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Chap. 3 Common Diode Application

3.1 Transformers

3.2 Half-wave rectifiers

3.3 Full-wave rectifiers

3.4 Full-wave bridge rectifiers

Basic power supply block diagram

Power supply A group of circuits used to convert ac to dc. Rectifier A circuit that converts ac to pulsating dc. Filter A circuit that reduces the variations in the output of a rectifier. Voltage regulator A circuit designed to maintain a constant power supply output Transformer A circuit connecting dc power supply to the ac input

3.1 Transformers

Transformers are made up of inductors that are in close proximity to each other, yet not electrically connected.

A transformers provides ac coupling from primary to secondary while providing physical isolation between the two circuits.

3.1 Transformers

Three type of transformer Step-up transformer (승압형) : provides a secondary voltage that is greater than the primary voltage Step-down transformer (감압형) : provides a secondary voltage that is less than the primary voltage Isolation transformer (분리형) : provides an out voltage that is equal to the input voltage

3.1 Transformers

turns ratio (권선비) : the ratio of the number of turns in the primary to the number of turns in the secondary.

turns ratio is equal to the voltage ratio of the component.

When the turns ratio an primary voltage are known, the secondary voltage can be found as :

3.1.2 calculating secondary current

Ideally, transformer transfers 100 % of it’s power to the secondary.

Since power equals the product of voltage and current ,

SP P PS S P P S P P

S P S S

VI V NV I V I I I I

I V V N

Example 3.1

Assume that the fuse limits the value of Ip to 1 A. What is the limit on the value of the secondary current ?

1 250

4

PS P

S

NI I A mA

N

If the secondary current tries to exceed the 250 mA limit, the primary current will exceed its limit and blow the fuse.

3.2 Half-wave rectifiers

Three type of rectifier circuits: Half-wave rectifier Full-wave rectifier Bridge rectifier

Half-wave rectifier circuit : a diode is connected in series with a transformer and a load resistance

3.2.1 Basic circuit operation of the positive half-wave rectifier

Positive half-wave rectifier ~ the negative half-cycle of the input to the rectifier is eliminated by the one-way conduction of the diode

During the positive half-cycle of the input, D1 is forward biased and provides a current path

When D1 is reverse biased, there is no current flow and no voltage drop through RL

Ideal model VD1 VL

Forward bias 0 V VS

Reverse bias VS 0 V

3.2.2 Negative Half-wave rectifiers

Positive & Negative Half-wave rectifiers

Basic Transformer

Small power transformer

The basic transformer is formed from two coils that are usually wound

on a common core to provide a path for the magnetic field lines.

Air core Ferrite core Iron core

Direction of windings

The direction of the windings determines the polarity of the voltage across the secondary winding

with respect to the voltage across the primary. Phase dots are sometimes used to indicate polarities.

In phase Out of phase

secondary prisec

primary pri sec

N IVn

N V I

3.2.3 Calculating load voltage & current

When we take VF into account (i.e. introducing practical model), the peak load voltage, VL(pk) is found as

where VS(pk) is the peak secondary voltage of transformer:

with VP(pk) = the peak transformer primary voltage

Example 3.2

Q. The input voltage is 120 Vac. Determine the peak load voltage

The secondary peak voltage is obtained from the primary value.

Example 3.4

Q. The input voltage is 120 Vac. Determine the peak load current.

( )

( )

( ) ( )

( ) ( )

( )

( )

120170

0.707 0.707

17056.7

3

56.7 0.7 56

565.6

10

P rms

P pk

SS pk P pk

P

L pk S pk F

L pk

L pk

L

V VV V

N VV V V

N

V V V V V V

V VI mA

R k

3.2.4 Average voltage

The value of an average voltage (Vave) is a dc voltage value equivalent to the ac waveform after an average over one period.

Vave

0

sin 2pk aveV d V

0.318ave pkV V

3.2.4 Average current

The value of an average current (Iave) is a dc current value equivalent to the ac waveform after an average over one period.

Example 3.6 Assume an input voltage of 120 Vac. Determine the value of Iave

( )

( )

( ) ( )

( ) ( )

( )

120170

0.707 0.707

17085

2

85 0.7 84.3

84.326.8

26.81.34

20

P rms

P pk

SS pk P pk

P

L pk S pk F

L pk

ave

aveave

L

V VV V

N VV V V

N

V V V V V V

V VV V

V VI mA

R k

( )

( )

( )

84.34.22

20

1.34

L pk

L pk

L

L pk

ave

V VI mA

R k

II mA

3.3 Full-wave rectifiers

Center-tapped transformer: the voltage from the center tap to each terminal of the outer winding coils is equal to one half of the secondary voltage

3.3.1 Basic circuit operation

During an positive half-cycle of the input, D1 is forward-biased and D2 is reverse- biased

When D2 is forward-biased, D1 is reverse-biased. But, the direction of the load current is the same as before.

