Chapter 10 : Power Circuits and Systemssvbit-it.weebly.com/uploads/2/4/8/5/2485214/chap-10.pdf[ N 1 / N 2] 2 = 9 N 1 N 2 = 3 …Ans. Ex. 10.9.4 : For a class B push-pull amplifier

Basic Electronics (GTU) 10-1 Power Circuits and Systems

Chapter 10 : Power Circuits and Systems

Section 10.9 :

Ex. 10.9.2 : A class B push-pull amplifier supplies power to a loud speaker of 8 . The output

transformer has a turns ratio N1 : N2 of 4 : 1 and efficiency of 95%. Calculate the

following :

1. Maximum power output. 2. Maximum power dissipation in each transistor.

3. Maximum base currents for each transistor.

Assume hfe = 15 and VCC = 24 Volts. Page No. 10-43.

Soln. :

Given : [ N1 / N2 ] = 4, of transformer = 0.95, VCC = 24 V, RL = 8 , hfe = 15

To calculate Pac ( max ) :

The maximum output power can be obtained by assuming Vm = VCC .

Pac ( max ) = V

2

m

2 RL

= V

2

CC

2 RL

...(1)

But R L = [ N1 / N2 ]

2 RL = ( 4 )

2 8 = 128 ...(2)

Pac ( max ) = ( 24 )

2

2 128 = 2.25 Watt ...(3)

But since the transformer efficiency = 0.95 the actual output power is given by,

Po = Pac ( max ) = 2.25 0.95 = 2.137 W ...Ans.

To calculate the power dissipation :

Maximum power dissipation = 2 V

2

CC

2 R

L

= 2 (24 )

2

2 128

= 0.912 W.

But this power dissipation corresponds to two transistors. Hence the maximum power dissipation

per transistor is given by,

Pd ( max ) / Transistor = 0.912

2 = 0.456 W ...Ans.

To calculate Im :

We know that, Pac ( max ) = Vm Im

2 =

VCC Im

2 2.25 =

24 Im

2

Im = 2.25 2

24 = 0.1875 Amp ...(4)

To calculate the maximum base current :

Ib ( max ) = Im

hfe

= 0.1875

15 = 12.5 mA ...Ans.

Ex. 10.9.3 : The maximum ac power supplied to a 4 loud speaker by a class B push-pull amplifier is

2 W. Calculate the turns ratio of the transformer if VCC = 12 V and the transformer is ideal.

Page No. 10-43.


Soln. :

Given : VCC = 12 V, Pac ( max ) = 2 W, RL = 4

We know that the maximum ac power is obtained when Vm = VCC .

Pac ( max ) = V

2

m

2 RL

= V

2

CC

2 RL

R L =

V2

CC

2 Pac ( max ) =

( 12 )2

2 2 = 36 ...(1)

But R L = [ N1 / N2 ]

2 RL

36 = [ N1 / N2 ]2 4

[ N1 / N2 ]2

= 9

N1

N2

= 3 …Ans.

Ex. 10.9.4 : For a class B push-pull amplifier operating on a 24 V supply, the collector voltage of each

transistor swings from 24 V i.e. VCC to 4 V, with application of the input signal. Each

transistor has the maximum collector dissipation rating of 10 W. Calculate :

1. Output power 2. DC input power

3. Collector circuit efficiency. Assume [ N1 / N2 ] = 1. Page No. 10-43.

Soln. :

Given : VCC = 24 V, Vmax = 24 V, Vmin = 4 V, Pd ( max ) / transistor = 10 W.

To calculate the ac output power and dc power, it is necessary to calculate R L .

Step 1 : Calculate the value of R L :

Pd ( max ) = 2 Pd ( max ) / transistor = 2 10 = 20 W

But, Pd ( max ) = 2 V

2

CC

2 R

L

20 = 2 ( 24 )

2

2 R

L

, R L = 5.84 ...(1)

Step 2 : Calculate the output power ( Pac ) :

Pac = V

2

m

2 RL

= ( )Vmax – Vmin

2

2 RL

= ( )24 – 4

2

2 5.84 = 34.12 Watt ...Ans.

Step 3 : Calculate the dc input power ( Pdc ) :

Pdc = VCC 2 Im

But, Im = Vm

RL

= ( 24 – 4 )

5.84 = 3.42 Amp.

Pdc = 24 2 3.42

= 52.32 Watt ...Ans.


Step 4 : To calculate the % efficiency :

% = Pac

Pdc

100 = 34.12

52.32 100 = 65.22 % ...Ans.

Ex. 10.9.7 : A single transistor operates as an ideal class B amplifier. The dc current drawn by the

transistor from the supply is 20 mA. Calculate the amount of ac power delivered to the load

if load resistance is 1 k. Page No. 10-48.

Soln. :

Given : Idc = 20 mA, RL = 1 k

Steps to be followed :

Step 1 : Calculate the value of Im from Idc as Idc = ( Im / ) .

Step 2 : Then calculate the output power Pac = I

2

m RL

2

Fig. P. 10.9.7: Supply current waveform

Step 1 : Calculate the value of Im :

Referring to Fig. P. 10.9.7 we can write the expression for the average value of the source current

as,

Idc = 1

2

0

Im sin t dt = – Im

2 [ ] cos t

0

= – Im

2 [ ] cos – cos 0 =

– Im

( ) –1 – 1

Idc = Im / ...(1)

Im = Idc = 20 mA = 62.63 mA ...(2)

Step 2 : Calculate the output power :

Pac = I2

L ( rms ) RL ...(3)

From the current waveform of Fig. P.10.9.7, we can calculate IL rms as follows :

IL rms =

1

2

0 I

2

m sin

2 t dt

1 / 2

=

I

2

m

2

0 1 – cos 2t

2 dt

1 / 2

=

I

2

m

4

0 ( ) 1 – cos 2t dt

1 / 2

= Im

2

1

( ) – 0 = Im / 2 …(4)


Substitute Equation (4) into (3) to get,

Pac = ( Im / 2 )2 RL = I

2

m RL / 4

= ( ) 62.83 10

–3

2 1 10

3

4 = 0.985 W ...Ans.

Ex. 10.9.8 : For an ideal class B push-pull amplifier show that the maximum output power can be as

high as five times the maximum power dissipation in each transistor. Page No. 10-48.

Soln. : We are expected to prove the following relation :

Pac ( max ) = 5 Pd ( max ) per transistor

Referring to Equation (10.9.14) we can write that,

Pac ( max ) = V

2

CC

2 RL

...(1)

Referring to Equation (10.9.15) we can write that,

Pd ( max ) per transistor = 2

2 Pac ( max )

Pac ( max ) =

2

2 Pd ( max ) per transistor = 4.934 Pd ( max ) per transistor

Pac ( max ) 5 Pd ( max ) per transistor

Hence proved.

Section 10.12 :

Ex. 10.12.1 : A complementary push-pull amplifier has capacitive coupled load RL = 8, supply voltage

12 V calculate :

1. Pac( max ) 2. PD of each transistor 3. Efficiency Page No. 10-53.

Soln. :

Step 1 : To calculate Pac ( max ) :

Pac ( max ) = V

2

m

2 RL

Substitute Vm = VCC to get,

Pac ( max ) = V

2

CC

2 RL

= ( 12 )

2

2 8 = 9 Watt ...Ans.

Step 2 : To calculate Im :

Im = Vm

RL

VCC

RL

= 12

8 …(1)

= 1.5 Amp

Step 3 : To calculate Pdc :

Pdc = 2 VCC Im

= 2 12 1.5

= 11.46 W Fig. P. 10.12.1 : Given circuit


Step 4 : To calculate efficiency :

% = Pac

Pdc

100 = 9

11.46 100 = 78.5 % ...Ans.

Step 5 : To calculate Pd of each transistor :

Total Power dissipation Pd = Pdc – Pac

= 11.46 – 9 = 2.46 W ...Ans.

