Consider the circuit shown in the figure below:

$The cut-in voltage V_{B E (O N)} = 0.7 V, then the value of stability factor S (I_{C O}) is equal to _______.$
91

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Solution

The correct option is A 91
$I_{C} = β I_{B} + (1 + β) I_{C O}$

$Differentiate with respect to I_{C}$

$1 = \frac{β d I_{B}}{d I_{C}} + (1 + β) \frac{d I_{C O}}{d I_{C}}$

$1 = \frac{β d I_{B}}{d I_{C}} + (1 + β) \frac{1}{S (I_{C O})}$

$\frac{\partial I_{C}}{\partial I_{C O}} = \frac{1 + β}{1 - β \frac{d I_{B}}{d I_{C}}}$

$now, \frac{d I_{B}}{d I_{C}} = 0$

$S (I_{C O}) = 1 + β = 90 + 1 = 91$

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