A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V
Question:
A transistor is connected in common emitter circuit configuration, the collector supply voltage is $10 \mathrm{~V}$ and the voltage drop across a resistor of $1000 \Omega$ in the collector circuit is $0.6 \mathrm{~V}$. If the current gain factor $(\beta)$ is 24 , then the base current is________$\mu \mathrm{A} .$ (Round off to the Nearest Integer)
Solution:
$\beta=\frac{I_{C}}{I_{B}}=24 ; \quad R_{C}=1000$
$\Delta \mathrm{V}=0.6$
$\mathrm{I}_{\mathrm{C}}=\frac{0.6}{1000}$
$\mathrm{I}_{\mathrm{C}}=6 \times 10^{-4}$
$\mathrm{I}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{C}}}{\beta}=\frac{6 \times 10^{-4}}{24}=25 \mu \mathrm{A}$
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