In an n-p-n common emitter (CE) transistor,th | vedclass

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

Question

(A) The current gain $(\beta)$ in a common emitter $(CE)$ transistor is defined as the ratio of the change in collector current $(\Delta I_C)$ to the

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(A) The current gain $(\beta)$ in a common emitter $(CE)$ transistor is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.Given:$\Delta I_C = 16\,mA - 5\,mA = 11\,mA = 11 	imes 10^{-3}\,A$$\Delta I_B = 200\,\mu A - 100\,\mu A = 100\,\mu A = 100 	imes 10^{-6}\,A$Using the formula:$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{11 	imes 10^{-3}\,A}{100 	imes 10^{-6}\,A} = \frac{11 	imes 10^{-3}}{10^{-4}} = 11 	imes 10^1 = 110$Thus,the current gain of the transistor is $110$.

In an $n-p-n$ common emitter $(CE)$ transistor,the collector current changes from $5\,mA$ to $16\,mA$ for a change in base current from $100\,\mu A$ to $200\,\mu A$,respectively. The current gain of the transistor is:

Similar Questions

In the given circuit, a voltmeter $V$ is connected across a bulb $B$. If the value of the resistor $R$ is increased, what happens?

When $n-p-n$ transistor is used as an amplifier :

In a transistor,the collector current is always less than the emitter current because:

In a common emitter transistor amplifier,the load resistance is $6.5 \text{ k}\Omega$ and the input resistance is $1.3 \text{ k}\Omega$. If the current gain is $78$,the voltage gain is:

The thinnest part of a transistor is . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module