In an NPN transistor,10^{10} electrons enter | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The emitter current $I_E$ is given by $I_E = \frac{Q}{t} = \frac{n \cdot e}{t}$.Given $n = 10^{10}$,$e = 1.6 	imes 10^{-19} \, C$,and $t = 10^{-6

Vedclass · Accepted Answer

$1.57 \, mA, 49$

Vedclass · Answer

(A) The emitter current $I_E$ is given by $I_E = \frac{Q}{t} = \frac{n \cdot e}{t}$.Given $n = 10^{10}$,$e = 1.6 	imes 10^{-19} \, C$,and $t = 10^{-6} \, s$.$I_E = \frac{10^{10} 	imes 1.6 	imes 10^{-19}}{10^{-6}} = 1.6 	imes 10^{-3} \, A = 1.6 \, mA$.Since $2 \%$ of electrons are lost in the base,the current gain $\alpha$ is $1 - 0.02 = 0.98$.The collector current $I_C$ is given by $I_C = \alpha I_E = 0.98 	imes 1.6 \, mA = 1.568 \, mA \approx 1.57 \, mA$.The current amplification factor $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.$\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.

In an $NPN$ transistor,$10^{10}$ electrons enter the emitter region in $10^{-6} \, s$. If $2 \%$ of electrons are lost in the base region,then the collector current and the current amplification factor $(\beta)$ are,respectively:

Similar Questions

In the following circuit, a voltmeter $V$ is connected across a lamp $L$. What change would occur in the voltmeter reading if the resistance $R$ is reduced in value?

When the transistor is used as an amplifier,

$NPN$ transistors are preferred to $PNP$ transistors because they have

An oscillator is nothing but an amplifier with

For a transistor,$\beta = 100$. The value of $\alpha$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module