The input resistance of a silicon transistor is 100 . Base current is changed by 40 μ A which results in a change in collector current by 2 ~mA . This transistor is used as a common-emitter amplifier with a load

Correct answer: A. 2000 We know V_g=β (R_L)/(R_i) =(Δ I_C)/(Δ I_B) (R_L)/(R_i) ( β=(Δ I_C)/(Δ I_B) ) On putting given values aligned & V_g= 2 × 10^ -3 40 × 10^ -6 × (4 × 10^3)/(100) & V_g=2000 aligned

NEET 2012

Exam: NEET
Year: 2012
Subject: Physics
Chapter: Semiconductors

NEET 2012 · Physics · Semiconductors

The input resistance of a silicon transistor is 100 . Base current...

The input resistance of a silicon transistor is

100 \Omega

. Base current is changed by

40 \mu \mathrm{A}

which results in a change in collector current by

2 \mathrm{~mA}

. This transistor is used as a common-emitter amplifier with a load resistance of

4 \mathrm{k} \Omega

. The voltage gain of the amplifier is

Options

A2000
B3000
C4000
D1000

Correct answer

A. 2000

Step-by-step solution

We know V_g=\beta

\frac{R_L}{R_i} =\frac{\Delta I_C}{\Delta I_B} \frac{R_L}{R_i}

\quad

\left(\because \beta=\frac{\Delta I_C}{\Delta I_B}

\right)

On putting given values

\begin{aligned} & V_g=\frac{2 \times 10^{-3}}{40 \times 10^{-6}} \times \frac{4 \times 10^3}{100} \\ & V_g=2000 \end{aligned}

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