Pd per transistor = 2.46

2 = 1.23 W …Ans.

Ex. 10.12.2 : The circuit shown in Fig. P. 10.12.2

operates with a sinusoidal input.

Calculate :

1. Maximum A.C. power output.

2. Power dissipation in each

transistor.

3. Conversion efficiency at

maximum power output.

Page No. 10-53.

Fig. P. 10.12.2

Soln. :

As only a single supply is being used, we have to use all the expressions derived for the

complementary symmetry with a substitution of VCC = VCC / 2 .

Step 1 : Maximum A.C. power output Pac ( max ) :

Pac = V

2

m

2 RL

To obtain Pac ( max ) we substitute Vm = VCC

But here we will substitute Vm = VCC / 2

Pac ( max ) = ( )VCC / 2

2

2 RL

= V

2

CC

8 RL

= ( 20 )

2

8 8 = 6.25 Watt ...Ans.

Step 2 : Power dissipation in each transistor :

We know that,


2 Pac ( max )


2 6.25 = 1.26 W ...Ans.

Total power dissipation = 1.26 2 = 2.53 W

Step 3 : Efficiency :

= Pac ( max )

Pac ( max ) + Pd

= 6.25

6.25 + 2.53 = 0.7115 = 71.15% ...Ans.


Ex. 10.12.3 : An ideal class B complementary symmetry push pull amplifier operates with VCC = 12 V

and RL = 5 . If the input is sinusoidal, calculate :

1. Maximum power output.

2. Power dissipation in both transistors.

3. Power dissipation in each transistor.

4. Conversion efficiency for maximum output. Page No. 10-54.

Soln. : This example is similar to Ex. 10.12.2.

1. Maximum ac power Pac ( max ) :

Pac ( max ) = V

2

CC

8 RL

= ( 12 )

2

8 5 = 3.6 W ...Ans.

2. Maximum power dissipation in both the transistors :

Pd ( max ) = 4

2 Pac ( max ) =

4

2 3.6 = 1.46 W ...Ans.

3. Maximum power dissipation per transistor :

Pd ( max ) / transistor = 1.46 / 2 = 0.73 W ...Ans.

4. Efficiency :

% = Pac ( max )

Pac ( max ) + Pd ( max )

100 = 3.6

5.06 100 = 71.15% ...Ans.

Ex. 10.12.4 : A series fed class A amplifier shown in Fig. P. 10.12.4, operates from D.C. source and

applied sinusoidal input signal generates peak base current 9 mA. Calculate

1. Quiescent current ICQ 2. Quiescent voltage VCEQ

3. D.C. input power PDC 4. A.C. output power PAC

5. Efficiency.

Assume = 50 and VBE = 0.7 V. Page No. 10-54.

Soln. :

1. IBQ = VCC – VBE

RB

= 20 – 0.7

1.5 103

= 12.867 mA

2. ICQ = IBQ

= 50 12.867

= 643.33 mA ...Ans.

Fig. P. 10.12.4 : Given circuit

3. VCEQ = VCC – ICQ RC = 20 – (0.6433 16 ) = 9.7 ...Ans.


4. DC input power PDC = VCC ICQ = 20 0.6433 = 12.866 W ...Ans.

5. AC output power PAC = I2

C rms RC

Ib peak = 9 mA = 50 9 = 540 mA

IC rms = IC ( peak )

2 = 381.83 m

PAC = (0.3818)2 16 = 2.3328 W ...Ans.

6. Efficiency = PAC / PDC = 2.3328

12.866 = 0.1813 or 18.13% ...Ans.

Ex. 10.12.5 : A class B push-pull amplifier supplies power to a resistive load of 12 . The output

transformer has a turns ratio of 3 : 1 and efficiency of 78.5%. Obtain :

1. Maximum power output.

2. Maximum power dissipation in each transistor.

3. Maximum base and collector current for each transistor.

Assume hfe = 25 and VCC = 20 V. Page No. 10-54.

Soln. :

Given : [ N1 : N2 ] = 4, of transformer = 0.785

VCC = 20 V, RL = 12 and hfe = 25

Step 1 : Calculate Pac ( max ) :

The maximum output power can be obtained by assuming Vm = VCC .

Pac ( max ) = V

2

m

2 RL

= V

2

CC

2 RL

But R L = [N1 / N2 ]

2 RL = (3)

2 12 = 108

Pac ( max ) = ( 20 )

2

108 = 3.7 W ...Ans.

Step 2 : Maximum power dissipation :

Pd ( max ) = 2 V

2

CC

2 R

L

= 2 ( 20 )

2

2 108

= 0.75 W.

Pd ( max ) per transistor = 0.75 / 2 = 0.375 W ...Ans.

Step 3 : Calculate Im :

We know that Pac ( max ) = Vm Im

2 =

VCC Im

2

3.7 = 20 Im

2 Im = 0.37 Amp

Step 4 : Calculate the maximum base current :


Ib max = Im

hfe

= 0.37

25 = 14.8 mA ...Ans.

Ex. 10.12.6 : For the transformer coupled class A circuit shown in Fig. P. 10.12.6 calculate the

maximum efficiency. Assume 1. the device to be ideal and 2. RB is adjusted for maximum

input swings. Page No. 10-54.

Fig. P. 10.12.6

Soln. : Please Refer section 10.7.1.

Ex. 10.12.7 : In a class B amplifier VCE ( min ) = 1 Volt and the supply voltage VCC = 18 V. Calculate the

collector efficiency. Page No. 10-54.

Soln. : As proved in section 10.9.2,

Efficiency = Pac / Pdc = Vm Im / 2

2 VCC Im / =

Vm

4 VCC

But Vm = VCC – VCE ( min ) = 18 – 1 = 17 V

= 17

4 18 = 0.7417 or 74.17 % ...Ans.

Ex. 10.12.9 : A single transistor amplifier with transformer coupled load produces harmonic amplitudes

in the output as,

B0 = 1.5 mA B1 = 120 mA

B2 = 10 mA B3 = 4 mA

B4 = 2 mA B5 = 1 mA

1. Determine the percentage total harmonic distortion

2. Assume a second identical transistor is used alongwith a suitable transformer to

provide push-pull operation.

Use the above harmonic amplitudes to determine the new total harmonic distortion.

Page No. 10-55.

Soln. :

Part (i)

Second harmonic distortion D2 = | B2 |

| B1 | =

10

120 = 0.0833


Third harmonic distortion D3 = | B3 |

| B1 | =

4

120 = 0.0333

Fourth harmonic distortion D4 = | B4 |

| B1 | =

2

120 = 0.0166

Fifth harmonic distortion D5 = | B5 |

| B1 | =

1

120 = 8.33 10

–3

Hence percentage total harmonic distortion

D = D2

2 + D

2

3 + D

2

4 + D

2

5 100

= ( )0.08332 + ( )0.0333

2 + ( )0.0166

2 + ( )8.33 10

–3 2 100

= 0.09165 100 = 9.165 % ...Ans.

Part (ii)

When a second transistor is used, all the even harmonic components are automatically cancelled

out. So the THD is given by,

D = D2

3 + D

2

5

100

= ( )0.03332 + ( )8.33 10

–3 2 100 = 3.4326 % ...Ans.

Ex. 10.12.10: A transformer coupled class A, AF power amplifier uses an ideal output transformer

having a load resistance of 1.5 on its secondary side. If the amplifier is to deliver

maximum power output of 3 W to the load when the collector supply voltage is 15V,

determine :

1. Turns ratio of output transformer. 2. Power transistor ratings required.

3. DC power input to the amplifier. Page No. 10-55.

Soln. :

Given : RL = 1.5 , Pac = 3 W, VCC = 15 V.

