In the case of class A amplifier, we have observed that the transistor conducts for the full cycle of the input signal i.e. the conduction angle is 180◦. Although the transistor conducts for the full cycle of the input signal, the power conversion efficiency is poor in class A amplifier. In addition to that, a great deal of distortion is introduced by the nonlinearity in the dynamic transfer characteristic of the transistor. The power conversion efficiency can be improved by biasing the transistor at cut off point on VCE axis and a great deal of the distortion due to nonlinearity in dynamic transfer characteristic may be eliminated by the push-pull configuration of the transistor as discussed in next section

1. Unit-2 : Large Signal Amplifiers
Vikas R. Gupta
Assistant Professor
Department of Electronics Engineering
Shri Ramdeobaba College of Engineering and Management, Nagpur-13.
guptavr1@rknec.edu
February 2, 2017

2. Chapter 1
Amplifiers
1

3. Chapter 2
Large Signal Amplifier
2.1 Introduction:
In case of class A amplifier, we have observed that the transistor conducts for
the full cycle of the input signal i.e. the conduction angle is 180◦
. Although
the transistor conducts for full cycle of the input signal, the power conversion
efficiency is poor in class A amplifier. In addition to that, a great deal of
distortion is introduced by the nonlinearity in dynamic transfer characteristic
of the transistor. The power conversion efficiency can be improved by biasing
the transistor at cut off point on VCE axis and a great deal of the distortion
due to nonlinearity in dynamic transfer characteristic may be eliminated by
the push-pull configuration of the transistor as discussed in next section.
2.2 Push-Pull Configuration:
In ideal conditions, the static output characteristics are equidistant for equal
increment of input excitation and hence the dynamic transfer characteristic
is assumed to be linear, because of this assumption the output waveform will
resemble the input waveform.
But in practice this assumption is not valid. Hence it may introduce
distortions in the output signal. The distortion introduced by nonlinearity
in dynamic transfer characteristic may be eliminated by the circuit shown in
figure 2.1, known as push-pull configuration .
In this circuit the input signal is introduced through a center-tapped
transformer where two equal voltages which differ in the phase by 180◦
is
produced across the secondary winding. Thus, when the signal at base ter-
minal of the transistor Q1 is positive, the signal at base terminal of the
transistor Q2 is negative by an equal amount. Basically the input center-
2

4. Large Signal Amplifier Unit-2
Figure 2.1: Two transistors in a push-pull arrangement.
tapped transformer is used here as a phase splitter. Therefore any other
circuit that provides two equal voltages which differ in phase by 180◦
may
be used in place of the input transformer.
Let us consider an input signal of the form
ib1 = Ibmcosωt
applied to transistor Q1. We know that the output current (i.e. total instan-
taneous collector current) of the transistor is given by
i1 = iC1 = IC + B0 + B1cosωt + B2cos2
ωt + B3cos3
ωt + B4cos4
ωt + ... (2.1)
Where B0, B1, B2, B3, B4, ... are constants determined by the nonlinearity of
the transistor. From equation 2.1, it is clear that apart from the fundamen-
tal frequency (i.e. input frequency) ω, certain higher order terms given by
2ω, 3ω, 4ω, ..., are also available in the output signal iC1 with respect to the
input signal. this type of distortion is called as harmonic distortion and this
should be eliminated.
The corresponding inp ut signal to Q2 is
ib2 = −ib1 = Ibmcos(ωt + π) (2.2)
Thus, the output current of the transistor Q2 is obtained by replacing ωt by
(ωt + π) in equation 2.1. That is,
iC2 (ωt) = iC1 (ωt + π) (2.3)
hence,
i2 = iC2 = IC + B0 + B1cos(ωt + π) + B2cos2
(ωt + π) + B3cos3
(ωt + π) + ...
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 3

