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In this video, we will discuss pn junctions and the reverse bias.

This is by far the most interesting case for us because the pn junctions we

encountered in MOS transistors normally are under reverse bias, or at most around

their zero bias. So, here we again have a step junction.

I remind you this means that the n type material has a fixed uniform doping

throughout, and the p type material again has a fixed uniform doping throughout.

The donor concentrations are denoted by N sub D and the acceptor concentrations by N

sub A. Just like on the forward bias, the

depletion region got narrower. When you apply a reverse bias, the

depletion region gets wider. There is a potential drop, c1, across the

depletion region part on the n-side. And there's a potential c2 across the

depletion range on all the p-side. And the total potential from the n-side to

the p-side is C sub C. Also, we will have a need for the depth of

the depletion region the n-side, d1 and the depth d2 on the p-side.

Because the structure is over on neutron, the total number of depleted acceptor

atoms here is equal to the total number of donor atoms on the end site.

So if you happen to have a more heavily dubbed n region than p region, then you

only need to uncover the donor atoms over a cellular portion of the n-side.

Now we are in non-equilibrium, just like we were in non-equlibrium when we

discussed the structure under forward bias, and we're in non-equilibrium because

we can exchange energy with external world Represented by this battery with reverse

bias V sub R. Leakage currents can flow because we have

electron hole generation even inside the depletion region, it's not a pure

depletion region after all. Although we do make this simplifying

approximation. And those generated whole electron pairs

move under the influence of the electric field in such a direction as to give rise

to a leakage current. The energy band situation now looks like

this. Because we have connected the external

voltage source in this way with the negative terminal connected to the p-side,

you are actually increasing the electron energy on the p-side relative to the

n-side. So, this is why the energies here are

lifted, so to speak, corresponding to those over here.

In the, when we discuss the forward bias case, we saw that the forward bias

subtracts from the total built in potential.

Here you have the opposite case where it adds.

So just like with zero bias, we had a total potential equal to phi bi from the

n-side to the p-side, now we have phi bi plus VR.

Another way to see this is to go around the loop from here all the way back to

there. And you're going to find the built in

potential that depends on the first and last material.

In other words, the n type and the p type. But also in the loop, we have an extra

voltage we call to VR. So the total potential difference is now

phi bi plus VR. Yeah, where phi bi is the built in

potential. Correspondingly, the total bad banding for

the energy bands will be Q times this change in potential, Q times phi bi plus

VR. So we have increased the barrier that

electrons have to face when they go from the n-side to the p-side by an amount

equal to QVR. We have made it more difficult for the

electrons to move. Now, the total charge in the structure is

still zero. Q1 plus Q2 is equal to 0.

Q1 is positive, q2 is negative, and they have the same absolute value.

The ratio of this distance to this distance depends on the relative dopings

of the n-side and the p-side. For example, if here you have twice as

much doping concentration on the n-side as you have on the p-side, then you only need

to uncover the donor atoms over half as much difference as on the p-side.

So therefore, d1 over d2 is equal to NA divided by ND.

And very often, we will see the case where ND will be assumed to be much, much larger

than NA. In which case, d1 will be a very small

distance compared to d2. And practically, the entire width of the

depletion region will be equal to d2. As I mentioned already, leakage current

flows in this case. We will for now neglect the leakage

current. Now, we come to a more detailed picture in

terms of charge density electric field and the electrostatic potential.

We take the equations I showed you that relate density to field, and field to

potential. And we develop after working through the

math, this picture. So, let me just go through it.

Although I expect that you will derive the details in piece step by step at your own

pace. Here, we are plotting versus distance y,

the density role of y. We assume a step junction where we have a

constant doping concentration throughout the end region.

And only part of the n region is depleted, and this part here has a density which is

N sub D. These charges here have a charge density

which corresponds to the donor concentration per unit volume.

So then, the charge density goes positive, as you can see here.

And then, immediately after that, it goes negative, it goes like that.

And in the neutral areas it is zero, so here and here, you have neutrality.

Now, the amount of charge per unit volume here is the amount of donor atoms per unit

volume, N sub D, times the charge plus Q. And over here, you have the corresponding

situation for the p-side. If you integrate this curve under here,

the area is equal to Q1 prime. In other words, it is the charge Q1 that

you have in here, divided by the area of the structure as you look at it from

above. And correspondingly, this area here is Q2

prime, which is the total charge Q2 that we have in here, divided by the cross

sectional area A, as you look at it again from above.

Now to find the electric field, as you may recall from a previous video, we will

integrate the chart's density and divide by the permittivity.

So, as we integrate the constant here, you get a straight line over here.

And then, we integrate a constant which is negative so you get another straight line

over here. As you expect, the field is 0 over here,

and over there. In other words, in the neutral regions.

But, it grows like this and it becomes maximum at this point, which is right at

the boundary between the two regions. Finally, we have to integrate the field to

get the potential but I remind there was also a minus sign involved.

So, as you integrate this straight line, you get a quadratic.

And with a minus sign, it goes down over here.

And then, as you integrate this, you get another quadratic that goes like that.

The total change in potential from this point to this point, we call C sub C, and

it is the sum of the potential across the n-type part of the diffusion region, plus

p2 which is the potential across the p-type part of the depletion region.

I am now going to assume an n plus p junction, where n plus indicates that the

inside is heavily doped. This is called a one-sided junction.

So, we already have assumed a step junction.

This is a one-sided step junction. Then, a detailed analysis gives the

following results. First of all, d1 is much less than d2.

We already said why this is the case. The case is because this is, this area is

so densely doped that you only need to uncover it over a very small distance to

uncover as many charges as you have over here for overall charged neutrality.

Because of the way we developed the potential here, it turns out that in the

case we're discussing, this case here, C1 is much smaller than C2.

So, we have this fact over here. That means that C sub C is essentially the

same as C2. And the total dope, dopal potential from

one side to the n-side before we've seen in the previous slide was phi bi plus VR.

So basically, C2 is approximately equal to phi bi plus VR.

I will now show you a couple of important results.

I will not go through the, the math derivations but I assume that you will go

through them at home or you will read them in, In the book.

When you calculate d sub 2 from this analysis, you find that it is given by

this where epsilon sub s is the permittivity of silicon, N sub A is the

doping or the p-side, C sub C is the total potential across the depletion region.

And d2 is the width of the depletion region on the p-side.

Finally, to find the charge per unit area, Q2 prime, we take the total charge on the

p-side, Q2, and divide by the cross-sectional area as we look at it from

above, A, and we get this result. Again, I encourage you to either go

through the derivation of this equation and convince yourself that you agree with

it or follow the derivation in the book. In the next video, we will conclude our

discussion of the reverse bias pn junction by talking about an important quantity

called the junction small signal capacitance.