In a class A amplifier shown in fig.1, the dc source VCC must supply direct current to the voltage divider and the collector circuit.
Fig. 1
Assuming a stiff voltage divider circuit, the dc current drain of the voltage divider circuit is
I 1 = V CC / (R 1 +R 2 )
In the collector circuit, the dc current drain is
I 2 = I CQ
In a class A amplifier, the sinusoidal variations in collector current averages to zero. Therefore, whether the ac signal is present or not, the dc source must supply an average current of
I S = I 1 + I 2.
This is the total dc current drain. The dc source voltage multiplied by the dc current drain gives the ac power supplied to an amplifier.
P S = V CC IS
Therefore, efficiency of the amplifier, h = (P L (max) / P S ) * 100 %
Where,, P L (max) = maximum ac load line power. In class A amplifier, there is a wastage of power in resistor R C and R E i.e. ICQ 2 * (R C + R E ).
To reduce this wastage of power R C and R E should be made zero. R E cannot be made zero because this will give rise to bias stability problem. R C can also not be made zero because effective load resistance gets shorted. This results in more current and no power transfer to the load R L. The R C resistance can, however, be replaced by an inductance whose dc resistance is zero and there is no dc voltage drop across the choke as shown in fig. 1.
Since in most application the load is loudspeaker, therefore power amplifier drives the loudspeaker, and the maximum power transfer takes place only when load impedence is equal to the source impedence. If it is not, the loud speaker gets less power. The impedence matching is done with the help of transformer, as shown in fig. 2.
Fig. 2
The ratio of number of turns is so selected that the impedence referred to primary side can be matched with the output impedence of the amplifier.
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