Question on GM from Razavis book

[nibo_mmx]

a basic question on gm

When I am reading Razavi's book "Design of Analog CMOS IC"

I saw:

Id1=gm1*Vgs1, I don't understand why.

Because, gm=µ*Cox*W/L*(Vgs-Vth)
and, Id=1/2*µ*Cox*W/L*(Vgs-Vth)²

Then, How could Id1=gm1*Vgs1? I am really confused. I know I must confused some fundmental concept. But I don't know where.

I hope some one help me make it clear. Thank you very much.

 

[aryajur]

Re: a basic question on gm

Well yes, you have got one fundamental concept wrong here. The equations:

Ids = 1/2*un*Cox*W/L*(Vgs - Vth)²
gm = un*Cox*W/L*(Vgs-Vth)

these equations are large signal equations. Which means once you set the DC biasing, i.e. the Vgs of your transistor you can calculate the Ids(Large Signal, i.e. biasing current) and gm(small signal) using these equations. Now once you get the value of gm according to your biasing point you can use this gm in the small signal model of the transistor and find the small signal current Id1 = gm*Vgs1 where Vgs1 is the small signal Gate source voltage.
I know razavi does not clearly differentiate between Large Signal Analysis and Small Signal Analysis, but thats what are the 2 parts to any circuit problem.

 

[nibo_mmx]

Re: a basic question on gm

Thanks a lot for your reply.

Can I understand your explanation in this way:
you said:

the small signal current Id1 = gm*Vgs1 where Vgs1 is the small signal Gate source voltage.

so does the small signal current Id1 mean a small variant current on the biasing current Id. We can wirte it as ΔId?

Say, if the biasing Vgs is 1V and the basing current is 1mA.
When the small signal Gate source voltage is 10mV, then the small signal current is gm*10mA.

Then we have Vgs1=10mV and Id1=gm*10mA. that's different from Vgs and Id

Am I right?

 

[aryajur]

Re: a basic question on gm

Yes you are right. The small signal analysis assumes that the variation of your signals is so small that we can appoximate all the graphs as linear for the whole signal swing, and thats why we calculate the slopes of the graphd for example gm is the slope of the Id-Vgs graph at the biasing point, so that becomes just a linear line for our small signal with the slope equal to the slop at our biasing point.
So the small signals sit over your biasing values.

 

[nibo_mmx]

Re: a basic question on gm

Okay,

I understand what small signal means in the equation.

But based on the above discussion, I cannot understand the following statement in Razavi's book.

I draw the figure as below:

in Razavi's book (P122), he wrote:

We can write Id1=gm1*(Vin,CM-Vp), Id2=gm2*(Vin,CM-Vp)
Then (Id1+Id2)*R2=Vp

Since Id1 and Id2 stand for the small variat of the biasing currents on M0 and M1, How can they sum up to the voltage cross the resistor?

Does he mean Vp is also a small variant? But I dont think so because he use the smae Vp in Id1=gm1*(Vin,CM-Vp).

Thanks a lot,

 

[aryajur]

Re: a basic question on gm

Yes he means Vp is also small signal. ANd that is why it is also used in the equation:

Id1 = gm1*(VinCM - Vp)

where VinCM-Vp is the small signal Vgs of the transistor and VinCM and Vp are both small signal voltages.
Just remember this small helpful tip. Any manipulation you do on small signals will always give you a small signal, you can never get a large signal value from a small signal analysis. While SMall signals are always derived from the large signal values or by operations on other small signal values.
In this problem he is trying to find the Common mode to differential mode gain in case there is a mismatch between the transistors. All these gains are defined for small signals, bacause our signals of interest are the small signals for this case.

 

[nibo_mmx]

Re: a basic question on gm

Thank you very much.

My foundation is really poor, but I try to make clear everthing I don't understand.

Thanks