Physics - Semiconductors Question with Solution | TestHub

For given electrons

Emitter current$I_e=\frac{n_{e}e}{t}$

Where$n_e$is the number of electrons enters into the emitter,$e$is the charge on electron and$t$is time.

$I_e=\frac{10^{10}\times \left(1.6\times 10^{-19}\right)}{10^{-6}}$

$I_e=1.6\times 10^{-3}$

$I_e=1.6 \mathrm{mA}$

$2\%$electrons are lost in base region hence base current$I_b=2\%\times 1.6\times 10^{-3}$, therefore$I_b=0.032 \mathrm{mA}$

Therefore$I_c=I_e-I_b$

$I_c=1.6-0.032$

$I_c=1.57 \mathrm{mA}$

and the current amplification factor

$\beta =\frac{I_c}{I_b}=\frac{1.568}{0.032}=49$