Chapter 20
Power Amplifier
This chapter deals with
• Differences between voltage and power amplifiers
• Classification of Power Amplifiers
• Harmonic Distortion analysis
• Class-A Power Amplifier
- Series fed Class-A Power Amplifier
- Inductor coupled Class-A Power Amplifier
- Transformer coupled Class-A Power Amplifier
• Push-pull Amplifier
- Class-A push-pull amplifier
• Class-B Power Amplifier
- Direct coupled push-pull class-B amplifier
- Complementary Symmetry push-pull class-B amplifier
•
•
•
•
•
Class-AB push-pull Power Amplifier
Class-C Power Amplifier
Class-D Power Amplifier
Class-S Power Amplifier
Heat transfer in semiconductor devices
20.1 Introduction
Power amplifiers are designed to deliver a large amount of power to drive loads like
loud speaker, CRT, servo motor etc. Power amplifiers are meant for raising the power
level of the input signal to a large extent. Hence the input to the power amplifiers
should be large. That is why in an electronic system a voltage amplifier precedes a
power amplifier to provide a large input signal. Hence the power amplifiers are also
known as large signal amplifiers. A large signal amplifier must operate efficiently
and should be capable of handling power ranging from a few Watts to few 100 Watts.
These amplifiers are characterized by high efficiency, maximum power handling
capacity and good impedance matching.
20.2
Electron Devices and Circuits
A power amplifier actually draws power from the DC power supply and
converts it into AC signal. That is why the efficiency of a power amplifier is defined as
conversion efficiency. Hence a power amplifier may be defined as an amplifier that
converts DC power into AC power and whose action is controlled by the input signal.
Let us now discuss in detail these amplifiers in this chapter.
20.2 Difference between a Voltage and Power Amplifier
The main function of a voltage amplifier is to amplify a low level signal to a high
level signal. The voltage gain of the amplifier is given in general by
AV =
Ai Z0
Zi
(20.1)
Where Ai is current gain, Z0 is the output impedance and Zi is the input
impedance of the amplifier.
Hence to obtain a high voltage gain, we need
1. Transistors with high β value.
2. The input impedance should be less than the output impedance (generally
Z0 = RC = 1KΩ − 2KΩ).
3. Collector resistance RC should be larger to obtain high voltage gain. The
collector current of voltage amplifier is low.
4. Collector current IC is low, the power dissipation is low and hence BJTs with
low power handling capacity can be used.
Let us now tabulate the differences between voltage and power amplifiers.
Parameters
Current gain, β
Collector resistance, RC
Input voltage, Vin
Collector current, IC
Output power
Power dissipation
Output Impedance
Method of coupling
Voltage Amplifier
High (> 100)
High
Low (a few mV)
Low (≈ 1mA)
Low (few mW)
Low
High
RC coupling
Power Amplifier
Low (20 − 50)
Low
High (2 − 4 Volts)
High (>100mA)
High (few Watts − few 100 W)
High
Low
Transformer coupling
Since a power amplifier needs to handle large signals, its output current is high.
Due to high currents, the power is also high. Therefore, to obtain high power
amplification, we need:
1. A transistor which can dissipate the power. Hence large sized power transistors
mounted on heat sinks are used, which dissipate the heat generated during the
operation.
Power Amplifier
20.3
2. The β value of the transistors should be low.
3. The collector current IC is large; the collector resistance RC should be low.
Major applications of power amplifiers
1.
2.
3.
4.
5.
6.
Used in the last stage of a transmitter in a communication system
Used in radio and TV receivers
Used in electro-mechanical control system to drive motors
Used in computer hard disk driver stage
Used in motorized valves
Used in robotics and process control instruments.
20.3 Classification of Amplifiers
An amplifier can be classified based on the choice of Q-point on the transfer
characteristics. They are
1. Linear amplifier
(a)
(b)
(c)
(d)
Class-A power amplifier
Class-B power amplifier
Class-AB power amplifier
Class-C power amplifier
2. Non-linear amplifier
(a) Class-D power amplifier
(b) Class-S power amplifier
Choice of the Q-point
Consider the transfer characteristics of a transistor.
Output current (mA)
QA
IAQ
QAB
QC
QB
Input Voltage / Current
VAQ/IAQ
Fig. 20.1 Transfer characteristics: Q-point for various power amplifiers
20.4
Electron Devices and Circuits
20.3.1 Class-A amplifier
For a class-A amplifier, the quiescent point QA is chosen in the linear region of the
transfer characteristics as shown in Fig.20.1. In the linear region, the current and
voltage vary linearly around the Q-point QA . At the Q-point of class-A amplifier,
there exists a current in the circuit even after VAQ reduces to a very small value.
Thus an amplifier in which the current does not become zero for any part of the cycle
is known as class-A amplifier.
iC
ICQ
DC load
t
Fig. 20.2 Collector current of class-A amplifier
20.3.2 Class-B amplifier
The Q-point QB for a class-B amplifier is chosen at cut-off, where the output current
is zero. Hence for a sinusoidal input, during the positive half cycle, the transistor is
brought out of cut-off as it gets a forward bias and it conducts. During the negative
half cycle, the transistor gets a negative bias and it is taken deep into cut-off and so
the transistor does not conduct. Thus an amplifier in which the output current flows
for only one half cycle is called as class-B amplifier.
The waveform indicates that the collector current flows for only one half cycle.
iC
ICQ = 0
0
p
2p
3p
t
Fig. 20.3 Collector current of class-B amplifier
20.3.3 Class-AB amplifier
The Q-point QAB for a class-AB amplifier is chosen above cut-off and below linear
region, that is, between QA and QB . Hence for a sinusoidal input, the transistor
conducts for a full positive half cycle and for a part of the negative half cycle. When
Power Amplifier
20.5
the negative input becomes very high, the transistor is taken into cut-off and there is
no output. Thus for a class-AB amplifier, the current flows for full positive half cycle
and part of the negative half cycle.
The waveform indicates that the output is for more than half a cycle and less than
full cycle.
iC
ICQ > 0
DC load
0
p
2p
3p
4p
t
Fig. 20.4 Collector current of class-AB amplifier
20.3.4 Class-C amplifier
The Q-point QC for a class-C amplifier is chosen much beyond cut-off. For a
sinusoidal input, even during the positive half cycle, the output current flows only
when the input signal exceeds the minimum current required to take the transistor
into conduction. During the negative half cycle, the transistor is driven deeper into
cut-off and hence there is no output. Thus an amplifier in which the output current
flows for less than half a cycle is known as class-C amplifier.
The waveform indicates that the output is for less than half a cycle.
iC
ICQ < 0
0
p
t
2p
Fig. 20.5 Collector current of class-C amplifier
The amplifiers discussed above are linear amplifiers whereas class-D and class-S
are non-linear amplifiers.
20.3.5 Class-D amplifier
A class-D amplifier operates for pulsed inputs and hence the transistor is ON for
a short interval and OFF for a long interval. Hence the drop across the transistor
is negligible and so maximum power is delivered to the load which increases the
efficiency to 100%.
20.6
Electron Devices and Circuits
20.3.6 Class-S amplifier
The class-D amplifier does not respond to variation in input signal, hence it cannot be
used for amplitude modulated signals. Hence a class-S amplifier finds its application
in such cases.
20.4 Important Terms used in the Performance Analysis
of Power Amplifier
20.4.1 Efficiency
A power amplifier converts DC power into AC power. Hence conversion efficiency is
defined as the ability of the transistor to convert the DC power of the supply into AC
power delivered to the load. This is also called as theoretical efficiency or collector
circuit efficiency.
