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:
Ex=1ε∫ρxdx
E
x
1
ε
x
ρ
x
(1)
We
could write down an expression for
ρx
ρ
x
and then formally integrate it to get
Ex
E
x
but we can also just do it graphically, which is a lot
easier, and gives us a much more intuitive feeling for what is
going on. Let's start doing our integral at [x equals -infinity] Whenever we perform an integral such as
Equation 1, we've got to remember to add a constant to
our answer. Since we can not have an electric field which
extends to infinity (either plus or minus) however, we can
safely assume
E-infinity=0
E
-infinity
0
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|.
In the n-side of the depletion region
ρx=+q
N
d
=ε∂∂xE
ρ
x
q
N
d
ε
x
E
(2)
and so we plot
Ex
E
x
with a (positive) slope of
q
N
d
ε
q
N
d
ε
, starting at
Ex=-Emax
E
x
-Emax
at
x = 0.
This line continues with this positive slope until it reaches a value of 0 at x =
xn
xn
. We know that E(x) must equal 0 at x =
xn
xn
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 everything
about this problem, if we just knew what
xn
xn
was. Note that since we know the slope of the triangle now, we can find -Emax
-Emax
in terms of the slope and xn
xn
. We can derive an expression for
xn
xn
, if we remember that the integral of
the electric field over a distance is the potential drop across that distance. What is the potential
drop in going from the p-side to the n-side of the diode?
As a reminder, Figure 3 shows the junction band
diagram again. The potential drop must just be
Vbi
Vbi
the "built-in" potential of the
junction. Obviously
Vbi
Vbi
can not be greater than 1.1 V, the
band-gap potential. On the other hand, by looking at Figure 3, and remembering that the bandgap in silicon
is 1.1 eV, it will not be some value like 0.2 or 0.4 volts
either. Let's make life easy for ourselves, and say
V
bi
=1Volt
V
bi
1
Volt
. This will not be too far off, and as you will see
shortly, the answer is not very sensitive to the
exact value of
Vbi
Vbi
anyway.
The integral of
Ex
E
x
is now just the area of the triangle in
Figure 2. Getting the area is easy:
area=
1
2
base
×
height
=1/2
x
n
q
N
d
x
n
ε=q
N
d
x
n
22ε=
V
bi
area
1
2
base
×
height
12
x
n
q
N
d
x
n
ε
q
N
d
x
n
2
2
ε
V
bi
(3)
We can simply turn
Equation 3 around and solve for
xn
xn
.
x
n
=2ε
V
bi
q
N
d
x
n
2
ε
V
bi
q
N
d
(4)
As we said, for silicon,
ε
Si
=1.1×10-12
ε
Si
1.1-12
. Let's let
N
d
=1016cm-3
N
d
10
16
cm
-3
donors. As we already know from before,
q=1.6×10-19
q
1.6-19
Coulombs. This makes
x
n
=3.7×10-5cm
x
n
3.7-5
cm
or 0.37 μm long. Not a very wide depletion region!
How big is
|
E
max
|
E
max
? Plugging in
E
max
=q
N
d
x
n
ε
E
max
q
N
d
x
n
ε
(5)
We find
|
E
max
|=53,000Vcm
E
max
53,000
V
cm
! Why such a big electric field? Well, we've got to
shift the potential by about a volt, and we do not have much
distance 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 exercise
was 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 things
out!
"This course offers an introduction to solid state device including field effect and bipolar transistors. Properties of transmission lines and propagating E&M waves are also presented. It is […]"