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CLASS B AND CLASS AB PUSH-PULL AMPLIFIERS
When an amplifier is biased at cut-off such that it operates in the linear region for 180o of the
input cycle and is in cut-off for 180o it is a class B amplifier. Class AB amplifiers are biased
to conduct for slightly more than 180o. The primary advantage of a class B or class AB
amplifier over a class A amplifier is that they are more efficient; you can get more output
power for a given amount of input power. A disadvantage of class B or class AB is that it is
more difficult to implement the circuit in order to get a linear reproduction of the input
waveform. As you will see in this section, the term push-pull refers to a common type of class
B or class AB amplifier circuit in which the input wave shape is reproduced at the output.
After completing this section, you should be able to
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Explain and analyse the operation of class B and class AB amplifiers
Explain class B operation
Discuss Q-point location for class B amplifiers
Describe class B push-pull operation
Explain crossover distortion and its cause
Explain class AB operation
Analyse class AB push-pull amplifiers
Calculate maximum output power
Calculate dc input power
Determine class B maximum efficiency
Find the input resistance
Analyse a Darlington push-pull amplifier
Discuss methods of driving a push-pull amplifier
Class B Operation
The class B operation is illustrated in Figure 18, where the output waveform relative to the
input in terms of time (t).
The Q-Point is at Cut-off The class B amplifier is biased at the cut-off ICQ and VCEQ = VCE(cutoff). It is brought out of cut-off and operates in its when the input signal drives the transistor
into conduction. This is illustration 19 with an emitter-follower circuit where, obviously, the
output is not a input. To get a sufficiently good reproduction of the input waveform, a
configuration, known as a push-pull amplifier, is necessary.
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Push-Pull Class B Operation Figure 20 shows one type of push-pull class B amplifier using
two emitter-followers. This is called a complementary amplifier because one emitter-follower
uses an npn transistor and the other a matched pnp,the transistors conduct on opposite
alternations of the input cycle. A matched complementary pair of transistors have identical
characteristics, except one is an npn and the other a pnp. The 2N3904 and 2N3906 are
examples. Notice that there is no dc base bias voltage (VB = 0); thus, only the signal voltage
drives the transistors into conduction. Q1 conducts during the positive half of the input cycle,
and Q2 conducts during the negative half
Crossover Distortion When the dc base voltage is zero, both transistors are off and the input
signal voltage must exceed VBE before a transistor conducts. Because of this, there is a time
interval between the positive and negative alternations of the input when neither transistor is
conducting, as shown in Figure 21. The resulting distortion in the output waveform is quite
common and is called crossover distortion.
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Class AB Operation
To eliminate crossover distortion, both transistors in the push-pull arrangement must be
biased slightly above cut-off when there is no signal. This variation of the class B push-pull
amplifier is designated as class AB. Class AB biasing can be done with a voltage-divider
arrangement, as shown in Figure 22(a). It is, however, difficult to maintain a stable bias point
with this circuit due to changes in VBE with temperature. The requirement for dual-polarity
power supplies is eliminated when RL is capacitively coupled more suitable arrangement is
shown in Figure 22(b). When the diode characteristic D1 and D2 are closely matched to the
transconductance characteristics of the transistor stable bias can be maintained over
temperature. This stabilisation can also be accomplished by using the base-emitter junctions
of two additional matched transistors instead of D1 and D2. Although technically incorrect,
class AB amplifiers are often referred t class B in common practice.
The dc equivalent circuit of the push-pull amplifier is shown in Figure 23. Resistors R1 and R2
are of equal value; therefore the voltage at point A between the two diodes
is VCC/2. Assuming that the transconductance characteristics of the diodes and the transistors
are identical, the drop across D1 equals the VBE of Q1 and the drop across D2 equals the VBE
of Q2. As a result, the voltage at the emitters is also Vcc/2, and therefore, VCEQ1 = VCEQ2 =
VCC/2, as indicated. Because both transistors are biased near cut-off,
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Example 7 Determine the dc voltages at the bases and emitters of the matched
complementary transistors Qt and Q2 in Figure 24. Also determine VCEQ for each transistor.
Assume VD1 = VD2 = VBE = 0.7 V and ICQ = 0.
Solution The equivalent for the bias circuit is shown in Figure 25(a) where VD1 and VD2
represent the barrier potential of the diodes. The total current through the bias circuit (R1, D1,
D2, and R2) is
IT
VCC - VD1 - VD2 20V - l.4V

 93mA
200
200
VB1  Vcc I T R 1  20 V - (93 mA)(l00 )  10.7 V
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and
VB2  VB1 - VD1 - VD2  10.7 V - 1.4 V  9.3 V
VE1  VE2  10.7 V - 0.7 V  10 V
Therefore,
VCEQ1  VCEQ2 
Vcc 20V

 10 V
2
2
These values are shown in figure 25(b)
AC Operation Under maximum conditions, transistors Q1 and Q2 in an amplifier are
alternately driven from near cut-off to near saturation. During the positive alternation of the
input signal, the Q1 emitter is driven from its Q-point value of VCC/2 near Vcc, producing a
positive peak voltage approximately equal to VCEQ. At the time, the Q1 current swings from
its Q-point value near zero to near-saturation value as shown in Figure 26(a).
During the negative alternation of the input signal, the Q2 emitter is driven from its Q-point
value of Vcc /2 to near zero, producing a negative peak voltage approximately equal to V CEQ.
Also, the Q2 current swings from near zero to near-saturation value as shown in Figure 26(b).
In terms of the ac load line operation, the Vce of both transistors swings from VCC/2 to near
zero, and the current swings from zero to IC(SAT) as shown in Figure 26 c. Because the peak
voltage across each transistor is VCEQ, the ac collector saturation current is
I C(SAT ) 
VCEQ
RL
Since I E  I C and the output current is the emitter current, the peak output current is also
VCEQ /RL.
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Figure 26 Ideal ac push-pull operation for maximum signal operation
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Determine the maximum ideal peak values for the output voltage and current in Figure 27.
Solution The maximum peak output voltage is
Vout(peak)  VCEQ 
Vcc 20V

