A common emitter low_frequency amplifier has an effective input resistance of 1 kΩ. The collector load resistance is 5 kΩ. On applying a signal, the_base current changes by 10 μA and the collector current changes by 1 mA. what is the power gain of this amplifier?

Asked by Topperlearning User | 10th Jul, 2014, 03:03: PM

Expert Answer:

The power gain or power amplification (Ap) is the product of the current gain βac and the voltage gain Av:

Ap = βac x Av

Here βac = ΔIC/ΔIB = (1 x 10-3)/(10 x 10-6) = 100 and

Av = -βac x RL/RiWhere RL is the load resistance and Ri is the input resistance. The negative sign just indicates that in the common emitter configuration the output  signal is phase shifted by 180o.

[We use the load resistance RL instead of the output resistance Ro(of the amplifier stage) in the above expression since the load resistance is small compared to the transistor output resistance. (The output resistance of the amplifier stage is the parallel combined value of RL and the transistor output resistance)].

Ignoring the negative sign, Av = 100 x 5/1 = 500

Therefore, power gain Ap = βac x Av = 100 x 500 = 5000.

Answered by  | 10th Jul, 2014, 05:03: PM