SATHEE: Electronic Devices Question 251

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:

$I_{e} = 10^{10} \times 1.6 \times 10^{- 19} \times \frac{1}{10^{- 6}} = 1.6 m A$

$(∵ I = \frac{Q}{t})$ Since 2% electrons are absorbed by base, hence 98% electrons reaches the collector i.e.

a = 0.98

Therefore $I_{c} = α I_{e} = 0.98 \times 1.6 = 1.568 m A \approx 1.57 m A$ Also current amplification factor $β = \frac{α}{1 - α} = \frac{0.98}{0.02} = 49$

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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:

Answer:

Solution:

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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