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?
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?
Potential barrier = 0.2 V
(a) The minimum kinetic energy of the hole should be equal to the band gap of the material.
Band gap = eV
KE = Potential difference × e = 0.2 eV
(b) In forward biassing,
Kinetic energy = Effective potential of the barrier
∴ Kinetic energy = Potential under unbiased condition - Applied voltage
⇒ KE + Ve = 0.2 eV
Here, V is the applied voltage.
⇒ KE = 0.2 − 0.1 = 0.1 eV
(c) In reverse biassing,
Kinetic energy = Effective potential of the barrier
∴ Kinetic energy = Potential under unbiased condition + Applied voltage
⇒ KE − Ve = 0.2 eV
Here, V is the applied voltage.
⇒ KE = 0.2 + 0.1 = 0.3 eV
Potential barrier = 0.2 V
(a) The minimum kinetic energy of the hole should be equal to the band gap of the material.
Band gap = eV
KE = Potential difference × e = 0.2 eV
(b) In forward biassing,
Kinetic energy = Effective potential of the barrier
∴ Kinetic energy = Potential under unbiased condition - Applied voltage
⇒ KE + Ve = 0.2 eV
Here, V is the applied voltage.
⇒ KE = 0.2 − 0.1 = 0.1 eV
(c) In reverse biassing,
Kinetic energy = Effective potential of the barrier
∴ Kinetic energy = Potential under unbiased condition + Applied voltage
⇒ KE − Ve = 0.2 eV
Here, V is the applied voltage.
⇒ KE = 0.2 + 0.1 = 0.3 eV
