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Q.
For a transistor $\frac{I_C}{I_E}=0.96$, then current gain for common emitter is
AIPMTAIPMT 2002Semiconductor Electronics: Materials Devices and Simple Circuits
Solution:
The current gain of a common emitter transistor $(\beta)$ is defined as the ratio of collector current $\left(I_{C}\right)$ to the base current $\left(I_{B}\right)$.
Also, $I_{E}=I_{B}+I_{C} ; I_{C} / I_{E}=0.96$ (given)
$\therefore \beta=\frac{I_{C}}{I_{B}}=\frac{I_{C}}{I_{E}-I_{C}}$
Now, $\frac{I_{E}}{I_{C}}=\frac{1}{0.96} .$
$ \therefore \frac{I_{E}-I_{C}}{I_{C}}=\frac{1}{0.96}-1=\frac{0.04}{0.96}$
$\therefore \beta=\frac{I_{C}}{I_{E}-I_{C}}=\frac{0.96}{0.04}=24 \text {. }$