Class B Power Amplifier
Class B Amplifier Operation
Class B
The amplifier conducts through 180 of the input.
The Q-point is set at the cutoff point.
Output only conducts for 180 or one-half of the AC input signal.
The Q-point is at 0V on the load line, so that the AC signal can only swing for one-half cycle.
To obtain output for the full cycle of signal, it is necessary to use
two transistors.
Since one part of the circuit pushes the signal high during one half-
cycle and the other part pulls the signal low during the other half-
cycle. The circuit is referred to as a push-pull circuit.
A pnp transistor that provides the positive half.
An npn transistor that provides the negative half of the AC cycle
Class B Amplifier Efficiency
The maximum efficiency of class B amplifier is 78.5%.
The power supplied to the load is drawn from the power supply,
then
where Idc is the average current expressed as
The power delivered to the load can be calculated by either one of
these equations
( ) dcdci CCP V I=
Peak value of output current
( )( )
2 rmsac L
o
L
VP
R= ( )
( ) ( )2 2p-p pac
8 2
L Lo
L L
V VP
R R= =
)(
2pIV LCC
)(2
pII Ldc
Efficiency
Power dissipated by output resistors
( ) ( )2 dc acQ i oP P P= -
Power dissipated by the two output power transistors
2
2
Q
Q
PP =
Power dissipated by each output power transistor
%100)(
4%100
)(
)(% 0
CC
L
i V
pV
dcP
acP
Example
For a class B amplifier providing a 20V peak signal to a 16 load
and a power supply of VCC = 30V, determine the input power,
output power, and circuit efficiency.
A peak of load current is
The dc value of the current drawn from the power supply is
( )pLV LR
( )( )p
p LL
L
VI
R= = A
V25.1
16
20
AApII Ldc 796.0)25.1(2
)(2
The input power delivered by the supply voltage is
The output power delivered to the load is
Hence, the efficiency of this circuit is
( ) dcdci CCP V I= ( )( )30V 0.796A 23.9W= =
( )( )2 p
ac2
Lo
L
VP
R= =
%100)(
)(% 0
dcP
acP
i
WV
5.12)16(2
)20( 2
%3.52%1009.23
5.12
W
W
Maximum power consideration
The maximum output power is when VL(p) = VCC
The corresponding peak ac current IL(p) is then
( )( )2 p
ac2
Lo
L
VP
R= ( )
2
maximum ac2
CCo
L
VP
R=
( )p CCL
L
VI
R=
( ) dcdci CCP V I=
L
CC
L
CCCCi
R
V
R
VVdcP
222)( maximum
L
CCL
R
VpII
2)(
2 maximum dc
The maximum circuit efficiency for class B is
The maximum power dissipation of class B is found to be
%100/2
2/
%100)(
)(%
2
2
0
LCC
LCC
i
RV
RV
dcP
acP
%54.78%1004
L
CCQ
R
VP
2
2
2
2 maximum
The maximum efficiency of a class B amplifier can also be
expressed as
Example VCC = 24V
VL(p) = 22V
VL(p) = 6V
( )( )2 p
ac2
Lo
L
VP
R=
)(
2)( pIVdcP LCCi
CC
L
i V
pV
dcP
acP )(%)54.78(%100
)(
)(% 0
%7224
22%)54.78(%
V
V
%6.1924
6%)54.78(%
V
V
Example
For a class B amplifier using a power supply of VCC = 30V and
driving a load of 16 load, determine the maximum input power,
output power and transistor dissipation.
The maximum output power is
The maximum input power is
( )
2
maximum ac2
CCo
L
VP
R= W
V125.28
)16(2
)30( 2
L
CCi
R
VdcP
22)( maximum W
V81.35
)16(
)30(2 2
The maximum circuit efficiency is
The maximum power dissipation of class B is found to be
%100)(
)(% 0
dcP
acP
i
%54.78%10081.35
125.28
W
W
L
CCQ
QR
VPP
2
22 2
5.02
maximum maximum
WV
7.5)16(
)30(25.0
2
2
Transformer-Coupled Push-Pull Circuit
The center-tapped transformer on the input produces opposite
polarity signals to the two transistor inputs.
The center-tapped transformer on the output combines the two
halves of the AC waveform together.
Complementary-Symmetry Push-Pull Circuit
During the positive half-cycle of the AC input, transistor Q1 (npn)
is conducting and Q2 (pnp) is off.
During the negative half-cycle of the AC input, transistor Q2 (pnp)
is conducting and Q1 (npn) is off.
Crossover distortion
If the transistors Q1 and Q2 do not turn on and off at exactly the
same time, then there is a gap in the output voltage.
Quasi-Complementary
Push-Pull Amplifier
A Darlington pair and a feedback pair combination perform the
push-pull operation. This increases the output power capability.
Example
Calculate the input power, output power, power handled by each
transistor, and circuit efficiency for an input of 12 Vrms.
For 12V rms input signal, the peak input voltage thus is
By assuming that the amplifier has ideally a voltage gain of unity,
then
and the output power across load is
( )p 17 VLV =
( )( )2 p
ac2
Lo
L
VP
R=
VVrmsVpV ii 17)12(2)(2)(
WV
125.36)4(2
)17( 2
In addition, the peak load current is
from which the dc current of the supply is
Hence, the input power is
AV
R
pVpI
L
LL 25.4
4
17)()(
AApII Ldc 71.225.42
)(2
WAVpIVIVdcP LCCdcCCi 75.67)71.2)(25()(2
)(
The power dissipated by each transistor thus is
The circuit efficiency for the 12 V rms input voltage is then
( ) ( )2 dc ac
2 2
Q i oQ
P P PP
-= =
( )67.75 36.125 W15.8 W
2
-= =
%100)(
)(% 0
dcP
acP
i
%3.53%10075.67
125.36
W
W
Homework
Determine (a) Maximum Po(ac), (b) maximum Pi(dc), (c)
Maximum %, (d) Maximum power dissipated by both transistors
Quiz
Calculate (a) Po(ac), (b) Pi(dc), (c) %, (d) power dissipated by both transistors
Quiz
(a)
(b)
(c)
(d)
( ) dcdci CCP V I=
( ) ( )2 dc ac 40.5WQ i oP P P= - =
WVrms
R
rmsVacP
L
Lo 5.40
8
)18()()(
22
WV
VR
pVV
L
LCC 81
8
2182)40(
)(2
%100)(
)(% 0
dcP
acP
i
%50%10081
5.40
W
W
Homework
(a)
(b)
(c)
(d)
( )
2
maximum ac2
CCo
L
VP
R= W
V25.56
)8(2
)30( 2
L
CCi
R
VdcP
22)( maximum W
V62.71
)8(
)30(2 2
%54.78%10062.71
25.56%
W
W
WV
R
VP
L
CCQ 8.22
)8(
)30(22 maximum
2
2
2
2
2