Turns ratio of the output transformer :

The maximum output power is given by,

Pac ( max ) = ( )V1m

2

2 RL

where V1m = Maximum primary voltage = VCC

R L = Reflected load resistance on the primary side.


Pac ( max ) = V

2

CC

2 RL

R L =

( )152

2 3 = 37.5 ...(1)

We know that, R

L

RL

=

N1

N2

2

N1

N2

2

= 37.5

1.5 = 25

The turns ratio N1

N2

= 5 …Ans.

Power dissipation of the transistor :

Pd ( max ) = VCC ICQ But ICQ = VCC

R L

= 0.4

Pd ( max ) = 15 0.4 = 6.0 W ...Ans.

DC input power to the amplifier :

PDC = VCC ICQ = 15 0.4 = 6 W ...Ans.

Ex. 10.12.11 : A push-pull class B, AF power amplifier uses ideal output transformer whose primary has

a total of 160 turns and whose secondary has 40 turns. It must be capable of delivering

40W to 8 . load under maximum power conditions. What is the minimum possible

value VCC ? Page No. 10-55.

Soln. :

Given : Turns ratio N1

N2

= 4, Pac ( max ) = 40W, RL = 8

We know that for a push-pull class B amplifier,

Pac ( max ) = V

2

m

2 RL

where Vm = VCC and R L = ( N1 / N2 )

2 RL

Pac ( max ) = V

2

CC

2 ( N1 / N2 )2

RL

V2

CC = 2 (4)

2 8 40 = 10240

VCC = 101.19 Volts ...Ans.

Thus the minimum required VCC is 101.19 Volts.


Ex. 10.12.12 : Derive the expression for power dissipation rating of a transistor operated as class A and

class B amplifier and show that class B can deliver 5 times the output power than class

A. A complimentary symmetry amplifier supplies output to a load of 3 from supply

voltage of 20V. Calculate maximum power output and rating of transistors.

Page No. 10-55.

Soln. :

Part I :

1. We know that for a class A amplifier

Pac ( max ) = 0.5 VCC ICQ ...(1)

and the maximum power dissipation is given by,

Pd ( max ) = VCC ICQ ...(2)

As there are two transistors in a class A push-pull amplifier.

Maximum power dissipation/transistor = 0.5 VCC ICQ = 0.5 Pd ( max ) . Therefore, for a class A

amplifier,

Pac ( max ) = Maximum power dissipation per transistor ...(3)

2. For a class B push pull amplifier


2 Pac ( max )

Pac ( max ) =

2

2 Pd ( max ) per transistor = 4.934 Pd ( max ) per transistor

Pac ( max ) 5 Pd ( max ) per transistor ...(4)

But, from Equation (3), Pd ( max ) per transistor = Pac ( max ) of class A amplifier hence Equation (4)

gets modified as :

Pac ( max ) of class B = 5 Pac ( max ) of class A ...Proved.

Part II : Given : A complementary symmetry amplifier, RL = 3 , VCC = 20V.

Maximum output power :

Pac ( max ) = V

2

CC

2 RL

= ( )20

2

2 3 = 66.66 W ...Ans.

Maximum power rating of transistor :

Im = Vm

RL

= VCC

RL

= 20

3 = 6.66 Amp.

Pdc = 2 VCC Im

RL

= 2 20 6.66

= 84.88 W.

Total power dissipation = Pdc – Pac = 84.88 – 66.66 = 18.22 W

Power dissipation per transistor = 18.22

2 = 9.11 W ...Ans.


Ex. 10.12.13 : The reflected load resistance of a push-pull amplifier is 4. It draws a current of 3.25

Amp from 24 V d.c. power supply with sinusoidal input.

Calculate : 1. a.c. power output. 2. Conversion efficiency

3. Minimum dissipation rating of each transistor. Page No. 10-55.

Soln. :

Given : R L = 4 , Idc = 3.25 Amp., VCC = 24 V.

AC output power Pac :

Pac = I2

1rms R

L ...(1)

But Idc = 2 I1m

Peak primary current I1m =

2 Idc =

2 3.25 = 5.1 Amp.

RMS primary current I1 rms = I1m

2 = 3.6 Amp.

Substituting this value into Equation (1) we get,

Pac = (3.6)2 4 = 52.12 Watts ...Ans.

Conversion efficiency ( ) :

DC input power PDC = VCC Idc = 24 3.25 = 78 W

Efficiency = Pac

PDC

= 52.123

78 = 0.6682 or 66.82% ...Ans.

Minimum dissipation rating of each transistor :

Total power dissipation Pd = PDC – Pac = 78 – 52.12 = 25.88 W

Minimum dissipation rating per transistor = 25.88

2 = 12.94 W ...Ans.

Ex. 10.12.14 : Explain three point method to obtain second harmonic distortion in a power amplifier. A

sinusoidal signal Vs = 1.75 sin 600t is fed to a power amplifier. The resulting output

current is :

Io = 15 sin 600t + 1.5 sin 1200t + 1.2 sin 1800t + 0.5 sin 2400t. Calculate percentage

increase in power due to distortion. Draw circuit diagram of complementary symmetry

power amplifier with a driver circuit using transistors and explain its operation in brief.

Page No. 10-56.

Soln. :

1. The output current is given by,

Io = 15 sin 600t + 1.5 sin 1200t + 1.2 sin 1800t + 0.5 sin 2400t. Comparing this expression with

the following ,

Io = A1 sin t + A2 sin 2t + A3 sin 3t + A4 sin 4t.


We get the amplitudes of various harmonic components A1, A2, A3 and A4 as :

A1 = 15, A2 = 1.5, A3 = 1.2, A4 = 0.5.

2. The percentage nth

harmonic distortion is given by,

% nth

harmonic distortion = % Dn = | An |

| A1 | 100

Second order distortion, D2 = | A2 |

| A1 | =

1.5

15 = 0.1

Third order distortion, D3 = | A3 |

| A1 | =

1.2

15 = 0.08

Fourth order distortion, D4 = | A4 |

| A1 | =

0.5

15 = 0.03333.

3. Let the power delivered to the load without distortion be P. Hence the load power when distortion

is considered is given by,

P = P ( 1 + D2 )

where D = Total harmonic distortion.

D = D2

2 + D

2

3 + D

2

4 + …..

P = P (1 + D2

2 + D

2

3 + D

2

4 ) = P [ ]1 + ( )0.1

2 + ( )0.08

2 + ( )0.0333

2

P = [ 1.0175] P

4. Percent increase in load power = P – P

P 100 =

1.0175 P – P

P 100 = 1.75% ...Ans.

For complementary symmetry power amplifier with a driver circuit refer section 10.9.7.

Ex. 10.12.15 : A single stage common-emitter A.F. power amplifier, working in class-A condition,

supplies 2 W to 4 k load. The zero signal D.C. collector current is 35 mA and D.C.

collector current with signal is 39 mA. Determine the percentage of second harmonic

distortion in the output. Page No. 10-56.

Soln. :

Given : ICQ = 35 mA .…when the signal is absent.

1. When the signal is present

Ic = ICQ + Ao = 39 mA

Ao = 39 – 35 = 4 mA = A2 = 4 mA

2. Now Pac =

A

2

1

2 +

A2

2

2 RL =

1

2 RL ( A

2

1 + A

2

2 )

2 = 1

2 4 10

3 [A

2

1 + ( 4 10

–3 )

2 ]


A2

1 = 9.84 10

–4 A1 = 0.0313 Amp or 31.368 mA.

3. Therefore the second harmonic distortion is given by,

% D2 = | A2 |

| A1 | 100 =

4

31.368 100 = 12.7% ...Ans.

Ex. 10.12.16 : For a class A amplifier shown in the Fig. P. 10.12.16(a) transformer turns ratio 8 : 1. The

transformer has an efficiency of 90% and d.c. resistance of 10 for the primary.

