Lecture - 31: Class B Power Amplifiers

# Cross over distortion:

Fig. 1 shows the ac equivalent circuit of a class B push pull amplifier. Suppose that no bias is applied to the emitter diodes. Then the incoming voltage has to rise to about 0.7 V to overcome the barrier potential. Because of this no current flows through Q, when the signal is less than 0.7 V. The action is similar on the other half cycle no current flows in Q2 until ac voltage is more negative the – 0.7 V. If no bias is applied the output of class B amplifier looks like as shown in fig. 1.

Fig. 1

The signal output is distorted. Because of clipping action between half cycles, it no longer is a sine wave. Since the clipping occurs between the time one transistor cuts off and the time the other comes on, it is called cross over distortion. To eliminate cross over distortion, the slight forward bias must be applied to each emitter diode. This means locating the Q-point slightly above cut off as shown in fig. 2. In fact, this is class AB operation. This means that collector current flows for more than 180 degrees but less than 360°.

Fig. 2

Class A amplifier introduces non-linear distortion in input wave means elongates one half cycle and compresses one half cycle. This can be reduced by swamping. In this case it can be further reduced because both half cycles are identical in shape, is given by non-linear distortion is much less than class A.

Since the ac output compliance equals the peak-to-peak voltage, the maximum load power is

Where, I1 = current through biasing resistance. When no signal is present I2 = ICQ and the current drain is small. But when a signal is present, the current drain increase because the upper collector current becomes large.

If the entire ac load line is used, then the upper transistor has a half sine wave of current through it with a peak value of

IC(sat) = VCEQ / RL

The average value of half sine wave is given by

The dc power is supplied to the circuit is PS = VCC is under no signal conditions, the dc power is small because the current drain is minimum. But when a signal uses the entire ac load line, the dc power supplied to the circuit reaches a maximum.

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