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pn Junctions – Zero and Forward Bias
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来自 Columbia University 的课程
MOS 晶体管
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与讲师见面
Yannis TsividisCharles Batchelor Professor of Electrical Engineering
Fu Foundation School of Engineering and Applied Science
In this video, we will discuss an important combination of materials for us,
the pn junction, a junction between a p and an n region.
We start with pn junctions with zero and forward biased.
Let's start with zero biased. To biased the junction with 0 volts, we
connect an external wire, as you can see here, between n and p.
All of this region is n. And all of the regions is p.
I'm assuming that this is a Step junction, which means that we have uniformly doped n
region and a uniformly doped p region, and there is a sharp boundary between them.
Now, because you have plenty of electrons here, electron, and very few electrons
here in the p region, electrons tend to diffuse downwards and at the same time,
holes are in plenty of supply in the p region and they tend to diffuse upwards.
When electrons diffuse downwards, they leave behind positively charged donor
atoms and when holes diffuse upwards, they leave behind negatively charged acceptor
atoms. The region here, consisting of positively
charged donor atoms and negatively charged acceptor atoms, is called the depletion
region. We will make the simplifying assumption
that the edges of this depletion region are sharply defined, as shown in the
figure. Overall, the structure is neutral, because
these pluses cancel these minuses. Now, why doesn't this process continue?
Why don't even more electrons diffuse down and holes diffuse up?
The reason is that you have now developed an electric field in this direction, which
tends to prevent holes from moving up and also tends to prevent electrons form
moving down. You have a charged density in the
depletion region and when you integrate it twice, you end up with a potential across
the depletion region. This would be called the built-in
potential of the junction. Now, how is it that we have a potential
here despite the fact that we have 0 potential difference between the contact
of the junction? Well, the reason is understood if you use
the material in the, the contact potential lecture, there is not only a contact
potential between n and p, but there is a contact potential between n and this
material here, that I used to contact n. There is a contact potential from here to
there, to the y, similarly here. And if you start this point and end up
with this point, the sum of all contact potential is 0, which is consistent with
the fact that you have sorted the junction.
If you break the wire here, the situation remains the same as when you have the
junction sorted. Now, let's see how the energy bands look
like in this structure that is in equilibrium.
Since it is in equilibrium, we have a single Fermi level throughout.
And locally, in the inside, by the way, this is the energy band that you get when
you take this structure and you rotate it 90 degrees in this direction.
So, n comes to the left and p comes to the right.
This is the n-side and this is the p-side. Now, let us look at the situation on the n
and p sides, away from the diffusion region.
Locally, on the n -side, the situation looks like I have shown to you for
equilibrium and type material in the past. And luckily over here, this usually, this
situation looks like it should for p-type material.
For this to be consistent with the fact that the Fermi level is constant
throughout, the bands have to bend, as shown like this.
And as we know, the bottom of the conduction band represents potential
energy for electrons. And if it goes up, it means that the
corresponding potential goes down because the charge of an electron is negative.
Energy per unit energy divided by the negative charge, will give you a delta C
that will be the opposite from delta EC, the change in the conduction band level.
The total change in potential is the built-in potential.
It's this quantity over here. And therefore, in the energy band,
multiply by 2 and you get the corresponding energy band bending here.
We know already that the contact potential of this region with respect to this region
is going to be the difference between the corresponding work functions, but an
easier way to see what happens here is this.
The total band bending here is the sum of this distance here which is q phi Fn.
Q is the electron charged band [unknown] and phi Fn is the Fermi potential for the
n-type region, which we have discussed before.
And the total distance here is correspondingly q phi Fp.
Phi Fp being the Fermi potential for the p-side.
So then, the total bending, the total bending for this or for this level, is the
sum of this and that. But because they have opposite signs, you
have to take this one and subtract this negative quantity to get the total.
So phi Bi is the built-in potential of the junction and depends, of course, on the
doping concentration because the Fermi potentials depend on doping concentration.
It can be several tenths of one volt, close to a volt depending on doppler
concentration. Let us know look at the same junction but
forward biased. So, we apply a voltage V between the
p-side and the n-side. Now, this positive terminal here has a
tendency to make holes moving in that direction and the negative terminal here
has a tendency to make the electrons moving the opposite direction.
What I'm saying is just the fundamental logical way of explaining things to help
you remember things. The, the actual physics of this are rather
complicated. The net result is that now, the depletion
region width becomes less than before. And because we are pushing through the,
the, the supplication of potential, we are pushing holes go up and electrons go down,
there's a net positive current that flows through the junction.
And if you plot I verses U, you find the well-known exponential relation for pn
junctions or diode. Now, let's look at the energy band because
we are a non-equilibrium. And why are we non-equilibrium?
Because now, the structure can exchange energy with the external world.
The external world being the, the voltage source that we have applied.
The energy bands now look this. Far away from the divisional region and
the n material we still have the same situation as before.
And far away from the depletion region, we still have on the p-sides, the same
situation as before. Because of the polarity of the externally
voltage which has the plus connected to the p and the minus connected to the n
there is a higher energy on the n-side, for the electrons there.
And therefore the quasi-Fermi energy level e sub Fn on the n-side is higher than the
corresponding quasi-Fermi level is e sub Fp on the p-side.
For this to happen, the bands must bend. You will recall in the previous slide when
we had zero bias, that bending was q times phi Bi, the built-in potential.
So now, they will bend, but they will bend less by an amount corresponding to the
difference of the two Fermi levels. And that difference is basically V, the
externally applied voltage. So, instead of q phi Bi, you now have q
times phi Bi minus V. That means that the corresponding change
in potential versus distance is now phi Bi minus V, which shows that the total band
bending decreases by an amount which depends on the externally applied voltage,
V. Large currents can flow and one way to
describe this effect is to say that before the band bending was larger, which means
that from the n-side, the electron phase, the energy barrier which was large enough
to prevent many of them from flowing. But now, the energy barrier has been
reduced, the band bending is less, so more electrons can make it to the other side.
Correspondingly, more poles can make it to the left.
So, in this video, we have seen pn junctions under zero biased and under
forward biased. In the next video, we will talk about
reverse biased pn junctions, which is actually the case of most [unknown].
