 D1

R1>R

a) Polarized

R1 C

R

D

LsR1>R

b) Reverse Polarized

C

R

LsR1>R

T1T2

c) Unpolarized

L

RL

Rs Cs

T1+

-When switch is closed

L

RL

+

-

Vs

Rs



0max

==⎟⎠⎞

⎜⎝⎛ tat

LV

dtdi s

- (1)

Voltage across SCR, iRa s .=

dtdiR

dtdv

sa .=

( )2maxmax

−⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

dtdiR

dtdv

sa

L

VR s

s .=

( )3max

−⎟⎠⎞

⎜⎝⎛=

dtdv

VLR a

ss

ss c

LR ρ2= where, ρ = damping ratio = LcR s

2

( )4.22

−⎟⎟⎠

⎞⎜⎜⎝

⎛= L

RC

ss

ρ

When SCR is turned on capacitor Cs will discharge a maximum current of s

s

RV

and total current

through thyristor will be ⎟⎟⎠

⎞⎜⎜⎝

⎛+

L

s

s

s

RV

RV

. It should be less than peak current rating of thyristor.

Q. Following are the specification of a thyristors operating from a peak supply of 500V and

repetitive peak current sec/200sec,/60250maxmax

VdtdvA

dtdiAI p =⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛= . Design the snubber

circuit, if the minimum load resistance is 20 Ω . Take 65.0=ρ . Take a factor of safety for the given data. Solution: For a factor of safety 2, permitted values are:

sec/302

601252

250

max

AdtdiI p ==⎟

⎠⎞

⎜⎝⎛==

sVdtdv µsec100

2200

max

==⎟⎠⎞

⎜⎝⎛

Now,

H

dtdiV

L s µ67.1630500

max

==⎟⎠⎞

⎜⎝⎛

=

max⎟⎠⎞

⎜⎝⎛

=

dtdiV

L s



Ω=×

=⎟⎠⎞

⎜⎝⎛= 33.3

50010067.16

maxdtdv

VLR

ss

When thyristor is turned on, the peak current through thyristor is

ARV

RV

L

s

s

s 15.17520

50033.3

500=+=+

This peak current is more than permissible peak current of 125A. The value of sR must be increased.

Take Ω= 7sR , peak current through thyristor is A42.9620

5007

500=+ , which is also less than

permissible peak current. Also,

max

⎟⎠⎞

⎜⎝⎛=

dtdv

VLR

ss

H

dtdv

VRL ss µ35

1005007.

max

=×

=⎟⎠⎞

⎜⎝⎛

=

( ) FLR

Cs

s µρ 20.1357

65.02.2 22

=×⎟⎠⎞

⎜⎝⎛ ×

==

Take FCs µ1=

Reverse Recovery Transients: Due to the reverse recovery time rrt and reverse current RI and amount of energy is trapped in the circuit inductance and as a result transient voltage appears across the device. In addition to

dtdv protect, the snubber limits the peak transient voltage across the device. The equivalent circuit for

a circuit arrangement is shown in the figure. The energy stored in the inductor L which is transferred to the snubber capacitance C is dissipated mostly in snubber resistor. Supply and Load side transient:

15aH

20Ω

R=7-2 C=1uF

+

-

Vs

IR

+

-

C

RDivice underrecovery

L



A transformer is normally connected to the input side of converter, under steady state condition and amount of energy is stored in the magnetizing inductance mL of the transformer and switching

off the supply provides a transient voltage to the input of converter, a capacitor may be connected across the primary or secondary of the transformer to limit the transient voltage.

When load is disconnected the transient voltage are produced due to the energy stored in inductance. Voltage Protection by Selenium Diodes and Metal Oxide Varistors:

The selenium diode may be used for protection against transient over voltage. The characteristics of selenium diode are shown in the figure.

Normally, the operating point lies before knee of the characteristics curve and draws very small current from the circuit. However, when an over voltage appears, the knee point is crossed and the reverse current flow through the selenium increases suddenly, thereby, limiting the transient voltage.

Varistors are non-linear variable impedance device consisting of metal oxide particles. As the applied voltage is increased, the film become conductive an the current is increased. The current is expressed as αKVI = where,

K = constant and V = applied voltage. The value of α varies between 30° and 40°.

