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8th February 2010, 14:20 #1
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Miller's theorem and virtual ground
There are many misinterpretations of such a simple theorem, including textbooks by Razavi, Sedra, this forum, etc. For example, Razavi writes that Miller's theorem can give wrong gain. Sedra does not suggest finding the output impedance after applying Miller's theorem. A number of people on this forum say that zeroes of the gain are lost after applying Miller's theorem.
All the above is caused by the misuse of a virtual ground. Miller's concept is actually based on the substitution of an impedance Zµ with two series ones, Zµin and Zµo, in such a way that their total impedance equals Zµ, and the potential at their interconnection equal zero, which is a virtual ground. If one does not connect the virtual ground to the real ground, he or she has no problems with gain, zeroes, etc. This is so because the current entering the virtual ground from Zµin should continue to flow through Zµo. This does not happen when the virtual ground and the real ground are short circuted. Short circuit a virtual ground and the real ground in any circuit and it will puzzle you, so why to do it in the case of Miller?
Regarding the zeroes of the gain, consider a CE amplifier. In this amplifier, the Miller gain, µ=[1+gm*(Rcro)]/[1+j*ω*(Rcro)*Cµ]+1, can easily be found from the following equation with one unknown: µ=gm*[Rcro1/(j*ω*Cµ*(11/µ)]. Look at µ, does it become zeroes for an ω?

8th February 2010, 14:20

1st March 2010, 11:36 #2
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Re: Miller's theorem and virtual ground
hi Jasmin ur question is not clear can you reform??

1st March 2010, 11:36

1st March 2010, 18:12 #3
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Re: Miller's theorem and virtual ground
Hi jasmin !
Interesting contribution!
However, there is one question from my side:
Don't you think that application of equ. 63 from the Razavi excerpt leads to Z2=R2 only (instead of the parallel connection as shown in Fig. 63) ?
And  what is the consequence on the gain for the example (voltage divider) ?
Added somewhat later: JASMIN, , please forgot the above question. I know my mistake now. Sorry!

1st March 2010, 18:12

3rd March 2010, 17:53 #4
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Re: Miller's theorem and virtual ground
HI jasmin!
thank you again for your contribution because it directs our attention to a formulation from RAZAVI (explanation to his Fig. 6.3) which sounds a bit "strange".
I think the purpose and the motivation of Miller`s theorem is to allocate a current flowing through an element and which is driven by TWO sources to only ONE single source. Thus, the parmeter "input impedance" can be computet.
With other words, the Miller theorem can and should be applied only to a signal path between two nodes (X resp. Y) which are connected to TWO signals sources.
Therefore, it makes really no sense (and does not simplify any calculations) to apply ("misuse") the theorem to a simple voltage divider.
Even if (perhaps by accident) this may lead to some correct results (input resistance and "gain"), it is  as mentioned by you  a misuse of the sentence (what about the output resistance for the voltage divider example?).
Question to the community: Does anybody has a counter example?
That means: Any circuitry with only one source which can be analyzed in a simplified manner using the Miller theorem ?
Thanks
LvW
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