A transistor is connected in common emitter ( | vedclass

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

Question

(C) The current gain $\beta$ in common emitter configuration is related to $\alpha$ by the formula:$\beta = \frac{\alpha}{1 - \alpha} = \frac{0.96}{1

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(C) The current gain $\beta$ in common emitter configuration is related to $\alpha$ by the formula:$\beta = \frac{\alpha}{1 - \alpha} = \frac{0.96}{1 - 0.96} = \frac{0.96}{0.04} = 24$.The collector current $I_C$ is determined by the voltage drop across the collector resistor $R_C = 800\,\Omega$:$I_C = \frac{V_{drop}}{R_C} = \frac{0.5\ V}{800\,\Omega} = 0.000625\ A = 0.625 	imes 10^{-3}\ A$.The base current $I_B$ is related to the collector current by $I_C = \beta I_B$,so:$I_B = \frac{I_C}{\beta} = \frac{0.625 	imes 10^{-3}\ A}{24} \approx 0.02604 	imes 10^{-3}\ A$.$I_B \approx 26 	imes 10^{-6}\ A = 26\,\mu A$.

$A$ transistor is connected in common emitter $(CE)$ configuration. The collector supply is $8\ V$ and the voltage drop across a resistor of $800\,\Omega$ in the collector circuit is $0.5\ V$. If the current gain factor $\alpha$ is $0.96$,find the base current in $\mu A$.

Similar Questions

In a common base amplifier,the phase difference between the input and output voltage is .......

In a transistor,if $\frac{I_C}{I_E} = \alpha$ and $\frac{I_C}{I_B} = \beta$. If $\alpha$ varies between $\frac{20}{21}$ and $\frac{100}{101}$,then the value of $\beta$ lies between:

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

$A$ common emitter transistor amplifier has a current gain of $50$. If the load resistance is $9 \, k\Omega$ and the input resistance is $500 \, \Omega$,the voltage gain of the amplifier will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module