Physics - Semiconductors Question with Solution | TestHub

Given that $90\%$ of the electrons emitted reach the collector, the collector current $I_C$ is $90\%$ of the emitter current $I_E$.

$I_C=0.9I_E$

Given $I_C=9\text{ mA}$, the emitter current $I_E$ is:

$I_E=\frac{9}{0.9}=10\text{ mA}$

The base current $I_B$ is given by:

$I_B=I_E-I_C=10-9=1\text{ mA}$

The common-base current amplification factor $\alpha$ is:

$\alpha =\frac{I_C}{I_E}=\frac{9}{10}=0.9$

The common-emitter current amplification factor $\beta$ is:

$\beta =\frac{I_C}{I_B}=\frac{9}{1}=9$

Answer: $\alpha =0.9,\beta =9$