A transistor is used as an amplifier in CB mo | vedclass

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

Question

(A) In common base $(CB)$ mode,the current gain $\alpha = 0.98$,load resistance $R_L = 5 \, k\Omega = 5000 \, \Omega$,and input resistance $R_{in} = 7

Vedclass · Accepted Answer

$70, 68.6$

Vedclass · Answer

(A) In common base $(CB)$ mode,the current gain $\alpha = 0.98$,load resistance $R_L = 5 \, k\Omega = 5000 \, \Omega$,and input resistance $R_{in} = 70 \, \Omega$.The voltage gain $(A_v)$ is given by the formula: $A_v = \alpha 	imes \frac{R_L}{R_{in}}$.Substituting the values: $A_v = 0.98 	imes \frac{5000}{70} = 0.98 	imes 71.428 \approx 70$.The power gain $(A_p)$ is given by the product of current gain and voltage gain: $A_p = \alpha 	imes A_v$.Substituting the values: $A_p = 0.98 	imes 70 = 68.6$.Thus,the voltage gain is $70$ and the power gain is $68.6$.

$A$ transistor is used as an amplifier in $CB$ mode with a load resistance of $5 \, k\Omega$. The current gain of the amplifier is $0.98$ and the input resistance is $70 \, \Omega$. The voltage gain and power gain respectively are:

Similar Questions

$A$ certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid,then:

$A$ plane $\pi$ makes intercepts $3$ and $4$ respectively on the $z$-axis and $x$-axis. If $\pi$ is parallel to the $y$-axis,then its equation is:

The shield-shaped cotyledon seen in monocot seeds is known as:

Bulk modulus of water is $2 \times 10^9 \ N/m^2$. The pressure required to increase the volume of water by $0.1 \%$ in $N/m^2$ is:

If $\int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = \frac{A \sqrt{x}}{\sqrt{1-x}} + \frac{B}{\sqrt{1-x}} + C$,where $C$ is a real constant,then $A+B$ equals to

Vedclass Test Series

Exam Paper Generator

Online Exam Module