Current Protection: The power converter may develop short circuit of faults and the fault current must be clear quickly. Fast acting fuses are normally used to protect the semi-conductor device. As the fault current increases, fuse opens and clears a fault current in few milliseconds. Fusing:

a) Circuit Diagram

+

-

Vp

S1

Np NsR

C

+

-

Vo

i I0

Lm +

-

R

CVo

b) Equivalent Circuit during turn off

L

LOAD

C

R

c) Equivalent Circuit due to load disconnection

Clamping Vz Voltage

o v

i

a) V-I characteristics

+

_

v

b) Symbol



The semi-conductor device may be protected by carefully choosing the location of fuse. When the fault current rises, the fuse temperature also rises and fuse melts and arcs are developed across the fuse. Due to the arc, the impedance of fuse as increased thereby reducing current. Crowbar Protection Circuit:

Transistors can be protected by a crowbar circuit. A crowbar is used for protecting circuits or equipment under fault condition where the amount of energy involved is to high and the normal protection circuit cannot be used. A crowbar consists of thyristor with voltage or current sensitive firing circuit. The crowbar thyristor is placed across the converter circuit to be protected. If the fault condition are sensed and crowbar thyristor cT is fired. A virtual short circuit is created and fuse link

1F is blown thereby relieving the converter circuit from over current. Fault Current with AC source and DC source: The fault current and fuse clearing time will be dependent on time constant of fault circuit. The fuse manufacture specify the current-time characteristics for ac circuit and there is no equivalent curve for dc circuit for a circuit operating from dc-voltage, the voltage rating of fuse should be typically 1.5 times the equivalent ac rms voltage. The fuse protection of dc circuit require more careful design then that of ac circuit.

T1 T3

F1 F3

T4 T2

F2 F2 R

L

ac supply

Fig : Controlled Rectifier

R

L

Vs

F

Tc

Dm

θ1

+

_

Fig : Protection by crowbar



Chapter 7 Power Transistors

Introduction: Power transistor have controlled turn ON and turn OFF characteristics. The power transistors can be classified broadly into four categories:

i) By-polar junction transistor (BJT) ii) Metal-oxide semi-conductor filed effect transistor (MOSFET) iii) Static-induction transistor (SIT) iv) Insulated gate by-polar transistor (IGBT)

i) BJT: There are three operating region of transistor circuit active and saturation. In circuit off region, the transistor is off or the base current is not enough to turn it ON. In active region, the transistor acts amplifier by a gain. In the saturation region the base current is sufficiently high so that the collection emitter voltage CEV is low and the transistor acts as switch.

Fig : a) NPN transistor

B

IE

IB

IcC

E

n

p

n

Base

emitter

co llector

B

IE

IB

IcC

E

n

p

n

Base

collector

emitter Fig : b) NPN transistor

RB

IB

VBE

VCE

+

_

IE

RC

+_

Ic

VB

IBVCF1 VCE1

Fig a) Circuit Diagram

IC

Active region

IB3IB2IB1IB=0

VCE

Active region

VCE

Cut off Saturation

VCE

VCC

Fig c) Output Characterisitcs Fig d) Transfer Characterisitcs



The equation neglecting the current is ( )1−+= Bce III

Current gain,

( )2−=B

c

II

β

( )3−+= CEOBC III β where, =CEOI collector to emitter leakage current

From (1) and (3), we have ( ) ( )41 −++= CEOBE III β

)1( β+≈ BI

( )5111 −⎟⎟⎠

⎞⎜⎜⎝

⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛+≈

ββ

β CCE III I

The collector current can be expanded as: ( )6−≈ EC II α

Where, ( )71

−+

=β

βα

Or, ( )81

−−

=α

αβ

Let us consider the circuit of fig 1(a) where transistor is operated as switch.

( )9−−

=B

BEBB R

VVI

( )10−−== ceCCCEC RIVVV

CBCC RIV β−=

( )BEBB

CCC VV

RR

V −−=β

( )11−+= BECBCE VVV

BECECB VVV −=

The maximum collector current in active region which can be obtained by setting isVVandV CEBECB == 0

( )12−−

=−

=C

BECC

C

CECCCM R

VVR

VVI

And the corresponding value of base current ( )13−=βCM

BMI

I

The transistor saturation may be defined as the point above which any increase in base current does not increase collector current significantly. In the saturation, the collector current remains almost constant

( ) ( )14−−

=C

CECCCsS R

satVVI



And the corresponding base current ( )15−=βCS

BSI

I

Normally, the circuit is designed so that BI is higher than BSI .

