An amplifier circuit employs a transistor in common-emitter mode, with a load resistor of 2 k connected in the collector branch. The transistor has a current gain β = 50 and an input resistance of 0.50 k . When the

Correct answer: C. 25 mV Given: Input resistance, R_ in = 0.50 k = 500 Change in input current, Δ I_B = 50 = 50 × 10^ -6 A The change in input voltage is given by Ohm's law applied to the input circuit: Δ V_ in = Δ I_B × R_ in Δ V_ in = (50 × 10^ -6 A) × (500 ) Δ V_ in = 25000 × 10^ -6 V Δ V_ in = 25 × 10^ -3 V = 25 mV

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Semiconductors and Semiconductor Devices

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Semiconductors and Semiconductor Devices

An amplifier circuit employs a transistor in common-emitter mode,...

An amplifier circuit employs a transistor in common-emitter mode, with a load resistor of

2 \text{ k}\Omega

connected in the collector branch. The transistor has a current gain

\beta = 50

and an input resistance of

0.50 \text{ k}\Omega

. When the input current is altered by

50 \mu\text{A}

, determine the change in the input voltage.

Options

A $10 \text{ mV}$
B $50 \text{ mV}$
C $25 \text{ mV}$
D $2.5 \text{ mV}$

Correct answer

C. $25 \text{ mV}$

Step-by-step solution

Given:
Input resistance,

R_{in} = 0.50 \text{ k}\Omega = 500 \Omega

Change in input current,

\Delta I_B = 50 \mu\text{A} = 50 \times 10^{-6} \text{ A}

The change in input voltage is given by Ohm's law applied to the input circuit:

\Delta V_{in} = \Delta I_B \times R_{in}

\Delta V_{in} = (50 \times 10^{-6} \text{ A}) \times (500 \Omega)

\Delta V_{in} = 25000 \times 10^{-6} \text{ V}

\Delta V_{in} = 25 \times 10^{-3} \text{ V} = 25 \text{ mV}

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