c power a class-a power amplifier · class-a power amplifier: in class-a amplifier, the transistor...

P a g e | 1 Unit 5| Power Amplifiers

2018-19

Syllabus:

Power dissipations in transistors, Harmonic distortion, Amplifiers Classification (Class-A, Class-B,

Class-C, Class-AB) Efficiency, Push-pull and complementary Push-pull amplifiers, Cross-over

distortion.

Give the classification of power amplifiers. (S-14/6m)

CLASSIFICATION OF POWER AMPLIFIER: The power amplifiers can be classified on the basis of biasing voltage level into Class-A, Class-B,

Class-AB & Class-C.

CLASS-A POWER AMPLIFIER: In Class-A amplifier, the transistor is biased such that the Q-point is located at the center of load

line due to which output current flows for the full cycle of the input signal (i.e. 360°).

CLASS-B POWER AMPLIFIER In Class-B amplifier, the transistor is biased such that the Q-point is located on the intersection

point of the voltage axis and the load line (i.e. on cutoff point) due to which output current flows

for only half cycle of the input signal (i.e. 180°).


2018-19

CLASS-AB POWER AMPLIFIER In Class-AB amplifier, the transistor is biased such that the Q-point is located between mid-point of

load line and cutoff point due to which output current flows for more than half cycle but less than

complete cycle of the input signal (i.e. 180° < ∠ < 360°).

CLASS-C POWER AMPLIFIER In Class-C amplifier, the transistor is biased such that the Q-point is located just below the voltage

axis due to which output current flows for less than half cycle of the input signal (i.e. ∠ < 180°).


2018-19

CLASS-A DIRECT COUPLED POWER AMPLIFIER

RL

RE CE

R1

R2

Vin

Ci

VCC

Figure shows circuit diagram of single stage class A

direct coupled resistive load power amplifier. &

form voltage divider network, & provide

stabilization to the amplifier. The transistor is biased

such that the Q point is located at the center of load

line.

The rms value of the output voltage and current are

given as,

=

√2 & =

√2


2018-19

= −

2

= −

2

= −

2√2

V = −

2√2

The output AC power is,

= ×

= −

2√2 −

2√2

=( − )( − )

8

also

=

= 0

= 2

= 0

= × 2

8

=4

The input DC power supplied by the source VCC is

=

The collector efficiency is given by,

=

× 100

=4

× 100 =

1

4× 100 = 0.25 × 100

= %

This is the maximum efficiency of class A amplifier with direct coupled resistive load.


2018-19

Show that the efficiency of transformer coupled power amplifier in class A is 50%. (W-13/6m) (W-14/6m) (S-17/6m) Explain transformer coupled class A power amplifier and show that its efficiency is 50%. (W-16/6m)

CLASS-A TRANSFORMER COUPLED POWER AMPLIFIER

R1

R2

Vin

Cin

VCC

RL

RE CE

Figure shows circuit diagram of class A transformer coupled power amplifier. & form self-

bias, & provides thermal stabilization to the amplifier. The transistor is biased such that the

Q point is located at the center of load line.

Assuming the inductive reactance of transformer primary and emitter resistance very small, the

voltage-drop across them will be very small as compared to , thus, neglecting them, and almost

all the supply voltage will come across .

=

The rms value of the output voltage and current are given as,

=

√2 & =

√2

= −

2

= −

2

= −

2√2

= −

2√2

The output AC power is,

= ×

= −

2√2 −

2√2

=( − )( − )

8

Also

=

= 2

= 0

= 2

= 0

=2 × 2

8

=2

The input DC power supplied by the source VCC

is

=

The collector efficiency is given by,


2018-19

=

× 100

=2

× 100 =

1

2× 100 = 0.50 × 100

= %

This is the maximum efficiency of a

transformer coupled class A amplifier.

