In NPN transistor, 1010 electrons enter in emitter region in 10-6 s. If 2 % electrons are lost in base region then collector current and current amplification factor (β) respectively are :-

Question

In NPN transistor, 1010 electrons enter in emitter region in 10-6 s. If 2 % electrons are lost in base region then collector current and current amplification factor (β) respectively are :- (A) 1.57 mA , 49 (B) 1.92 mA , 70 (C) 2 mA , 25 (D) 2.25 mA , 100

Tardigrade Admin · Accepted Answer

i e =( N e / t )=(1010 × 1.6 × 10-19/10-6)=1.6 mA i b =2 % of i e so i c =98 % of i e i c =(98/100)=1.6 mA ≃ 1.57 mA β=( i c / i b ) or (α/1-α)

Q. In NPN transistor, $1 0^{10}$ electrons enter in emitter region in $1 0^{- 6} s$ . If $2%$ electrons are lost in base region then collector current and current amplification factor $(β)$ respectively are :-

1.57 mA , 49

1.92 mA , 70

2 mA , 25

2.25 mA , 100

$i_{e} = \frac{N _{e}}{t} = \frac{1 0 ^{10} \times 1.6 \times 1 0 ^{- 19}}{1 0 ^{- 6}} = 1.6 m A$
$i_{b} = 2%$ of $i_{e}$ so $i_{c} = 98%$ of $i_{e}$
$i_{c} = \frac{98}{100} = 1.6 m A ≃ 1.57 m A$
$β = \frac{i _{c}}{i _{b}}$
or $\frac{α}{1 - α}$

Q. In NPN transistor, 1010 electrons enter in emitter region in 10−6s. If 2% electrons are lost in base region then collector current and current amplification factor (β) respectively are :-