04/03/2024
Power Amplifiers
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Power Amplifiers (PA)
Small
Signal
Amplifier
In small-signal amplifiers the
main factors are:
• Linearity
• Gain
Power
Amplifier
Since large-signal, or power, amplifiers handle
relatively large voltage signals and current
levels, the main factors are:
• Efficiency
• Maximum power capability
• Impedance matching to the output device
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Classification of Amplifiers
According to use:
1. Voltage amplifier (Small signal amplifier)
2. Power amplifier – to deliver power efficiently to a load.
According to coupling methods:
1. RC coupled
2. Transformer coupled – for impedance matching
According to mode of operation:
1. Class A
2. Class B
3. Class AB
4. Class C
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Comparison of Amplifier
Classes
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Power Amplifiers
Class A
Class B
Portion of input signal during which collector current is supposed to flow.
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RC Coupled Class A PA
VCC
Differs from small signal amplifier by:
1. Size of transistor is considerably
larger
Rb
RL
2. Smaller value of amplification factor
3. Output power ~ 1-10 W
4. Lower output resistance ~Ohms
Vi
0
0
For small signal amplifiers power consideration is not important, but for
power amplifier collector efficiency is very important.
Power conversion – transformation from dc power to ac power!!
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RC Coupled Class A PA
Power considerations:
Input power – power drawn from dc supply
Pi
Idc
VCC
RL
Rb
Vcc  Idc
Pdc
Where Idc – operating point of the device
i
dc
cc dc
Idc
IQ
 
0
Vi
0
 Vcc  Vbe 

 Rb 
Ib << Ic
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RC Coupled Class A PA
Vcc
i
2RL
Vcc/RL
IL
Ic
i
Operating point
Ic = β Ib = Vcc/ 2 RL
For best performance
Vo
Due to ac signal
Vp
Vcc
Vce
Vp
Vcc
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RC Coupled Class A PA
VCC
Output power – ac power delivered to the load:
Po.transistor.ac
Rb
RL
Ic
IQ  i
Vcc  Vce Ic
Po.ac.load
Ic
Vce  Ic
IQ  i
Consists of ac signal
Vi
0
0
If the output voltage, Vo, is known across the load, the output power
c Q
can be defined by:
2
Vo.rms
Po.ac
RL
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RC Coupled Class A PA
Maximum output voltage swing is given by:
L
Vcc
Vp Vce.p
2
Maximum output current
L swing:
i
Ic.p
Vcc
2RL
The maximum output power:
Po.max Vo.max.rms Io.max.rms
Po.max
Vcc
Vcc

2 2 2 2 RL
Vcc
2
8 RL
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RC Coupled Class A PA
The maximum input dc power is given by:

Vcc 

2RL

Pi.max.dc Vcc  Idc
Pi.max.dc
Vcc

2
2RL
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RC Coupled Class A PA
The efficiency of the amplifier is described as the ratio of the ac power
delivered to the load to the power drawn fromLthe dc supply

 max
Po
Pi
P i.max
V cc
 max
Pi.max.dc
P o.max
Po.max
2
2RL
Vcc
2
8 RL
2
8R L
1
2
4
V cc
Vcc
2R L
25 %
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Transformer coupled Class A PA
Usually the resistance in the collector loop is greater than a load resistance
which must be connected to the amplifier (loudspeakers).
In order to transfer maximum power to low resistance loads, a transformer may
be used in the output for impedance matching.
100% power delivery is achieved from input to output of transformers.
N1
P 1
N2
I1  V 1
V1
V2
R1
R2
Primary
Secondary
Coil
Coil
P 2
I2  V 2
V2
N 2
V1
N 1
R 2
R 1
 N 2

 N 1
I2
N 1
I1
N 2



2
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Transformer coupled Class A PA
VCC
The dc resistance of a
transformer is ideally ZERO
RL
RT.dc
Rb
0
Hence
Vce.Q
Vcc
Vi
The base current and operating current
Ib
IQ
Vcc  Vbe
Rb
  Ib
The impedance of the transformer is:
2
 N1 
RT.ac 
  RL
 N2 
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The load line for transformer
coupled PA is obtained graphically.
dc load line
VceQ = Vcc
Ic
Zero signal current is:
IQ
IQ
Vcc
  Ib
RT.ac
Vce
Vcc
1. Draw the dc load line for VceQ = Vcc
2. Determine the operating point current
3. Draw the ac load line having gradient -1/RTac through the operating
point.
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Transformer coupled Class A PA
Power considerations.
Input power drawn form the dc supply:
Pdc
Pi
Pi
Vcc  IQ
Vcc
Vcc 
RT.ac
Vcc
2
IQ
Vcc
RT.ac
RT.ac
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Transformer coupled Class A PA
Ac power delivered to the input of the transformer:
Vo
Vp
Po
Po
Vo.rms Io.rms
Vp
2

