Question

A $n-p-n$ transistor is connected to common emitter configuration in a given amplifier. A load resistance of $800$ is connected in the collector circuit and the voltage drop across it is $0.8V$. If the current amplification factor is $0.96$ and the input resistance of the circuit is $192\Omega$, the voltage gain and the power gain of the amplifier will respectively be

• A. 3.69, 3.84
• B. 4 , 4
• C. 4, 3.69
• D. 4, 3.84

Answer 1

Answer: (D)

Given $\alpha =0.96$
so, $\beta =\frac{\alpha }{1-\alpha }=\frac{0.96}{0.04}\Rightarrow \beta =24$
Voltage gain for common emitter configuration
$A_V=\beta .\frac{R_L}{R_1}=24\times \frac{800}{192}=100$
Power gain for common emitter configuration
$P_V=\beta A_V=24\times 100=2400$
Voltage gain for common base configuration
$A_V=\alpha .\frac{R_L}{R_P}=0.96\times \frac{800}{192}=4$
Power gain for common base configuration
$P_V=A_V\alpha =4\times 0.96=3.84$
* In the question it is asked about common emitter configuration but we got above answer for common base configuration.