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A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of $800 Ω$ is connected in the collector circuit and the voltage drop across it is $0.8 V$ . If the current amplification factor is $0.96$ and the input resistance of the circuit is $192 Ω$ , the voltage gain and the power gain of the amplifier will respectively be
$3.69, 3.84$
$4, 3.84$
$4, 4$
$4, 3.69$

A

4, 3.84

B

3.69, 3.84

C

4, 4

D

4, 3.69

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Solution

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

Given:
$R_{c} = 800 Ω, V_{L} = 0.8 V$
$R_{B} = 192 Ω$
Current amplification factor =0.96
Hence, $\frac{I_{c}}{I_{B}} = 0.96$
$β = 0.96$

Assumption:
Transistor is ideal and working in forward active region (Junction BE is forward biased and Junction BC is reverse biased).
$⟹ V_{B E} = 0$

$V_{E} = 0 ⟹ V_{B} = 0$

Collector current is given by $I_{C} = V_{L} / R_{c} = 0.8 / 800 = 10^{- 3} A$
Base current is given by, $I_{B} = I_{C} / β = \frac{10^{- 3}}{0.96} A$
Applying KVL on Base circuit, input voltage $V_{B B} = I_{B} R_{B}$
$V_{B B} = \frac{10^{- 3}}{0.96} \times 192 = 0.2 V$

Voltage Gain, $A_{v} = \frac{V_{o u t}}{V_{i n}} = \frac{V_{L}}{V_{B B}} = \frac{0.8}{0.2} = 4$
Current Gain, $A_{I} = \frac{I_{o u t}}{I_{i n}} = \frac{I_{C}}{I_{B}} = β = 0.96$
Power Gain, $A_{P} = A_{V} A_{I} = 4 \times 0.96 = 3.84$

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