A transistor is connected in common emitter c | vedclass

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(C) Given:Collector resistance $R_C = 1000 \, \Omega$Voltage drop across collector resistor $\Delta V_C = 0.6 \, V$Current gain $\beta = 24$The collec

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$25$

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(C) Given:Collector resistance $R_C = 1000 \, \Omega$Voltage drop across collector resistor $\Delta V_C = 0.6 \, V$Current gain $\beta = 24$The collector current $I_C$ is given by Ohm's law:$I_C = \frac{\Delta V_C}{R_C} = \frac{0.6 \, V}{1000 \, \Omega} = 0.6 	imes 10^{-3} \, A = 6 	imes 10^{-4} \, A$The relationship between collector current $I_C$ and base current $I_B$ is:$I_C = \beta 	imes I_B$Therefore,the base current $I_B$ is:$I_B = \frac{I_C}{\beta} = \frac{6 	imes 10^{-4} \, A}{24} = 0.25 	imes 10^{-4} \, A$Converting to microamperes $(\mu A)$:$I_B = 0.25 	imes 10^{-4} 	imes 10^6 \, \mu A = 25 \, \mu A$.

Similar Questions

In the study of a transistor as an amplifier,if $\alpha = \frac{I_C}{I_E} = 0.98$ and $\beta = \frac{I_C}{I_B} = 49$,where $I_C, I_B,$ and $I_E$ are collector,base,and emitter currents respectively,then $\left(\frac{1}{\alpha} - \frac{1}{\beta}\right)$ is equal to:

The base current is $100 \mu A$ and the collector current is $3 \text{ mA}$. $A$ change of $20 \mu A$ in the base current results in a change of $0.5 \text{ mA}$ in the collector current. The value of $\beta_{ac}$ is:

The given transistor operates in the saturation region. What should be the value of $V_{BB}$ (in $Volt$)?
$(R_{out} = 200 \, \Omega, R_{in} = 100 \, k\Omega, V_{CC} = 3 \, V, V_{BE} = 0.7 \, V, \beta = 200)$

For a transistor,$\alpha$ and $\beta$ are given as $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$. Then the correct relation between $\alpha$ and $\beta$ is:

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