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Magnetic Flux and Faraday's Law

The transfer ...

Question

The transfer characteristic curve for a based transistor is common-emitter configuration is shown in figure where $v_{0}$ represents output voltage and $v_{1}$ is input voltage. For the action of transistor as amplifier.

Region I of graph is used

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Region II of graph is used

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Region III of graph is used

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Both regions I and III are used

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Solution

The correct option is B Region II of graph is used
Region (I) is the cut-off region, region (II) is the active region and region (III) is the saturation region.
The region (II) of transfer characteristic is an active region and hence it is used as an amplifier. So, option (B) is correct.

(If we use a transistor as a switch, the cut off region (I) is used for OFF state and the saturation region (III) is used for ON state)

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Similar questions

(V_{0})

(V_{1})

(V_{0})

(V_{i})

(r_{i})

β

(R_{i}

= {[\frac{Change in emitter base voltage (△ V_{B E})}{Change in base current (△ I_{B})}]}_{at constant V_{C E}}

R_{i} = {(\frac{△ V_{B E}}{△ I_{B}})}_{V_{C E}}

(β) = {(\frac{△ I_{C}}{△ I_{B}})}_{V_{C E}}

\frac{△ I_{C}}{△ I_{E} - △ I_{C}}

\frac{\frac{△ I_{C}}{△ I_{E}}}{\frac{△ I_{C}}{△ I_{E}} - \frac{△ I_{C}}{△ I_{E}}}

\frac{α}{1 - α}

\Rightarrow β = \frac{α}{1 - α}

I_{B}

I_{C}

V_{C E}

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