A change of 0.04 V takes place between the b | vedclass

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

Question

(B) Given: $\Delta I_b = 20 \ \mu A = 20 	imes 10^{-6} \ A$,$\Delta I_c = 2 \ mA = 2 	imes 10^{-3} \ A$,and $\Delta V_{BE} = 0.04 \ V$.The input res

Vedclass · Accepted Answer

$2 \ k\Omega \ \& \ 100$

Vedclass · Answer

(B) Given: $\Delta I_b = 20 \ \mu A = 20 	imes 10^{-6} \ A$,$\Delta I_c = 2 \ mA = 2 	imes 10^{-3} \ A$,and $\Delta V_{BE} = 0.04 \ V$.The input resistance $R_{	ext{input}}$ is defined as the ratio of the change in base-emitter voltage to the change in base current:$R_{	ext{input}} = \frac{\Delta V_{BE}}{\Delta I_b} = \frac{0.04}{20 	imes 10^{-6}} = \frac{0.04 	imes 10^6}{20} = 2 	imes 10^3 \ \Omega = 2 \ k\Omega$.The $AC$ current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:$\beta = \frac{\Delta I_c}{\Delta I_b} = \frac{2 	imes 10^{-3}}{20 	imes 10^{-6}} = \frac{2000}{20} = 100$.Thus,the input resistance is $2 \ k\Omega$ and the $AC$ current gain is $100$.

Similar Questions

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

In a $p-n-p$ transistor working as a common base amplifier, when the current gain is $0.96$ and emitter current is $7.2 \,mA$, the base current is (in $\,mA$)

In common emitter mode of a transistor, the current gain is $8$. The input impedance is $25 \, k\Omega$ and load resistance is $75 \, k\Omega$. The power gain is

$A$ transistor having a $\beta$ equal to $80$ has a change in base current of $250 \mu A$. What is the change in collector current?

Vedclass Test Series

Exam Paper Generator

Online Exam Module