A class A transformer coupled power amplifier is to deliver maximum 5W to a 4 Ω load. The quiescent point is adjusted for symmetrical clipping and the collector supply voltage is = . Find (i) What is transformer turns ratio = /? (ii) What is the peak collector current and ? (iii) What is the quiescent operating point

and ? (iv) What is the collector circuit efficiency? (W-15/7m) (S-14/6m) (S-16/7m)

Solution:

Given:

= 5 ; = 4 Ω; = 20

Part (1):

Maximum Power at secondary of transformer is

= I =

√2

√2=(/

) × 2

=

2

Output power will be maximum only when = = 20

=

2 =

20

2

From above equation will be

=

20

2=

400

2 × 5=400

10= 40Ω

We know another formula to calculate as

=

× =

×

From this formula we can calculate the turns ratio as

= =

4

40=

1

10

= 1

10=

1

3.162= 0.316

Part (2):

The peak collector current is

=


2018-19

For maximum power, = = 20

=20

40=1

2= 0.5

Part (3):

The quiescent operating point is ( , ). It is a coordinate of a point on the load line in output

characteristics.

It is given that; the quiescent point is adjusted for symmetrical clipping. This means that the Q point

is at center of load line. i.e. Class A Amplifier with transformer coupled.

For Class A Amplifier with transformer coupled, = & =

Therefore, = 20 & = 0.5 i.e. (20 ,0.5)

Part (4):

= 5

= = 20 × 0.5 = 10

% =

=5

10= 50%

A class A amplifier shown in figure operates from = , draws no-signal current of 5A and feeds a load of 40 Ω through a step-down transformer of / = . . Find (i) Whether the amplifier is properly matched for maximum power transfer, (ii) Maximum ac signal power output, (iii) Conversion efficiency at maximum signal input.

R1

R2

Vin

CE

VCC=20V

RL=40

1:3.16

Solution:

Part 1:

We know for impedance matching the input impedance of transformer must be same as that of

output of amplifier.


2018-19

Let & be the input voltage and current and & be the output voltage and current of

transformer.

RL=40ΩV2V1

I1 I2

R’L

We know,

=

∴ =

×

We know,

=

∴ =

×

=

=

×

× =

×=

× = 1

3.16

× 40

= --- (1)

Where, is the load impedance appearing at transformer’s primary.

The output impedance of amplifier is /

Given:

= 20 = 5

=

=20

5

= --- (2)

From equations (1) & (2) we can say that the amplifier is properly matched with the transformer

for maximum power transfer.

Part 2:

() = =

√2×

√2=2

=2

=20 × 5

2

() = 50

Part 3:

=()

()=2

=

2

=1

2

= 50 %


2018-19

A transformer coupled class A power amplifier draws a current of 200 mA from a collector supply of 10V when no-signal is applied to it. Determine, (i) Maximum output power, (ii) Maximum collector efficiency, (iii) Power rating of the transformer

If the load connected across the transformer secondary is of 2 Ω and transformer turn ratio is :, comment on the impedance matching.

Solution:

Given:

= 200

= 10

= 2 Ω

= 5:1 =5

1

Part 1: Maximum output power

() = =

√2×

√2=2

=

2=10 × 200

2= 5 × 200

() = 1

Part 2: Maximum collector efficiency

=()

()=2

=

2=1

2

= 50 %

Part 3: Power rating of transformer

Let & be the input voltage and current and & be the output voltage and current of

transformer.

RL=2ΩV2V1

I1 I2

R’L

The power rating of transformer is the rated output power of transformer and it is given by,

() = ×

We know,

==

=1

5


2018-19

=1

5 &

=1

5

=5=5

=10

5

=

= 5 × = 5 × = 5 × 200

=

The power rating of the transformer is

= 2

Part 4:

We know for impedance matching the input impedance of transformer must be same as that of

output of amplifier. From the figure shown above we can write,

We know,

=

∴ =

×

We know,

=

∴ =

×

=

=

×

× =

×

=

× = (5) × 2

= 50 Ω --- (1)

The output impedance of amplifier is /

Given:

= 10 = 200

=

=10

200

= 50 Ω --- (2)

From equations (1) & (2) we can say that the amplifier is matched with the transformer for

maximum power transfer.


