In the shown common emitter amplifier circuit | vedclass

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

Question

(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:$8.7 - I_B(200 	imes 10^3) - V_{BE} = 0$$8.7 - I_B(200 	imes 10^3) - 0.7 = 0$$8 = I_B

Vedclass · Accepted Answer

$2.5\,k\Omega$

Vedclass · Answer

(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:$8.7 - I_B(200 	imes 10^3) - V_{BE} = 0$$8.7 - I_B(200 	imes 10^3) - 0.7 = 0$$8 = I_B(200 	imes 10^3)$$I_B = \frac{8}{200 	imes 10^3} = 4 	imes 10^{-5}\,A = 40\,\mu A$Given $\beta = 80$,we find the collector current $I_C$:$I_C = \beta I_B = 80 	imes 4 	imes 10^{-5} = 320 	imes 10^{-5} = 3.2 	imes 10^{-3}\,A = 3.2\,mA$Applying $KVL$ to the output loop:$11 - I_C R_C - V_{CE} = 0$$11 - (3.2 	imes 10^{-3}) R_C - 3 = 0$$8 = (3.2 	imes 10^{-3}) R_C$$R_C = \frac{8}{3.2 	imes 10^{-3}} = 2.5 	imes 10^3\,\Omega = 2.5\,k\Omega$

In the shown common emitter amplifier circuit,$\beta = 80$ and $V_{BE} = 0.7\,V$. The value of $R_C$ is:

Similar Questions

In a common emitter amplifier, a change of $0.2 \text{ mA}$ in the base current causes a change of $5 \text{ mA}$ in the collector current. If input resistance is $2 \text{ k}\Omega$ and voltage gain is $75$, the load resistance used in the circuit is

In the following transistor amplifier circuit,$\beta = 50$. The $V_{CE}$ of the transistor is .......... $V$.

What is the basic constitutional unit in the electronic circuit?

The load resistance and the base-emitter voltage of a transistor in $CE$ configuration are $5 \ k\Omega$ and $0.02 \ V$ respectively. If the collector current is $2 \ mA$,then its voltage gain is

The power gain and voltage gain of a transistor connected in common emitter configuration are $1800$ and $60$ respectively. If the change in the emitter current is $0.62 \text{ mA}$, then the change in the collector current is: (in $\text{ mA}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module