Signal power delivered to the load
× 100%
DC power drawn from the supply
Pac
%η =
× 100%
Pdc
%η =
(20.2)
20.4.2 Input power
The power into a power amplifier is given by the supply or bias. With no input signal
to the circuit, the DC power drawn is the product of the collector bias current ICQ
and the maximum drop across the transistor VCE .
Pi(dc) = VCE ICQ
(20.3)
20.4.3 Output power
The output power is calculated as the AC power delivered to the load given an AC
input signal. It can be calculated either in terms of RMS value, peak value or peakto-peak value.
In terms of RMS value, the output AC power is given by
Po(ac) = VCE(rms) IC(rms)
Po(ac) =
2
VCE(rms)
RC
2
RC
Po(ac) = IC(rms)
(20.4)
(20.5)
(20.6)
In terms of maximum value or peak value, the output AC power is given by
Po(ac) =
VCE(p) IC(p)
VCE(p) IC(p)
√
√ =
2
2
2
(20.7)
Power Amplifier
Po(ac) =
Po(ac) =
2
VCE(p)
2RC
2
IC(p)
2
RC
20.7
(20.8)
(20.9)
In terms of peak-to-peak value, the output AC power is given by
Po(ac) =
Po(ac) =
Po(ac) =
VCE(pp) IC(pp)
VCE(pp) IC(pp)
√
√ =
8
2 2 2 2
2
VCE(pp)
8RC
2
IC(pp)
8
RC
(20.10)
(20.11)
(20.12)
20.4.4 Power dissipation
This is defined as its ability to dissipate the heat developed within it. As already
discussed, power amplifiers handle high currents and hence get heated up during
operation. This heat has to be dissipated; if not the transistor will get damaged. So
the power dissipation capacity of the transistors should be increased by using suitable
provisions like heat sinks.
20.4.5 Distortion
This is defined as the change in output from the input in terms of its shape, frequency
and phase. Accordingly it is called as amplitude distortion, frequency distortion and
phase distortion. In the case of power amplifiers due to the large input signal, the
transistor operation becomes non-linear and hence results in amplitude distortion.
For good performance, the distortion is expected to be low.
20.5 Class-A Power Amplifier
A class-A amplifier has its operating point chosen in the linear or active region.
Hence for a given AC input, the transistor remains forward biased and conducts for
both the half cycles. This results in an output current, which flows for both the half
cycles. Since the output current flows for both the cycles, the power loss in the circuit
is more and the power transferred to the load is less. This reduces the efficiency of the
class-A amplifier. Now let us discuss the operation of following class-A amplifiers
and derive their efficiency.
1. Series fed class-A amplifier
2. Inductor coupled class-A amplifier
3. Transformer coupled class-A amplifier
20.8
Electron Devices and Circuits
20.5.1 Series fed class-A amplifier
The circuit diagram of a series fed class-A amplifier is shown in Fig.20.6. Here
resistor RB is used to provide a fixed bias to the transistor to operate it in the linear
region. The collector resistor RC acts as the load and it decides the power dissipation
of circuit. Since the load is connected in series with the transistor output, the circuit
is known as series fed class-A amplifier.
+VCC
RC
RB
IC
C
Vi
+
–
+
VBE
–
Fig. 20.6 Series fed class-A amplifier
Since we have a resistive load, we have a maximum power loss across it. Hence
the efficiency is low as shown in the calculation below.
DC load line
The DC load line is the line passing through output characteristics of the transistor
connecting the extreme conditions of operation as shown in Fig.20.7.
IC(mA)
VCC
RC
Q
ICQ
IB =0
VCEQ
Fig. 20.7 DC load line
VCC
VCE (volts )
Power Amplifier
20.9
To determine IC(max) and VCE(max)
Apply KVL to output loop of fixed bias circuit,
VCE = VCC − IC RC
(20.13)
When IB = IB(min) and IC = IC(min)
Then,
VCE(max) = VCC − IC(min) RC
Neglecting IC(min) RC as IC(min) ≈ 0, we get
VCE(max) = VCC
(20.14)
When IB = IB(max) and IC = IC(max)
Then,
VCE(min) = VCC − IC(max) RC
Neglecting VCE(min) as VCE(min) ≈ 0, we get
IC(max) =
VCC
RC
(20.15)
Thus the DC load line is drawn on the output characteristics joining points
obtained from IC(max) = VRCC
on y-axis and VCE(max) = VCC on x-axis. The
C
1
slope of the line will be /RC . Similarly, the AC load line also passes through
same points with a slope RC 1||RL . The point of intersection of the load line with the
characteristics in the active region is chosen to be the operating point. Operating the
amplifier for this quiescent conditions results in an output for one full cycle.
Consider the load line characteristics shown in Fig.20.8. Here quiescent
collector current ICQ and quiescent collector-emitter voltage VCEQ indicate the
operating current and voltage when the AC input is zero. When the AC signal is
applied, the operating point shifts above and below causing collector current IC and
collector-emitter voltage VCE to vary about their maximum and minimum values.
From the waveform, it is clear that the output voltage varies between 0 and VCC to
give an output swing of VCC .
Let us now see how the input DC power is distributed in an amplifier. The total
input power from DC supply is given by
Pi(dc) = VCC ICQ
(20.16)
20.10
Electron Devices and Circuits
IC(mA)
VCC
RC
ICQ
Q
VCEQ
O
VCC VCE – (volts )
Fig. 20.8 Load line characteristics
This DC power is divided across the collector resistor and transistor, that is,
1. The power dissipated across collector resistor RC is given by
2
RC
PR = ICQ
(20.17)
2. The power dissipated to transistor is given by,
PT = Pi(dc) − PR
(20.18)
Substitute equations (20.16) and (20.17) in equation (20.18)
2
RC
PT = VCC ICQ − ICQ
(20.19)
The power dissipated to transistor PT is further subdivided as
(a) The power across the load
(b) The power dissipation across the transistor
3. The power across the load in terms of RMS value is given by
Po =
2
IC(rms)
RC
=
2
VCE(rms)
RC
(20.20)
The power across the load in terms of peak-to-peak value is given by,
Po =
VCE(pp) IC(pp)
8
(20.21)
4. Power dissipation across the transistor is in the form of heat.
Let us now calculate the conversion efficiency of a series fed class-A amplifier.
Power Amplifier
20.11
The conversion efficiency is defined as
signal power delivered to load
DC power drawn from supply
Po(ac)
η=
Pdc
η=
(20.22)
DC power drawn from supply is given by
Pdc = VCC ICQ
(20.23)
From the load line graph, it is clear that,
ICQ =
IC(max)
VCC
=
2
2RC
Substitute equation (20.24) in equation (20.23)
V2
VCC
= CC
Pdc = VCC
2RC
2RC
(20.24)
(20.25)
The AC power delivered to the load is given by
VCE(pp) IC(pp)
8
VCE(max) − VCE(min) IC(max) − IC(min)
=
8
Po(ac) =
(20.26)
Po(ac)
(20.27)
From the load line graph, we get
IC(max) =
VCC
RC
(20.28a)
IC(min) = 0
(20.28b)
VCE(max) = VCC
(20.28c)
VCE(min) = 0
(20.28d)
Substitute equation (20.28) in equation (20.27)
VCC VRCC
V2
C
= CC
Po(ac) =
8
8RC
(20.29)
20.12
Electron Devices and Circuits
Substitute equations (20.25) and (20.29) in equation (20.22)
Po(ac)
η=
=
Pdc
2
VCC
8RC
2
VCC
2RC
=
1
= 0.25
4
η(%) = 25%
(20.30)
This is the maximum efficiency of a series fed class-A power amplifier.
Generally, the maximum efficiency occurs only under ideal conditions and so
a practical series fed class-A amplifier has an efficiency of less than 25%.
Solved Problem 20.1
A class-A series fed amplifier has VCE(max) = 25 V, VCE(min) = 5 V, IC(max) =
10 mA and IC(min) = 2 mA. Determine the RMS value of current and voltage.
