Question

In an $\text{NPN}$ transistor ${10}^{10}$ electrons enter the emitter in ${10}^{-6}\text{s}.$ If $2\%$ of the electrons are lost in the base. The current transfer ratio of the transistor $\frac{98}{n}.$ The value of $n$ is .(integer only)

Answer 1

The emitter current $\left(I_E\right)$ is given by,

$I_E=\frac{Ne}{t}=\frac{{10}^{10}\times (1.6\times {10}^{-19})}{{10}^{-6}}=1.6\text{mA}$

The base current $\left(I_B\right)$ is given by,

$I_B=\frac{2}{100}\times \frac{{10}^{10}\times (1.6\times {10}^{-19})}{{10}^{-6}}=0.032\text{mA}$

In a transistor, $I_E=I_B+I_c$

$\Rightarrow I_c=I_E-I_B=1.6-0.032=1.568\text{mA}$

$\text{The current transfer ratio}=\frac{I_c}{I_E}=\frac{1.568}{1.6}=0.98=\frac{98}{100}$

$\therefore n=100$