Voltage gain for Common Source Degenerative MOSFET

[Renshai]

Homework Statement

We are going over gain, Gm, Ro, and Rin for MOSFETS. We are told to know and memorize certain equations and how to get these equations. In this particular problem we are told to find voltage gain by analyzing small signal. My professor gave this as the answer:

(-gm* Rd)/(1+gm*Rs),

to the gain to a common source with degeneracy... basically a common source MOSFET with a resistor on the source side, but I am not seeing how he got it. I know it comes from the basic common source MOSFET voltage gain: (-gm* Rd). I am not sure how he got the denominator (1+gm*Rs). I know (or think anyway) that he used KCL to get the equation in terms of gm. I am somehow missing that link. I put a picture of the MOSFET below.

Homework Equations

The equation we are to use to find gain is Vo/Vin

The Attempt at a Solution

This is the equation I keep coming up with, but I know it is wrong, but I am having a heck of a time figuring out the correct relationship:

(Vo/Rd) + gm(Vin -It*Rs) = 0

This is what I am struggling with. It seems that I am missing something or doing the equation totally wrong (I am becoming aware, due to this problem, of my lack of KVL and KCL skills. It has been a while since I have had to use them, but I am getting back into it, but this is tripping me up). Any help and or guidance will be extremely appreciated.

 

[Quinzio]

The circuit you linked is not the one your professor used to give (-gm* Rd)/(1+gm*Rs).
The circuit you have to use is slightly different.
Can you draw the correct circuit ?

That circuit you posted is useful to find the output resistance (if somehow).
One equation of that circuit is Vgs + Vgs*gm*R = 0
That gives Vgs=0
or
R=-1/gm

both R and gm are fixed parameters, that last equation is not possible, so you see how it is pretty unuseful and that there's something wrong.

 

[Renshai]

Thank you for your post and help. Here is the changed circuit. All I did was add a small signal voltage to the input. I hope that is right.

 

[Quinzio]

Wow, ok. Good job. Now you should be able to get
Vout/Vin = (-gm* Rd)/(1+gm*Rs)

 

[Renshai]

Wow, ok. Good job. Now you should be able to get
Vout/Vin = (-gm* Rd)/(1+gm*Rs)


Thanks for the Kudos, but this is where I get into trouble. I really do not understand how that circuit is getting the (1+gm*Rs) part. I have played with it and this is where my KCL and KVL is really lacking. If you could try to explain that to me I would really appreciate it.

 

[The Electrician]

Keeping in mind that we are dealing with only small signal voltages and currents, we have:

Vgs = voltage between gate and source
Vg = voltage from gate to ground
Vin = Vg
Vd = voltage from drain to ground
Vout = Vd
Vs = voltage from source to ground (this is the voltage across Rs)
Is = source current
Id = drain current

Starting with the most basic relationship:

Id = -gm Vgs

Then:

Eq 1) Vd = Vout = -gm Vgs Rd

Also, for a FET, Is = -Id = gm Vgs (small signal, remember)
So:

Vs = Rs Is = Rs gm Vgs
Vgs = Vg - Vs = Vg - Rs gm Vgs

So:

Eq 2) Vg = Vgs(1 + gm Rs)

divide Eq 1 by Eq 2:

Vd/Vg = Vout/Vin = (-gm Vgs Rd)/(Vgs(1 + gm Rs)) = (-gm Rd)/(1 + gm Rs)

 

[Quinzio]

Yeah, thank you The electrician.
It's important to remember that this is a dynamic circuit, all symbols are small signals superimposed to static levels and should be indicated with lower case letter.

 

[Renshai]

I absolutely get it now. Thank you for your help.

 

[The Electrician]

Yeah, thank you The electrician.
It's important to remember that this is a dynamic circuit, all symbols are small signals superimposed to static levels and should be indicated with lower case letter.

The lower case convention is a good idea, but when I'm writing in text as I was here, I find that Vgs looks like the "gs" part is a subscript, but with "vgs", the "gs" part doesn't look like a subscript. I preferred the appearance of Vgs, and others, with an admonition that the variables are small signal.

I suppose I could have written everything in Tex, but I wasn't willing to do that much work. :-(

However, when turning in hand written work in class or homework, it's good to follow the convention.