3.3.2 Calculating load voltage & current

With assuming the practical diode model, the peak load voltage (VL(pk)) for full-wave rectifier is found as

Average load voltage (Vave) for a full-wave rectifier is found as

full-wave rectifier

half-wave rectifier

Example 3.9

Q. Determine the dc load voltage

3.3.3 Peak inverse voltage (PIV)

From practical model, D1 is forward-biased and exhibits a voltage drop of 0.7 V

Because D1 is in series with D2, the PIV across D2 is reduced by the voltage drop (VF) across D1

Note that PIV = VS(pk) from the ideal diode model

3.3.4 Negative full-wave rectifier

3.4 Full-wave bridge rectifiers

1. The bridge rectifier does not require the use of a center-tapped transformer, or the transformer itself. It can be connected directly to the AC input even without the transformer. 2. When it is connected to a transformer with the same secondary voltage, it produces nearly double the peak output voltage of the full-wave center-tapped rectifier.

3.4.1 Basic circuit operation

During the positive half-cycle of the input, D1 and D3 are switched “on”, while D2 and D4 are “off”.

During the negative half-cycle of the input, D2 and D4 are switched “on”, while D1 and D3 are “off”.

3.4.2 Load voltage & current

Since there are two conducting diodes in series with the load resistance, the peak load voltage is founds as:

Example 3.11 Determine the dc load voltage and current

( )

( ) ( )

( )

1217

0.707

1.4 15.6

2 2(15.6 )9.93

9.9382.8

120

S pk

L pk S pk

L pk

ave

aveave

L

VV V

V V V V

V VV V

V VI mA

R

3.4.3 Bridge vs. full-wave rectifier

Example 3.12 Assume a full-wave rectifier with an input voltage of 12 Vac and a load resistance of RL = 120 . Determine Vave and Iave.

12 120

( )

( )

( )

( )

1217

0.707

170.7 0.7 7.8

2 2

2 2(7.8 )4.97

4.9741.4

120

acS pk

S pk

L pk

L pk

ave

aveave

L

VV V

V VV V V V

V VV V

V VI mA

R

( )

( ) ( )

( )

1217

0.707

1.4 15.6

2 2(15.6 )9.93

9.9382.8

120

S pk

L pk S pk

L pk

ave

aveave

L

VV V

V V V V

V VV V

V VI mA

R

Bridge rectifier results Center-tapped rectifier results

Double

efficiency !

3.4.4 Peak Inverse Voltage

The PIV across each diode in the bridge rectifier is approximately equal to the peak secondary voltage, VS(pk). Counting VF results in PIV = VS(pk) – 0.7

Chap. 3 Common Diode Application

3.5 Filters

3. 6 Zener voltage regulators

Basic power supply block diagram

Power supply A group of circuits used to convert ac to dc. Rectifier A circuit that converts ac to pulsating dc. Filter A circuit that reduces the variations in the output of a rectifier. Voltage regulator A circuit designed to maintain a constant power supply output Transformer A circuit connecting dc power supply to the ac input

3.6 Filters

The effects of filtering on the output of a half-wave rectifier.

There are still voltage variations after filtering, but, the amount of variation is severely reduced.

Ripple voltage (Vr): The variation in the output voltage of a filter ~ depends on the rectifier used, the filter component value, and the load resistance.

ripple voltage

3.6.1 Basic capacitive filter

The capacitive filter is the most basic filter type and the most commonly used.

The filtering action is based on the charge/discharge action of capacitor

During the positive half-cycle, D1 conducts and capacitor charge rapidly.

When the output voltage of the rectifier becomes smaller than the capacitor voltage, the capacitor acts as the voltage source for the load resistance.

3.6.1 Basic capacitive filter : time constant

For an RC circuit, the time constant for charging or

discharging is t = RC [seconds] ~ exponential curve 100%

80%

60%

40%

20%

00 1t 2t 3t 4t 5t

99%98%

95%

86%

63%

37%

14%

5%2% 1%

Number of time constants

Perc

ent o

f final v

alu

e

C

R

C

R

It takes five time constants for a capacitor to charge or discharge fully

T = 5RC

total capacitor charge time (T)

total capacitor discharge time (T)

resistance of diode under forward bias = 5

Note that the charging time is much shorter than the discharging time!

