What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifier

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skamthey

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What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifier

Hi All,
What will be the expression of Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifier.
Here Gm is trans-conductance of the amplifier.

If over all gain Av = (gm * Rs ) /(1+gm*Rs) & Rout = Rs || (1/gm) = Rs /(1+gm*Rs) then
Gm =gm .
But Gm = f(Vin) while gm =f(Vgs).
Please correct me.

 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

The correct formula is

Av = (gm * Rs ) /(1+gm*Rs)

Rout is the output resistance as seen from outside the device and, thus, does not appear as part of the gain formula.
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Hi LvW,
Please specify the Gm & Rout in the gain formula.
Are you saying Rout = Rs || (1/gm) = Rs /(1+gm*Rs) is not correct ??
& Gm = (gm)/(1+gm*Rs) is correct ?

Please reply.
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Please specify the Gm & Rout in the gain formula.
Are you saying Rout = Rs || (1/gm) = Rs /(1+gm*Rs) is not correct ??
& Gm = (gm)/(1+gm*Rs) is correct ?

No - that`s not what I am saying. Of course, it is correct that r,out=Rs||(1/gm). (Please note that I`ve used small letters for r,out because it is a dynamic resistance and NOT a static one).
More than that, if you like you can use the abbreviation Gm=gm/(1+gm*Rs). But in post#1 you say Gm=gm which proves not to be correct.
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Yes, So can I say
if r,out=Rs||(1/gm) & Gm = gm/(1+gm*Rs)
Then Gain Av= Gm*r,out
Av = gm/(1+gm*Rs) * Rs||(1/gm) = gm/(1+gm*Rs) * Rs/(1+gm*Rs) = ( gm*Rs)/(1+gm*Rs)^2
But CD stage Gain should be Av = (gm*Rs)/(1+gm*Rs)

Please correct me.
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Then Gain Av= Gm*r,out

Who told you this? It`s wrong.
Correct: Av=Gm*Rs

(In my post#2 I have mentioned that Rout is the output resistance as seen from outside the device. It does not appear as part of the gain formula.)
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Hi LvW,
Thanks for your explanation.
Considering your answer "Av=Gm*Rs" is correct. I am getting one more doubt. So I would like to ask same question in terms of CS stage amplifier with pmos current source as a load.




Considering ro1=ro2=ro
What I am thinking :

Here Gain Av = Gm * rout = gm1 * (ro1|| ro2) = gm1 *( ro/2)
Here Gm =gm1 & Rout = rout = (ro1|| ro2) .

According to your explanation : Av = Gm *ro2 = gm1 * ro2

Please correct me.
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Gain: Av=gm1*(r,load||ro1); with r,load=ro2 we have Av=gm1*ro/2

Thus, your result is correct.
(Comment: Here we have to take ro1 into account because it has the same value as the load resistor. For common drain stages (your post#1) we can assume that Rs<<ro)
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Hi LvW,
So In case of CS stage Av= Gm * rout Is it correct ??? If yes then
Av = gm1 * (r,load||ro1); with r,load=ro2 we have Av=gm1*ro/2

Here my point is : rout = (ro2||ro1), is used in gain formula. Av= Gm * rout

Who told you this? It`s wrong.
Correct: Av=Gm*Rs

Your point is correct that if ro2 <<<ro1, we can neglect ro1 and rout will be ro2, but i am looking in generic expression.

If above is correct then, in case of CD stage too Av = Gm * rout should be correct. Say if Rs=ro then ...??
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

I am afraid you are mixing gm and Gm.
For common drain (for example): Av=gm*Rs/(1+gm*Rs)=gm*rout with rout=Rs/(1+gm*Rs)=1/(1/Rs + gm)=Rs||(1/gm).
But, in general, I would not use the term rout in a gain formula.
 
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Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Hi LvW,
Yes, certainly I am missing some point about CD stage amplr. That`s why I am discussing it here.
Thanks for being patience with my doubts.
So now for CD stage if you are Saying rout=Rs/(1+gm*Rs)=1/(1/Rs + gm)=Rs||(1/gm). Then according to gain formula Gm should be equal to gm.
But in case of CD stage Vin is different than Vgs as gm = f(vgs) not vin. While Gm is function of Vin.
See load resistance Rs will act as a source degeneration resistance as well so Gm < gm.
I think irrespective of the node at which I am checking output (Drain in case of CS-stage or Source in case of CD stage), the source degeneration will change the Gm value from gm to gm/(1+gmRs)

So if you are getting my point ...I can see difference in the Gain formula Av =Gm * rout and Cd stage Gain = (gm*Rs)/(1+gm*Rs)

Please correct me .
 

Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Two comments:

1) "But in case of CD stage Vin is different than Vgs as gm = f(vgs) not vin. While Gm is function of Vin."

It seems you are mixing signals with DC biasing. The transconductance gm is fixed by the DC voltage VGS. And it does NOT depend on any input signal.

2.) I recommend NOT to use Gm. I never have seen this term, up to now, for considering feedback. I know what you mean with Gm - however, the problem is that sometimes Gm=gm and sometimes not.
Thus, misunderstandings cannot be avoided.
 
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Re: What will be the Gm, Rout & Av= Gm * Rout of a single-stage Common Drain amplifie

Okay Lvw,
I will think on your comments (should give some time on CD stage).
Thanks for your time and consideration.
 

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