Common source with Ideal current source load-short channel

Hi everybody,

Here is a question for you. suppose you Have an amplifier with common source configuration with ideal current source load.
The gain of this amplifier is therefore the intrinsic gain = gm*ro.
(assume this is a device with short channel lenght e.g., 0.18um).

Since the output resistance ,ro, is a function of Vds of the transistor (Channel length modulation, DIBL, hot electrons)...then the intrinsic gain is a function of Vds!
i.e., gain=gm*ro=f(Vds)
Assuming that you want gain of 20[V/V] this means that the output swing needs to be confined from one min voltage to output voltage.

The question: If the gain is changing with respect to the output swing, doesnt it means that the amplifier is "very very non-linear"?

(assuming very small input signal which dont excite the non-linearity due to the input).

Attached file is a figure from EE240 taught last year.

Regards,

Re: Common source with Ideal current source load

yes, the gain is a function of output voltage, and this is called gain-nonlinearity. this effect will cause distortion. one solution method is to reduce the gain-nonlinearity, using long-length transistor and working the signal in linear region; the other method is to improve the dc gain of circuit to reduce this effect.

hope it's helpful

Re: Common source with Ideal current source load

But usually when you deal with the nonlinearity of the common source it depends on the input signal:

Ids=k'(W/L)(Vgs,dc+vgs,ac-Vt)^2

when you expand this relation you will get the coeffiecient for the nonlinearity.
But this doesnt depends on the output voltage or on the output resistance.

I've never seen calculation of nonlinearity of an amplifier where the nonlinearity due to output voltage is taken into accound due to the output resistance change.

Re: Common source with Ideal current source load

surely, when you consider the current it's be affected in first-order approximately, but the output voltage vout=i*ro is influenced.

in a closed-loop amplifier, vout=vin*A(s)/(1+f*A(s)), where A(s)=-A(vout)/(1+s/w) is amplifier open-loop transfer function, the output voltage is gain-dependent.

Re: Common source with Ideal current source load-short chann

Then why In calculation of CS nonlinearity you dont see the output resistance taken into account?
You can design theoretically high IIP3 amplifier with hand calculation but when you simulate it, the results will be a disaster because of the output resistance change vs. Vds (and not because of the V-I nonlinear relationship!!)

Therefore, If this is true, the design procedure for linearity in amplifiers is very complicated (and the theory and literature is missleading) and the 1st order approximation (assuming constant rout) is irrelevant.

I'm attaching a paper by Razavi where he mentiones that the output resistance contributes to the nonlinearitys:

Razavi paper:
============
"Another troublesome effect is the output resistance of shortchannel
MOS transistors, and in particular its variation with
the drain-source voltage even in the saturation region. Shown
in Fig. 3, this phenomenon causes the intrinsic gain gM*ro to
depend on the output potential, thereby creating nonlinearity
in amplifiers."

Re: Common source with Ideal current source load-short chann

the gain is a function of output voltage A(vout), and when output voltage change, the output resistance change, so the output resistance influences the gain. in design, we set the output transistors' length be long in order to reduce the channel length modulation, and we also hold the output transistors in deep saturation to make the gain constant as possible. in general, make the Vds of output transistors twice the saturation voltage Vds,sat.

it's impossible to cancel the nonlinearity completely incured by output resistance, so how much cost we should pay to reduce the effect depends on the application.