Transistor has = 20 and VBE = 0.5 V. Determine :

(a) Maximum power delivered to the load. (b) Circuit efficiency. Page No. 10-56.

Fig. P. 10.12.16(a)

Soln. :

Step 1 : Draw the Thevenin's equivalent circuit :

The Thevenin's equivalent circuit is as shown in Fig. P. 10.12.16(b).

VTH = R2

R1 + R2

VCC

= 100

100 + 1000 25

= 2.27 V

RB = R1 || R2

= 100 || 1000 = 91

Step 2 : Calculate IB :

Applying KVL to the base loop we get,

VTH = IB RB + VBE + IE RE

Fig. P. 10.12.6(b) : Thevenin's equivalent


VTH = ( IB 91) + 0.5 + [ ( 1 + ) IB 10 ] 2.27 = IB ( 91 + 210 ) + 0 .5

IB = 5.88 mA ...(1)

Step 3 : Calculate IE and VB :

IE = ( 1 + ) IB = 21 + 5.88 mA = 123.48 mA or 0.12348 A ...(2)

The base voltage VB is given by,

VB = VBE + IE RE = 0.5 + ( 0.12348 10 ) = 1.7348 V ...(3)

Step 4 : Calculate current I1 through R1 and IC :

I1 = VCC – VB

R1

= 25 – 1.7348

1 k = 23.2652 mA ...(4)

and IC = IB = 20 5.88 mA = 117.6 mA ...(5)

Step 5 : Calculate the dc input power :

Total dc input power PDC = ( IC + I1 ) VCC

= ( 23.2652 + 117.6) mA 25 = 3.5216 W ...(6)

Step 6 : Calculate maximum power delivered to the load :

The reflected load resistance to the primary is

R L =

RL

n2 = RL

N1

N2

2

= 5

8

1

2

= 320

For maximum power delivered to the load, calculate the maximum value of primary voltage.

V1 m = VCEQ

To calculate VCEQ apply KVL to the collector loop of Fig. P. 10.12.6(b).

VCC = ( IC 10 ) + VCEQ + 10 IE

VCEQ = 25 – (0.1176 10) – (0.12348 10) = 22.59 volts

Maximum primary power

P1 ( max ) = 1

2

V2

1m

R L

= 1

2

( )22.592

320 = 0.7974 W

The maximum load power is given by,

Pac ( max ) = Transformer P1 ( max ) = 0.9 0.7974 = 0.7176 W


Step 7 : Calculate efficiency :

% = Pac (max)

PDC

100 = 0.7176

3.5216 100 = 20.38 % ...Ans.

Ex. 10.12.17 : The collector characteristics of a power transistor are as follows :

VCE IC (Amp) IC (Amp) IC (Amp)

volts for Ib = 5 mA for Ib = 10 mA for Ib = 20 mA

1 0.32 0.62 1.25

5 0.38 0.70 1.38

10 0.40 0.75 1.48

15 0.42 0.82 1.58

20 0.48 0.90 -

This transistor is used in a transformer coupled class B push-pull amplifier with

VCC = 20 V, reflected load RL = 15 ohms. If a sinusoidal input signal of 20 mA peak is

applied to the transistor amplifier calculate :

(a) AC power output (b) Collector circuit efficiency.

(c) % Third harmonic distortion. Page No. 10-56.

Soln. :

Given : VCC = 20 V, R L = 15 , input current = 20 mA peak

Type of amplifier : Transformer coupled class B.

Step 1 : Draw the characteristics of power transistor :

The characteristics of power transistor is shown in Fig. P. 10.12.17.

The operation is class B, so the co-ordinates of Q point are (20 V, 0 A) and it is on the x-axis.

The dc load line is a vertical line passing through the Q point. Thus line is parallel to y-axis.

Draw the ac load line. The ac load line passes through the Q point and its slope is ( – 1/ R L ). The

equation for ac load line is,

IC = – 1

R L VCE + C ...(1)

At Q point, IC = 0 and VCE = VCC = 20 V, whereas R L = 15 .

0 = – 20

15 + C C = 1.33 ...(2)

At VCE = 0,

IC = – 0 + 1.33 IC = 1.33 mA ...(3)


So the second point through which the ac load line passes through is (0, 1.33 A).

The input signal is Ib = 20 mA (peak). The point of intersection of the characteristics for

Ib = 20 mA and the ac load line is obtained as point A. Its co-ordinates are (1.2 V, 1.26 A) from

the graph.

IC corresponding to point A is 1.26 A and VCE is 2.2 volts.

So as shown in Fig. P. 10.12.17 maximum output voltage is

Vm = 20 – 1.2 = 18.8 volts and Im = 1.26 A

Fig. P. 10.12.17

Step 2 : Calculate the third harmonic distortion :

We will use the five point method for calculating the third harmonic distortion.

D3 = | B3 |

| B1 |

For a class B operation, the even order harmonics B2 and B4 are absent and the dc value B0 and

ICQ is zero.

Substituting these values we get,


Imax = ICQ + B0 + B1 + B2 + B3 + B4 = B1 + B3 ...(1)

and I1 / 2 = ICQ + B0 + 0.5 B1 – 0.5 B2 – B3 – 0.5 B4 = 0.5 B1 – B3 ...(2)

Adding Equations (1) and (2) we get,

Imax + I1/ 2 = 1.5 B1 B1 =

Imax + I1 / 2

1.5 =

1.26 + 0.7

1.5 = 1.3

And from Equation (1) B3 = Imax – B1 = 1.26 – 1.3 = – 0.04

So the third harmonic distortion is given by,

D3 = | B3 |

| B1 | =

0.04

1.3 = 0.03076 or 3.076 %

Step 3 : Calculate ac power :

Pdc = Vm Im

2 =

18. 8 1.26

2 = 11.844 W ...Ans.

Step 4 : Calculate dc power and efficiency :

PDC = VCC Idc = VCC 2 Im

= 20

2 1.26

= 16.042 W ...Ans.

= Pdc

PDC

100 = 11.844

16.042 100 = 73.83% ...Ans.

Ex. 10.12.18 : In the complementary symmetry amplifier shown in Fig. P. 10.12.18, the transistors and

diodes are of silicon. Calculate :

(a) Base to ground d.c., voltages of Q1 and Q2.

(b) Power delivered to load under maximum signal conditions.

(c) Conversion efficiency under maximum signal.

(d) Approximate value of C, if the circuit is used at signal frequency down to 30 Hz.

Page No. 10-57.

Fig. P. 10.12.18


Soln. :

Step 1 : Calculate the base voltages of the two transistors :

Assume the base currents of both the transistors to be small and apply KVL to the input loop to get,

VCC = (1 k I ) + VD1 + VD2 + (1 k I )

I = VCC – VD1 – VD2

2k =

20 – 14

2 k = 9.3 mA

Base voltage of Q2 = VB2 = I 1 k = 9.3 1 = 9.3 V ...Ans.

Base voltage of Q1 = VB1 = VB2 + VD1 + VD2 = 9.3 + 1.4 = 10.7 V ...Ans.

Step 2 : Calculate maximum load power :

This circuit operates on a single polarity supply. So the new value of VCC to be used for amplifier

calculations is VCC

= VCC / 2 = 10 V.

Pac ( max ) = 1

2 V

2

L =

1

2

( 10 )2

10 = 5 W ...Ans.

Step 3 : DC power and conversion efficiency :

PDC = VCC

ICQ = VCC

2 Im

= V

CC

2 V

CC

RL

= 2

( VCC

)2

RL

= 2

( 10 )2

10 = 6.3661 W

% = Pac (max )

PDC

100 = 5

6.3661 100 = 78.5398 % ...Ans.

Step 4 : Value of C :

f = 1

2 RL C

C = 1

2 f RL

= 1

2 30 10 = 530.5 F ...Ans.

