The dc common emitter current gain of an n-p- | vedclass

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

Question

(A) Given: Current gain $\beta = 50$,$V_{CE} = 2 \ V$,and $R_{C} = 4 \ k\Omega$.The collector current $I_{C}$ is determined by the output circuit loop

Vedclass · Accepted Answer

$I_{B} = 10 \ \mu A, I_{C} = 0.5 \ mA$

Vedclass · Answer

(A) Given: Current gain $\beta = 50$,$V_{CE} = 2 \ V$,and $R_{C} = 4 \ k\Omega$.The collector current $I_{C}$ is determined by the output circuit loop: $I_{C} = \frac{V_{CE}}{R_{C}} = \frac{2 \ V}{4 	imes 10^{3} \ \Omega} = 0.5 	imes 10^{-3} \ A = 0.5 \ mA$.The base current $I_{B}$ is related to the collector current by the current gain formula: $\beta = \frac{I_{C}}{I_{B}}$.Rearranging for $I_{B}$: $I_{B} = \frac{I_{C}}{\beta} = \frac{0.5 	imes 10^{-3} \ A}{50} = 0.01 	imes 10^{-3} \ A = 10 	imes 10^{-6} \ A = 10 \ \mu A$.Thus,$I_{B} = 10 \ \mu A$ and $I_{C} = 0.5 \ mA$.

Similar Questions

The current amplification factor of a transistor is $50$. The input resistance when used in common emitter mode is $1 \text{ k}\Omega$. The peak value of the collector current for an a.c. input voltage of $0.01 \text{ V}$ peak is:

$A$ common emitter amplifier has a voltage gain of $50$,an input impedance of $100\; \Omega$,and an output impedance of $200\; \Omega$. The power gain of the amplifier is:

In the following common emitter circuit, if $\beta=100$, $V_{CE}=7 \, V$, $V_{BE}$ is negligible, and $R_C=2 \, k\Omega$, then $I_B=$ (in $\, mA$)?

In the case of a common emitter transistor amplifier,the ratio of collector current to emitter current is $0.96$. Then,the current gain $(\beta)$ of the amplifier is:

In an $NPN$ transistor,the collector current is $24 \, mA$. If $80\%$ of electrons reach the collector,its base current in $mA$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module