5. Large Signal Amplifier Unit-2
which reduces is
i2 = iC2 = IC + B0 − B1cosωt + B2cos2
ωt − B3cos3
ωt + B4cos4
ωt + ... (2.4)
As shown in figure 2.1, the current i1 and i2 flow in opposite direction
through the primary winding of the output transformer. Therefore the to-
tal output current i is proportional to the difference between the collector
currents in the two transistors. That is,
i = k(i1 − i2) = k(iC1 − iC2 ) = 2k(B1cosωt + B3cos3
ωt + ...) (2.5)
The above equation 2.5 shows that a push-pull circuit will eliminate all even
harmonics in the output and will leave the third harmonic term as the prin-
cipal source of distortion. In order to achieve this both the transistor must
be identical.
Since the output current contains no even harmonic terms the push-pull
system is said to have ”half-wave,” or ”mirror,” symmetry in addition to the
zero axis symmetry. The condition for mirror symmetry is mathematically
given by the following relation
i(ωt) = −i(ωt + π) (2.6)
Advantages
1. As no even harmonics are present in the output of a push-pull amplifier,
such a circuit provides less distortion for a given power output per
transistor.
2. The dc components of the collector current oppose each other magneti-
cally in the transformer core, thereby eliminates any tendency towards
core saturation which leads to nonlinear distortion.
3. The effects of ripple voltages contained in the power supply because of
inadequate filtering will be balanced out in push-pull configuration.
.
Disadvantages
1. The power supply hum will not be eliminated by the push-pull circuit.
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 4

6. Large Signal Amplifier Unit-2
2.3 Class B Amplifier:
In a class B amplifier the transistor is biased almost at cut-off (i.e. the
operating point is selected at cut-off point), so that the transistor will conduct
only for half cycle of the input signal. Hence its conduction angle is 180◦
.
The circuit shown in figure 2.1 operates in class B mode if R2 = 0. To bias
the transistor in cut-off region, the base and emitter terminals of transistor
are shorted (i.e. VBE = 0). In Class B amplifier, it is possible to obtain
greater power output, higher efficiency, and negligible power loss at no input
signal. For these reasons class B amplifier is employed
• in applications where the power supply is limited, say, operating from
solar cells or battery.
• as a power stage (or output stage) in the audio power amplifiers.
In class B push-pull amplifier circuit transistor Q1 conducts during the
positive half cycle of the input signal and current i1 flows through the primary
winding of the output center-tapped transformer. Whereas in the negative
half cycle of the input signal transistor Q2 conducts and current i2 flows
through the primary winding of the output center-tapped transformer. Since
the current i1 and i2 flows in opposite direction in the primary winding of the
output center-tapped transformer, the output current in the secondary wind-
ing of the output center-tapped transformer is proportional to the algebraic
sum of the two currents. That is,
i = k(i1 − i2)
where, k is the turns ratio of the transformer.
.
Power Consideration and Derivation of Efficiency
To investigate the conversion efficiency of Class B amplifier, let us assume
that:
1. The output characteristics are equally spaced for equal increments in
the input excitation.
2. The dynamic transfer curve is a straight line.
3. The minimum collector current is zero due to transistor biasing at cut-
off point.
4. The two transistors Q1 and Q2 are identical.
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 5

7. Large Signal Amplifier Unit-2
Figure 2.2: Graphical construction for determining the output waveforms of
a single class B transistor stage.
.
The Input Power Pdc:
Pdc = VCCIdc (2.7)
Since the transistor is biased at cut-off point, the dc collector current due
to the power supply VCC will be zero. But due to the rectification of the
input signal in each transistor there will be the flow of some dc current and
it will be equal to the average value of the half sine loop shown in figure 2.4.
Therefore total dc current will be given by
Idc = Iavg1 + Iavg2
where, Iavg1 and Iavg2 is the average value of the output collector current in
transistor Q1 and Q2 respectively.
Iavg1 = Iavg2 =
Im
π
Therefore,
Idc = 2
Im
π
Thus, the input power is
Pdc = 2
ImVCC
π
(2.8)
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 6

8. Large Signal Amplifier Unit-2
.
The Output PowerPac:
Pac =
ImVm
2
=
Im
2
(VCC − Vmin) (2.9)
.
The Collector Circuit Efficiency η :
η =
Pac
Pdc
(2.10)
=
VmIm
2
/
2ImVCC
π
(2.11)
=
π
4
Vm
VCC
(2.12)
=
π
4
VCC − Vmin
VCC
Vm = VCC − Vmin (2.13)
∴ % η =
π
4
1 −
Vmin
VCC
× 100% (2.14)
.
Maximum Collector Circuit Efficiency ηmax :
The equation 2.14 shows that the maximum efficiency can be obtained when
Vmin << VCC, and therefore the maximum efficiency will be
% ηmax =
π
4
× 100% = 25π = 78.5% (2.15)
This large value of η results from the fact that there is no current in class B
amplifier circuit if there is no input signal (i.e. excitation), whereas in class
A amplifier circuit the dc current ICQ drawn from the power supply flows
through the collector circuit even if the input signal is zero.
.
The Power Dissipation PC (in both transistors) :
It is the difference between the input power to the collector circuit and
the power delivered to the load.
PC = Pdc − Pac (2.16)
=
2ImVCC
π
−
VmIm
2
(2.17)
=
2VmVCC
πRL
−
V 2
m
2RL
Im =
Vm
RL
(2.18)
The equation 2.18 shows that the power dissipation in both the transistor
is zero at no signal (Vm = 0), The power dissipation increases as Vm increases.
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 7

