Enggineering Technology

Class B Push-Pull Amplifier

Class B Push-Pull Amplifier

Though the efficiency of class B power amplifier is higher than class A, as only one-half cycle of the input is used, the distortion is high. Also, the input power is not completely utilized. In order to compensate these problems, the push-pull configuration is introduced in class B amplifier.

Construction

The circuit of a push-pull class B power amplifier consists of two identical transistors T1 and T2 whose bases are connected to the secondary of the center-tapped input transformer Tr1. The emitters are shorted and the collectors are given the VCC supply through the primary of the output transformer Tr2.

The circuit arrangement of class B push-pull amplifier is same as that of class A push-pull amplifier except that the transistors are biased at cut off, instead of using the biasing resistors. The figure below gives the detailing of the construction of a push-pull class B power amplifier.

Push-Pull construction

Operation

The circuit of class B push-pull amplifier shown in the above figure clears that both the transformers are center-tapped. When no signal is applied at the input, the transistors T1 and T2 are in the cut off condition and hence no collector currents flow. As no current is drawn from VCC, no power is wasted.

When the input signal is given, it is applied to the input transformer Tr1 which splits the signal into two signals that are 180o out of phase with each other. These two signals are given to the two identical transistors T1 and T2. For the positive half cycle, the base of the transistor T1 becomes positive and collector current flows. At the same time, the transistor T2 has the negative half cycle, which throws the transistor T2 into the cutoff condition and hence no collector current flows. The waveform is produced as shown in the following figure.

Diagram for push-pull operation

For the next half cycle, the transistor T1 gets into cut off condition and the transistor T2 gets into conduction, to contribute the output. Hence for both the cycles, each transistor conducts alternately. The output transformer Tr3 serves to join the two currents producing an almost undistorted output waveform.

Power Efficiency of Class B Push-Pull Amplifier

The current in each transistor is the average value of half sine loop.

For half sine loop, Idc is given by

Idc=(IC)maxπIdc=(IC)maxπ

Therefore,

(pin)dc=2×[(IC)maxπ×VCC](pin)dc=2×[(IC)maxπ×VCC]

Here factor 2 is introduced as there are two transistors in push-pull amplifier.

R.M.S. value of collector current = (IC)max/2–√(IC)max/2

R.M.S. value of output voltage = VCC/2–√VCC/2

Under ideal conditions of maximum power

Therefore,

(PO)ac=(IC)max2–√×VCC2–√=(IC)max×VCC2(PO)ac=(IC)max2×VCC2=(IC)max×VCC2

Now overall maximum efficiency

ηoverall=(PO)ac(Pin)dcηoverall=(PO)ac(Pin)dc
=(IC)max×VCC2×π2(IC)max×VCC=(IC)max×VCC2×π2(IC)max×VCC
=π4=0.785=78.5%=π4=0.785=78.5%

The collector efficiency would be the same.

Hence the class B push-pull amplifier improves the efficiency than the class A push-pull amplifier.

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