Also determine the AC power and conversion efficiency given VCC = 25 V and
ICQ = 5 mA.
Solution:
Pdc = VCC ICQ = 25 × 5m = 125mW
VCE(rms) =
VCE(rms) =
IC(rms) =
IC(rms) =
Po(ac) =
Po(ac) =
η(%) =
VCC(max) − VCC(min)
VCC(pp)
√
√
=
2 2
2 2
20
25 − 5
√ = √ = 7.071V
2 2
2 2
IC(max) − IC(min)
IC(pp)
√ =
√
2 2
2 2
8
10 − 2
√ = √ = 2.824 mA
2 2
2 2
VCE(max) − VCE(min) IC(max) − IC(min)
8
20 × 8m
= 20mW
8
Po(ac)
20mW
× 100 = 16%
=
Pdc
125mW
Solved Problem 20.2
For the given class-A amplifier, find a) input power b) output power delivered to
transistor c) power lost across the transistor d) power rating of transistor.
Power Amplifier
20.13
+20V
2 kW
50W
b =20
IB(max)=12mA
Vi
Solution:
IC(max) =
VCC
20
= 0.4A = 400 mA
=
RC
50
VCE(max) = 20V
IBQ =
VCC − VBE
20 − 0.7
= 9.65 mA
=
RB
2K
ICQ = βIBQ = 20 × 9.65m = 193 mA
VCEQ = VCC − ICQ RC = 20 − 193m × 50 = 10.35V
For the above data, DC load line is drawn and Q-point is fixed for ICQ = 193 mA
and VCEQ = 10.35V .
IC(mA)
400mA
ICQ = 193mA
Q
VCE=10.35V
20V VCE (volts )
20.5.2 Inductor coupled class-A amplifier
A class-A power amplifier with a resistive load has an efficiency of 25%, since most
of DC power is lost in the circuit across the resistor and transistor. The wasted DC
power, therefore does not contribute to the useful AC output. Apart from this it is
20.14
Electron Devices and Circuits
preferable to avoid DC current passing to an output load such as a loud speaker.
This is not possible in a circuit with resistive load as it allows both AC and DC to
pass through it. Hence to overcome these problems, an inductor can be used as load.
Alternatively the resistive load can be connected to the amplifier through transformer.
This arrangement also solves the problem with an additional advantage of providing
good impedance matching.
+VCC
L
RB
CO
+
Ci
IB
+
RL
VO
Vi
–
–
Fig. 20.9 Inductor coupled class-A amplifier
Let us now consider a class-A amplifier with an inductor load at the collector as
shown in Fig.20.9. The resistor RB biases the transistor in the active region. Since
the reactance of the inductor XL = 2πf L is zero for DC, there is no DC drop across
inductor. Hence this DC power which is not wasted is converted into useful AC
power and transferred to an output. This improves the efficiency of the circuit to a
maximum value of 50% which is double of a class-A amplifier with resistive load.
Now let us discuss AC and DC load lines and determine the collector efficiency of
the circuit.
DC load line
For simplifying the analysis, neglect the series resistance of inductor. With this
assumption, the entire DC drop is across the transistor and so VCE(max) is equal
to VCC . The maximum current through the circuit tends to infinity since IC = VXCC
L
where f = 0, and XL = 0. Therefore the DC load line is a vertical line from
VCE = VCC on the x-axis to infinity.
AC load line
The AC load line passes through the Q-point with a slope of −1/RL . From the plot
shown in Fig.20.10, it is obvious that the maximum output collector current swing is
0 to 2VCC , which is twice that of a series fed class A amplifiers.
Power Amplifier
20.15
IC(mA)
2ICQ
dc load line
ICQ
Q
ac load line
O
2VCC V (volts )
CE
VCC
Fig. 20.10 Load line characteristic of inductor coupled class-A amplifier
Now let us discuss why the output swing is doubled?
For AC analysis, since the reactance of an inductor is large, no AC current passes
through it. Hence it can be replaced by a constant current source of value ICQ . Since
the reactance of the capacitor is very small, the AC drop across it is negligible. Hence
it can be replaced by a voltage source VCC , the voltage to which it would be charged
in the absence of that AC input. With these assumptions, we get the circuit as shown
in Fig.20.11.
VCC
ICQ
a
VCC
+ –
iC
+
iL
VCE
¯
RL
Fig. 20.11 Simplified (with assumption) inductor coupled class-A amplifier
20.16
Electron Devices and Circuits
Given an AC input, consider for an instant when ic = 0.
Apply KCL at node ‘a’
IcQ = ic + iL
(20.31)
IcQ = iL
(20.32)
Since ic = 0,
Solving the collector-emitter loop, we get,
Since ICQ =
VCE = VCC + iL RL
(20.33)
VCE = VCC + IcQ RL
(20.34)
VCC
RL
VCE = VCC +
VCC
RL
RL = 2VCC
(20.35)
When the input signal polarity is reversed,
ic = 2IcQ
(20.36)
Substitute equation (20.36) in equation (20.31)
IcQ = 2IcQ + iL
Therefore,
IcQ = −iL
(20.37)
Substitute equation (20.37) in equation (20.34)
VCE = VCC − ICQ RL
Since ICQ =
VCC
RL
VCE = VCC −
VCC
RL
(20.38)
RL = 0
(20.39)
Thus it is clear that the voltage swings between 0 to 2VCC .
Efficiency Calculation
As already defined,
η=
Signal power delivered to load
DC power drawn from supply
η=
Po(ac)
Pdc
(20.40)
Power Amplifier
20.17
DC power drawn from supply is given by
Pdc = VCC ICQ
(20.41)
From the load line graph, it is clear that,
VCC
RC
Substitute equation (20.42) in equation (20.41)
V2
VCC
= CC
Pdc = VCC
RC
RC
ICQ =
(20.42)
(20.43)
The AC power delivered to the load is given by
VCE(pp) IC(pp)
8
VCE(max) − VCE(min) IC(max) − IC(min)
Po(ac) =
8
From the load line graph, we get
Po(ac) =
IC(max) = 2ICQ =
2VCC
RC
IC(min) = 0
(20.44)
(20.45)
(20.46a)
(20.46b)
VCE(max) = 2VCC
(20.46c)
VCE(min) = 0
(20.46d)
Substitute equation (20.46) in equation (20.45)
2VCC 2VRCC
4V 2
V2
C
= CC = CC
Po(ac) =
8
8RC
2RC
Substitute equations (20.45) and (20.47) in equation (20.40)
Po(ac)
=
η=
Pdc
2
VCC
2RC
2
VCC
RC
=
η(%) = 50%
(20.47)
1
= 0.5
2
(20.48)
Thus from the above derivation, it is clear that the maximum efficiency is 50%
for an inductor coupled class-A power amplifier.
Figure of merit
It is defined as the ratio of maximum power dissipation
at the collector PC(max) to
the maximum power dissipation in the load PL(max) .
F =
PC(max)
PL(max)
(20.49)
20.18
Electron Devices and Circuits
Maximum power dissipation at the collector of the transistor is given by
VCC
(20.50)
PC(max) = VCEQ ICQ = VCC
RL
V2
PC(max) = CC
(20.51)
RL
Maximum power dissipation in the load,
2 IC(p)
VCC 2 RL
RL =
(20.52)
PL(max) =
2
RL
2
2
VCC
2RL
Substitute equations (20.51) and (20.53) in equation (20.49)
PL(max) =
PC(max)
=
F =
PL(max)
2
VCC
RL
2
VCC
2RL
(20.53)
=2
(20.54)
20.5.3 Transformer coupled class-A amplifier
In order to avoid DC power loss across the collector load and to improve the
collector efficiency, we go in for a transformer coupled class-A amplifier as shown
in Fig.20.12. Here also the resistor RB is used to bias the transistor in the active
region. A transformer is connected at the collector through which the load resistor
RL is connected.
