Question

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 1

Solution:

The potential barrier is 0.2 V

(a) The hole’s minimum kinetic energy must be equal to the material’s band gap.

Gap’s band = eV

KE = e x potential difference = 0.2 eV

(b) In forward biasing,

Kinetic energy = Effective potential of the barrier

Therefore, Kinetic energy = – Applied voltage + Potential with unbiased state

⇒ Ve + KE = 0.2 eV

Here, the applied voltage is denoted as V

⇒ KE = − 0.1 + 0.2 = 0.1 eV

(c) In reverse biasing,

Kinetic energy = Barrier’s effective potential  

Therefore, Kinetic energy = Applied voltage + Unbiased condition’s potential

⇒ KE − Ve = 0.2 eV

Here, the applied voltage is V

⇒ KE = 0.1 + 0.2 = 0.3 eV

Answer 2

Answer:

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Explanation:

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