Overdrive factor,

( )16−=BS

B

II

ODF

Forced ( )17−==B

CS

II

fββ

The total power less in two junctions is ( )18−+= CCEBBET IVIVP

Q. The bipolar transistor shown in the figure is specified to have β in the range 8 to 40. The load resistance is Ω= 11CR . The dc supply voltage VVCC 200= and the input voltage to the base circuit is

VN B 10= . If ( ) ( ) VVVV satBEsatCE 5.1&1 == . Find a) Value of 3R that results in saturation with overdrive factor (ODF) of 5. b) Force Fβ c) Power loss tP in the transistor.

Solution: 51020011 ===Ω= ODFVVVVR BCCC ( ) ( ) 4085.11 maxmin ==== ββVVVV satBEsatCE a) ?=BR

( )

B

satBEBB I

VVR

−=

( ) ARVV

IC

satCECCCS 1.18

111200

=−

=−

=

AI

I CSBS 26.2

81.18

min

===β

BS

B

II

ODF =

AIODFI BSB 31.1126.25 =×=×=

Ω=−

= 7514.031.1

5.110BR

b) 6.131.111.18

===B

CSs I

Iβ

WIVIVP CCEBBET 07.351.18131.115.1 =×+×=+=

RB

VBEIE

R

+

_

VB+

_

VCC

Ic



Switching Characteristics: When the input voltage is reverse from 1V to 2V− and the base current is also changed to 2BI− the

collector current does not change for a time st called the storage time. sT is required to remove the

saturating charge from the base. Due to the change in polarity of BV from 21 VtoV − , the reverse

current 2BI− helps to discharge base current .The turn on time oNT is the sum of delay time dt and

rise time rt .

rdoN ttT +=

Q. The wave form of the transistor switch is shown in the figure. The parameters are

( ) ( ) 3,5,1,5.0,100,2.8,3,250 ========= fsrdCSsatCEBsatBECC TsTsTsTAIVVAIVVVV µµµ.10KHzfand s = The duty cycle is 50%, the collector to emitter leakage current is mAIceo 3= .

Determine the power loss due to collector current a) During turn on (ton = td + tr ) b) During conduction period (tn) c) During turn off (Toff = ts + tf) d) During off time to e) Total average power loss (PT)

t

VBV1

-V2

KT

t

VB

t

VBV1

-V2

KT (1-K)T

t

VBIB

-IB

totftstntd

0.1Ics

0.9cs

Ics

Fig : Switching times of bipolar transistor



Solution:

sf

Ts

µ1001==

ststststK fsdr µµµµ 3,5,5.0,1,5.0 ===== stttKT nrd µ50=++= stn µ5.485.0150 =−−=

( ) stttTK ofs µ501 =++=− sto µ425350 =−−=

a) Due to delay time, dtt ≤≤0 ( ) CEOc Iti =

( ) CCCE VtV = Instantaneous power due to collector current, ( ) ( ) ( ) WVItVtitP CCCEOCECC 75.0250105. 3 =××=== −

Average power loss during delay time is

( ) mWtfdttPT

P ds

t

Cdd 75.375.01

0=××== ∫

During rise time,

( ) ( )CEO

r

CEOCSc I

ttII

ti ++

=

( ) ( )CC

r

satCECCCE V

tVV

tV +⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

VCE

o

VCC

VCC(Sat)

IC

ton toff

Ics

ICCo

iB td tr tr ts tf to

t

t

t

KT (1-K)T

T=1/fc

IBs

iB

o

VBE(Sat)

o



( ) ( ) ( )TVTITP CECC =

Average Power during rise time,

( ) WdtVtt

Vt

tIT

dttPT

P rr t

CCr

CC

r

CSt

Cr 33.42..11

00=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−== ∫∫

Total power loss during turn on time WWmWPPP rdon 33.4232.4275.3 =+=+=

Total power WPT 05.274= Switching Limits: i) Secondary Breakdown:

Secondary (second) break down is caused by localize thermal runaway, resulting from high current concentration. The current concentration may be caused by defects in transistor structure. ii) Forward Bias Safe Operating Area (FBSOA):

FBSOA indicates CEC VI − limits of X’sistor during turn off and for reliable operation, the

transistor must not be subjected to greater power dissipation, then that shown by FBSOA curve. iii) Reverse Bias Safe Operating Area (RBSOA): The manufacture provide CEC VI − I limit during reverse bias turn off, as RBSOA

iv) Power Derating: The ambient temperature and thermal resistance must be consider when interpreting the rating of

device. The manufacturer shows the derating curve for thermal derating and second breakdown derating. Base Drive Control: Ton can be reduced by allowing base current picking during turn .onT and offT and be reduce by

reducing base current and allowing base current picking turning turn off. The commonly used technique for optimizing the base drive of transistor are:

