Electronic Devices Question 251
Question: In NPN transistor, 1010 electrons enters in emitter region in ${10^-6}$ sec. If 2% electrons are lost in base region then collector current and current amplification factor (b) respectively are
Options:
A) 1.57 mA, 49
B) 1.92 mA, 70
C) 2 mA, 25
D) 2.25 mA, 100
Show Answer
Answer:
Correct Answer: A
Solution:
$ I _{e}=10^{10}\times 1.6\times {{10}^{-19}}\times \frac{1}{{{10}^{-6}}}=1.6mA $
$ ( \because I=\frac{Q}{t} ) $ Since 2% electrons are absorbed by base, hence 98% electrons reaches the collector i.e.
a = 0.98
Therefore $ I _{c}=\alpha I _{e}=0.98\times 1.6=1.568mA\approx 1.57mA $ Also current amplification factor $ \beta =\frac{\alpha }{1-\alpha }=\frac{0.98}{0.02}=49 $