+VCC
+
NP:NS
RB
Ci
Vi
–
R L VO
–
+
Fig. 20.12 Transformer coupled class-A amplifier
The reflected value of resistor connected at the secondary into the primary side
of transformer is given by
2
Np
rL =
RL
(20.55)
Ns
Power Amplifier
20.19
Where, NP is the primary turns ratio, NS is the secondary turns ratio and RL is
load resistance. Since we have the primary coil connected at the collector, the circuit
analysis is similar to an inductor coupled class-A power amplifier and hence the AC
and DC load line graph remains the same as shown in Fig.20.10. The output swings
between 0 to 2VCC which gives a maximum efficiency of 50%.
Efficiency calculation
As already defined,
η=
signal power delivered to load
DC power drawn from supply
η=
Po(ac)
Pdc
(20.56)
DC power drawn from supply is given by
Pdc = VCC ICQ
(20.57)
From the load line graph, it is clear that,
ICQ =
VCEQ
rL
(20.58)
Substitute equation (20.58) in equation (20.57)
VCEQ
Pdc = VCC
rL
Since VCEQ = VCC
Pdc =
2
VCC
rL
(20.59)
The AC power delivered to the load is given by
Po(ac) =
2
Vsy(pp)
8RL
Ns
Vsy(pp) = Vpy(pp)
Np
Ns
Vsy(pp) = 2VCC
Np
Substitute equation (20.61) in equation (20.60)
2
2
Ns
Ns
2
2VCC N
4V
CC
Np
p
Po(ac) =
=
8RL
8RL
2
2
V
Ns
Po(ac) = CC
2RL Np
(20.60)
(20.61)
(20.62)
20.20
Electron Devices and Circuits
Substitute equations (20.62) and (20.59) in equation (20.56)
2
2
VCC
Ns
Po(ac)
2RL Np
=
η=
2
VCC
Pdc
(20.63)
rL
Substitute (20.55) in equation (20.63)
η=
Po(ac)
=
Pdc
2
VCC
2RL
Ns
Np
2
V2
CC
Np 2
RL
N
=
1
= 0.5
2
s
η(%) = 50%
(20.64)
Thus from the above derivation, it is proved that a transformer coupled class-A
power amplifier has a maximum efficiency of 50%.
Figure of merit
It is defined as the ratio of maximum power dissipation
at the collector PC(max) to
the maximum power dissipation in the load PL(max) .
F =
PC(max)
PL(max)
(20.65)
Maximum transistor collector dissipation is given by
VCC
PC(max) = VCEQ ICQ = VCC
RL
2
VCC
PC(max) =
RL
Maximum power dissipation in the load,
2 IC(p)
VCC 2 RL
RL =
PL(max) =
2
RL
2
PL(max) =
2
VCC
2RL
(20.66)
(20.67)
(20.68)
(20.69)
Substitute equations (20.67) and (20.69) in equation (20.65)
PC(max)
=
F =
PL(max)
2
VCC
RL
2
VCC
2RL
=2
(20.70)
Since class-B, AB and C amplifiers conduct for less than 360◦ , they result in
harmonic distortion. Let us now discuss the harmonic distortion in power amplifiers.
Power Amplifier
20.21
20.6 Harmonic Distortion in Power Amplifier
In general, the dynamic characteristic of an amplifier is not linear. The non-linearity
arises if the static output characteristics are not equidistant straight lines for constant
increments of input excitation. Due to this non-linearity of the dynamic characteristic
curve the shape of the output waveform differs from that of the input. This type of
distortion is known as non-linear/ amplitude/ harmonic distortion.
20.6.1 Second order harmonic distortion
When non-linear distortion is present the output waveform contains lots of harmonic
components, that is, second, third and higher harmonic components. Out of these,
the second harmonic component contributes more to the distortion. Hence let us now
determine the magnitude of second harmonic distortion. In a transistor, generally,
collector current ic is related to base current ib by a constant. This equation
holds-good if the dynamic transistor characteristics is a straight line. Due to the
non-linearity of the transistor the dynamic transfer characteristic is a parabola and so
ic = G1 ib + G2 i2b
(20.71)
Where G1 and G2 are constants.
Let the input be a sinusoidal signal. Then
ib = Ibm cos ωt
(20.72)
Substitute equation (20.72) in equation (20.71)
ic = G1 (Ibm cos ωt) + G2 (Ibm cos ωt)2
Since, cos2 θ =
Therefore,
1+cos 2θ
2
ic = G1 Ibm cos ωt +
2
G2 Ibm
ic = G1 Ibm cos ωt +
2
G2 Ibm
2
(20.73)
1 + cos 2ωt
2
2 cos 2ωt
G2 Ibm
+
2
(20.74)
Now the total instantaneous collector current will be due to both AC and DC
inputs. Hence,
iC = IC + ic
(20.75)
Substitute equation (20.74) in equation (20.75)
2
2 cos 2ωt
G2 Ibm
G2 Ibm
+
2
2
2 cos 2ωt
G2 Ibm
+ G1 Ibm cos ωt +
2
iC = IC + G1 Ibm cos ωt +
iC = IC +
2
G2 Ibm
2
(20.76)
20.22
Electron Devices and Circuits
The equation (20.76) can be rewritten as,
iC = IC + B0 + B1 cos ωt + B2 cos 2ωt
Where,
(20.77)
2
2
G2 Ibm
G2 Ibm
= B0 ; G1 Ibm = B1 ;
= B2
2
2
From equation (20.77), it is clear that the output contains a constant current B0 ,
a term of the same frequency as the input and a second harmonic term along with a
DC current IC . Thus the total DC at the output is IC + B0 .
Let us now determine the magnitude of B0 , B1 and B2 . Consider the output
characteristics of an amplifier that shows the quiescent, maximum and minimum
values of collector current as shown in Fig.20.13.
IC(mA)
Icmax
Im
t
1
Q
0
p
2
p
wt
ICQ
2
Icmin
VCC VCE (volts )
Vm
0
VC
p
2
wt
Fig. 20.13 Output characteristic of a push-pull amplifier
From the Fig.20.15,
When ωt = 0; iC = Imax
When ωt = π2 ; iC = IC
When ωt = π; iC = Imin
Substitute these conditions in equation (20.77),
When ωt = 0; iC = Imax
iC = Imax = IC + B0 + B1 cos 0 + B2 cos 0
Power Amplifier
20.23
Since cos 0 = 1
Imax = IC + B0 + B1 + B2
When ωt =
(20.78)
π
; iC = IC
2
iC = IC = IC + B0 + B1 cos π/2 + B2 cos π
Since cos π/2 = 0 and cos π = −1
IC = IC + B0 − B2
(20.79)
When ωt = π, iC = Imin
iC = Imin = IC + B0 + B1 cos (π) + B2 cos 2π
Since cos π = −1 and cos 2π = 1
Imin = IC + B0 − B1 + B2
(20.80)
From equation (20.79), we get
B0 = B2
(20.81)
Subtracting equation (20.80) from equation (20.78)
Imax − Imin = 2B1
Imax − Imin
B1 =
2
Substituting equations (20.81) and (20.82) in equation (20.78)
Imax = IC + 2B0 +
(20.82)
Imax − Imin
2
2Imax = 2IC + 4B0 + Imax − Imin
Imax + Imin − 2IC
= B2
4
The second harmonic distortion D2 is defined as,
B2 D2 = B1
B0 =
(20.83)
(20.84)
Knowing B1 and B2 , D2 is determined.
