Let me pick from where we left in the last video.
So we were discussing the basics of p-n junction diode.
And I had derived these two relationships.
So, they relate the depletion width on the p side, and the depletion width on the
n side, and these are the two variables, and these are the two equations.
So this is equation number one. And equation number two.
And I can solve
for these two equations and I can express these
depletion widths on the n and p side regions.
And I can express them in term of these known numbers, such as the doping concentration
of acceptors on the p side and the doping concentration for donors on the n side.
And this other term which is the built-in
voltage, which I'll talk more about in this video.
So
I get these two relationships, and I get these relationships for xn and xp.
And then I can further essentially express my overall total depletion width,
which is the sum of the depletion width on the n side
and the sum of the depletion width on the p side.
And I get, this formula for, for my depletion width.
Where it's, related to the doping, in, in, this
form.
So one thing which I can immediately recognize is that my repletion width
is essentially inversely related, to the doping I have in that, region.
And let me illustrate that by using this equation.
So this says that the overall charge neutrality is has to be maintained.
So the depletion width, multiplied by the doping density,
has to be equal on both the p and the
n side. So now, let me consider this case.
So this, let's say this junction was p-n
junction with equal doping of p and n types.
But I'll consider a asymmetrically doped p-n junction.
So I'll Essentially denote this by p+ meaning
a much higher doping on the p side.
And then, this is going to form a junction
with with with this n-type semiconductor, which has
a much lower doping concentration as compared to this p type.
So what I see over here is that my Nd, which is related
to the doping of the n region Is much smaller as compared to Na.
So I have this relationship over here.
So what I have over here is Na is much larger
as compared to Nd, and what that would result in is
that Xn would be much larger as compared to Xp,
because the multiple of these two has to be the same.
So what is indicating is that if I have this p+-n
junction, most of my depletion width would essentially lie in
the n region, because my my depletion width in the n
region is going to be much larger as compared to the depletion width
in the p region. So I get this relationship,
where my Xn is going to be much larger than Xp.
And I can observe that using the individual
equations for the depletion widths in the n and p
regions as well, so what I've stated is that Na is much larger than Nd, so
the division of this division of the two or Na by Nd has
to be much larger than one, and I see Na by Nd occurring
over here, so my Xn is going to be very large.
The same time, I have this term in reverse over here, so I have Nd by
Na which is going to be very small, so Xp is going to be very small.
So, what a, a good statement to, you know, always, keep at the back of your
mind, is that, depletion, most of the depletion
width always lies in the lightly doped region.
In the lightly doped region.
So, if you have asymmetric junction, most of your depletion region is going to be
occurring in the lightly doped region, which
happened to be n-type region in this case.
And, this is something which is used you know, used
very often in other devices which make use of p-n diodes.
So the only other term which is I need to, you know,
investigate a little more is what is this Vbi, and how is
it related to the doping in the p and n regions?
So let me make use of you know, some band diagrams over here.
So I'll draw essentially a p-type semiconductor and
in my p-type semiconductor, my Fermi level is close to the valence band.
And then I have this n-type semiconductor in which my Fermi level
is close to the conduction band. This is
my p-type semiconductor and this is my n-type semiconductor.
Also, I know that for reference, also for reference, I draw the
intrinsic energy level, which approximately lies in the middle of your bandgap.
So this is my conduction band.
Let me buy it write in the same color.
So this is my conduction band. This is my valence band.
Similarly over here.
This is my conduction band, this is my valence band.
This is my Fermi level, this dotted blue line.
And then this dotted yellow line is my intrinsic energy level.
So, when I'm in equilibrium I can, essentially I have no current flowing
so I can connect the Fermi level between my p and n regions.
So, this would be connected. And then
I can, you know, invert my potential profile and then I can
connect the p and n regions using this continuity in the band.
So, I'll I'll just the invert the shape of my potential profile.
And that would be the same shape of this intrinsic energy level as well.
So now the concentration in my p and n semiconductors are related to
the, how far is this Fermi level from my from my intrinsic energy level.
So I know that for n-type semiconductor, the number of holes is
essentially related to this this difference in the position of the Fermi level.
Let me call it Phi fn and it's related to
essentially difference between Ef and Ei by this relationship where it
varies exponentially, depending upon how far is my
Fermi level from my intrinsic energy level? So I know this relationship.
And then I also know that, for my, for my p type semiconductors,
I'll call this, Phi p The number of holes
is essentially related to
Is, related to, again, the intrinsic, carrier concentration.
And it depends upon how below is my Fermi level from my intrinsic energy level.
So it's related by, again this this relationship.
And these these you know, potentials, my built-in potential
in my p-n junction is essentially the total
potential difference between this p and n region.
So this is my built-in potential, and I can,
you know, take that reference, instead of taking that with
a reference referencing that to the conduction band, I
can reference that same built-in potential to this Fermi level.
And I see that this built-in potential is equal to, this built-in
potential is equal to this Phi p. That is, you know,
the difference between the Fermi level and the intrinsic energy level on the p side,
plus this phi n, which is equal to the difference in
the Fermi level and the intrinsic energy level on the n side.
And these are related to you know these are related to the carrier concentration
using these two formulas.
And I also know that my carrier concentration
on the n-type region is equal to Nd.
So this would be equal to Nd and similarly this concentration of
holes on my p-type region, this would be equal to Na.
So then I can take a log of these two equations and that basically expressed,
gives me an expression for this phi n and phi p.
So this phi n, so let me correct this mistake, this says phi n, so this phi
n on the n-type semiconductor, it depends upon
the doping doping density in the n-type semiconductor.
Similarly this phi p it depends upon the
concentration of acceptors in the p-type semicondutor and
this built in voltage is basically It's equal to the sum of this phi p and phi n.
And that depends upon the product of the doping density of the product on the n side and the
produce of the doping density on the p side, and it's given by this formula.