Very high frequency operation, sometimes called the radio frequency or RF

operation, necessitates considerations in addition to those we have given in the

previous videos. These considerations are summarized in

this video. In order to proper model the transistor

for very high-frequency operation, we need the non quasi-static small signal

equivalent circuit for the intrinsic part, which we have already derived.

And a model for the extrinsic part of the transistor, which has already been

discussed back when we considered the dynamic operation of the, the large

signal dynamic operation of the transistor.

So I will not repeat the discussion of these elements.

for hand analysis, this model is too complicated and sometimes the two source

resistances are lumped into one. The same is done for the drain

resistances and the four body resistances are lumped into one.

And you end up with this model. In addition, if you have interconnects

connecting the device to other elements, you need to model the inductance of those

interconnects. But this is supposed to be outside the

main transistor, so we don't show such inductances here.

I would like now to discuss the transition or cut-off frequency of the

device, including extrinsic parasitics. I remind you we have already discussed

the transition frequency for the intrinsic part of a transistor.

Now we will do so, for the complete device.

So here we have a device assumed to have source and body connected together.

A simplified model for it, is shown here. gate to source capacitance, gate body

capacitance. And gate drain capacitance are assumed to

be in parallel because we assume that the drain is grounded for small signals.

This simply means that there is a DC source, VDS, which does not have small

signal variations in it. Therefore it is represented here by a

short. So this allows us to use a single gate

capacitance which is the sum of Cgs, Cgb, and Cgd, and this is the source, excuse

me, the gate resistance. I'm driving the device with a current,

and I would like to look at the output current and consider the current gate.

The ratio of this current Io to this current, ii.

Now, the current ii goes through rge, the same current appears here.

And then it will create the voltage here which would be the current times the

impedance you see at this node. And the impedance you see at this node is

1 over j omega cg, where cg is the total gate capacitance.

So this is then the voltage at the internal gate note, the external note is

g, the internal node after the gate resistance, is g prime.

Notice that because the resistor is in series with ii, this voltage does not

depend on the resistor. So this current simply goes through the

resistor, comes out of here, and creates this voltage strong.

Now, to find this current here, I'm going to ignore the small current through Cgd

because we will not need to go to very high frequencies where this current would

be very large. And then the current you see is just gm

Vg prime s for Io. If for, instead of Vgs prime.

Vg prime s, we use this quantity here. We end up with this result.

So, Io is equal something times II. This something is the current gain.

Now I would like to find the frequency at which the magnitude of the current gain

becomes one. Notice by the way, that there is no Rge

here, no gate resistance, for the reason I have already explained.

So, at what frequency does this have a magnitude of one.

If you set this equal to one and solve for a magnitude you find this frequency.

So, omega T, the transition or cut-off frequency, which is the frequency at

which the current gain is reduced to the value of 1, is simply Gm over Cg.

So the detailed expression will, will depend the particular region of operation

we're in and what assumptions we make in terms of channel length and so on.

Let me give you two examples. If we assume no velocity saturation then

gm is given by this expression. We have already derived this way back.