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Semiconductor Electronics: Materials, Devices and Simple Circuits

Npn transistor

An n-p-n transistor is used in common emitter configuration as

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An $n - p - n$ transistor is used in common emitter configuration as an amplifier with 1 $k Ω$ load resistance. Signal voltage of $10 mV$ is applied across the base-emitter. This produces a $3 mA$ change in the collector current and $15 μ A$ change in the base current of the amplifier. The input resistance and voltage gain are
(a) $0.67 k Ω, 200$
(b) $0.33 k Ω, 1.5$
(c) $0.67 k Ω, 300$
(d) $0.33 k Ω, 300$

Views: 5,412 students

Updated on: Jan 18, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

Found 6 tutors discussing this question

Luna Discussed

n - p - n

transistor is used in common emitter configuration as an amplifier with 1

k Ω

load resistance. Signal voltage of

10 mV

is applied across the base-emitter. This produces a

3 mA

change in the collector current and

15 μ A

change in the base current of the amplifier. The input resistance and voltage gain are
(a)

0.67 k Ω, 200

(b)

0.33 k Ω, 1.5

(c)

0.67 k Ω, 300

(d)

0.33 k Ω, 300

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

Exp. (c)
Given, load resistance,

R_{L} = 1 k Ω

Input voltage,

V_{in} = 10 mV = 10 \times 1 0^{- 3} V

Base current,

Δ_{B} = 15 μ A = 15 \times 1 0^{- 6} A

Collector current,

Δ I_{C} = 3 mA

Input resistance,

R_{in} = \frac{V _{in}}{Δ _{B}} = \frac{10 \times 1 0 ^{- 3}}{15 \times 1 0 ^{- 6}} = 0.67 k Ω

and voltage gain

= β \times \frac{R _{L}}{R _{in}}

= \frac{Δ _{C} \times R _{L}}{Δ _{B} \times R _{in}} = (\frac{3 mA}{15 μ A}) \times (\frac{1 k Ω}{0.67 k Ω})

= (\frac{3 \times 1 0 ^{- 3}}{15 \times 1 0 ^{- 6}}) (\frac{1 \times 1 0 ^{3}}{0.67 \times 1 0 ^{3}})

= \frac{1000 \times 3 \times 3}{15 \times 2} = 300 (∵ 0.67 ≅ 2/3)

Alternate Solution
Voltage gain

= \frac{V _{o u tp u t}}{V _{in p u t}}

= \frac{R _{L} \times Δ _{C}}{V _{in}}

= \frac{1 \times 1 0 ^{3} \times 3 \times 1 0 ^{- 3}}{10 \times 1 0 ^{- 3}} = 300

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Question Text	An $n - p - n$ transistor is used in common emitter configuration as an amplifier with 1 $k Ω$ load resistance. Signal voltage of $10 mV$ is applied across the base-emitter. This produces a $3 mA$ change in the collector current and $15 μ A$ change in the base current of the amplifier. The input resistance and voltage gain are (a) $0.67 k Ω, 200$ (b) $0.33 k Ω, 1.5$ (c) $0.67 k Ω, 300$ (d) $0.33 k Ω, 300$
Updated On	Jan 18, 2023
Topic	Semiconductor Electronics: Materials, Devices and Simple Circuits
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 3
Upvotes	320
Avg. Video Duration	10 min

Text solutionVerified

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