Ex. 10.12.19 : A complementary-symmetry class-AB, A. F. power amplifier uses two matched

transistors and a dual power supply of 30 V to feed a common load of 8 . If input

voltage of this amplifier is 8 V( rms ), calculate :

1. D.C. power input 2. A.C. power output

3. Maximum possible a.c. power output 4. Efficiency

5. Power dissipation of each transistor.

Assume that the two transistors are connected in an ideal C-C configuration.

Page No. 10-57.

Soln. :

Given : VCC = 30 V, RL = 8 , Vin = 8 V (rms)

Type of amplifier : Complementary symmetry class AB.

Step 1 : Calculate peak load current Im :


For C.C. configuration, voltage gain = 1

Vo ( rms ) = Vin ( rms ) = 8 V

Vom = 2 Vo ( rms ) = 2 8 = 11.3137 V

Im = Vom

RL

= 11.3137

8 = 1.4142 Amp.

Step 2 : Calculate DC input power :

PDC = VCC IDC = VCC 2 Im

=

30 2 1.4142

= 27 W ...Ans.

Step 3 : AC output power :

Po (AC) = 1

2

V2

om

RL

= 1

2

( 11.3137 )2

8 = 8 W ...Ans.

Step 4 : Maximum possible ac output power :

Po AC ( max ) = 1

2

V2

CC

RL

= 1

2

( 30 )2

8 = 56.25 W ...Ans.

Step 5 : Efficiency = PAC

PDC

= 8 W

27 W = 0.2962 OR 29.62 % ...Ans.

% = 29.62 %

Step 6 : Power dissipation per transistor :

PD = PDC – PAC

2 =

27 – 8 W

2 = 9.5 W …Ans.

Ex. 10.12.20 : An ideal class B power amplifier is shown in Fig. P. 10.12.20. Assume that the

characteristics of the transistors used are linear.

For this circuit, derive the following. Make suitable assumptions if necessary :

1. Power output derived to load ( Pout )

2. DC input power drawn from the supply ( Pcc )

3. Efficiency of the class B amplifier ( )

4. Maximum efficiency of the amplifier ( max )

5. Power dissipated in each transistor. Page No. 10-57.


Fig. P. 10.12.20

Soln. :

Let the turns ratio be N1

N2

R L =

Vm

Im

=

N1

N2

2

RL

iprimary = ic1 – ic2

Step I : To find power output delivered to load ( Pout ) :

VL rms = Vm

2 IL rms =

Im

2

Pout = VL rms IL rms = Vm

2

Im

2 =

Vm Im

2

Step II : To find DC input power drawn from the supply ( PCC ) :

The average value of supply current is,

Idc = 2 Im

PCC = VCC Idc = VCC

2 Im

=

2 VCC Im

Step III : To find efficiency of class B amplifier ( ) :

= Pac

Pdc

= Pout

PCC

=

Vm Im

2

2 VCC Im

=

4

Vm

VCC

Step IV : To find maximum efficiency of the amplifier ( max ) :

% =

4

Vm

VCC

100 = max when Vm = VCC

% max =

4

VCC

VCC

100 = 78.5 %

Step V : To find power dissipated in each transistor :


Pd = Pdc – Pac = 2 VCC Im

–

Vm Im

2

But Im = Vm

R L Pd =

2 VCC Vm

R L

– V

2

m

2 RL

Ex. 10.12.21 : For the circuit shown in Fig. P. 10.12.20 in the previous example show that

1. Power dissipated in the transistor is maximum when load voltage Vm = 2 VCC

2. Maximum collector power dissipation for both transistors combined as 2 V2

CC /

2 R

L

Page No. 10-58.

Soln. :

Part I : We know that, PD = 2

VmVCC

RL

– V

2

m

2 RL

Differentiating w.r.t. Vm

d PD

d Vm

= 2

VCC

R L

– 2 Vm

2 RL

0 = 2

VCC

R L

– Vm

R L

2 VCC

R L

= Vm

R L

Vm = 2 VCC

Thus, maximum PD occurs when Vm = 2 VCC

Part II : We know that, PD = 2

Vm

R L

VCC – 1

2

V

2

m

R L

Maximum PD occurs when Vm = 2 VCC

max PD = 2

VCC

R L

2 VCC

–

1

2 RL

2 VCC

2

max PD = 4 V

2

CC

2 R

L –

2

2

V

2

CC

R L =

V

2

CC

2 R

L [ 4 – 2 ] =

2 V

2

CC

2 R

L

Ex. 10.12.22 : The dynamic transfer characteristic curve for a given transistor is :

ic (in mA) = 50 Ib + 1000 i2

b where ib ( in mA ) = 10 cos 2 (100 )t

Calculate the percent harmonic distortion. Page No. 10-58.

Soln. :

Given : G1 = 50, G2 = 1000, Ibm = 10, cos t = cos 2 (100 )t

Step I : To find B0 and B1 :


B0 = B2 = G2 I2

bm =

1

2 1000 (10)

2 10

–6 = 0.05

B1 = G1 = 50

Step II : To find % harmonic distortion :

% D = B2

B1

100 % = 0.05

50 100 = 0.1 % ...Ans.

Ex. 10.12.23 : A sinusoidal signal Vs = 1.95 sin 400t is applied to a power amplifier. The resulting

current is given by

io = 12 sin 400t + 1.2 sin 800t + 0.9 sin 1200t + 0.4 sin 1600t

Calculate : 1. Total harmonic distortion

2. Percentage increase in power because of distortion. Page No. 10-58.

Soln. :

Step 1 : Obtain the amplitudes of Fourier components :

The standard expression for io is as follows :

io = B0 + B1 sin t + B2 sin 2t + B3 sin 3t + B4 sin 4t

Comparing with the given equation we get,

B0 = 0, B1 = 12, B2 = 1.2, B3 = 0.9, B4 = 0.4

Step 2 : Various distortions and THD :

Second harmonic distortion D2 = B2

B1

= 1.2

12 = 0.1

Third harmonic distortion D3 = B3

B1

= 0.9

12 = 0.075

Fourth harmonic distortion D4 = B4

B1

= 0.4

12 = 0.0333

Total harmonic distortion. THD = D2

2 + D

2

3 + D

2

4 100

= ( )0.12 + ( )0.075

2 + ( )0.0333

2 100 = 12.9359 % ...Ans.

Step 3 : Percent increase in power due to distortion :

Output power including the distortion is

P = ( 1 + D2 ) P1

where D = Total harmonic distortion P1 = Fundamental power

P = [ 1 + (0.1294)2 ] P1 = 1.01673 P1

So % increase = ( P – P1 ) 100 = ( 1.01673 P1 – P1 ) 100 = 1.6733 % ...Ans.


Section 10.13 :

Ex. 10.13.1 : For a regulated dc power supply, the output voltage varies from 12 V to 11.6 V when the

load current is varied from 0 to 100 mA which is the maximum value of IL. If the ac line

voltage and temperature are constant, calculate the load regulation, % load regulation

and output resistance of the power supply. Page No. 10-64.

Soln. :

Given : VNL = 12 V, VFL = 11.6 V, IL max = 100 mA.

1. The load regulation, L.R. = VNL – VFL = 12 – 11.6 = 0.4 V ...Ans.

2. % load regulation = VNL–VFL

VFL

100 = 0.4

11.6 100 = 3.45% ...Ans.

3. Output resistance Ro = Vo

IL

= 0.4 V

100mA = 4 ...Ans.

Ex. 10.13.2 : For a regulated dc power supply, the output voltage changes in the range of 12 V ± 0.2 V

when the ac line voltage fluctuates in the range of 230V ± 10%. Calculate the source

regulation (SR) and percent source regulation. Page No. 10-64.

Soln. :

1. Source regulation, S.R. = VLH – VLL

VLH = 12 + 0.2 = 12.2 V and VLL = 12 – 0.2 = 11.8 V

S.R. = 12.2 – 11.8 = 0.4 V ...Ans.