2018-19

A power transistor is operating in class A to deliver a maximum of 5 W power to a load of 4 Ω. The collector supply voltage is Vcc = 20 V. (i) What is the turn ratio n2/n1? (ii) peak collector current (iii) Quiescent operating point (iv) Collector circuit efficiency. (W-15/7m) (S-14/6m)


2018-19

What is harmonic distortion in power amplifiers and how it can be reduced? (S-15/6m)

HARMONIC DISTORTION An amplifier is a circuit that gives enlarged form of input and also provide a faithful reproduction

of the input waveform. The harmonic distortion results from the non-linearity of the active device

used for amplification. The non-linear characteristics may be as shown below,

The positive and negative portion of the input signal is not equally amplified by the active device.

It happens when the input signal is so large that it is applied over non-linear region of the transistor

characteristics. Due to this, additional frequency components appear in the amplifier’s output

other than the fundamental frequency component. These additional frequency components are

known as Harmonics. Let an input applied to Class A amplifier with fundamental frequency be

= sin

Due to nonlinear nature, the harmonic frequency components of frequencies ,2,3,… will

appear at the output of amplifier as

= + sin+ sin2+ sin3…

Where,

→ DC Component

sin → First Harmonic or Fundamental frequency component

sin2 → Second Harmonic

2 = =

3 = =

Let P1 be the output power due to fundamental frequency component or first harmonic,

= =

√2

=

√2

=2

Let P2 be the output power due to second harmonic,


2018-19

=2

Let P3 be the output power due to third harmonic,

=2

The total power due to all the harmonic components at the output is,

= + + …

=2

+2

+2

+ ⋯

= 1 +

+

+ ⋯ 2

= [1 + +

+ ⋯ ]

The distortion factor or total distortion D is given by

= +

+ ⋯

Thus, the total power including all the harmonic distortion is

= +

The harmonic distortion can be reduced by a factor of 1 + , with the help of negative feedback.

It can also be reduced by restricting the signal over linear region of amplifier.

Calculate harmonic distortion for an output signal of a power amplifier having fundamental amplitude of 2.5 V, second harmonic amplitude of 0.25 V, third of 0.15 V and fourth harmonic of 0.05 V. (S-15/7m)

Solution:

Given:

= 2.5 ; = 0.25 ; = 0.15 ; = 0.05

The harmonic distortion is given by

= +

+ + ⋯

Second harmonic distortion, = / = 0.25/2.5 = 0.1

Third harmonic distortion, = / = 0.15/2.5 = 0.06

Fourth harmonic distortion, = / = 0.05/2.5 = 0.02

Total harmonic distortion, = +

+ = √0.1 + 0.06 + 0.02 = 0.1183

Percentage harmonic distortion, % = 11.83%


2018-19

A transistor supplies 0.85 Watts to 4KΩ load. A zero-signal dc collector current is 31 mA and the dc collector current with signal is 34 mA. Determine the percentage second harmonic distortion.

Solution:

Given:

= 0.85

= 4Ω

ℎ = = 31

ℎ = () = 34

=?

Following figure shows the collector current without input signal at 31mA (black line) and average

collector current with signal at 34mA. The average current with signal should appear at 31mA but

it is appearing at 34mA, this increase in current is due to second order harmonics (neglecting higher

harmonics). Thus, magnitude of second harmonic current I2 is given by

= () − = 34 − 31 = 3

We know,

=2

+2

+

=2

=

2

−

= 2

−

= 2(0.85)

4− (3 ) = 20.39

The second harmonic distortion D2 is

=× 100 =

3

20.39× 100 = 0.147× 100 = . %


2018-19

Explain the working of Class-A push pull amplifier and show its effect on harmonic distortion. (W-13/7m) Draw the circuit and explain the working of class A push pull amplifier with input and output transformer. (W-15/7m) (S-14/6m)

CLASS-A PUSH PULL AMPLIFIER Circuit Description:

Both the transistors Q1 and

Q2 are used in common

emitter configuration.

