Question

Consider the circuit shown in the figure below:
The cut-in voltage $V_{BE\left(ON\right)}$ = 0.7 V, then the value of stability factor $S\left(I_{CO}\right)$ is equal to _______

• A. 91

Answer 1

The correct option is A 91
$I_C=\beta I_B+(1+\beta )I_{CO}$

$1=\frac{\beta d{I}_B}{d{I}_C}+(1+\beta )\frac{d{I}_{CO}}{d{I}_C}$

$\frac{\partial I_C}{\partial I_{CO}}=\frac{1+\beta }{1-\beta \frac{d{I}_B}{d{I}_C}}$

now, $\frac{d{I}_B}{d{I}_C}=0$

$S\left(I_{CO}\right)=1+\beta$

$=90+1=91$

$(\therefore \frac{\partial I_C}{\partial I_{CO}}=S(I_{CO}\left)\right)$