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?

Class 12th HC Verma - Concepts of Physics Part 2 45. Semiconductors and Semiconductor Devices

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

K.E._min = (0.2+0.1) × e

= 0.3eV

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