Resistances R1 & R2 form

voltage divider network

such that voltage across R2

forward biases both the

transistors. When no-

signal is applied to the

primary of transformer T1,

constant collector currents

(IC1 & IC2) flows in the

output circuit of both the

transistor. As both these

currents are ideally equal in magnitude and flow through transformer T2 in opposite direction, flux

saturation of the transformer is prevented. In this case, when no-signal is applied to it, current IL is

zero.

Circuit Operation:

During positive half cycle of the input, transistor Q1 is more forward bias while transistor Q2 is less

forward bias. Thus, magnitude of current IC1 will be larger than that of IC2 and their directions are

opposite. When such currents flow through transformer T2, the output current IL will be

proportional to ( − ).

. . ∝ ( − )

In this case current IL flows in forward direction

During negative half cycle of the input, transistor Q2 is more forward bias while transistor Q1 is less

forward bias. Thus, magnitude of current IC2 will be larger than that of IC1 and their directions are

opposite. When such currents flow through transformer T2, the output current IL will be

proportional to ( − ).

. . ∝ ( − )

∝ −( − )

In this case current IL flows in reverse direction.

During positive half cycle transistor Q1 push the current into the load and during negative half cycle

transistor Q2 pull the current out of the load. Both the transistors are biased such that Q point is

located at the center of the load line. Thus, it is known as Class-A Push Pull Amplifier.


2018-19

Draw the schematic diagram of Class-B push pull amplifier and show that output consists of odd harmonics only. (W-14/7m) Explain class B push pull power amplifier. Calculate its conversion efficiency. (S-15/6m) (S-16/7m) Show that the efficiency of class B push pull amplifier is 78.5%. Draw circuit diagram of class B push pull amplifier. (W-16/7m)

CLASS-B PUSH PULL AMPLIFIER

Q1

Q2

T1

Vin

T2

RL

IC1

IC2

IL

IB1

IB1

Circuit Description:

Both the transistors Q1 and Q2 are used in common emitter configuration. The transistors T1 and T2

are biased (base is connected to ground i.e. zero volts) such that the Q point is located on cutoff

point. When no-signal is applied to the primary of transformer T1, both the transistors will remain

in cutoff state. In this case, no collector currents will flow in output circuit and thus output current

IL is zero.

Circuit Operation:

During positive half cycle, transistor T1 will be forward biased and transistor T2 will remain in cutoff

state. Thus, only current IC1 will flow in output circuit and output current IL will be proportional to

only IC1.

. . ∝

In this case current IL flows in forward direction

During negative half cycle, transistor T2 will be forward biased and transistor T1 will remain in cutoff

state. Thus, only current IC2 will flow in output circuit and output current IL will be proportional to

only IC2.

. . ∝

In this case current IL flows in reverse direction.


2018-19



located at the cutoff point on the load line. Thus, it is known as Class-B Push Pull Amplifier.

In class B, each transistor works for only 180° (half cycle). Thus, if one transistor gives output of half

cycle, second will give output of another half cycle, which is phase shifted by 180°.

The collector current of transistor Q1 is

= + cos+ cos2+ cos3+ I cos4…

The collector current of transistor Q2 is

= + cos(+ ) + cos2(+ ) + cos3(+ ) + cos4(+ )…

∵ cos(+ ) = −cos

where, = ,3,5,…

∴ = − cos+ cos2− cos3+ I cos4…

Output current is

= −

= ( + cos+ cos2+ cos3+ I cos4)

− ( − cos+ cos2− cos3+ I cos4)

= 2 × cos+ 2 × cos3+ 2 × I cos5+ ⋯

Thus, it is proved that output of class B push pull amplifier contains odd harmonics only. It will

have only odd harmonic distortion such as , and so on.

EFFICIENCY OF CLASS-B PUSH PULL AMPLIFIER The input dc power to the amplifier is given by,

() =

Where, is the average current delivered to the amplifier.

=1

2 sin

=2

() =2

∵ =

() =2

The output ac power supplied by the amplifier is given by,

() =


2018-19

Where & are the RMS values of collector voltage and current respectively.

