The current gain beta of a transistor is 50. | vedclass

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

Question

(D) Given: Current gain $\beta = 50$, Input resistance $R_i = 1\,k\Omega = 1000\,\Omega$, Input peak voltage $V_i = 0.01\,V$.First, calculate the peak

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(D) Given: Current gain $\beta = 50$, Input resistance $R_i = 1\,k\Omega = 1000\,\Omega$, Input peak voltage $V_i = 0.01\,V$.First, calculate the peak input base current $(I_b)$:$I_b = \frac{V_i}{R_i} = \frac{0.01\,V}{1000\,\Omega} = 10^{-5}\,A$.Using the relation for current gain in common emitter configuration:$\beta = \frac{I_c}{I_b}$Therefore, the peak collector current $(I_c)$ is:$I_c = \beta 	imes I_b = 50 	imes 10^{-5}\,A$.Converting to microamperes $(\mu A)$:$I_c = 50 	imes 10^{-5} 	imes 10^6\,\mu A = 500\,\mu A$.

The current gain $\beta$ of a transistor is $50$. The input resistance of the transistor when used in the common emitter mode is $1\,k\Omega$. The peak value of the collector current for an input peak voltage of $0.01\,V$ is ...... $\mu A$.

Similar Questions

In a transistor,a change of $8.0\, mA$ in the emitter current produces a change of $7.8\, mA$ in the collector current. What change in the base current is necessary to produce the same change in the collector current? (in $\mu A$)

$A$ transistor has $3$ impurity regions of different doping levels. In the order of increasing doping level,the regions are

The current amplification factor of a transistor is $50$. In a $CE$ amplifier circuit, the collector resistance is $5 \, k\Omega$ and the input resistance is $1 \, k\Omega$. If the input voltage is $0.01 \, V$, the output voltage is ........ $V$.

In a transistor circuit,the emitter-base circuit of a $n-p-n$ transistor is always

In an $n-p-n$ transistor,$200$ electrons enter the emitter in $10^{-8} \ s$. If $1 \%$ of electrons are lost in the base,then the current that enters the emitter and the current amplification factor are respectively $\left[e=1.6 \times 10^{-19} \ C\right]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module