Question

An npn BJT having reverse saturation current $I_s={10}^{-15}$ A is biased in the forward active region with $V_{BE}=700mV$. The thermal voltage $\left(V_T\right)$ is $25mV$ and the current gain $\left(\beta \right)$ may vary from $50$ to $150$ due to manufactuing variations. The maximum emitter current (in $\mu A$) is

Answer 1

$I_C=I_s{e}^{V_{BE}/V_T}$
$I_E=\frac{\beta +1}{\beta }{I}_C$
$=\frac{\beta +1}{\beta }{I}_s.e^{V_{BE}/V_T}$
$I_E$ will be maximum when $\beta$ is 50
$=\left(1.02\right)({10}^{-9}\times {10}^{-6})e^{\frac{700\times {10}^{-3}}{25\times {10}^{-3}}}$
$=1475\mu A$