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Semiconductor Electronics: Materials, Devices and Simple Circuits

Common emitter transistor

For a common emitter configuration, if and have their usual

Question

Medium

Solving time: 3 mins

For a common emitter configuration, if $α$ and $β$ have their usual meanings, the incorrect relationship between $α$ and $β$ is
(a) $\frac{1}{α} = \frac{1}{β} + 1$
(b) $α = \frac{β}{1 - β}$
(c) $α = \frac{β}{1 + β}$
(d) $α = \frac{β ^{2}}{1 + β ^{2}}$

Views: 5,301 students

Book:

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

Exam:

JEE Mains 2016

Found 7 tutors discussing this question

Amelia Discussed

For a common emitter configuration, if

α

and

β

have their usual meanings, the incorrect relationship between

α

and

β

is
(a)

\frac{1}{α} = \frac{1}{β} + 1

(b)

α = \frac{β}{1 - β}

(c)

α = \frac{β}{1 + β}

(d)

α = \frac{β ^{2}}{1 + β ^{2}}

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

Exp.

(a, c)

As, we know, In case of a common-emitter configuration, DC current gain,

α = \frac{l _{c}}{I _{e}}

.
Where,

I_{c}

is collector current and

I_{e}

is emitter current
and AC current gain,

β = \frac{l _{c}}{I _{b}}

.
where,

I_{b}

is base current.
Also,

I_{e} = I_{b} + I_{c}

Dividing whole equation by

I_{c}

, we get

\Rightarrow \frac{l _{e}}{l _{c}} = \frac{I _{b}}{l _{c}} + 1 \Rightarrow \frac{1}{α} = \frac{1}{β} + 1 \Rightarrow α = \frac{β}{1 + β}

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Question Text	For a common emitter configuration, if $α$ and $β$ have their usual meanings, the incorrect relationship between $α$ and $β$ is (a) $\frac{1}{α} = \frac{1}{β} + 1$ (b) $α = \frac{β}{1 - β}$ (c) $α = \frac{β}{1 + β}$ (d) $α = \frac{β ^{2}}{1 + β ^{2}}$
Topic	Semiconductor Electronics: Materials, Devices and Simple Circuits
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	44

For a common emitter configuration, if $α$ and $β$ have their usual meanings, the incorrect relationship between $α$ and $β$ is
(a) $\frac{1}{α} = \frac{1}{β} + 1$
(b) $α = \frac{β}{1 - β}$
(c) $α = \frac{β}{1 + β}$
(d) $α = \frac{β ^{2}}{1 + β ^{2}}$

Text solutionVerified

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For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between α and β is(a) α1​=β1​+1(b) α=1−ββ​(c) α=1+ββ​(d) α=1+β2β2​

Text solutionVerified

Questions from JEE Mains 2016 - PYQs

Practice questions from Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

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Previous year question papers

For a common emitter configuration, if $α$ and $β$ have their usual meanings, the incorrect relationship between $α$ and $β$ is
(a) $\frac{1}{α} = \frac{1}{β} + 1$
(b) $α = \frac{β}{1 - β}$
(c) $α = \frac{β}{1 + β}$
(d) $α = \frac{β ^{2}}{1 + β ^{2}}$