Download CLASS B POWER AMPLIFIER and more Essays (university) Electronics in PDF only on Docsity!
Class B Power Amplifiers:
- We know the Q point of a class B amplifier is adjusted
to be on the X-axis. i.e. the. transistor is biased at
cut-off. Due to this, a transistor can conduct only
during the positive half cycle of the ac input signal.
- That means the conduction takes place for only 180
0
or a half cycle period. Lot of distortion will therefore be
present at the output.
- In order to overcome this problem, two transistors
are used instead of one. These transistors conduct in
the alternate half cycles of the input signal to produce
the full ac signal.
Types of class B power amplifier:
The class B amplifier using two transistors can be
of two types:
- Class B Push pull amplifier.
- Class B Complementary symmetry amplifier.
Let us discuss these circuits one by one.
Operation of the circuit:
The operation of the circuit can be divided into two parts as follows:
Operation in the positive half cycle of input (0≤ ωt ≤ π):t ≤ π):):
Equivalent circuit for the positive half cycle of the input
As shown in the
equivalent circuit
shown in fig, in the
positive half cycle of the
input, secondary
voltage of the driver
transformer V AO
is
positive and V
BO
is
negative.
This voltage polarity will
forward bias the BE
junction of transistor Q
1
only. BE junction of Q
2
Thus a sinusoidal base
current for Q 1
i.e, i b 1
will
flow. But the base
current for Q
2
i.e. 1
b
=
Hence the collector
current I
C
will flow
through the upper
half of the primary
winding of the
output transformer
and 1
C
= 0 as Q
2
does not conduct.
A positive
sinusoidal voltage
will appear across
the load as shown in
the waveforms of Fig.
The amplitude of the
secondary load
voltage is dependent
on the primary to
secondary turns ratio
of the output
transformer.
- Operation in the negative half cycle of input during –ve half cycle: (π ≤ ωt t
≤ 2π):
cycle of the input
voltage Vs, the polarity
of induced voltage in
the secondary winding
of the driver
transformer gets
reversed i.e. V
AO
is
negative and V
BO
is
positive as shown in Fig.
This will forward bias
the BE junction of
transistor Q
2
and
reverse bias the BE
junction of transistor
Q
1
.
Equivalent circuit for the negative half cycle of the input
The base current i
b
=
0 and i
b
will flow
through transistor Q
2
.
Thus in this half cycle
Q
1
is OFF and Q
2
is
ON.
The collector current
for Q 2
i.e. I C
flows
through the lower
half of the primary
winding of the
output transformer
as shown in Fig.
The amplitude of this secondary induced voltage is
dependent on the turns ratio of the output
transformer.
As the direction of
primary current in
the output
transformer has
reversed, the
induced voltage on
the secondary side
will reverse its
polarities. Hence a
negative sinusoidal
voltage will appear
across the load as
shown in Fig.
Analysis of Class B Push Pull Amplifier:
The analysis is carried out by following the steps given
below:
Steps to be followed:
Step 1: Obtain the Q-point and dc input power(P
dc
Step 2: Obtain the expression for the ac output power
(P
ac
Step 3: Get the expression for efficiency (η). ).
Step 4: Find the expression for maximum efficiency
max
Step 1: Q point of the circuit
- For the class B operation the Q point is positioned on the X-axis
in order to bias the transistors at cut-off. That means the co-
ordinates of the Q point are ( V
CC
, 0 ) as V
CEQ
= V
CC
and I
CQ
int ^ ^ , 0 ^ 1
CEQ
Q po V
- Thus in absence of the ac
input signal, the..power
drawn from the source is
zero.
Step 2: AC output power:
. 4
ac Lrms Lrms
P V I
ssuming the turns ratio to be 1 : 1: we get,
I RMSvalueof load current
Lrms
where V RMSvalueof load voltage
Lrms
m
Lrms
m
Lrms
I
and I
V
V
5
m m m m
ac
V I V I
P
Alternate expressions for P
ac
For deriving the alternate expressions for P
ac
, we must consider the
transferred resistance R
L
’
from the secondary to primary of the output
transformer.
6
L L
R
N
N
R
7
L
L rms
ac Lrms L
R
V
P I R
Step 4: Maximum efficiency (% η).
max
The maximum efficiency corresponds to the maximum swing of the
load voltage, and the maximum output swing V
m
= V
CC
Substituting this value of V
m
in Equation (8) we get,
100 78. 5 %
4
100
4
%
CC
CC
CC
m
V
V
V
V
This is the maximum possible efficiency under the ideal operating
conditions.
But practically the value of η).
max
will be between 65 to 70 %.
Step 5 : Power Dissipation:
To find the total power dissipation taking place in the two transistors is
given by,
d dc ac
P P P
CC m m m
V I V I
10
'
( 6 ),
L
m
m
R
V
But from equ I
11
. 2.
2
'
2
'
L
m
L
CC m
d
R
V
R
V V
P
This equation shows that the power dissipation taking place in the transistor
under no signal condition is,
P
d
= 0 as V
m
= 0 when there is no input signal.
Advantages of the Push-Pull amplifier are:
Due to Class B operation, their collector efficiency is quite
high.
- They give more AC output power per device.
- Distortion free output is obtained.
- Output Transformer used for push-pull amplifier circuit are
lighter, smaller and less expensive than the transformers of
comparable quality that are used in single-ended circuits.
In the core of the output transformer, the DC components for
output currents of the two devices oppose each other
magnetically which eliminates the tendency of the core to
saturate.
Push Pull Amplifier Disadvantages:
Following are the disadvantages of a Push-Pull
amplifier:
- It requires two equal and opposite voltages at the
input, therefore, driver stage has to be employed.
- Unequal amplification of the two halves of the signal
introduces more distortion.
Two identical transistors are required.
