Question

A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of $800\Omega$ 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. $4,3.84$
• B. $3.69,3.84$
• C. $4,4$
• D. $4,3.69$

Answer 1

Answer: (A)

$4,3.84$

Transistor is working in common emitter configuration. Hence, input terminal is at base and output terminal is at collector.

Given:

$R_c=800\Omega ,V_L=0.8V$

$R_B=192\Omega$

Current amplification factor =0.96

Hence, $\frac{I_c}{I_B}=0.96$

$\beta =0.96$

Assumption:

Transistor is ideal and working in forward active region (Junction BE is forward biased and Junction BC is reverse biased).

$⟹V_{BE}=0$

$V_E=0⟹V_B=0$

Collector current is given by $I_C=V_L/R_c=0.8/800={10}^{-3}A$

Base current is given by, $I_B=I_C/\beta =\frac{{10}^{-3}}{0.96}A$

Applying KVL on Base circuit, input voltage $V_{BB}=I_B{R}_B$

$V_{BB}=\frac{{10}^{-3}}{0.96}\times 192=0.2V$

Voltage Gain, $A_v=\frac{V_{out}}{V_{in}}=\frac{V_L}{V_{BB}}=\frac{0.8}{0.2}=4$

Current Gain, $A_I=\frac{I_{out}}{I_{in}}=\frac{I_C}{I_B}=\beta =0.96$

Power Gain, $A_P=A_V{A}_I=4\times 0.96=3.84$