Class B Complementary Symmetry Power Amplifier
Class B Complementary Symmetry Power Amplifier
Class B Complementary Symmetry Power Amplifier
Let us analyze the circuit operation in both the half cycles of the ac input signal.
All the results obtained for class B push-pull amplifier are applicable to this
circuit replace RL’ in all those expressions by RL for the complementary
symmetry circuit.
Analysis of Complementary Symmetry Class B Push Pull Amplifier:
The analysis is carried out by following the steps given below:
Steps to be followed:
• 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 ( VCC, 0 ) as VCEQ = VCC and ICQ = 0.
1
I dc I m sin t dt
0
Im
cos t 0 Im
1 1 2 I m 2
I dc
Im • Therefore the total dc power input is given by,
cos cos 0
2VCC I m
Pdc VCC .I dc 3
Step 2: AC output power: 2
VLrms
Pac V Lrms . I Lrms 4 I Lrms RL VLrms I Lrms RL
2
RL
VLrms
where VLrms RMS value of load voltage & I Lrms
RL
I Lrms RMS value of load current
Assuming the turns ratio to be 1 : 1: we get,
Vm Im
V Lrms and I Lrms
2 2
Vm I m Vm I m
Pac . 5
2 2 2
2 2
VLrms I Lrms RL
2 RL 2
Step 3: Efficiency (η):
Pac
Pdc
Vm I m 2VCC I m
Let , Pac , and Pdc
2
Vm I m
2 Vm
% 100 4 V 100 6
2VCC I m CC
Step 4: Maximum efficiency (% ηmax):
Vm VCC
% 100 100 78.5%
4 VCC 4 VCC
This is the maximum possible efficiency under the ideal operating conditions.
To find the total power dissipation taking place in the two transistors is given by,
Pd Pdc Pac
2VCC I m Vm I m
8
2
Vm
But from equ (6), I m
RL
2VCCVm Vm2
Pd 9
.RL
'
2.RL
This equation shows that the power dissipation taking place in the transistors
under no signal condition is,
Pd = 0 as Vm = 0 when there is no input signal.
Step 6: Maximum power dissipation (Pd):
• Transformers are not being used. This makes the circuit less expensive and
less bulky.
• Even harmonics are automatically balanced out. Hence only the odd
harmonics are present.
Disadvantages of Complementary Symmetry Amplifier: