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In this video, I will give you a short summary of

the so-called Quantum-Mechanical Effects notably

including the ones that give rise to Gate Current.

Let me begin with Quantum-Mechanical Effects.

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The Fermi potential is fixed and

let us consider just the two terminal structure for simplicity.

And the variable tension in the bulk is below E sub i,

as it befits a pit type region.

But the bands bent at near the surface, the firming potential is above E sub i,

which is the situation for a end pipe semi-conductor.

This is because the surface has been inverted.

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Now, if you have very thin oxides and very high substrate doping,

you end up with very high fields and then the bending is very sharp.

Now the so-called potential well in which the electrons must find themselves,

which is defined by the conduction band edge E sub c, and

the oxide wall here becomes very narrow.

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And because of this, the so called physics of tightly confined

particles start becoming evident and you have quantum mechanical effects.

Now, it is totally beyond the scope of this course to discuss

the nature of such effects.

So all I will do is describe those effects have on device characteristics.

There are two main effects that take place.

One is that the electron density instead of being maximum next to the interface

between semiconductor and oxide becomes maximum a little into the semiconductor.

In other words, if you pluck electron density per unit volume versus

distance away from the interface, you get a maximum which is not

right next to the interface but away from it by a distance called d sub m here.

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This has an effect of increasing your effective oxide thickness.

In fact, the increase in oxide thickness is of the same type as we have seen when

we discuss depletion.

It is equal to this distance d sub m weighted

by the ratio of the permittivities of the oxide and the semiconductor.

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The second effect is that the allowable energy levels

split into discrete sub-bands.

So rather than having a continuum of energy levels that are allowed in

the conduction band, right here where inversion layer takes Occurs.

You have certain energy levels above Ec that can no longer be occupied.

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This effect is similar to increasing the energy gap between the valance band and

the conduction band.

If the gap increases,

you can expect to have a lower n sub i, the intrinsic carrier concentration.

And if you go back to the formula for

Fermi potential, this means a higher phi sub F, a higher Fermi potential.

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Because of this, you need a larger surface potential to have inversion.

Effectively then, instead of having a surface potential phi sub 0 In strong

inversion, you have phi subzero glass, and amount in delta phi sub-zero and

the net effect is you have a larger threshold voltage.

So, phi sub-m,

the distance of the peak of the inversal of the density of the interface and

delta sub zero the increase in the potential can impressively be calculated,

but you need Poisson's equation and Schrodinger's equation.

It is an involved calculation.

Some results, not proofs, are given in the text if you're interested.

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The result is this.

Because you have an increase of the effective

oxide thickness you have a decrease in the capacitance of,

the CV curve goes down as you can see here, here, and there.

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And the total capacitance in inversion is less than you would have otherwise.

And in addition because V0 now became V0 plus delta V0 and

the threshold has increased, there is a shift of curve to the right.

These are the main effects of the quantum mechanical effects I just described.

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We assume polositical gate, so you can define a balance conduction band,

both in the gate and the semiconductor.

The bands as we show that are already bent in the semiconductor and

the inversion layer is very close to the interface.

And this dot here indicates one electron being there, just schematically.

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above it's conduction band here, and this is called Fowler Nordheim tunneling.

Normally, this effect is small.

But there's another effect which says that an electron can find itself

from the immersion layer to the other side of the oxide without having to go over

the barrier like this, and this is the main form of tunneling that worries us,

it's called direct tunneling, and it is dominant for very thin oxides.

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The direct tunneling current increases drastically for thinner and

thinner oxides.

In fact, half a nanometer of change in the oxide thickness can result

in a current that is 100 times larger.

And I'm talking about the gate current that is possible because of the direct

tunneling here.

We will later see a set of curves that show this.

By the way, holes can tunnel too in the same fashion.

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So the effect of direct tunneling is a current through the oxide

which manifests itself as the gate current.

So here I show ou a transistor with a thin oxide.

There is tunneling current that goes from the gate to the channel,

effectively electrons negatively charged go from the channel to the gate,

that's the equivalent of positive current going from the gate to the channel.

And part of that current ends up going to the source and

part of it goes to the gate.

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There is also tunneling between the gate and

the end source through this thin oxide.

Over the so-called overlap region between the gate and the source.

And this is called I GS OV.

And of course one near the gate we have I GD OV.

So the gate current is the sum of all of these currents.

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Also you can have direct tunneling from the substrate to the gate,

especially if the gate is negatively charged, and

all of these currents contribute to a gate current that in some cases can

be significant and it can interfere with proper device operation.

