[Neuroscience] Re: Op Amps - The Voltage Follower Circuit
(by connelly.bill from gmail.com)
Sun Feb 15 03:48:56 EST 2009
Yes, I see how without the G term in that equation, it breaks down to
output = input/2, i.e. halfway between those osscillation i described
And so this is the importance of the "infinite open-loop gain" that
the perfect op amp has.
On Feb 14, 4:02 pm, r norman <r_s_norman from _comcast.net> wrote:
> On Fri, 13 Feb 2009 17:16:27 -0800 (PST), Bill
> <connelly.b... from gmail.com> wrote:
> >I'm having a hard time understanding the classical Voltage Follower
> >Circuit made by a single op amp. I wont bother trying to draw it in
> >ASCII, just have a look here if you don't know what I'm talking about:
> >This is who I'm thinking about it (which I'm sure is wrong):
> >1) Lets imagine you have a 1mV input to the + input.
> >2) At the instant this is switched on the - input is 0, so the op amp
> >outputs 1mV
> >3) Now the + input still gets 1mV and - input gets 1mV, so the amp
> >outputs 0mV, sending us back 1)
> >I appriciate that the op amp works 'instantaneously' so you don't get
> >oscillations like I described, but I still don't know how one can
> >conceptualize op amps without getting into these kind of oscillations.
> >Thanks for anyone who can tell me the correct way to think about this
> You forget that the op amp is really an amplifier with a very high
> gain. Lets suppose it has a gain of one thousand. Actually it is
> normally higher than that but this will do for calculation.
> When you put 1 mv on the input, it responds by putting an output not
> of 1 mv, but only about 999 microvolts, 1/1000 less than the desired 1
> mv. Then there is 1 millivolt on the + input and 0.999 mv on the -
> input for a difference of 0.001 mv which, times the 1000 gain, would
> produce an output of 1 mv. Remember, I said "about" 999 microvolts.
> The difference is the approximation error.
> To be precise, if the gain is G, then the output is G times the
> difference between the + and - inputs. Let the output be y and the
> input x. Then y = G(x - y) since the output is applied to the -
> input. Solve that to get y = x * G/(1+G). If G is very large, then
> G/(1+G) is very close to one.
> Oscillations are an entirely different story. For that you have to
> describe the output in terms of the LaPlace (or Fourier) transform of
> its "transfer function". But control theory tells you exactly how to
> do it and to build a circuit that does not produce oscillations.
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