=

√2 & =

√2

() =2

=(/

)

2

() =

2

=()

()× 100

=

2

2

× 100

=

4

× 100

This is the equation for efficiency of Class-B Push Pull Amplifier. The maximum efficiency will be

when =

=

4× 100

= . %

With the load of 4 Ω, a push pull amplifier takes 3.25 A from the dc supply, with the sinusoidal signal. If VCC = 24 volts, find conversion efficiency. (W-13/6m) (S-15/7m)

Solution:

= 3.25

=2

=2

= (3.25)

2

= = (3.25)

2× 4 = 2(3.25)

=

4

× 100 =

42 × 3.25

24 × 100

= . %


2018-19

POWER DISSIPATION OF TRANSISTORS IN CLASS-B PUSH PULL

AMPLIFIER Power dissipation is the loss of power from transistors in the form of heat. It is given by the

difference of input power and output power.

= = () − ()

=

−

= 2

−2

= ×

−

This is power dissipation of the amplifier that takes place from the transistors. As both the

transistors are identical, power dissipation for each transistor will be half of total power dissipation.

=

=

−

CONDITION FOR MAXIMUM POWER DISSIPATION: Power dissipation is given by,

= = () − ()

=2

−

2

The condition for maximum power dissipation can be obtained by differentiating above equation

with respect to and equating with zero.

= 0

2

×

−1

2 ×

= 0

2

−22

= 0

2

=

=

This is the condition for maximum power dissipation. Therefore, maximum power dissipation can

be obtained by putting this condition in equation of power dissipation.

=2

2

−

2

2 =

4

−

4

2 =

4

−

2

=

×


2018-19

This is maximum power dissipation of the amplifier that takes place from the transistors. As both

the transistors are identical, power dissipation for each transistor will be half of total power

dissipation.

=

×

An ideal class B push pull amplifier with = , = , = and a transistor has = . For the maximum output signal, determine: (i) Output signal power (ii) Collector dissipation in each transistor (iii) Conversion efficiency η (iv) Power rating of each transistor (W-15/8m) (S-14/8m) (W-16/7m)

Solution:

Given: = 20;

= 2; = 20Ω; = 20

Part-1: Output Signal Power is given as

() =

2

For an ideal class B push pull amplifier =

() =

2

is given as

=

= 1

2

× 20 = 5Ω

() =

2 =

20

2 × 5= 40

Part-2: Collector dissipation in each transistor is the half of power dissipation of the amplifier and

is given as

=

−4 =

−4 =

20

520

−20

4 = .

Part-3: Conversion efficiency η

=()

()

() =2

=2

=2 × 20

× 5= .

∴ =()

()=

40

50.93= 0.7853

% = . %

Part-4: Power rating of each transistor is the maximum power available at the collector of a

transistor and is given as the product of &


2018-19

Power rating of each transistor = × = × =

As each transistor works for only half cycle in class B amplifier,

=2=

2

=2

=20

2 × 5= 2

Power rating of each transistor = = 20 × 2 =

The ideal class B push pull amplifier shown1 with = , = , the input signal is sinusoidal. Determine (i) Maximum output signal power (ii) Maximum dc input power (iii) The conversion efficiency (iv) What is the maximum dissipation of each transistor? (S-16/8m)

The ideal Class B push pull amplifier shown with = , = and the input signal. Determine (i) Maximum output signal power (ii) Maximum d.c. input power (iii) The conversion efficiency (iv) What is the maximum dissipation of each transistor and what is the efficiency under these conditions? (W-14/7m)

Solution:

Given:

= 15; = 5 Ω

Part (1):

() =

2=

2=

15

2 × 5= 22.5

Part (2):

() =2R

=2 × 15 × 15

× 5= 28.65

Part (3):

Conversion efficiency:

=()

()=

22.5

28.65= 0.785

% = 78.5%

Part (4):