Note:
if the input contains two frequencies ω1 and ω2 , then
a non-linear transistor will give an output which will consist of a DC term and sinusoidal components of frequencies
ω1 , ω2 , 2ω1 , 2ω2 , ω1 + ω2 and ω1 − ω2 . The sum and difference of frequencies in the output are known as inter-modulation or combination
frequencies.
20.24
Electron Devices and Circuits
20.6.2 Higher Order Harmonic Distortion
The analysis carried out in the previous section holds-good for small signal
amplifiers. For large signal amplifiers, the current is represented as,
ic = G1 ib + G2 i2b + G3 i3b + G4 i4b + ......
(20.85)
Where G1 , G2 , G3 , G4 are constants.
Let the input be a sinusoidal signal. Then
ib = Ibm cos ωt
(20.86)
Substitute equation (20.86) in equation (20.85)
ic = G1 (Ibm cos ωt) + G2 (Ibm cos ωt)2 + G3 (Ibm cos ωt)3 +
G4 (Ibm cos ωt)4 + · · ·
(20.87)
The total instantaneous collector current will be due to both AC and DC inputs.
Hence,
(20.88)
iC = IC + ic
Substitute equation (20.87) in equation (20.88) and simplify as in the previous
section and we get
iC = IC + B0 + B1 cos ωt + B2 cos 2ωt + B3 cos 3ωt + B4 cos 4ωt + ......
(20.89)
In the previous case, we had to determine B0 , B1 and B2 hence we defined three
points Imax , IC and Imin . Now in this case we have to determine five coefficients
B0 , B1 , B2 , B3 and B4 and hence we have to define five points Imax , I 1 , IC , I− 1 and
2
2
Imin . Hence this method is known as five-point method, compared to the two point
method discussed earlier.
Let us consider five points for the graphical analysis. From the graph as shown
in Fig.20.14 for a sinusoidal input ib = Ibm cos ωt, we get
When ωt = 0; iC = Imax
When ωt = π3 ; iC = I 1
2
When ωt = π2 ; iC = IC
When ωt =
2π
3 ; iC
= I− 1
2
When ωt = π; iC = Imin
Applying these conditions to equation (20.89), we get
iC = IC + B0 + B1 cos ωt + B2 cos 2ωt + B3 cos 3ωt + B4 cos 4ωt + · · ·
Power Amplifier
20.25
iC(A)
Icmax
I
1
2
IC
I– 1
2
Icmin
iB(mA)
Ibmax
p
2
p
2
ib
p
3
2p
3
p
wt
Fig. 20.14 Output characteristic of a push-pull amplifier
When ωt = 0; iC = Imax
Imax = IC + B0 + B1 + B2 + B3 + B4
When ωt =
π
; iC = I 1
2
3
I 1 = IC + B0 +
2
When ωt =
B1 B2
B4
−
− B3 −
2
2
2
(20.91)
π
; iC = IC
2
IC = IC + B0 − B2 + B4
When ωt =
(20.90)
(20.92)
2π
; iC = I− 1
2
3
I− 1 = IC + B0 −
2
B1 B2
B4
−
− B3 −
2
2
2
(20.93)
20.26
Electron Devices and Circuits
When ωt = π; iC = Imin
Imin = IC + B0 − B1 + B2 − B3 + B4
(20.94)
Solving the above equations, we get
B0 =
B1 =
B2 =
B3 =
B4 =
1
Imax + 2I 1 + 2I− 1 + Imin − IC
2
2
6
1
Imax + I 1 − I− 1 − Imin
2
2
3
1
(Imax − 2IC + Imin )
4
1
Imax − 2I 1 + 2I− 1 − Imin
2
2
6
1 Imax − 4I 1 + 6IC − 4I− 1 + Imin
2
2
12
Now the harmonic distortions are defined as,
B2 D2 = B
1
B3 D3 = B
1
B4 D4 = B1
In general, the nth harmonic is defined as,
Bn Dn = B1
The total distortion or distortion factor is defined as,
D = D22 + D32 + D42 + D52 + · · · Dn2
(20.95)
The power output of a distorted signal at the fundamental frequency is given by
P1 =
B12 RL
2
Similarly, the power at harmonic frequencies are,
P2 =
B22 RL
;
2
P3 =
B32 RL
etc.,
2
Power Amplifier
20.27
Hence the total output power is given by
B 2 RL
B12 RL B22 RL B32 RL B42 RL
+
+
+
+ ··· n
2
2
2
2
2
RL 2
B1 + B22 + B32 + B42 + · · · Bn2
=
2
2 2 2
2 B2
B3
B4
Bn
= P1 1 +
+
+
+ ···
B1
B1
B1
B1
= P1 1 + D12 + D22 + D32 + · · · Dn2
= P1 1 + D2
(20.96)
P =
P
P
P
P
where
D2 = D12 + D22 + D32 + · · · Dn2
20.7 Push-Pull Amplifier
In a single ended power amplifier due to the non-linearity of the transfer
characteristics of the transistor, the output is appreciably distorted. Such distortions
are greatly reduced by using push-pull configurations where two transistors are used
and made to conduct alternatively. These amplifiers are extensively used in transmitters, receivers, tape recorders, public addressing system etc.
20.7.1 Working principle
The push-pull amplifier has a centre tapped transformer both at the input and output.
The input transformer splits the signal with equal amplitude and opposite phase and
feeds it to the base of the transistors. Since the input to the transistor is out of phase,
their outputs are also out of phase. The output transformer combines these outputs to
provide a sinusoidal output. The push-pull amplifier is operated in class-A, class-B
and class-AB mode. Let us now discuss these amplifiers in detail.
20.8 Class-A Push-Pull Amplifier
The circuit of a class-A push-pull amplifier is shown in Fig.20.15. The input
transformer takes in the input signal and provides it at the bases of the transistors
with equal amplitude and opposite phase. The transistors Q1 and Q2 are identical, so
that they can conduct in the same manner for a given input. The resistors R1 and R2
form a voltage divider bias and they bias the transistors in active or linear region.
The output or collector current of the transistors flow in opposition in the output
transformer. This results in an alternating voltage across the load. Since the currents
flow in opposition, the transformer core saturation also does not occur.
20.28
Electron Devices and Circuits
Ic1
Q1
RS
T1
Vi
T2
R1
+
~
R2
–
IL
+
RL VO
VCC
–
Q2
Ic2
Fig. 20.15 Class-A push-pull amplifier
Circuit operation
During the positive half cycle, the base bias to transistor Q1 is more compared to
transistor Q2 and so transistor Q1 conducts more heavily than transistor Q2 , that
is, transistor Q2 conducts less. Hence collector current Ic1 of transistor Q1 is greater
than collector current Ic2 of transistor Q2 . We know that load current is the difference
between the collector current, that is, IL = IC1 − IC2 . During the positive half cycle,
the load current is dominated by collector current Ic1 of transistor Q1 .
During the negative half cycle, the base bias to transistor Q2 is more compared
to transistor Q1 and so transistor Q2 conducts more heavily than transistor Q1 , that
is, transistor Q1 conducts less. Hence collector current Ic2 of transistor Q2 is greater
than collector current Ic1 of transistor Q1 . We know that load current is the difference
between the collector current, that is, IL = IC1 −IC2 . During the negative half cycle,
the load current is dominated by collector current Ic2 of transistor Q2 .
Thus during every half cycle, either of the transistor conducts more and the other
conducts less resulting in a push-pull operation. As already said, since Ic1 and Ic2
are out of phase with each other, the voltage developed across load RL is alternating.
Advantages
1. The even harmonics are cancelled in the output. Hence a push-pull amplifier
gives a larger output per active transistor for a given amount of distortion.
2. A push-pull amplifier is also used to obtain a reduced distortion for a given
output power per transistor.
3. Since the collector current flows in opposition, there is no transformer core
saturation.
