10^{10} electrons enter the emitter of a junc | vedclass

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

Question

(D) The emitter current $I_E$ is given by the rate of flow of charge: $I_E = \frac{q}{t} = \frac{n \cdot e}{t}$.Given $n = 10^{10}$, $e = 1.6 	imes 1

Vedclass · Accepted Answer

$3.8$

Vedclass · Answer

(D) The emitter current $I_E$ is given by the rate of flow of charge: $I_E = \frac{q}{t} = \frac{n \cdot e}{t}$.Given $n = 10^{10}$, $e = 1.6 	imes 10^{-19} 	ext{ C}$, and $t = 0.4 	imes 10^{-6} 	ext{ s}$.$I_E = \frac{10^{10} 	imes 1.6 	imes 10^{-19}}{0.4 	imes 10^{-6}} = \frac{1.6 	imes 10^{-9}}{0.4 	imes 10^{-6}} = 4 	imes 10^{-3} 	ext{ A} = 4 	ext{ mA}$.Since $5 \%$ of the electrons are lost in the base, the collector current $I_C$ is $95 \%$ of the emitter current $I_E$.$I_C = 0.95 	imes I_E = 0.95 	imes 4 	ext{ mA} = 3.8 	ext{ mA}$.

$10^{10}$ electrons enter the emitter of a junction transistor in a time of $0.4 \mu s$. If $5 \%$ of the electrons are lost in the base, then the collector current is (in $\text{ mA}$)

Similar Questions

For a transistor,current gain $(\beta) = 50$. To change the collector current by $350 \mu A$,the base current should be changed by:

In a transistor amplifier, $AC$ current gain is $64$, the load resistance is $5400 \ \Omega$ and the input resistance of the transistor is $540 \ \Omega$. The voltage gain is

$A$ transistor connected in common emitter mode has a load resistance of $5 \, k\Omega$. If the input peak voltage is $5 \, mV$ and the current gain $(\beta)$ is $50$, find the voltage gain.

The correct relationship between $\alpha$ and $\beta$ for a transistor is ........

In a transistor,the base is very lightly doped as compared to the emitter because by doing so:

Vedclass Test Series

Exam Paper Generator

Online Exam Module