P-N Junction: Part II2.142000/08/042007/08/14 11:46:13.031 GMT-5BillWilsonwlw@madriver.netBillWilsonwlw@madriver.netLiqunWangliqun@rice.eduElizabethGregoryelizabeth.gregory@gmail.comJeffreyMSilvermanJSilverman@astro.berkeley.eduGerardWysockigerardw@rice.edudepletion regionpn-junctionAdvanced discussion about P-N junction, especially focus on the affect of the depletion region in energy band diagram.
If you look closely at these pictures, you will notice
something. As we remove more and more electrons and holes, we
are starting to "uncover" the fixed charges associated with the
donors and acceptors. We are making what is known as a
depletion region, so named because it is
depleted of mobile carriers (holes and
electrons). The uncovered net charge in the depletion region is
separated, with negative charge in the p-region, and positive
charge in the n-region. What will such a charge separation give
rise to? Why, an electric field! Of course! Which way will the
field point? The electric field which arises from a separation
of charges always goes from the positive charge, towards the
negative charge. This is shown in .
The pn-junction with the resultant built-in electric field
What effect will this field have on our device? It will have
the tendency to push the holes back into the p-region and the
electrons into the n-region. This is just what we need to
counteract the recombination which has been going on, and
hopefully bring it to a stop.
Now try to think through what effect this field could have on
our energy band diagram. The band diagram is for electrons, so
if an electron moves from the right hand side of the device (the
n-region) towards the left hand side (the p-region), it will
have to move through an electric field which is opposing its
motion. This means it has do some work, or in other words, the
potential energy for the electron must go up. We can show this
on the band diagram by simply shifting the bands on the left
hand side upward, to indicate that there is a shift in potential
energy as electrons move from right to left across the junction.
Energy band diagram for a p-n junction at equilibrium
The shift of the bands, which is just the difference between the
location of the Fermi level in the n-region and the Fermi level
is the p-region, is called the built-in potential,
Vbi. This built-in potential keeps the
majority of holes in the p-region, and the electrons in the
n-region. It provides a potential barrier, which prevents
current flow across the junction. (On the band diagram we have
to multiply the built-in potential
Vbi by the charge of an electron,
q, so that we can represent the
shift in energy in terms of electron volts, the
unit of potential energy used in band diagrams.)
How big is Vbi? This is not too hard to figure out.
Let's look at a little more
carefully. Remember, we know from this equation and this equation
that since
nNd in the n-region and
pNa
in the p-region, we can relate the distance of the Fermi level
from
Ec and
Ef by
EcEfkTNcNd
and
EfEvkTNvNa
Look at and see if you can agree
that
qVBIEgEcEfEfEvEgkTNcNdkTNcNdEgkTNcNvNdNa
Where Nd and
Na are the doping densities in the n and
p sinc respectively. Remember,
kT140eV0.025eV,
Eg1.1eV and Nc and
Nv are both
≈1019. Thus,
qVBI1.1eV0.025eV1038NdNa Here the q in front of
the VBI and the
e in
eV are both the charge of 1
electron and they cancel out making
VBI1.10.0251038NdNavolts
Suppose both Nd and Na are both about [10 to the 15th] - not uncommon values. How big would the built-in potential be in this case?
It turns out that we can actually derive some specific details
about the depletion region if we make only a coupled of
simplifying (and often justified) assumptions. In order to make
the math easier, and also because many p-n
junctions are built this way, we will consider what is known as
a one sided junction. is
a picture of such a beast: In this diode, one side is much more heavily doped
than the other. In this particular example, the p-side is heavily doped, and the n-side is relatively lightly doped. We can not show the
true picture here, because typically, the more heavily doped
side will be doped several orders of
magnitude greater than the lightly doped
side. Typical values might be
Na1019 and
Nd1016. Regardless of how big the difference is however,
there must be exactly the same amount of "uncovered" charge on
both side of the junction. Why? Because each time a hole and
electron recombine to form the depletion region, they each leave
behind either a donor or an acceptor. A careful count of the
exposed charge in shows that I was careful enough to draw my figure accurately for you. We do not need to have a one-sided
diode to do the analysis that will follow, but the equations are
easier to solve if we do.
An example of a one-sided diode
In order to proceed from here, the first thing we do is make a
plot of the charge density
ρx as we move through the junction. Naturally, in the
bulk, since the holes and the acceptors (in the p-side), or the
electrons and the donors (in the n-side) just equal one another,
the net charge density is zero. In the depletion region, the
charge density is -
qNa on the p-side and
qNd on the donor side. (All the mobile carriers are gone,
and we are left with just the charged acceptors or donors.) We
will make the assumption that on the n-side, the depletion
extends a distance
xn from the junction. On the p-side, the acceptor charge
density is so large, that we will treat it is a
δ-function, with essentially no width. The areas of the
two boxes must be the same (equal amount of positive and
negative charge) and hence, the tall thin box actually has a
width of
NdNaxn, which, since
Na is several orders of magnitude greater than
Nd, means that the tall box has a very very small width
compared to the lower, wider one, which is
qNd tall, and has a width of
xn.