Milller theorem fails

[luckypearl25]

I used the miller's theorem to calculate the transfer function of a common source amplfier which has a capacitance between gate and drain .First calculated without the miller theorem i get a zero at gm/C
but when i implement miller's theorem i don get any zero.and the poles have changed a little .why does the miller theorem fail here .are there any restrictions on its usage..

 

[Humungus]

the Miller theorem is ony valid when the forward element does not influences the original transfer. Otherwise you must calculate the transfer with the forward element by traditional methods.

 

[deepa]

i dont understand,,wat do u mean by forward element .
IF u meant the capacitor ,then whenver we change the capacitor's postion it will always affect the transfer function ,right?Isnt that effect only taken care by the miller theorem

 

[pixel]

Miller theorem does not exactly say in which case is it valid.
How much is your amplifier´s gain and are you sure that it does not depend on C?

 

[jasmin_123]

To apply Miller's theorem and replace Cm by Cm_in and Cm_out, you have to find the gain, G(jw)=v1/v2(jw). To find G(jw) in a simple way, you need Cm_out.
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Conclusion: you use an approximate G(dc), and Cm_out=Cm.
Such an approximation is good only if G(jw)<<-1 and the input loop pole is dominant.

Added after 2 minutes:

Miller theorem does not exactly say in which case is it valid.

Miller's theorem is always valid; the issue is that we do not always apply it correctly.

 

[deepa]

the file contains two figures.The gain of the circuit id gm/gds.
hence the Cin =C(1+gm/gds) and Cout=C(1+gds/gm).for now lets omit CL and Cgs from the figures,,
if i find the transfer function with C i get a zero whereas with Cin and cout i get 2 poles..
am i doing a mistake anywhere here?

 

[jasmin_123]

Hi, deepa,

It does not matter for you to compare the exact solution against an approximation?

Jasmine

 

[10kangstroms]

If I understand you correctly, I think you are assuming the output resistance of your amplifier is zero ohms. Finite output resistance will give you a zero in the frequency response.

 

[deepa]

no the output resistance is Gds ,hence the gain i have taken is gm/gds.

It does not matter for you to compare the exact solution against an approximation?

Jasmine

Well jasmine does that mean miller's is an approximation and so it will not be able to find the zeroes n the poles exactly..

 

[jasmin_123]

Well jasmine does that mean miller's is an approximation and so it will not be able to find the zeroes n the poles exactly..

If you do not know the exact gain and use an approximate one, while applying Miller's theorem, then your solution is an approximation.
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Doing approximations, you have to be ready to drop a part of the exact solution.

 

[10kangstroms]

Here is a quote from a presentation called **broken link removed**.

B. Feed forward zero.
With any capacitor which is connected from output to input of
an amplification stage, there is an alternative path for the signal
from input to output of the stage, and which means there is
potentially a zero in the system transfer function. If the gain
stage is an inverting stage (such as a miller amplifier) then the
zero will be in the right half plane which will cause
degeneration of the phase margin. When the Miller gain stage is
made up of a single stage, the position of the zero can be
approximated as sz=+gm/(Cmiller+Cload) where the
gm is the transconductance of the miller stage.

The point I was making in my previous post is that if the amplifier output impedance is zero, this RHP zero disappears. However, this may not be the reason you do not see the zero in your simplified model. I think the Miller theorem simply ignores this zero.

 

[deepa]

If you do not know the exact gain and use an approximate one, while applying Miller's theorem, then your solution is an approximation

Well isnt my gain Gm/Gds .SO since i know the gain exaclty shouldn my result be same,or is it tat on adding the capacitance the gain of the common source has changed .

 

[jasmin_123]

Gm/Gds is not the exact gain. The exact gain is a function of Co!

 

[aomeen]

Hello

Miller theory is an approximate theory to calculate the poles of the system only. In case you want to get the zero, you will either have to solve the problem exactly which is a bit lengthy, or you can use the signal path splitting method introduced by Razavi.

Assume a transconductance Gm with a shunting cap C. a zero in the transfer function will occur at a "complex" frequency for which the output will be zero, as if connected to ground. This means that the curents through the two paths "through transistor transcunductor and through the shunting cap, must cancell out.

That is : Vin x Gm = Vin / (1/sc) -- > s = Gm/C..

Actually any signal splitting will result in the rise of a zero...

 

[10kangstroms]

Why are you people nitpicking about exact gain? That has little bearing on the presence of a zero in the transfer function that is not accounted for in Miller's theorem.

Added after 1 minutes:

Hello

Miller theory is an approximate theory to calculate the poles of the system only. In case you want to get the zero, you will either have to solve the problem exactly which is a bit lengthy, or you can use the signal path splitting method introduced by Razavi.

Assume a transconductance Gm with a shunting cap C. a zero in the transfer function will occur at a "complex" frequency for which the output will be zero, as if connected to ground. This means that the curents through the two paths "through transistor transcunductor and through the shunting cap, must cancell out.

That is : Vin x Gm = Vin / (1/sc) -- > s = Gm/C..

Actually any signal splitting will result in the rise of a zero...

Excellent explanation!

 

[jasmin_123]

Why are you people nitpicking about exact gain?

Because if you use the exact gain in Miller's theorem, you get both the poles and zeroes.
Miller's theorem is not approximate theory; it gives you a 100% equivalent circuit!

It is a different issue that you do not always know the exact gain and use approximations.

By the way, applying Miller's theorem for currents usually causes no problems because in most cases you do know the exact current gain.

 

[10kangstroms]

Why are you people nitpicking about exact gain?

Because if you use the exact gain in Miller's theorem, you get both the poles and zeroes.
Miller's theorem is not approximate theory; it gives you a 100% equivalent circuit!

It is a different issue that you do not always know the exact gain and use approximations.

By the way, applying Miller's theorem for currents usually causes no problems because in most cases you do know the exact current gain.

If you solve the transfer function by using network analysis, you will get the poles and the zero. The problem as I understand it is that Miller's theorem completely decouples the input and output circuits, ignoring the fact that there is an alternate path around the amplifier which creates the zero. Using exact gain in Miller's theorem does not magically reinstall the alternate path around the amplifier.
Here is **broken link removed** that explains the approximation of Miller's theorem, and shows how simple it is to do the exact analysis instead.

 

[jasmin_123]

Hi, 10kangstroms,

Do you like to say that applying Miller's theorem (in its exact form) does not yield a 100% equivalent circuit, with exactly the same transfer function?

Jasmine

 

[10kangstroms]

Hi, 10kangstroms,

Do you like to say that applying Miller's theorem (in its exact form) does not yield a 100% equivalent circuit, with exactly the same transfer function?

Jasmine

Please post the exact form of Miller's theorem, and a cite for it. All references I have found totally ignore the alternate path around the amplifier.

 

[jasmin_123]

Please post the exact form of Miller's theorem, and a cite for it. All references I have found totally ignore the alternate path around the amplifier.

Here you are:

1. Exact application of Meller's theorem.

For infinite f, the exact gain G =Vds/Vgs=1 (see the circuit of 21 Feb 2007 18:14); Co=C(1-1/G)=C(1-1/1)=0, and Vo=Vds=V1>0 (is not this effect of a zero?!).

2. Approximate application of Meller's theorem.

If you use an approximate (dc) gain: G<0, then Co=C(1-1/G)>0, and, for infinite f, Vo=0. This means no effect of zero.

Is not this simple?!

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Miller's theorem does not fail; APPROXIMATION DOES!