1. Turn on Control 2. Turn off Control 3. Proportional Control 4. Anti-saturation Control

IB1

t

IBS

-IB2

IB

FIg : Base Drive Current Waveform



Turn on Control: When an input voltage is turned on, the base current is limited by 1R and initial value of base current

is 1

1

RVV

I BEBO

−= and final value of base current is

21

11 RR

VVI BE

R +−

=

The capacitor C charges up to infinity

21

11. RR

RVVC +

≅

Turn off Control: If the input voltage is changed to –V2, during turn off, the capacitor voltage CEV is added to 2V

as a reverse voltage across the transistor. There will be base current picking during turn off. Proportional Base Control: If the collector current changes due to change in load demand, the base drive current is changed in proportional to collector current. Anti saturation Control: If the transistor is driven hard, the storage time which is proportional to the base current increases and switching speed is reduced. The storage time can be reduced by operating the transistor in sub-saturation rather than hard saturation. Power MOSFET: A power MOSFET is a voltage controlled device. The switching speed is very high. Switching times are of the order of nanosecond. MOSFET are of two types: 1) Diplation MOSFET 2) Enhancement MOSFET

t

t1 t2

VB

V1

-V2

C

R1

R2VB

+

_

IC

IE

Vcc

Rc

VGS

+

_

GD RD

VDDVGS

+

_

G

DRD

VDD

ID

a) n-channel b) p-channel



Switching Characteristics:

VGS

+

_

G

DRD

VDDSVGS

+

_

G

D RD

VDDS


Fig : Enhancement MOSFET ID

VP o VGS -ID

VPo

VGS


a) Depletion MOSFET

o VGSVr

VGS

Vr

-IDb) p-channel

a) n-channel

b) Enhancement MOSFET

VDS

Saturation region

VGS2VGS1

VGS = VG

Linearregion

Fig : Output Characteristics

VG

V1

O

VGS

V1VGSo

VT

t

td(on) tr td(off) tf



( ) timerise

delay timeon - turn=

=

r

d

tont

( ) delay time off turn =offtd timefall =ft

Gate Drive: Turn on delay time can be reduced by connecting RC circuit. SIT: A SIT is a high power, high frequency device. A SIT has a short channel length, loop gate series resistance, loop gate source capacitance and small thermal resistance. It has low noise, low distortion and high audio frequency power capacity. The turn on and turn off times are very small. Typically 0.25 sµ . The current rating of SIT can be output 300A, 1200V and switching speed can be as high as 100kHz. It is most suitable for high frequency applications. Example: audio, VHF, UHF and microwave amplifier. IGBT:

_

GVDD

RS R1

RG

RD

VG

+

G

D

S Symbol

B

C

E

Symbol



IGBT combines the advantes of BJT and MOSFET. IGBT has high input impedance like MOSFET and low ON-state conduction loss like BJT. IGBT is inherently fast then BJT. However, the switching speed of IJBT is inferior to that of MOSFET. The current rating of IGBT can be up to 400A, 1200V and switching frequency can be up to 20 kHz. IGBT’s are finding increasing application in medium power application such as DC and AC motor drive, power supplies. Series and Parallel Operation: Transistor may be operated in series to increase their voltage handling capability. It is very important that the series connected transistor are turn on and turn off. Simultaneously, the device should be match for gain, transconductance threshold voltage, on-state voltage, turn on time and turn off time. Even the gate or base drive characteristics should be identical. Voltage sharing n/w, similar to diode could be used. Transistor are connected in parallel if one device cannot handle load current demand for equal current sharing, the transistor should be matched for gain, transconductance, turn on time and turn off time. But, in practice, it is not always possible to meet this requirements, a reasonable amount of current sharing can be obtained by connecting resistor in series with emitter (or source) terminals as shown in figure. Q. Two MOSFET are connected in parallel to carry a total current of .20AIT = The drain to source voltage of MOSFET 1m is 11 5.2 VVDS = and that of MOSFET 2m is VVDS 32 = . Determine the drain current of each transistor (MOSFET).

( )iIII DDT −+= 21 ( )iiRIVRIV sDDssDDS −+=+ 222111 .

IE2

θ2 θ1

RC

Vcc

RC1RC2

IE1

IT

VDD

D D

RS1 RS2

RDIT = 20A

S S

M2M1

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