Page 187 - Engineering Electromagnetics, 8th Edition

CHAPTER 6 Capacitance 169

6.8 EXAMPLE OF THE SOLUTION
OF POISSON’S EQUATION: THE P-N
JUNCTION CAPACITANCE

To select a reasonably simple problem that might illustrate the application of Poisson’s
equation, we must assume that the volume charge density is specified. This is not
usually the case, however; in fact, it is often the quantity about which we are seeking
further information. The type of problem which we might encounter later would
begin with a knowledge only of the boundary values of the potential, the electric
field intensity, and the current density. From these we would have to apply Poisson’s
equation, the continuity equation, and some relationship expressing the forces on
the charged particles, such as the Lorentz force equation or the diffusion equation,
and solve the whole system of equations simultaneously. Such an ordeal is beyond
the scope of this text, and we will therefore assume a reasonably large amount of
information.
As an example, let us select a pn junction between two halves of a semiconductor
bar extending in the x direction. We will assume that the region for x < 0is doped p
type and that the region for x > 0is n type. The degree of doping is identical on each
side of the junction. To review some of the facts about the semiconductor junction,
we note that initially there are excess holes to the left of the junction and excess
electrons to the right. Each diffuses across the junction until an electric field is built
up in such a direction that the diffusion current drops to zero. Thus, to prevent more
holes from moving to the right, the electric field in the neighborhood of the junction
must be directed to the left; E x is negative there. This field must be produced by a net
positive charge to the right of the junction and a net negative charge to the left. Note
that the layer of positive charge consists of two parts—the holes which have crossed
the junction and the positive donor ions from which the electrons have departed.
The negative layer of charge is constituted in the opposite manner by electrons and
negative acceptor ions.
The type of charge distribution that results is shown in Figure 6.12a, and the
negative field which it produces is shown in Figure 6.12b. After looking at these two
figures, one might profitably read the previous paragraph again.
A charge distribution of this form may be approximated by many different
expressions. One of the simpler expressions is
x x
ρ ν = 2ρ ν0 sech tanh (44)
a a
which has a maximum charge density ρ v,max = ρ v0 that occurs at x = 0.881a. The
maximum charge density ρ v0 is related to the acceptor and donor concentrations N a
and N d by noting that all the donor and acceptor ions in this region (the depletion
layer) have been stripped of an electron or a hole, and thus
ρ v0 = eN a = eN d
We now solve Poisson’s equation,
2 ρ ν
∇ V =−