Now the gate to channel current,

this current here that is separated into Igs and

Igd is proportional to the inversion layer charge, not very surprisingly.

And also proportional to the tunneling probability at position x in the channel.

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The total gate channel current, which will be the sum of IGS plus IGD,

can be found if you take the density of the current, which is what this is here,

it's the density it's the current per unit area of the gate and

you integrate it by multiplying by W dx,

which is an element of the gate area and integrating over all the gate area.

This is equivalent to taking W outside the integral and integrating from 0 to L.

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Now some of this current will go towards the gate,

and some of it goes towards the source.

And it is not very surprising that the closer X is towards the drain,

the more that current goes towards the drain.

And the closer you are towards the source, the more the corresponding tunnel and

current can be Seem to go towards the source.

And it turns out that you can actually find those two currents in this way.

Let's talk about IGD.

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So now you can plot the current versus VDS and it turns out if you go through this

calculation, which actually is involved, you'll find IGS and IGD go like this.

IGD is less than IGS because as you increase VDS,

you increase the drained potential.

So, the gate to drain potential,

which is the potential across the oxide becomes less and less.

It is less than the gate to source potential,

which is the potential across the oxide over here.

So that's why GD is less than IGS.

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Here you see a plot of the absolute value of a gate current

versus VGS for two different values of VGS.

Because this X axis is logarithmic, we cannot show negative current values on it,

so what you see here is only the absolute value of the gate current.

But we distinguish two branches for each of these curves, for example,

this branch here.

Is labeled as being for positive current, and the corresponding branch here

is the absolute value of what actually is a negative current, and similarly for

the other current.

So let us first concentrate on VDS equals zero, for which we get this curve.

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Now if we have large values of VGS we have an inverted channel and

the electrons tunnel form channel to gate, they are negatively charged but we said

before this corresponds to a positive current from the gate to the channel and

this takes place through four paths, Igs, Igd,

Igs overlap and Igd overlap,

and the higher the value of Vgs, the more the tunneling current becomes.

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So this then becomes a current in this direction of negatively charged electrons,

which means it is a negative current.

This is why this current actually shows negative currents,

it shows the absolute value but actually you have negative currents.

Now somewhere in between the negative current region and

the positive current reason, the two cancel each other out.

And this curve goes to minus infinity,

indicating on the log axis that the current becomes zero.

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Now, if instead you have a positive drain voltage, VDS,

then let's take 1 case where VGS is equal to 0 and

VDS is equal to 1 corresponds to this point which shows a negative current.

Why is the current negative?

Because if the drain is positive,

more positive than the gate then you have tunneling of electrons from the gate.

To the drain, which corresponds to a negative current

through this overlap region, that's why this is shown as negative.

Eventually of course, if Vgs becomes high enough,

you end up with current similar to those that you have before, and

some were in between, there is cancellation, so if we had enough

resolution you would see this going all the way down as the other current did.

Let us now take a look at the current per unit of gate area.

This is the current density nms per square centimeter.

Each of these curves is similar to one of this curve, but

the barometer here is the oxide thickness.

So we start with a rather thick oxide by modern standards, 2.1 nm.

Go 1.9, 1.7, 1.5, and 1.3.

You can see clearly here the drastic increase of current density as you make

the oxide thinner and thinner.

So, for example, if you decrease the oxide thickness from 1.7 to 1.5,

which is only 0.2 nanometers, or 2 angstroms,

you see that the current goes up by more than an order of magnitude.

And of course it quickly goes out of hand.

Once you try to decrease the thickness even more.

So we can not make the oxide extremely thin.

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On the other hand, we would like to have a thin oxide,

because that gives us a large oxide combustance per unit area.

C ox prime and as you can recall C ox prime is a constant.

It's one of the quantities in the constant of proportionality in any current

equation of the device for example, IDS is W over L times u(greek letter)

times C ox Prime times a bunch of other terms.

So you would like to have a large CX prime in order to have a large car, and for

a given gate drive.

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On the other hand, you cannot arbitrarily increase CX prime by making the oxide

thinner because of the tunneling problems, as we have already mentioned.

So another possibility to increase C ox prime is to keep the oxide thicker,

but make its permittivity large.

So use a different type of insulating material that has a high k,

k meaning a high dielectric constant,

which would give you a large permittivity and it would allow you to obtain a large

oxide capacitance per unit area without having to go to very thin oxides.

In the next video,

we will talk about another form of leakage current that has to do with junctions.