Reply-To: "Kevin Aylward"
From: "Kevin Aylward"
Subject: Re: How to write the transfer function of the circuit?
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Sun, 22 Sep 2002 08:32:07 +0100
NNTP-Posting-Date: Sun, 22 Sep 2002 08:32:12 BST
"Artist" wrote in message
> Kevin Aylward wrote:
> >"Zhao" wrote in message
> >>what we want to know is as follow:
> >>The outpt function of above is as follow:
> >>Vo=(1/2)*Vcc*(1+R2/R1) - (R2/R1)*V1 ;
> >> If we use block diagram in a control system as follow, now the
> >>question is what the transfer function is ?
> >> |-----------|
> >>Vi _____| H(s) |____Vo
> >> | |
> >> |-----------|
> >>H(s) = Vo/Vi = ?
> >Transfer function, in this context, usually refers to the small
> >transfer function, in which case terms involving the power supply do
> >appear, i.e. no DC terms.
> >In this case, the ideal transfer function is:
> >H(s) = R2/R1
> >If you include the GBW of the opamp H(s) will then become a function
> This will explain all:
Not really. You post a question as if you don't know the answer, then
direct one to a paper, which seems, on the face of it a basic solution
to your problem, written by yourself. What was the intent of the
There is a small problem though regarding the effect of power supply
noise. The approach here is a standard treatment, however, it cannot be
used directly to actually obtain the real PSRR of a real amplifier. What
this analysis shows is that what ever the PSRR was, it will be reduced.
It does not show what the PSRR is in real terms. The reason for this is
that the analysis simple assumes that the PS noise (D) is applied
directly to a *summer* with the main output. This is not usually
correct. The PS noise reaches the output by an additional transfer
function i.e. it is not the main loop gain, and might typically be due
to the output capacitances of devices. Without knowing what this
transfer function feedthrough is, you cannot determine what the real
So, your final equation (15), showing that the supply noise appears
directly on the output for high frequencies, is incorrect in general.
You need to determine what the transfer function is from the supplies
point of view, not the input signal. It is not an ideal summer.
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