Physics - Semiconductors Question with Solution | TestHub

Given that, change in emitter current,

$∆I_E=8\mathrm{ }\mathrm{mA}$

And the change in collector current,

$∆I_C=7.9 \mathrm{mA}$

We know that,

$\alpha =\frac{∆I_C}{∆I_E}\Rightarrow \alpha =\frac{7.9}{8}\Rightarrow \alpha \approx 0.99$

Also, we know that

$\beta =\frac{\alpha }{1-\alpha }\Rightarrow \beta =\frac{\frac{7.9}{8}}{1-\frac{7.9}{8}}$

$\Rightarrow$$\beta =\frac{7.9}{8-7.9}\Rightarrow \beta =\frac{7.9}{0.1}=79$

Hence,$\alpha =0.99 \mathrm{a}\mathrm{n}\mathrm{d} \beta =79$