Transition frequency of source follower circuit

Why is Cgs / gm approximately equal to ωT = 2π*(fT) , where fT is the transition frequency ?

Extracted from page 193 of Razavi book "Design of Analog CMOS Integrated Circuits" 2nd edition

Why is Cgs / gm approximately equal to ωT

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Surely not right. Just by looking at the units, it must be

ωT = gm/Cgs [A/V * V/(As) = 1/s]

The pages are different in my first edition, you have apparently simplified the equations for an unloaded case. But the derivation is in the book.

The other way round, of course: ωT = 2π*(fT) ≈ gm / Cgs .

A valid approximation only for a source follower with input voltage control (Rs = 0) and output load C_L ≪ Cgs .

Extracted from page 179 of B. Razavi book "Design of Analog CMOS Integrated Circuits" International Edition 2001

I guess ωT = 2π*(fT) ≈ gm / Cgs is only dominant pole approximation with the condition of unloaded output.

How is dominant pole frequency equal to unity-gain frequency (Transition frequency) ?

Who say it's unity gain frequency? Source follower has unity gain at DC, thus gain at ωT is 0.7

Hi!
As other people pointed out ωT = 2π*(fT)≈ gm / Cgs. There is a mistake in the book.
fT is the frequency at which the current gain of the MOS transistor becomes unity (at which Iout/Isig=1, see the figure).

This parameter has the same value (≈gm/Cgs) regardless of the configuration: common gate, drain or source.

There is a mistake in the book.

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Is it so? The 1st edition doesn't even mention the term transition frequency or fT respectively ωT in this context. Not sure about the 2nd edition.

fT is the frequency at which the current gain of the MOS transistor becomes unity (at which Iout/Isig=1, see the figure).

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I see that gm/Cgs can be interpreted this way if Cgd << Cgs. Nevertheless it's also the cutoff frequency of source follower voltage gain.

Source follower has unity gain at DC, thus gain at ωT is 0.7

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I am a bit confused on the above statement. Could anyone elaborate ?

As ZoOneR pointed out, the term ωT is usually dedicated to current gain transition frequency. The discussed Razavi chapter is however discussing source follower voltage gain and it's cut-off frequency. Where do you see ωT in the book chapter, can you quote the respective formula or text paragraph?

Hello.

I said it is a mistake in the book, because at page 193 in "Design of Analog CMOS Integrated Circuits 2nd Edition" it is written: "Note that Cgs/gm is approximately equal to ωT = 2*pi*fT." and as you pointed, it should have been gm/Cgs.
The first book does mention the term "transit frequency" at page 40, exercise 2.13. This exercise is alos in the 2nd edition at page 41. Since Cgs >> Cgd because when the transistor is in saturation Cgd is ≈ Cgdo (the overlap capacitance), we can safely asume, with small error that ft≈gm/Cgs.

For the source follower Razavi approximates the first pole as: ωp1=gm/(Cgd*Rs*gm+CL+Cgs) (page 189 2nd edition). We see now that if CL+Cgd*Rs*gm << Cgs, ωp1≈ωT≈gm/Cgs. If we asume our source follower has a first order LPF characteristic, with the pole located at ωp1, then the voltage gain at fp1 is 1/sqrt(2)*A0=0.7*A0, where A0 is the dc gain. (This asumption is valid if ωp1<<ωp2). Since for a source follower the dc gain is 1(idealy), this means that the gain at fp1 = 0.7.

Hope it helps.