Q. In a common emitter configuration of a transistor, the voltage drop across a $500\Omega$ resistor in the collector circuit is $0.5\ V$ when the collector supply voltage is 0.96, the base current is
A) $\frac{1}{20}\mu A$
B) $\frac{1}{5}\mu A$
C) $\frac{1}{20}mA$
D) $\frac{1}{24}mA$
Explanation
(d) $\alpha = 0.96$
$\beta = \frac{\alpha}{1 - \alpha} = \frac{0.96}{1 - 0.96} = 24$
$I_{C} = \frac{V_{C}}{R} = \frac{0.5}{500} = 10^{- 3}A$
$I_{b} = \frac{I_{c}}{\beta} = \frac{10^{- 3}}{24} = \frac{1}{24}mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.