The charging time is not determined by RL but by the diode resistance in forward bias

3.6.1 Basic capacitive filter

The amplitude of the ripple voltage at the output of a filter varies inversely with the values of filter capacitance and load resister

RLC time for discharge increases

Larger capacitance value results in highly elongated discharge time and increased charging time as well.

3.6.2 Surge current

Surge current: high initial current in a power supply

For the first instant, the discharged capacitor acts as a short circuit ~ charging of positive and negative charge on both sides of capacitor

Isurge

Example 3.14

Surge limiting resistor

The resistor Rsurge helps to limit surge current, but it also limits the output voltage on RL

The other way to limit the surge current is to use small capacitance capacitor.

Smaller results in a shorter charging time of , but it also results in larger ripple voltage

Q V VQ CV I C t C

t t I

C t

.

3.6.3 Filter Output voltage

peak output voltage of rectifier

peak-to-peak ripple vol

2

tage

rdc pk

pk

r

V

V

V V

V

where dc load current,

t = interval time between charging peaks

C = capacitance [F

]

L

L

r

I tV

I

C

3.6.3 Filter Output voltage

1 116.7 ~ half-wave rectifier

60

1 18.3 ~ full-wave rectifier

120

half wave

full wave

t msf Hz

t msf Hz

, ,2L

r r half wave r full wave

I tV V V

C

EXAMPLE 3.15

EXAMPLE 3.16

( ) ( ) ( )

( )

( )

1

2 2 2

12

16.2816.16

1 0.007381

2

dcL

L

Lr

dcr Ldc L pk L pk L pk

L

dc L pk

L

L pk

dc

L

VI

R

I tV

C

VV I t tV V V V

C C R

tV V

CR

VV

t

CR

Chap. 3 Common Diode Application

3.7 Zener Voltage Regulators

Power supply : Regulators

3.7 Zener Voltage Regulators

Connection of zener diode :

• reverse bias operation

• constant voltage for IZK < I < IZM

• parallel circuit of a zener diode and a load resistor

• VL = VZ if IZK < ID < IZM

3.7.1 Total Circuit Current

3.7.4 Zener Current

Load current

Total current

, : current through a load resistor,

,

Zener curr

, en t

L L

ZL

L

RS in ZT L Z

S S

Z T L

I R

VI

R

V V VI I I

R R

I I I

EXAMPLE 3.17 & 3.19

9.10.91

10

4.95 0.91 4.04

ZL

L

Z T L

V VI mA

R k

I I I mA mA mA

3.7.5 Load Variations

What are the practical limit on the value of RL?

If RL = 0 (the load is shorted), IT = IL and IZ = 0 < IZK ~ the regulation is lost!

There exists the minimum value of RL for the voltage regulation.

To maintain zener regulation, the minimum zener current must be equal to IZK. Therefore,

Since IT = IZ + IL and IZ(min) = IZK,

IL(max) = IT – IZ(min) = IT – IZK

Since IL(max) occurs when the load resistance is at a minimum,

(min)

(max)

Z ZL

L T ZK

V VR

I I I

regulating till IZ < IZM

EXAMPLE 3.20

Q. Assume diode current rating of IZK = 3 mA and IZM = 100 mA. Determine the minimum RL.

(max)

(min)

(max)

20 3.316.7

1

16.7 3 13.7

3.3241

13.7

in ZT

S

L T ZK

ZL

L

V V V VI mA

R k

I I I mA mA mA

V VR

I mA

3.7.7 Zener Reduction of Ripple Voltage

How does the zener diode reduce the ripple voltage?

The equivalent circuit of a zener diode consists of a zener impedance (ZZ) and VZ.

Using the voltage divider rule, the ripple output from the regulator can be found as

(

( )

)

and ( || )

where

the ripple present at the regulator output,

( || ) the parallel combiantion

( ||

of and the load resistance

= the reg

)

( || )

ulator se

Z Lr out r

Z L

Z LZ L

Z L

r out

Z

S

L Z L

S

Z RZ R

Z R

V

Z R Z R

Z RV V

Z R

R

R

ries resistance

= the peak-to-peak ripple voltage present at the regualtor inputrV

EXAMPLE 3.21

Q. The filter output has a 1.5 Vpp as a ripple voltage. Determine the ripple voltage at the load resistor.

( )

5 120( || ) 4.8

5 120

( || ) 4.81.5 0.13

( || ) 4.8 51

Z LZ L

Z L

Z Lr out r pp pp

Z L S

Z RZ R

Z R

Z RV V V V

Z R R

3.7.8 Putting It All Together :