Q. 17

The potential barrier existing across an unbiased p-n junction is 0.2 volt. What minimum kinetic energy a hole should have to diffuse from the p-side to the n-side if

(a) the junction is unbiased,

(b) the junction is forward biased at 0.1 volt and

(c) the junction is reverse biased at 0.1 volt?

Answer :

Potential barrier across unbiased p-n junction = 0.2 Volts

Minimum kinetic energy required for diffusion of a hole from p-side to the n-side,


K.E.min = (Potential barrier - biasing voltage) × Charge on hole


(a) Biasing Voltage=0V


K.E.min = (0.2-0) × e


= 0.2eV


(b) Biasing Voltage for forward biased=+0.1V


K.E.min = (0.2-0.1) × e


= 0.1eV


(c) Biasing Voltage for reverse biased=-0.1


K.E.min = (0.2+0.1) × e


= 0.3eV


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