 10V
2
2
The maximum peak output current is
VCEQ
lOV
 1.25A
RL
8
Related Exercise Find the maximum peak values for the output voltage and current in Figure
27 if VCC is lowered to 15 V and the load resistance is changed to 16 .
I out(peak)  I C(sat) 

Maximum Output Power It has been shown that the maximum peak output current
approximately IC(sat) and the maximum peak output voltage is approximately VCEQ maximum
average output power is, therefore,
Pout  VOUT (rms)I out(rms)
Since
Vout(rms)  0.707 Vout(peak)  0.707VCEQ
and
I out(rms)  0.707 I out(rms)  0.707I C (sat )
then
Pout  0.5VCEQ I C(sat )
Substituting Vcc /2 for VCEQ,
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Pout  0.25VCC I C(sat )
DC Input Power The dc input power comes from the VCC supply and is
PDC  VCC I CC
Since each transistor draws current for a half-cycle, the current is a half-wave signal with an
average value of
I CC 
I C (SAT )

So,
PDC 
VCC I C( SAT)

Efficiency An advantage of push-pull class B and class AB amplifiers over class A is a much
higher efficiency. This advantage usually overrides the difficulty of biasing the class AB
push-pull amplifier to eliminate crossover distortion. Recall that efficiency is defined as the
ratio of ac output power to dc input power.
Efficiency 
Pout
PDC
The maximum efficiency for a class B amplifier (class AB is slightly less) is designated max
and is developed as follows.
Pout  0.25VCC I C(sat )

0.25VCC I C(sat )
Pout
 Pout 
 0.25
VCC I C( SAT)
PDC

 max  0.79
or, as a percentage,
max = 79%
Recall that the maximum efficiency for class A is 0.25 (25 percent).
Input Resistance The complementary push-pull configuration used in class B/class AB
amplifiers is, in effect, two emitter-followers. The input resistance is, therefore
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R in   ac (r e  R E )
Since RE = RL, the formula is
R in   ac (r e  R L )
Find the maximum ac output power and the dc input power of the amplifier in Figure 28.
Also, determine the input resistance assuming ac = 50 and r e  6
FIGURE 28
Solution IC(sat) for this same circuit was found to be 1.25 A in Example 8.
Pout  0.25Vcc I c(sat)  0.25(20 V)(1.25 A)  6.25W
PDC 
Vcc I c(sat)


(20 V)(1.25 A)
 7.96W

R in   ac (r e  R L )  50(6  8)  700
Related Exercise Determine the maximum ac output power and the dc input power
in Figure 28 for Vcc = 15 V and RL = 16 
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Darlinglon Class AB Amplifier
In many applications where the push-pull configuration is used, the load resistance is relatively small. For example, an 8 speaker is a common load for a class AB push-pull
amplifier.
As a result of low-resistance loading, push-pull amplifiers can present a quite low input
resistance to the preceding amplifier that drives it. Depending on the output resistance of the
preceding amplifier, the low push-pull input resistance can load it severely and significantly
reduce the voltage gain. As an example, if the complementary transistors in a push-pull
amplifier exhibit an ac beta of 50 and the load resistance is 8, the input resistance
(assuming re = 5 ) is
 R in (r e  R L )  50(5   8 )  650 
If the collector resistance of the driving amplifier is, for example, 1 k, the input resistance
of the push-pull amplifier reduces the effective collector resistance of the driving amplifier
(assuming a common-emitter) to Ro= Rc //Rin = 1 k ||650  = 394 . This drastically
reduces the voltage gain of the driving amplifier because its gain is Ro/ r/e.
In certain applications with low-resistance loads, a push-pull amplifier using Darlington
transistors can be used to increase the input resistance presented to the driving amplifier and
avoid severely reducing the voltage gain. The ac beta of a Darlington pair is generally in
excess of a thousand.
In the previous case, for example, if ac = 50 for each transistor in a Darlington pair, the
overall ac beta is ac = (50)(50) = 2500. The input resistance is greatly increased, as the
following calculation shows.
 R in (r e  R L )  2500(5   8 )  32.5 k
A Darlington class AB push-pull amplifier is shown in Figure 29. Four diodes are required in
the bias circuit to match the four base-emitter junctions of the two Darlington pairs.
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Push-Pull Amplifier with Class A Driver
Up to this point, the input signal to class B and class AB push-pull amplifiers has been
capacitively coupled from the input source to each of the two transistors. Another method is
to incorporate a class A stage as the driver, as shown in Figure 30. As you can see, the driver
is a voltage-divider-biased amplifier.
The dc bias voltages for the push-pull amplifier are established by the dc collector current of
the driver transistor Q3. When an input signal is coupled to the base of Q3, the resulting ac
collector current of Q3 produces signal voltages at the bases of Q1 and Q2, as shown in Figure
30. The ac voltage dropped across the dynamic resistances of the bias diodes is negligible, so
the Q1 and Q2 base signals are essentially equal. The reason for this is that the dynamic
resistances of the diodes are very small compared to R1 and, for all practical purposes, the
diodes can be considered shorts to the ac signal. This driver circuit has proved to be an
effective way to interface the push-pull amplifier with a preamplifier circuit that precedes it.
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