2. % S.R. = S.R.

VNom

100

Now VNom = 12 V

% S.R. = 0.4

12 100 = 3.33%

Ex. 10.13.3 : A DC power supply is known to have a ripple factor of 10%. If the DC output voltage is

10V, what is the rms value of the ripple voltage in the output ? Assuming this ripple to be

sinusoidal in nature, what is its peak to peak voltage ? Page No. 10-64.

Soln. : R.F. = 0.1, VLdc = 10V

1. Ripple factor = VRMS

VLdc

where VRMS = RMS value of ripple voltage.

0.1 = VRMS

10 VRMS = 0.1 10 = 1V ...Ans.

2. If the ripple is sinusoidal then the peak to peak ripple is given by,

Vp – p = 2 2 VRMS = 2 2 1 = 2.84 Volts. ...Ans.

Ex. 10.13.4 : A linear regulator has unregulated supply derived from FWR capacitor filter arrangement.

Ripple rejection of regulator is specified to be 60 dB. If input unregulated supply has a

ripple of 2 V peak-to-peak, what will be the output RMS ripple ? If the regulator has 12 V

output with full load current of 1 A, calculate percentage load regulation. Given-output

resistance of regulator = 200 milliohms. Page No. 10-64.


Soln. :

Given : Ripple rejection = 60 dB Peak to peak input ripple = 2V, Vo = 12 V, IFL = 1 A.

Ro = 200 m

To find : Rms ripple in output,

% load regulation

Step 1 : Rms ripple output :

Ripple rejection (dB) = 20 log10

Vripple (output)

Vripple (input)

– 60 = 20 log10

Vripple (output)

Vripple (input)

1 10– 3

= Vr (output)

2

Vr (output) = 2 10– 3

volts or 2 mV ...(1)

Note that this is the peak to peak value of output ripple voltage.

Assuming the output ripple to be sinusoidal we get

Rms Vr (output) = 2 mV

2 2 = 0.707 mV ...Ans.

Step 2 : Percent load regulation :

Referring to Fig. P. 10.13.4 we get,

Vo (NL) = 12 V

and Vo (FL) = 12 – (Ro IL)

= 12 – (200 10– 3

1)

= 11.8 volts Fig. P. 10.13.4

% load regulation = Vo (NL) – Vo (FL)

Vo (FL)

100 = 12 – 11.8

11.8 100 = 1.6949 % ...Ans.

Section 10.15 :

Ex. 10.15.1 : For a zener voltage regulator of

Fig. P. 10.15.1, if IZ min = 2 mA,

IZ max = 20 mA and VZ = 5V, determine

the range of input voltage over which

the output voltage remains constant.

Assume rZ = 0 . Page No. 10-68.

Fig. P. 10.15.1

Soln. :

Given : VZ = 5V, IZ min = 2 mA, IZ max = 20 mA., RS = RL = 1 k.

Step 1 : Calculate load current :

IL = VZ

RL

= 5V

1 k = 5 mA ...(1)


Step 2 : Obtain Vin (min) :

Vin (min) is the minimum input voltage which can keep the zener diode in the zener region. So

corresponding to Vin (min), the zener current IZ = IZ (min).

Source current I = IL + IZ (min)

I = (5 + 2) = 7 mA ...(2)

Vin (min) = VZ + Minimum voltage drop across RS.

= 5 + IRS = 5 + (7 10– 3

1 103) = 12 Volt ...Ans.

Step 3 : Obtain Vin (max) :

Vin (max) is the maximum input voltage which will ensure that IZ is less than or equal to IZ (max). So

corresponding to Vin (max), the zener current IZ = IZ (max).

Source current I = IL + IZ (max)

I = (5 + 20) = 25 mA ...(3)

Vin (max) = VZ + Maximum voltage drop across RS

= 5 + IRS = 5 + (25 10– 3

1 103) = 30 Volt ...Ans.

Thus the range of input voltage to have a constant output voltage is 12 to 30 V.

Ex. 10.15.2 : For a zener voltage regulator of the previous example if Vin = 30 V constant calculate the

range of load current over which the output voltage remains constant. Page No. 10-68.

Soln. :

Given : VZ = 5V, IZ (min) = 2 mA, IZ (max) = 20 mA, RS = 1 k, Vin = 30 V.

Step 1 : Calculate the source current :

Here due to constant Vin, the source current is going to remain constant.

Source current I = Vin – VZ

RS

= 30 – 5

1 k = 25 mA ...(1)

Step 2 : Calculate IL (min) :

IL(min) is the minimum value of load current which ensures that IZ is less than IZ (max). So

corresponding to IL (min), the zener current IZ = IZ (max).

Source current I = IL (min) + IZ (max)

IL (min) = I – IZ (max) = (25 – 20) = 5 mA ...Ans.

Step 3 : Calculate IL (max) :

IL (max) is the maximum value of load current which can keep the zener diode in the zener region.

So corresponding to IL (max) the zener current IZ = IZ (max).

I = IL (max) + IZ (min)

IL (max) = I – IZ (min) = 25 – 2 = 23 mA ...Ans.

Thus the range of load current over which output voltage remains constant is 5 mA to 23 mA.


Ex. 10.15.3 : Design a zener voltage regulator for the following specifications.

Vin = 20 2 Volt, Output voltage Vo = 6V, Load current IL = 50 mA, IZ (min) = 5 mA, Zener

wattage PZ = 0.5 Watt. Page No. 10-68.

Soln. :

Given : Vin (min) = 18V, Vin (max) = 22V, Vo = 6V, IL = 50 mA,

IZ (min) = 5 mA and PZ = 0.5 W.

Step 1 : Calculate IZ (max) :

IZ (max) =

PZ

VZ

= 0.5

6 = 83.33 mA ...(1)

Step 2 : Calculate RS (min) :

When RS = RS (min), the source current I can become maximum corresponding to Vin = Vin (max). As

the load current is constant, this increased source current will flow through the zener. Therefore the

value of RS (min) should be such that even when Vin = Vin (max), the zener current IZ IZ (max).

RS (min) = Vin (max) – VZ

[ ]IL + IZ (max)

...(2)

RS (min) = 22 – 6

[50 + 83.33] 10– 3 = 120 ...Ans.

Step 3 : Calculate RS (max) and select RS :

When RS = RS (max) the source current I can become minimum corresponding to Vin = Vin (min). As

IL is constant, care should be taken to ensure that IZ IZ (min).

RS (max) = Vin (min) – VZ

[ ]IL + IZ (min)

...(3)

RS (max) = 18 – 6

[50 + 5] 10– 3 = 218.18 ...Ans.

Thus RS 120 and RS 218.2 .

So, select RS = 180 .

Step 4 : Select zener diode :

Vo = 6 V

VZ = Vo = 6 V

Fig. P. 10.15.3 : Designed circuit

Select the zener diode with following specifications :

1. VZ = 6 V

2. IZ (max) 83.33 mA

3. PZ = 0.5 W.


Ex. 10.15.4 : Design and draw a zener regulator circuit to meet the following specifications. Load

voltage = 8V, source voltage = 30 V. IL = 50 mA. Assume IZ (min) = 5 mA, PZ = 1.0 Watt.

Page No. 10-68.

Soln. :

Given : VZ = 8 V, Vin = 30 V, IL = 50 mA, IZ (min) = 5 mA, PZ = 1W.

Step 1 : Calculate IZ (max) :

IZ (max) = PZ

VZ

= 1

8 = 125 mA ...(1)

Step 2 : Calculate RS (min) and RS (max) :

RS (max) = Vin – VZ

IL + IZ (min)

= 30 – 8

(50 + 5) 10– 3

RS (max) = 400 ...Ans.

