Question

Why is electric field is maximum at metallurgical junction of a pn junction

Answer 1

Let us imagine that a pn junction where p-type is on the left side and n-type is on the right side. Further, let the junction position be at the origin such that the width of depletion region is w = xp + xn where

xp is a distance of depletion region invading into the p-type (i.e., left side of the origin)

xn is a distance of depletion region invading into the n-type (i.e., right side of the origin)

........... -xp .............. 0 ............. xn ..........

Suppose that p-type has acceptor concentration of Na and n-type has donor concentration of Nd. Since the number of negative ions (due to ionized acceptors) and the number of positive ions (due to ionized donors) in the depletion region must be equal, we get

q A Na xp = q A Nd xn 

where q is the electron charge and A is the cross-sectional area of the junction.

It means that the larger part of depletion region is in the lower doping side. For instance, if Na > Nd, then xp < xn. Also, it will be shown later that the maximum electric field is NOT always at the center of depletion region if the doping levels of p- and n-type are not equal

To find the electric field (E) with the charge density already known, Poisson's equation is recalled. Two equations are consequently obtained:

For 0 < x < xn :      dE / dx = q Nd / epsilon      ......(1)

For -xp < x < 0 :     dE / dx = - q Na  / epsilon   ......(2)

Eq. (1) tells that E increases when x increases from 0 to xn and Eq. (2) indicates that E decreases when x increases from -xp to 0. With intuitive thinking, there must be a maximum electric field at some certain position within the depletion region. And it's done! That position is the junction (i.e., the origin).

Again, the maximum electric field is at the center of depletion region only if the doping levels of p- and n-type are equal.