A transistor is connected in common emitter CE configuration- The collector supply is 8V and the voltage drop across a resistor of 800 - in the collector circuit is 0-8 V- If the current gain factor - is 0-96- then the change in base current is

The correct option is A 124mA The collector current, I_C=V_CER_E #160; =8V800Ω #160; =1mA The current gain, β=I_CI_B Since, β ...

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Standard XII

Physics

P-N Junction

A transistor ...

Question

A transistor is connected in common emitter (CE) configuration. The collector supply is 8V and the voltage drop across a resistor of $800 Ω$ in the collector circuit is 0.8 V. If the current gain factor ( $α$ ) is 0.96, then the change in base current is

\frac{1}{24} m A

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\frac{1}{12} m A

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\frac{1}{6} m A

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\frac{1}{3} m A

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Solution

The correct option is A $\frac{1}{24} m A$
The collector current, $I_{C} = \frac{V_{C E}}{R_{E}}$ $= \frac{8 V}{800 Ω}$ $= 1 m A$
The current gain, $β = \frac{I_{C}}{I_{B}}$
Since, $β = \frac{α}{1 α} = \frac{0.96}{1 - 0.96} = 24$

Hence, $I_{B} = \frac{I_{C}}{β}$ $= \frac{1}{24} m A$

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A transistor is connected in common emitter (CE) configuration. The collector supply is 8V and the voltage drop across a resistor of 800Ω in the collector circuit is 0.8 V. If the current gain factor (α) is 0.96, then the change in base current is

The correct option is A 124mAThe collector current, IC=VCERE =8V800Ω =1mAThe current gain, β=ICIBSince, β=α1α=0.961−0.96=24Hence, IB=ICβ =124mA

A transistor is connected in common emitter (CE) configuration. The collector supply is 8V and the voltage drop across a resistor of $800 Ω$ in the collector circuit is 0.8 V. If the current gain factor ( $α$ ) is 0.96, then the change in base current is

The correct option is A $\frac{1}{24} m A$
The collector current, $I_{C} = \frac{V_{C E}}{R_{E}}$ $= \frac{8 V}{800 Ω}$ $= 1 m A$
The current gain, $β = \frac{I_{C}}{I_{B}}$
Since, $β = \frac{α}{1 α} = \frac{0.96}{1 - 0.96} = 24$

Hence, $I_{B} = \frac{I_{C}}{β}$ $= \frac{1}{24} m A$