In a transistor if $I_c/I_e=alpha$ and $I_c/I_b=beta$ , if $alpha$ varies between $(20)/(21)$ and $(100)/(101)$ , then the value of $beta$ lied between?

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Answer 1 · 2017-03-01T08:14:03

$20 and 100$

Representative circuit in Common emitter amplifier configuration is given below
![dnatechindia.com]( useruploads.socratic.org )

Given is the ratio of $I_c/I_b=beta$
And the ratio of $I_c/I_e= alpha$

From above expressions we have
$I_c=beta I_b=alphaI_e$ .....(1)

In this configuration, the current flowing out of the transistor must be equal to the currents flowing into the transistor. As such we have

$I_e = I_c + I_b$ .....(2)

To have a combined the expression for $α and β$ we divide both sides of (2) by $I_b$ and use (1) to get

$I_e/I_b = (I_c + I_b)/I_b$
$=>beta/alpha=beta+1$
$=>beta=alphabeta+alpha$

$=>beta=alpha/(1-alpha)$

For $alpha=20/21$

$beta=(20/21)/(1-20/21)=20$

Again for $alpha=100/101$
$beta=(100/101)/(1-100/101)=100$
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Note: value of $alpha$ will always be $<1$ .
and typically, $beta$ values lie between $20 and 200$ for most general purpose transistors.

In a transistor if I_c/I_e=alphaI_c/I_e=alpha and I_c/I_b=betaI_c/I_b=beta, if alphaalpha varies between (20)/(21)(20)/(21) and (100)/(101)(100)/(101), then the value of betabeta lied between?