Milller theorem fails

[10kangstroms]

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?!

---
Miller's theorem does not fail; APPROXIMATION DOES!

That is not a cite of Miller's theorem. By cite, I mean a scholarly reference, such as the one I provided. Can you provide a cite that defines it differently? Mind you, I'm not asking for Jasmin's theorem.
I also fail to see how your description defines the location of the zero, as defined in **broken link removed** I provided.

 

[jasmin_123]

As you see from your own reference,

https://paginas.fe.up.pt/~fff/eBook/MDA/Teo_Miller.html

to apply Miller's theorem is the same as to replace a circuit with its two-port equivalent.

If you insist that Miller's theorem fails, then you have to admit that two-port equivalents also fail. Are you ready to declare this???

 

[pixel]

Miller theorem does not fail. If you look it does not make something special, just transforms the circuit keeping same voltages and currents. Simply there are cases when it is much better to use some other method, because there is not always clear voltage dependance between nodes. It is specially useful for intuitive approach in the case of inverting amplifier with gain A with capacitance C between input and outut. Input capacitance is A+1 times higher than C, and output capacitance A+1 times smaller... (inverse for R)

 

[jasmin_123]

I second you, pixel,

and would just add that, even in the special case of a high and negative gain, Miller's theorem gives an accurate input impedance only if the input pole is dominant.

Added after 3 hours 18 minutes:

"If the impedance Z forms the only signal path between X and Y, then the [Miller's] conversion is often invalid." See p. 3.

 

[deepa]

this circuit shown below also implements miller's theorem ,but here it correctly calculates the gain n the poles ...how come miller's theorem is able to calculate it exactly in this case..how is this diff from the previous circuit

 

[pixel]

Now you have voltage amplifier with known voltage gain A (independant on C) and zero output resistance. In CS configuration, gain is affected with C.

 

[nandu]

Miller theorem cant be used to calculate the zeros of a system. It only finds the capacitive load seen at the input of the amplifier. This is because when we use miller theorem, we eliminate the feedforward path, and zeros can appear only if there are multiple (read more than one) paths from input to output

 

[jasmin_123]

Message is unavailable.