9. Large Signal Amplifier Unit-2
To find the maximum power dissipation we need to differentiate equation
2.18 with respect to Vm and equating it to zero.
∂PC
∂Vm
= 0 =
∂
∂Vm
2VmVCC
πRL
−
V 2
m
2RL
(2.19)
0 =
2VCC
πRL
−
2Vm
2RL
(2.20)
∴ Vm =
2VCC
π
(2.21)
If we substitute the value of Vm obtained in equation 2.21 into equation 2.18,
we will get the maximum power dissipation in class B amplifier,
PC(max) =
2V 2
CC
π2RL
(2.22)
Maximum Power Delivered to the Load Pac(max):
The maximum power which can be delivered to the load is obtained when
Vm = VCC(ifVmin = 0)
Pac(max) =
V 2
CC
2RL
(2.23)
Hence,
PC(max) =
4
π2
Pac(max) ≈ 0.4Pac(max) (2.24)
.
Distortion in Class B push-pull amplifier
From the derivation of total harmonic distortion, we have
B0 =
1
6
Imax + 2I1
2
+ 2I−1
2
+ Imin − IC (2.25)
B1 =
1
3
Imax + I1
2
− I−1
2
− Imin (2.26)
B2 =
1
4
(Imax − 2IC + Imin) (2.27)
B3 =
1
6
Imax − 2I1
2
+ 2I−1
2
− Imin (2.28)
B4 =
1
12
Imax − 4I1
2
+ 6IC − 4I−1
2
+ Imin (2.29)
We know that, the output of a push-pull configuration always possesses mir-
ror symmetry as explained in section 2.2, hence
IC = 0, Imax = −Imin, and I1
2
= −I−1
2
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 8

10. Large Signal Amplifier Unit-2
. Under these circumstances, equation 2.25 to 2.29 reduces to
B0 = B2 = B4 = 0 (2.30)
B1 =
2
3
Imax + I1
2
(2.31)
B3 =
1
3
Imax − 2I1
2
(2.32)
From equation 2.30 to 2.32 , we can note that there is no even-harmonic
distortion and the principal contribution to distortion is due to the third
harmonic i.e. B3. Therefore, distortion due to third harmonic is given by
D3 =
|B3|
|B1|
and the power due to fundamental component B1 is given by
P1 =
B2
1RL
2
and the power output, taking distortion into account is given by
P = 1 + D2
3
B2
1RL
2
Note: In order to find the values of Imax and I1
2
follow the procedure
as given below:
1. Draw a load line corresponding to RL = (N1/N2)2
RL on the
collector characteristics through the point IC = 0 and VCE =
VCC.
2. If the peak base current is IB then the intersection of the load line
with the IB curve will give Imax and with the IB/2 characteristics
is I1
2
, as shown in figure 2.4.
.
Advantages:
The advantages of class B push-pull amplifier as compared with class A
amplifier are:
1. It provides larger output power.
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 9

11. Large Signal Amplifier Unit-2
2. Efficiency of Class B amplifier is higher (ideally 78.5%) than class A
amplifier.
3. There is negligible power loss (as no collector current flows) at no input
signal.
4. Even harmonics are balanced out due to push pull configuration of
transistor.
5. Ripples in supply voltages are eliminated.
.
Disadvantages The disadvantages of class B push-pull amplifier are:
1. Use of center-tapped transformer at input as well as output makes the
circuit bulky and expensive.
2. Frequency response of class b amplifier is poor as compared to class A
amplifier.
3. The mismatch in the characteristics of the two transistors (i.e. Q1 and
Q2) and center-tapped transformer may create more severe harmonic
distortion.
4. Crossover distortion is introduced in the output signal due to nonlinear
input characteristics of the transistors (i.e. Q1 and Q2).
2.4 Special Circuits for Class B Amplifier:
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 10

12. Large Signal Amplifier Unit-2
Figure 2.3: A class B push-pull circuit which does not use an output trans-
former.
Figure 2.4: A push-pull circuit using transistors having complementary sym-
metry.
V. R. Gupta, Asst. Prof., Electronics Engg., RCOEM, Nagpur-13. 11