4. Since there is no core saturation, non-linear distortion due to transformer is
avoided.
5. Since the currents are flowing in opposition, the effective ripple voltages from
the supply are cancelled.
Power Amplifier
20.29
Disadvantages
1. Power supply hum is not eliminated.
2. Needs two identical transistors for equal amplification of the input for both half
cycles.
3. Needs an input stage that provides equal and opposite voltages for every half
cycle.
4. Uses transformers both at the input and output stages that increases the cost
and bulkiness of the circuit.
20.8.1 Proof for elimination of even harmonics
In a push - pull amplifier, the even harmonics are cancelled. Let the input base current
be
ib1 = Ibm cos ωt
(for transistor Q1 )
ib2 = −Ibm cos ωt
(for transistor Q2 )
The collector current ic1 of Q1 is given by
ic1 = ICQ + B0 + B1 cos ωt + B2 cos ωt + B3 cos 3ωt + · · ·
Since
ib2 = −ib1 = −Ibm cos ωt = Ibm cos(ωt + π)
The collector current ic2 of Q2 is given by
ic2 = ICQ + B0 + B1 cos(ωt + π) + B2 cos 2(ωt + π) + B3 cos 3(ωt + π) + · · ·
= ICQ + B0 − B1 cos ωt + B2 cos 2ωt − B3 cos 3ωt + · · ·
The load current is the algebraic sum of collector currents flowing due to transistors Q1 and Q2 , that is,
iL = ic1 − ic2
= ICQ + B0 + B1 cos 2ωt + B2 cos 2ωt + B3 cos 3ωt + · · ·
= −[ICQ + B0 − B1 cos 2ωt + B2 cos 2ωt − B3 cos 3ωt + · · · ]
= 2[B1 cos 2ωt + B2 cos 2ωt + · · · ]
Thus all the even harmonics are cancelled using push - pull configuration.
20.9 Class-B Power Amplifier
The practical circuit of a class-B push-pull power amplifier is shown in Fig.20.16.
The input signal is fed through a centre-tapped transformer. The output voltage is
collected across the transformer secondary.
In a class-B amplifier, the transistor is biased at cut-off and so the output current
flows for only one half-cycle as shown in Fig.20.17.
Electron Devices and Circuits
20.30
Ic1
RS
Vi
Q1
T2
T1
IL
+
RL V
~
O
–
Q2
Ic2
Fig. 20.16 Class-B push-pull power amplifier
Vi
VO
t
t
Fig. 20.17 Input-output voltage relation in class-B power amplifier
Since we obtain a partial output, class-B amplifiers are not useful in audio
applications. To obtain the full undistorted output, the class-B amplifiers are
connected in push-pull configuration. In the push-pull configuration, we have two
transistors biased at cut-off which conducts alternatively for every half cycle.
During the positive half cycle; transistor Q1 conducts and delivers the current
to the load in one direction (push) and during the negative half cycle; transistor Q2
conducts and delivers the current to the load in the opposite direction (pull).
Combining the individual signal output results in an undistorted output signal across
the load. The detailed analysis of push-pull amplifier is dealt in later sections.
Load line characteristic
As discussed for a class-A amplifier, the DC load line of class-B push-pull power
amplifier is shown in Fig.20.18.
The AC load line is the line connected between VCEQ = VCC and ICQ = VrCC
.
L
The quiescent operating point lies on the x− axis where the AC and DC load lines
intersect as shown in Fig.20.18. The slope of the DC load line is −1/rL , when rL is
Power Amplifier
20.31
IC
DC load line
VCC
rL
AC load line
VCEQ = VCC
VCE
Fig. 20.18 DC load line of class-B power amplifier
the reflected resistance into the primary and is given by
2
Np
RL
rL =
Ns
(20.97)
where, Np is the number of primary turns ratio and Ns is the number of secondary
turns ratio.
20.9.1 Efficiency calculation
DC power calculation
Since two transistors are conducting alternatively, the average value of current is
considered to determine the DC power obtained from the supply.
Pdc = VCC Idc
⎡
Pdc
Where, Idc
⎤
T/
⎥
⎢1 2
⎥
⎢
(ic1 (t) + ic2 (t)) dt⎥
= VCC ⎢
⎦
⎣T
T
− /2
T/
1 2
(ic1 (t) + ic2 (t)) dt
=
T T
− /2
(20.98)
(20.99)
20.32
Electron Devices and Circuits
IC(mA)
IC(p)
iC
iC2
iC1
p
t
2p
T
Fig. 20.19 Collector currents of class-B power amplifier
We know the average value of a unidirectional waveform is
maximum or peak value. Therefore,
Idc =
Ic(p) =
Where,
2Ic(p)
π
VCC
rL
2Im
π ,
where Im is the
(20.100)
(20.101)
Substitute equation (20.101) in equation (20.100)
Idc =
2VCC
πrL
(20.102)
Substitute equation (20.102) in equation (20.98)
2V 2
2VCC
= CC
Pdc = VCC
πrL
πrL
(20.103)
AC power calculation
The AC power transferred to the load is given by
Since Ic(p) =
Ic2
RL
2
2
Ic(p)
rL
=
2
Pac =
(20.104)
Pac
(20.105)
VCC
rL
Pac =
VCC
rL
2
2
rL =
2
VCC
2rL
(20.106)
Power Amplifier
20.33
Knowing Pdc and Pac , efficiency can be determined from the formula,
Pac
=
η=
Pdc
η=
2
VCC
2rL
2
2VCC
πrL
2
VCC
πrL
π
×
= = 78.5%
2
2rL
4
2VCC
(20.107)
From the equation (20.106), it is clear that the efficiency of a class-B push-pull
power amplifier is more than that of a class-A power amplifier.
20.9.2 Cross over distortions
In a class-B push-pull amplifier, the transistors are biased at cut-off. To make the
transistor conduct, the input at the base of the transistor should be greater than the
cut-in voltage of the base-emitter junction of the transistor. Thus only on exceeding
the cut-in voltage, a base current flows in the input junction to result in a collector
current at the output junction.
Vo
Q1ON
Q1ON
Q2ON
Q1ON
Q2ON
Cross over distortion
Fig. 20.20 Cross over distortion in class-B push pull amplifier
When an AC input is applied to a class-B push-pull amplifier and as long as
input voltage is less than cut-in voltage, the transistors do not conduct and there is
no output. When input voltage is greater than cut-in voltage, the transistors conduct
to respond to the input in an amplified form at the output. This results in a distorted
output. These distortions that occur at the zero cross-overs when input voltage is
less than cut-in voltage are known as cross-over distortions. These distortions can be
overcome if a small amount of current is allowed to flow through the transistor under
zero input conditions. In other words, the transistors have to be biased slightly above
cut-off, so that the transistors are already in the conducting state. Now if an AC input
is applied the transistors conduct for the full cycle and produce an undistorted output
waveform. This mode of operation where the transistors are biased above cut-off is
known as class-AB power amplifier, which is dealt in detail in the next section.
20.34
Electron Devices and Circuits
20.9.3 Direct coupled push-pull class-B amplifier
Fig. 20.21 Direct Coupled Push-pull class-B amplifier
The Fig.20.21 shows a direct coupled push pull class-B amplifier where the load
RL is directly connected to the circuit. The circuit shows an input transformer which
acts as a phase splitter.
Vi
+
+
+
t
–
–
–
V01
+
+
+
t
V02
t
–
–
–
V0
+
+
+
t
–
–
–
Fig. 20.22 Sequence of operation of direct coupled push-pull class-B amplifier
Power Amplifier
20.35
This transformer provides the input to the transistors with equal amplitude and
opposite phase. Hence the transistors conduct alternatively for every half cycle. The
transistors Q1 and Q2 are connected in such a way that the output is observed at the
emitter of Q1 and collector of Q2 . Hence when Q1 conducts the output is in phase
with the input and when Q2 conducts the output is 180◦ out of phase with the input.
