Question

The Width of a depletion region is 400 nm.The intensity of the electric field at the depletion region is 5*10⁵ V/m. Then calculate the following quantities : (1) The value of the potential barrier. (2) The minimum energy required by an electron to move from the N-type to the P-type region of the diode. [Ans: 0.2 V, 0.2 eV]

Answer 1

Answer:

N. A. = 2 × 10. 16 cm. −3 . Calculate the built-in potential V bi in eV and the total width of the depletion region W = xn0 + xp0 at zero.

Answer 2

Answer:

Explanation:

Width of a depletion region = 400nm

Intensity of the electric field at the depletion region = 5 × 10⁵ V/m

The potential difference is the the difference of an electrical potential between the two points. Thus,

P.d (v) = ED and w = qv

v =  5 × 10⁵ × 400 × 10`9

v = 20 × 10-²

v = 0.2 ev

Thus, the minimum energy required by an electron to move from the N-type to the P-type region of the diode is 0.2 ev