RS (min) = Vin – VZ

IL + IZ (max)

= 30 – 8

(50 + 125) 10– 3

RS (min) = 125.7 ...Ans.

Thus RS should be between 125.7 and 400 .

So select RS = 220 .

The circuit diagram is as shown in Fig. P. 10.15.4. Fig. P. 10.15.4 : Designed zener regulator

Step 3 : Select the zener diode :

Vo = 8 V VZ = 8 V

Select the zener diode having following specifications :

VZ = 8 V, IZ (max) = 125 mA, PZ = 1 W

Section 10.18 :

Ex. 10.18.2 : Design a series voltage regulator with an error amplifier as per the specifications given

below :

Specifications :

1. Unregulated d.c. input voltage Vin = 20 V

2. Regulated output voltage Vo = 12 V

3. Maximum load current IL max = 50 mA Page No. 10-80.

Soln. :

Refer Fig. 10.18.4. This is the regulator circuit to be designed.

Step 1 : Selection of zener diode :

Let VZ Vo / 2

VZ = 12/2 = 6 Volts

So select a zener diode having VZ = 6 Volts.


Step 2 : Calculate R :

Let IZT = 10 mA to operate the zener diode in the zener region.

IR = 10 mA

R = Vo – VZ

IR

= 12 – 6

10 10– 3 = 600

Use R = 560 which is a standard value.

IE1 (max) = IL (max) + IR = 50 mA + 10 mA = 60 mA

Step 3 : Selection of Q1 :

Q1 is the series pass transistor. The power dissipation in Q1 is given by,

PD (max) = IE1 (max) (Vin – Vo)

= 60 10– 3 (20 – 12) = 480 mW

VCE1 (max) = Vin = 20 V

IC1 (max) = IE1 (max) = 60 mA

Select Q1 as per these specifications.

Step 4 : Calculate R3 :

Let hFE (min) at IC = 60 mA of Q1 be 50.

IB1 (max) = IE1 (max)

hFE (min)

IB1 (max) = 60 mA

50 = 1.2 mA

Let IC2 >>> IB1 (max)

Let IC2 = 5 mA

R3 = Vin – VB1

IC2 + IB1 (max)

= 20 – (12 + 0.7)

(5 + 1.2) mA

R3 = 1.18 k

Select R3 = 1.2 k which is the standard value.


IZ = IE2 + IR = 5 mA + 10 mA = 15 mA

It is necessary that I2 >> IB2.

So let I2 = 1 mA.

R2 = VZ + VBE2

I2

= 6 + 0.7

1 mA

R2 = 6.7 k

Use 6.8 k which is a standard value.


R1 Vo – VR2

I2

= 12 – 6.7

1 mA = 5.1 k

Use 4.7 k or 5.1 k which is a standard value.


Section 10.19 :

Ex. 10.19.2 : For a transistorised series pass regulator shown in Fig. P. 10.19.2, calculate the output

voltage Vo and hence the temperature stability factor ST.

Given : Vz (0C) = 10.00 V, Vz (50C) = 10.25 V,

Q1 and Q2 are silicon transistors.

VBEQ2 (0C) = 0.650 V, VBEQ2 (50C) = 0.540 V

R1, R2, R3, R4 are stable resistors.

Assume IB2 << IR1, R2. Page No. 10-81.

Fig. P. 10.19.2

Soln. :

Steps to be followed :

Step 1 : Calculate Vo . Since IB2 << IR1, R2 , by voltage divider rule.

Vo = R1 + R2

R2

( VZ + VBE2 )

Step 2 : Calculate ST .

ST = Vo

T

Step 1 : Calculate Vo :

Vo =

R1 + R2

R1

(VBE2 + Vz)

at 0C Vo1 =

1 k + 1.2 k

1.2 k (0.650 + 10)

Vo1 = 19.525 Volts ...Ans.

at 50C Vo2 = 1.833 (0.540 + 10.25) = 19.78 Volts ...Ans.

Step 2 : Calculate stability factor ST :

ST = Vo

T =

Vo2 – Vo1

T2 – T1

= 19.78 – 19.525

50 – 0

ST = 5.13 mV/C ...Ans.


Section 10.21 :

Ex. 10.21.1 : For a series regulator shown in Fig. P. 10.21.1. Q1 and Q2 are silicon transistors with

hfe = 100 and VBE = 0.7. Calculate :

1. The range over which output voltage can be adjusted.

2. Collector current of Q2.

3. The zener diode current if Vz = 10 V. For 2. and 3. above assume, that output

voltage has been adjusted to 20 V. Page No. 10-84.

Fig. P. 10.21.1

Soln. :

Step I : To find range over which output voltage can be adjusted :

Case (i) : With RP at point A.

R1 = 1 k, R2 = 1.5 k + 0.5 k = 2 k

Vo =

1 +

R1

R2

( VBE2 + Vz )

Vo =

1 +

1

2 (0.7 + 10)

Vo = 1.5 10.7 = 16.05 V

Case (ii) : With RP at point B.

R1 = 1 k + 0.5 k = 1.5 k

R2 = 1.5 k

Vo =

1 +

R1

R2

(VBE2 + Vz)

Vo =

1 +

1.5 k

1.5 k (0.7 + 10)

Vo = (1 + 1) (10.7) = 21.4 V

The output voltage can be varied from 16 to 21 V.


Step II : To find collector current of Q2 :

Vout = 20 V ...given

IL = Vout

RL

= 20

40 = 0.5 A

Let the current through R1, RP and R2 be I such that I > IB2.

I = Vout

R1 + R2 + RP

= 20

1 k + 0.5 k + 1.5 k

I = 20

3000 = 6.67 mA ...Ans.

IE1 = I + IL, IB1 = IE1

1 + hfe1

IE1 = 6.67 10– 3

+ 0.5, IB1 = 506.67

1 + 100

IE1 = 0.5067 A. IB1 = 5.0165 mA

IE1 = 506.67 mA

VB1 = VBE1 + Vout = 0.7 + 20 = 20.7 V

IR3 = Vin – VB1

R3

= 30 – 20.7

1.5 103 = 6.2 mA

IC2 = IR3 – IB1 = 6.2 – 5 = 1.1834 mA ...Ans.

Step III : To find zener diode current :

IB2 = IC2

hfe

= 1.1834

100 = 11.8346 A

IE2 = IB2 + IC2

IE2 = 11.83 10– 6

+ 1.1834 10– 3

IE2 = 1.1952 mA

Zener current, IZ = IE2 + IR4

But IR4 = 30 – 10

2000

IR4 = 10 mA

IZ = 1.192 + 10 = 11.192 mA ...Ans.

Ex. 10.21.2 : In the circuit of Fig. P. 10.21.1 in the previous example, if the dynamic resistance of zener

diode RZ = 8 and hie = 0.5 k. hre = hoe = 0 for Q1 and Q2. Calculate the approximate

output impedance of the series pass regulator. Page No. 10-85.

Soln. :

Given : RZ = 8 , hie = 0.5 k.

To find : Ro

Ro = Vout

IL

Vin

temperature constant


Ro = R1 + R2

gm2 R2 hfe1

R1 = 1 k, R2 = 2 k

Step I : To calculate gm2 :

gm2 =

hfe R2

R1 + R2

[ R1 || R2 ] + [ hie2 + (1 + hfe2) RZ]

gm2 =

100 2

1 + 2

[ 1 || 2 ] + [ 0.5 + (1 + 100) 8 10– 3

]

gm2 = 66.67

0.6667 + [ 0.5 + 0.808 ]

gm2 = 33.763 mA/V

Step II : To find Ro :

Ro = (1 + 2) 10

3

33.763 10– 3

2 103 100

Ro = 0.4442 ...Ans.