2
Vp
Vp
2 RT.ac
2RT.ac
Vcc
t
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Transformer coupled Class A PA
The efficiency of the transformer coupled PA is given by:

Po
Pi
Vp

2
2RT.ac
2
Vcc
Po
1  Vp 


2 Vcc


2
Pi
RT.ac
Vp
2
2RT.ac
Vcc
2
RT.ac
Maximum efficiency occurs when Vp = Vcc; η = 50%
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RC Class A PA
Determine the supply voltage, base resistance and input power for an RC
coupled class A power amplifier for which power output in a load of 10 Ohm is
10 W. Consider β = 75.
T.ac
1.
Vcc
2
10W
8 RL
8 10 10
Vcc
2.
IQ
Ib
Rb
28V
Vcc
28
2 RL
2 10
IQ
1.4

75
Pi
1.4A
Vcc  IQ
28 1.4
39.2W
19mA
Vcc  Vbe
28  0.7
27.3
Ib
0.019
0.019
1.5k
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Class B Power Amplifier
In order to reduce power loss of the power amplifier, the system must
operate only when an input signal is applied to its input!!
=> Class B operation
Ic
In class B operation:
1. Zero signal current is ZERO
=> No bias is required!!
2. Operating point is located at
cut off voltage
3. Output is amplified half-wave
rectification.
Operating point
Vce
Vo
Output voltage
across load
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Class B Power Amplifier
T1
n
p
n
RL
Ip
V
Vi
Current across
transformer
Idc
Imean
Ip

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Class B Power Amplifier
In order to get a full AC cycle out of a class B amplifier, two
transistors are required:
• A transistor that provides the negative half of the AC cycle
• A transistor that provides the positive half.
Centre-tapped
transformer
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Push Pull Power Amplifier
1
n
p
n
Ip

V
RL
Vi
2
n
p
n
Ip

Average current drawn
Ip
from signal
Idc
2

1. During the first half of the cycle transistor 1 is properly biased by the ac
signal and the class B amplifier amplifies the first half wave.
2. During the second half of the cycle transistor 2 is properly biased by the ac
signal and the class B amplifier amplifies the second half wave.
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Push Pull Power Amplifier
Power considerations:
Input dc power
Vcc  Idc
Pdc
Pi
Pi
2 Ip
Vcc 

Ip
Pi
Vp
RT.ac
2 Vcc  Vp
  RT.ac
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Push Pull Power Amplifier
The output power delivered to the primary circuit of the transformer is
given by:
2
Po
Po
VL.rms
RT.ac
Vp
2
2RT.ac
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Push Pull Power Amplifier
Efficiency of the Push pull amplifier is given by:

Po
Pi
Vp

2
2RT.ac
2 Vcc Vp
 Vp

4 Vcc
  RT.ac
Maximum efficiency occurs when Vp = Vcc; η = π/4 = 78%
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Push Pull Power Amplifier
Maximum input and output power occurs when Vp = Vcc, therefore:
Pdc.max
Po.max


2 Vcc  Vp
2 Vcc
Vcc
  RT.ac
Vp
Vcc
2
  RT.ac
2
Vcc
2RT.ac
2
2RT.ac
Power dissipation in system:
P diss
P diss
Minimum dissipation for:
Pi  Po
2  V cc  V p
  R T.ac
d
P diss
dV p

Vp
2
2R T.ac
0
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Cross Over distortion
Vi
As there is no dc bias applied to the
base emitter junction, only when
the input ac signal reaches a peak
value of ±0.7V, the transistor is
open.
Vbe
Vo
Hence the output signal is
distorted.
Cross over distortion due to 2 half
cycles!!
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Elimination of distortion
To avoid cross over distortion, the
base emitter junction must be
prebiased by a very small voltage
~ 0.7V
R1
R2 = R3
R2
And
V
R1 = R4
RL
R2 and R1 must be chosen in
such a way that the potential drop
across R2 is 0.7V
Vi
R3
R4
This configuration is known as Class AB
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Elimination of distortion
R1
It is difficult to choose
appropriate resistors for the
input.
An easier way of biasing the
transistor is to use diodes of
similar material as the
transistors used.
When forward biased the
voltage drop across diode =
0.7V
0.7V
V
RL
Vi
0.7V
R4
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