2. z
CLASS B POWER AMPLIFIER
There are two amplifiers which operate in class B mode, 1. Class B push
pull amplifier, 2. Complimentary symmetry class B amplifier.
1. Class B push pull amplifier:
3. z
CONSTRUCTION:
The circuit diagram is as shown consisting of two identical transistors and two
transformers. The bases of transistor are connected to the secondary of
transformer. The i/p voltage applied to the transistor are 180 degree out of phase
using transformer Tr1. The collector of transistor are connected to the primary of
o/p transformer Tr2. As the two voltages applied to the bases of transistor are out of
phase, hence called as push pull amplifier. The Vcc supply voltage is connected to
the center tap of the transformer and bases of the transistor are grounded through
the center tap of transformer Tr1.
WORKING:
When the i/p signal is applied then during the positive half cycle of i/p signal, the
E/B junction of transistor T1 is forward biased. Hence, current Ic1 flows at the same
time E/B junction of transistor T2 becomes reverse biased. Hence, no current is
obtained i.e. Ic2 = 0. Similarly, during the negative half cycle of i/p signal, E/B
junction of transistor T1 becomes reverse biased. Hence, Ic1 = 0 and at the same
time, transistor T2 becomes forward biased. Hence, Ic2 flows.
4. z
In this way, the amplified o/p is obtained due to the combination of Ic1 and Ic2.
When the signal is not applied, then Vce = Vcc, Ic = 0. As the transformer is
ideal, primary resistance is 0. In this circuit, the transistor is biased at cutoff.
Hence, no current flows through both the transistors. When signal is applied
then the Q point exists at the cutoff. Hence, VcQ = Vcc.
Let R’L be the reflected resistance of secondary on primary implies R’L =
1/n2.RL.
From the load line analysis, slope of AC load line is given by RL’ =
Vccmax/Icmax.
Hence, AC o/p power i.e. Pac = I2RMS. R’L = (Icmax/2)2 . R’L = (Vccmax/ 2RL’)2
. R’L = Vccmax 2 / 4RL’.
The DC i/p power is Pdc = Vcc.IcQ = Vcc. Vcc /𝜋 RL’ = Vcc 2 / 𝜋 RL’.
Hence, efficiency 𝜂 = Pac/Pdc. 100% = (Vccmax 2 / 4RL’) / (Vcc 2 / 𝜋 RL’) . 100
%
5. z
Hence, 𝜂 = (𝜋/4). 100 = 78.5 %.
Hence, efficiency of class B push pull amplifier is found out to be 78.5 %.
ADVANTAGES:
1. It has high efficiency i.e. 78.5 %.
2. Harmonic distortion is avoided.
3. Distortion free o/p is obtained.
4. No loss of DC power.
DISADVANTAGES:
1. It requires two transformers which are bulky and expensive.
2. The driver transformer used affects the frequency response of amplifier.
3. The non linear behavior of core of transformer gives distortion.
4. There are copper losses and dissipation of a part of AC power in winding.