Ex. 10.21.3 : The emitter follower regulator is to supply a load current of 500 mA at 10.3 V. The

unregulated dc supply varies from 16 to 20 V. Use a zener 11 V which requires minimum

bias current of 2 mA for stable operation. The series pass transistor has parameters.

Given : hfe = 50, VBE = 0.7 and dVz

dIz = 20 , hie = 100 .

Determine :

1. Value of zener bias resistor Rb.

2. The wattage of Rb.

3. Power dissipation rating of zener.

4. Power dissipation rating of transistor.

5. Variation in Vo for variation in Vin from 16 to 20 V at IL = 500 mA constant.

6. Variation in Vo for load variation from 50 mA at Vin = 16 V constant.

Page No. 10-85.

Soln. :

Given : IL = 500 mA, Vo = 10.3 V

Vin = 16 to 20 V, Vz = 11 V

Iz min = 2 mA, hfe = 50, VBE = 0.7

Rz = 20 , hie = 100

Step I : To find Rb :

IL = IE = 500 mA

IB = IE

1 + =

500

1 + 50 = 9.8039 mA

IRb = IB + IZ = 9.89039 mA + 2 mA


IRb = 11.8039 mA

But IRb =

Vin – Vz

Rb

Rb = 16 – 11

11.80 10– 3 = 423.58 ...Ans.

Step II : To find wattage of Rb :

Wattage rating = (Vin – Vz)

2

Rb

Wattage rating = (20 – 11)

2

423.58 = 0.1912 W ...Ans.

Step III : To find PD of zener :

IRb = IB + Iz

Maximum IRb =

20 – 11

423.58 = 21.247 mA (when IL = 0, IB = 0 and IR

b = Iz max)

PD of zener = Vz Iz max

But Iz (max) = IRb (max)

PD of zener = 11 21.247 10– 3

= 0.23372 W ...Ans.

Step IV : To find PD of transistor :

VCE = VC – VE

VCE = 20 – 10.3 = 9.7 V

PD = VCE IE = 9.7 0.5 = 4.85 W ...Ans.

Step V : To find SV :

SV = Vo

Vin

IL

constant

SV = Rz

Rb + Rz

SV = 20

423.58 + 20 = 0.045 or 4.5 % ...Ans.

Step VI : To calculate Ro :

For the given circuit, Ro = Vo

IL

Ro = Rz + hie

1 + hfe

Ro = 20 + 100

1 + 50 = 2.3529 ...Ans.


Ex. 10.21.4 : A linear regulator has unregulated supply derived from FWR capacitor filter arrangement.

Ripple rejection of regulator is specified to be 60 dB. If input unregulated supply has a

ripple of 2 V peak-to-peak, what will be the output RMS ripple ? If the regulator has 12 V

output with full load current of 1 A, calculate percentage load regulation. Given-output

resistance of regulator = 200 milliohms. Page No. 10-85.

Soln. :

Given :

Ripple rejection = 60 dB

Peak to Peak input ripple = 2V, Vo = 12 V, IFL = 1 A.

Ro = 200 m

To find : Rms ripple in output

% load regulation

Step 1 : Rms ripple output :

Ripple rejection (dB) = 20 log10

Vripple (output)

Vripple (input)

– 60 = 20 log10

Vr (output)

2

1 10– 3

= Vr (output)

2

Vr (output) = 2 10– 3

volts or 2 mV ...(1)

Note that this is the peak to peak value of output ripple voltage.

Assuming the output ripple to be sinusoidal we get

Rms Vr (output) = 2 mV

2 2 = 0.707 mV ...Ans.

Step 2 : Percent load regulation :

Fig. P. 10.21.4

Referring to Fig. P. 10.21.4 we get

Vo (NL) = 12 V

and Vo (FL) = 10 – ( Ro IL )

= 12 – (200 10– 3

1)

= 11.8 volts

% load regualtion = Vo (NL) – Vo (FL)

Vo (FL)

100

= 12 – 11.8

11.8 100 = 1.6949 % ...Ans.


Ex. 10.21.5 : Design a Transistor Series Regulator to deliver 800 mA load current to load at + 12V DC.

The input voltage to regulator varies from 18 to 24 V DC. Assume suitable active and

passive devices (in terms of their specifications). Suggest practical values of resistors

from 5% tolerance series. For your design, calculate :

1. Worst case power dissipated by series pass transistor

2. Power dissipated by Zener diode

3. Maximum power delivered to load

4. Maximum power delivered by input unregulated supply

5. Efficiency of regulator circuit Page No. 10-85.

Soln. :

Given : IL = 800 mA, Vo = 12 V, Vin = 18 to 24 V

Step 1 : Draw the circuit diagram :

Fig. P. 10.21.5 shows the required circuit diagram.

Fig. P. 10.21.5 : A transistor series regulator

Step 2 : Design of the Regulator :

VZ = Vo + VBE = 12 + 0.7 = 12.7 V

Let PZ (max) = 1 W

IZ (max) = PZ (max)

VZ

= 1 W

12.7 = 78.74 mA

Therefore, Select the zener diode with the following specifications :

VZ = 12.7 V, PZ (max) = 1 W, IZ (max) = 80 mA

Now let us select the transistor.

IL = IE = 800 mA

Let of the transistor be 15.

typ = 15

IB = IL

1 + =

800

16 = 50 mA

IC = IB = 15 50 = 750 mA

Maximum power dissipation of the transistor corresponds to the short circuited output.


PD (max) = VCE (max) IC

= Vin (max) IC = 24 750 mA

PD (max) = 18 W

Select a transistor with following specifications.

VCE (max) = 36 V, IC (max) = 1.5 A, PD (max) = 54 W

Calculate RB :

RB = Vin –VZ

IRB

= 14 – 12.7

IB + IZ

RB (min) :

RB (min) should not be too small. Even at maximum value of Vin, the current through IZ should not

exceed IZ (max).

RB (min) = Vin (max) – VZ

IB + IZ (max)

RB (min) = 24 – 12.7

(50 + 78.78) mA = 87.74

RB (max) :

RB (max) should not be too large. Even at minimum Vin, the current through IZ should not be less

than IZ (min) i.e. 5 mA.

RB (max) = Vin (min) – VZ

IB + IZ (min)

RB (max) = 18 – 12.7

(50 + 5) mA = 96.36

The value of RB should be between RB (max) and RB (min).

RB = 92

Designed values :

RB = 92 , VZ = 12.7 V,

PZ (max) = 1 W, IZ (max) = 80 mA,

IZ (min) = 5 mA, = 15,

PD (max) = 54 W, IC (max) = 1.5 A,

VCE (max) = 36 V,

Calculations from the designed circuit :

1. Worst case power dissipation in the transistor :

PD (max) = Vin (max) IL

= 24 0.8 = 19.2 W ...Ans.

This corresponds to short circuited output terminals.


2. Power dissipated by zener :

Maximum zener current corresponds to Vin (max).

IB + IZ (max) = Vin (max) – VZ

RB

IB + IZ (max) = 24 – 12.7

92 = 122.83 mA

IZ (max) = 122.83 – 50 = 72.83 mA

PZ (max) = VZ IZ (max) = 12.7 72.83 10– 3

= 0.9248 W ...Ans.

3. Maximum power delivered to the load :

PL (max) = Vo IL = 12 0.8 = 9.6 W ...Ans.

4. Maximum power delivered by input supply :

Pin (max) = Vin (max) [ IC + IB + IZ (max) ]

= 24 [ 750 + 122.83] mA

= 20.95 W

5. Efficiency :

min = PL (max)

Pin (max)

= 9.6

20.95 = 0.4582 or 45.82% ...Ans.

Documents

Chapter 10 : Power Circuits and Systemssvbit-it.weebly.com/uploads/2/4/8/5/2485214/chap-10.pdf[ N 1 / N 2] 2 = 9 N 1 N 2 = 3 …Ans. Ex. 10.9.4 : For a class B push-pull amplifier