Get the electric field in the depletion region as a function of position.
We can now go back to the
charge density as a function of position
graph and easily find the electric field in the depletion
region as a function of position. If we integrate
Gauss' Law , we get
for the electric field:
We
could write down an expression for
and then formally integrate it to get
but we can also just do it graphically, which is a lot
easier, and gives us a much more intuitive feeling for what isgoing on. Let's start doing our integral at [x equals -infinity] Whenever we perform an integral such as
, we've got to remember to add a constant to
our answer. Since we can not have an electric field whichextends to infinity (either plus or minus) however, we can
safely assume
and remains at that value until we get to the edge of the depletion region at
(essentially) x equals zero.Since the charge density is zero all the way up to the edge of depletion region, Gauss tells us that the electric field can not change here either. When we get to x=0 we encounter the large negative delta-function of negative charge at the edge of the depletion region. If you can remember back to your calculus, when you integrate a delta function, you get a step. Since the charge in the p-side delta function is negative, when we integrate it, we get a negative step. Since we don't know (yet) how big the step will be, let's just call it -|Emax|.
Finding the electric field in the p-type regionFinishing the integral
In the n-side of the depletion region
and so we plot
with a (positive) slope of
, starting at
at
x = 0.This line continues with this positive slope until it reaches a value of 0 at x =
. We know that E(x) must equal 0 at x =
because there is no further charge outside of the depletion region and E must be 0 outside this region.
We are now done doing the integral. We would know everythingabout this problem, if we just knew what
was. Note that since we know the slope of the triangle now, we can find
in terms of the slope and
. We can derive an expression for
, if we remember that the integral of
the electric field over a distance is the potential drop across that distance. What is the potentialdrop in going from the p-side to the n-side of the diode?
As a reminder,
shows the junction band
diagram again. The potential drop must just be
the "built-in" potential of the
junction. Obviously
can not be greater than 1.1 V, the
band-gap potential. On the other hand, by looking at
, and remembering that the bandgap in silicon
is 1.1 eV, it will not be some value like 0.2 or 0.4 voltseither. Let's make life easy for ourselves, and say
. This will not be too far off, and as you will see
shortly, the answer is not very sensitive to the
exact value of
anyway.
Band diagram for a p-n junction The integral of
is now just the area of the triangle in
. Getting the area is easy:
We can simply turn
around and solve for
.
As we said, for silicon,
. Let's let
donors. As we already know from before,
Coulombs. This makes
or 0.37 μm long. Not a very wide depletion region!
How big is
? Plugging in
We find
! Why such a big electric field? Well, we've got to
shift the potential by about a volt, and we do not have muchdistance to do it in (less than a micron), and so there must be,
by default, a fairly large field in the depletion region.Remember, potential is electric field
times distance.
Enough p-n junction electrostatics. The point of this exercisewas two-fold;
A)
so you would know something about the details of what is
really going on in a p-n junction
B)
to show you that with just some very simple electrostatics
and a little thinking, it is not so hard to figure these thingsout!
