Question

A Bipolar Junction Transistor has $\alpha =0.98$, base current
is $I_B=25\mu A$ and $I_{CBO}=200nA$.
The emitter current is

• A. 1.33 mA
• B. 1.235 mA
• C. 1.05 mA
• D. 1.26 mA

Answer 1

The correct option is D 1.26 mA

Given,
Base current $I_B=25\mu A$
$I_{CBO}=200nA$
$\alpha =0.98$
where, $\beta =\frac{\alpha }{1-\alpha }=\frac{0.98}{1-0.98}=49$
$\text{Collector current},I_C=\beta I_B+(1+\beta )I_{CBO}$
$\therefore I_C=49\times 25\times {10}^{-6}+\left(50\right)\times 200\times {10}^{-9}$
$I_C=1.235\times {10}^{-3}A$
$\text{Emitter current}=I_E=I_C+I_B$
$=1.235mA+0.025mA$
$\therefore I_E=1.26mA$