The minimum potential difference between the base and emitter required to switch a silicon transistor $O N$ is approximately

1 V

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3 V

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5 V

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4.2 V

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Solution

The correct option is A $1 V$
To switch on the transistor the emitter-base junction of a transistor is forward biased while collector-base junction is reversed biased. The cut off voltage for silicon is $1 V,$ so to switch on a silicon transistor a potential difference of $1 V$ approximately is required between the base and emitter.

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I_{C} =

V_{C E} =

I_{B} =

V_{B B} =

V_{B E} =

β = 45

1 k Ω

5 V

β = 100

0.7 V

V_{C E}

(R = r_{1} = r_{2} = r_{3} = 1 Ω, E_{1} = 3 V, E_{2} = 2 V, E_{3} = 1 V)

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