Collector dissipation of each transistor is given by,

=1

×

=

1

15

5= 4.559

= . /

1 Figure was not given in question paper


2018-19

For Class B Push Pull amplifier = , = and = . (i) Find the efficiency, (ii) Find maximum power dissipated in transistor. (W-13/7m)

Solution: (do it, its procedure is same as above numericals)

A transistor power amplifier amplifies 10W power with efficiency of 45%. Find dc input power and power that must be dissipated as heat. (W-16/6m)

Solution:

Explain the working of complementary symmetry class B amplifier what are its advantages. (S-17/7m)

COMPLEMENTARY SYMMETRY (CLASS-B) PUSH-PULL AMPLIFIER

Q1

Q2

Vin

IC1

IC2

IB1

IB1

RL

VCC

VCC

Cin

IL

Circuit Description:

The circuit does not use any transformer. Thus, it uses complementary symmetry transistors i.e.

one transistor is npn while other is pnp, so that at a time one transistor is forward biased while

other is reverse biased. Complementary symmetry is only possible in class B amplifier.

Circuit Explanation:

During positive half cycle of the input, base emitter junction of transistor Q1 is forward biased and

that of Q2 is reverse biased. So the current flows through load resistor RL in forward direction is due

to transistor Q1 and is given by IC1. . . = .

During negative half cycle of the input, base emitter junction of transistor Q2 is forward biased and

that of Q1 is reverse biased. So the current flows through load resistor RL in reverse direction is due

to transistor Q2 and is given by IC2. . . = −.



located at the cutoff point on the load line. Thus, it is known as Complimentary Symmetry Class-B

Push Pull Amplifier.


2018-19

Advantages:

Circuit is light weight and cost worthy as it doesn’t uses the transformer. This improves it’s

frequency response.

Disadvantages:

The circuit requires two voltage supplies and output is cross-over distorted.

A complementary symmetry class AB audio frequency amplifier uses two matched BJTs

and dual power supply of ± 30V and feeds a common load of 8Ω. If the input voltage of

this amplifier is 8V (rms), calculate: (i) DC Power input (ii) A. C. Power output (iii)

Maximum possible a.c. power output (iv) Efficiency (v) Power dissipation by both the

transistors. (S-17/8m)

Explain crossover distortion in power amplifiers and how it can be eliminated? (W-13/6m) (W-14/6m) (W-15/6m) (S-16/6m) (S-14/2m) (S-17/5m)

CROSS OVER DISTORTION Ideally, transistor is assumed in cutoff when it is not biased and for every small value of biasing

voltage, it is assumed in linear region (active region). But practically, transistor remains in cutoff for

finite value of biasing voltage (VB). For silicon, biasing voltage is = 0.7 and for germanium it is

= 0.3. Due to this, the output of transistor remains zero upto the biasing voltage at the input.

This is shown in following figure.


2018-19

From the figure it is clear that the transistors Q1 & Q2 remain off when the input signal is crossing

over the zero line. Because of this, a dead-zone is created in the output between every two half

cycles. It the width of dead-zone for only Q1 is x, total width of dead-zone will be ‘2x’.

= =

= [−cos] = [−cos + cos0]= [1 − cos]= − cos

cos = −

cos = −

= cos −

Elimination of Crossover Distortion:

To eliminate crossover distortion, it is necessary to add a small amount of forward bias to take the

transistor to the average of conduction or slightly beyond. This slightly lowers the efficiency of the

circuit and there is a waste of standby power but it removes crossover distortion.

Technically, the operation of transistor lies between class A and class B mode. Therefore, the circuit

operation is often referred to as being class AB operation.

CLASS-AB PUSH PULL AMPLIFIER:

Q1

Q2

T1

Vin

T2

RL

IC1

IC2

IL

IB1

IB1

R1

D

In class AB push pull amplifier, the base emitter junction of both the transistors Q1 and Q2 are

slightly forward biased by the voltage across diode D. This method is used to minimize the crossover

distortion.

c power a class-a power amplifier · class-a power amplifier: in class-a amplifier, the transistor...

Documents