Question

The transistor shown below has $\beta$ = 50 and reverse saturation current is ${10}^{-16}$A. Then the base to emitter voltage of the transistor is



(Assume thermal voltage $V_T=26mV$)

• A. 0.745 V
• B. 0.635 V
• C. 0.844 V
• D. 0.623 V

Answer 1

The correct option is C 0.844 V

Given,
Reverse saturation current, $I_S={10}^{-16}A$
Base current, $I_B=250\mu A$
Collector current, $I_C=\beta I_B+(1+\beta )I_{\infty }=(50\times 250\times {10}^{-6}+51\times {10}^{-16})A$

$=12.5mA$

But, collector current, $I_C=I_B{e}^{\frac{V_{BE}}{V_2}}$

$12.5\times {10}^{-3}={10}^{-16}{e}^{\frac{V_{BE}}{26\times {10}^{-3}}}$

$\therefore$ Base to emitter voltage, $V_{BE}=(26\times {10}^{-3})In(12.5\times {10}^{13})=0.844V$