These two voltages when combined result in an alternating voltage across load RL .
The sequence of operation is explained in Fig.20.22.
During the positive half cycle, the transistor Q1 is forward biased and transistor
Q2 is reverse biased. Hence transistor Q1 conducts to reproduce the signal as such
at its emitter. During the negative half cycle, the transistor Q2 is forward biased and
transistor Q1 is reverse biased. Hence transistor Q2 conducts to reproduce the signal
with a phase of 180◦ at its collector. When both these outputs are combined we get a
sinusoidal output across the load.
Advantages
1. No output transformer is required
2. Due to the elimination of transformer, the cost of the circuit reduces drastically
3. Circuit become compact
Disadvantages
1. Need of two power supplies (±VCC )
2. It is not possible to obtain maximum power transfer to the load
3. Load isolation is not available
20.9.4 Complementary symmetry push-pull class-B amplifier
A complementary symmetry push-pull class-B amplifier overcomes the need for a
transformer both at the input and output. The circuit makes use of the two transistors
one npn and other pnp. The transistors have to be symmetrical which means they
have to be made of same semiconductor material and all the transistor parameters
have to be the same. Since the transistors are symmetrical and complementary of
each other the circuit is known as complementary symmetry push-pull amplifier.
Referring to the circuit shown in Fig.20.23, transistor Q1 is an npn type and
transistor Q2 is a pnp type. Both the transistors are operated in CC configuration
where the input is given to the base and output is observed at the emitter. The
collectors are connected to the supply voltage. Since the transistors are operated
in CC configuration, the voltage gain Av ∼
= 1 and hence the output at the emitter
follows the input at the base. That is why this circuit is also known as emitter
follower class-B push-pull amplifier. The resistors R1 and R2 act as a voltage
divider network to bias the transistor at cut-off.
The AC input is applied at the base of the transistors. During the positive half
cycle, the npn transistor Q1 is forward biased and pnp transistor Q2 is reverse biased.
20.36
Electron Devices and Circuits
+VCC
Ci1
R1
Q1
R2
RS
+
R2
VS ~
RL
Q2
Ci2
VO
R1
–
–VCC
Fig. 20.23 Complementary symmetry push-pull class-B amplifier
Hence transistor Q1 conducts above cut-in voltage to reproduce the positive half cycle
at the output. Similarly, during the negative half cycle, the pnp transistor Q2 is
forward biased and the npn transistor Q1 is reverse biased. Hence pnp transistor Q2
conducts above cut-in voltage and reproduces the negative half cycle at the output.
Thus the AC input is reproduced at the output with cross-over distortion which can
be over come by biasing the transistors slightly above the cut-off. The efficiency of
the class-B complementary symmetry circuit is 78.5%.
Advantages
1. Avoids the need for both input and output transformers. Hence the circuit is
cheap and compact.
2. Since either of the transistors conducts for every half cycle, the output
impedance of the circuit is low. Hence it finds use in low impedance circuit.
Example, the output stage of op-amp.
Disadvantages
1. Requires two biasing or supply voltage
2. If the transistors are not identical, even harmonics will not be cancelled. Hence
the distortion will be high.
20.10 Class-AB Push-Pull Amplifier
In a class-AB push-pull amplifier, a voltage divider network R1 and R2 is included
to provide the base bias for both the transistors so that they conduct under no signal
condition as shown in Fig.20.24.
Power Amplifier
20.37
Ic1
IL
T1
Q1
T2
+
R1
Vi
+
~
R2
–
R L VO
VCC
–
Q2
Ic2
Fig. 20.24 Class-AB power amplifier
The bias provided to the transistor keeps them just above cut-off so that the
transistors conduct for more than half a cycle and less than full cycle. That is why,
they are said to operate in class-AB mode. Since they conduct under zero signal
condition, they overcome the problem of cross-over distortion. The disadvantages of
a class-AB amplifier in comparison with a class-B amplifiers are
1. Wastage of standby power across the transistors
2. Low conversion efficiency compared to a class-B amplifier
20.10.1 Capacitor coupled class-AB push-pull amplifier
Fig. 20.25 Capacitor coupled class-AB power amplifier
20.38
Electron Devices and Circuits
The capacitor coupled class-AB push-pull amplifier is shown in Fig.20.24. The
input transformer coupling is replaced by the capacitor coupling. The resistors R1
and R2 form the voltage divider biasing network for both the transistors. Due to the
absence of input coupling transformer, the cost of the circuit reduces. The effective
area occupied by this configuration is relatively less than the transformer coupled
class-AB amplifier. The above configuration suffers from input coupling problem for
maximum signal transfer.
20.10.2 Phase splitter circuit for push-pull amplifier
One of the major disadvantages of a push-pull amplifier is the use of transformers
both at the input and output stages. The transformers increase the cost and bulkiness
of the circuit. We know that transformers are mainly used as phase splitters so that
the transistors are made to conduct alternatively for every half cycle at the input. The
simplest solution for this problem is to replace the transformer by a transistor phase
splitter circuit as shown in Fig.20.25.
+VCC
R1
RC
VO1
Ci
Vi
~
VO2
R2
RE
Fig. 20.26 Phase splitter circuit for push-pull amplifier
The circuit shows a single stage CE amplifier with a voltage divider bias. The
amplifier is designed to offer a voltage gain of unity by proper choice of collector
resistor RC and emitter resistor RE . The outputs are observed across the emitter and
collector. The output at the emitter will follow the input at the base and so it is a
in-phase signal, that is, during the positive half cycle of the input, the output at the
emitter is also positive and during the negative half cycle, the output at the emitter is
also negative.
The output at the collector is 180◦ out of phase with the input. For example,
during the positive half cycle the base bias is increased and so base current IB and
Power Amplifier
20.39
collector current IC increase. Since collector current IC increases, the collector
voltage VC = VCC − IC RC decreases. Thus V01 decreases with increase in
input during positive half cycle and hence it is out of phase with the input. Same
explanation holds good for negative half cycle. Thus the CE amplifier acts as a
phase splitter providing in phase voltage at its emitter and out of phase voltage at its
collector.
20.11 Class-C Power Amplifier
Circuit diagram
+VCC
L
C
+
RS
+
CO
RF choke
+
–
RL
VO
~
–
–VBB
–
Fig. 20.27
The Fig.20.27 shows a Class-C amplifier. For a Class-C amplifier, the transistor
is biased deep at cut-off so that it conducts only for a short interval of time.
Since the voltage drop across the load resistance (RL ) is not sinusoidal, the
output signal contains many harmonic frequencies. For a practical circuit, a
parallel resonance circuit must be connected at the collector as shown in the Fig.
L and C act as a parallel resonance circuit in the ac analysis. The LC-tuned circuit
removes the harmonic frequencies other than the fundamental frequency (fo ).
A Class-C amplifier is primarily used for high-power and high frequency
application like a radio-frequency transmitter. The principle advantage of Class-C
amplifier over other amplifiers is that, it has very high efficiency.
A negative bias (VBB ) is connected at the base of the transistor in order to drive
the Q-point deep into cut-off. The radio frequency choke (RFC) inductance presents a
high impedance to the high-frequency input and thereby prevents the DC component
from shorting the ac input signal.
Load line graph
The maximum collector current IC(max) =
bility for the transistor.
