Question

$\text{A Bipolar junction transistor has}\alpha =0.98$
$I_B=25\mu \text{A and}{I}_{CBO}=200nA$
The emitter current is

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

Answer 1

The correct option is A 1.26 mA

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