VCC
RL
< IP (max) is the peak current capa-
20.40
Electron Devices and Circuits
LC
V
IC = CC
RL
DC load line
(assuming inductor has zero
resistance)
AC load line (slope = –1/RL)
VCC
VCE
Q
Fig. 20.28
Signal analysis of Class-C amplifier
IS(t)
Iscoswot
t
IC(t)
IC,max
IC(t)
0
–t
t
t
T
+VCC
VC(t)
t
VC,max
0
t
Fig. 20.29
In a Class-C amplifier, the transistor conducts for a duration τ to +τ and
cut-off for the remaining period in a cycle. The period of conduction repeats with
time T = 1/fo .
Bias supply power calculation
The average power dissipated due to bias supply (VCC ) is given by,
Pdc
1
=
T
+τ
VCC IC (t)dt
−τ
(20.108)
Power Amplifier
20.41
The collector current can be represented by a general equation,
Ic(t) = a cos ωo t + b
(20.109)
Where ‘a’ is the amplitude of the collector current and b is variation with respect
to input signal Is (t).
Apply the boundary condition to the equation (20.109).
Boundary condition-1
At t = 0, IC (t) = IC(max)
then
IC(max) = a + b
(20.110)
At t = τ, IC (t) = 0
then
0 = a cos ωo τ + b
From equation (20.110),
b = a cos ωo τ
(20.111)
Substituting the equation (20.111) in the equation (20.110), we get
IC(max) = a − a cos ωo τ = a(1 − cos ωo τ )
IC(max)
Therefore,
a=
(1 − cos ωO τ )
(20.112)
Substitute the equation (20.1) in the equation (20.111),
b=
−IC(max)
cos ωo τ
(1 − cos ωo τ )
(20.113)
Substitute the equation (20.113) and the equation (20.112) into the equation (20.8),
then,
IC(max)
IC,(max)
IC (t) =
cos ωo t −
cos ωo τ
(1 − cos ωo τ )
(1 − cos ωo τ )
IC(max)
[cos ωo t − cos ωo τ ] · · · τ ≤ (t + nτo ) ≤ τ
IC t =
(20.114)
(1 − cos ωo τ )
Substitute the equation (20.114) into the equation (20.108) and we get
Pdc
1
=
T
+τ
VCC IC(max)
−τ
cos ωo t − cos ωo τ
dt
(1 − cos ωo τ )
(20.115)
20.42
Electron Devices and Circuits
Using Taylor’s series, expand cos ωo τ
(ωo τ )2 (ωo τ )4 (ωo τ )6
+
+
+ ···
2!
4!
6!
The higher order is neglected as it is too small. Therefore,
cos ωo τ = 1 −
cos ωo τ ≈ 1 −
(ωo τ )2
2!
(20.116)
for |t| < τ ,
(ωo τ )2
(20.117)
2
Substitute the equation (20.116) and the equation (20.117) into the equation (20.115).
We get
cos ωo τ ≈
Pdc =
Pdc =
Pdc =
Pdc
VCC IC(max)
T
VCC IC(max)
T
VCC IC(max)
T
+τ
1−
−τ
+τ
−τ
+τ
−τ
(ωo t)2
2
−1+
1−1+
ωo τ 2
2
(ωo τ )2
2
dt
τ 2 − t2
dt
τ2
2 t
1−
dt
τ
4τ
VCC IC(max)
=
3T
(20.118)
Output power calculation
The Fourier expansion of collector current is given by,
IC (t) = Idc + In cos ωo nt
(20.119)
The fundamental component can be evaluated by setting n = 1 in the equation
(20.119). Other harmonic components are alternated by the parallel tank circuit.
IC (t) = Idc + I1 cos ωo t
Where,
1
I1 =
T /2
(20.120)
+T
/2
IC (t) cos ωo tdt
(20.121)
−T /2
Substitute the equation (20.114) into the equation (20.121),
1
I1 =
T /2
+T
/2
IC(max)
−T /2
cos ωo t − cos ωo τ
cos ωo tdt
1 − cos ωo τ
(20.122)
Power Amplifier
20.43
Substitute the equation (20.116) and the equation (20.117) into the equation
(20.122) and we get
2IC(max)
I1 =
T
+T
/2
(ωt)2
1−
dt
2
−T /2
On integration and simplification above equation is reduced to
2IC(max) 4τ
I1 ≈
T
3
The power delivered to the load is given by,
2IC(max) 4τ
RL
I12 RL
=
Pload =
2
T
3
2
32 τ 2 2
Pload =
IC (max) RL
9 T
(20.123)
Efficiency calculation
The efficiency,
η=
Pload
Pdc
32 τ 2
η = 49 Tτ 3
IC2 (max) RL
IC2 (max) VCC
2
32 τ 2 VC (max)
T
9
T
R
η = V2 L
4
τ
C (max)
3
η=
T
RL
8 τ VC(max)
3 T
VCC
VCC
(20.124)
If VC(max) = VCC ; τ = T /8
Then,
η = 33%
VC(max) = VCC ; τ = T /3
Then,
η = 88.9%
As η increases, the efficiency is also increases if VC(max) = VCC
Collector power calculation
1
PC =
T
+τ
IC (t)VCE (t)dt
−τ
(20.125)
Electron Devices and Circuits
20.44
From figure,
Vce (t) =
VCC − VC(max)
cos ωo t−cos ωo τ
(1−cos ωo τ )
· · · |t + nTo | < τo
(20.126)
VCC
Substitute the equation (20.126) into the equation (20.125). We get
1
PC =
T
1
PC =
T
Pdc =
Pdc
1
T
1
=
T
+τ
VCC IC(max)
−τ
+τ
IC(max)
−τ
cos ωo t − cos ωo τ
1 − cos ωo τ
2
2
+τ
cos ωo t − cos ωo τ
τ − t2
1
IC(max) VC(max)
dt
dt −
1 − cos ωO τ
T
τ2
−τ
+τ
IC(max)
−τ
cos ωo t − cos ωo τ
1 − cos ωo τ
+τ
IC(max) VC(max)
−τ
cos ωo t − cos ωo τ
VCC − VC(max)
dt
1 − cos ωo τ
τ 2 − t2
τ2
2
dt
2
dt
16 τ IC,(max) VC(max)
PC = Pdc −
15
T
4
16
τ
τ
VCC IC(max) −
IC(max) VC(max)
PC =
3
T
15
T
VC(max)
4
4
τ
PC =
VCC −
VC(max)
3
T
RL
5
(20.127)
Figure of merit
F =
PC(max)
dPC
dVC
PC(max)
PL
dPC
=
=0
dVC(max)
τ VC C
τ
4 2VC(max)
4
4
=
−
3
T RL
3
T
5
RL
8
VC(max) = 0
VCC −
5
5VCC
VC(max) =
8
Substitute the equation (12.128) into the equation (12.127),
2
5 τ VCC
then
PC(max) =
12 T RL
(20.128)
(20.129)
Power Amplifier
20.45
Then, figure of merit is given by,
2 5/ τ/ VCC
12
T
2 RL
=
F =
Pdc
τ/ 2 VCC
32/
T
9
RL
15 T
/τ
F =
128
PC(max)
For τ = T /8, then F = 15/16 = 0.9375
τ = T /3, then F = 0.35
This shows that higher the efficiency of the amplifier, the figure of merit reduces.
20.12 Class-D Amplifier
The Class-D amplifier is also a non-linear power amplifier like that of a Class-C
amplifier. The transistor operates as a switch and consequently the power dissipation
is almost zero.
In the case of Class-D amplifier, the input signal is a series of pulses having
widths which correspond to the amplitude of the input signal as shown in the Fig.20.30
which in turn is a typical operation of a pulse-width modulator system.
Saw-tooth
generator
Vi
Modulator
VO
t
Fig. 20.30
For the low level signal, the pulse width is small and for the high level signal, the
pulse width is large. The amplitude of the saw-tooth must exceed the maximum input
signal amplitude.
The average power of the output signal changes with the input signal. Therefore,
average is high for large signal and it is small for small signal.