[Neuroscience] Re: Op Amps - The Voltage Follower Circuit
J.A.Legris
via neur-sci%40net.bio.net
(by jalegris from sympatico.ca)
Sun Feb 15 08:51:58 EST 2009
On Feb 15, 3:48 am, Bill <connelly.b... from gmail.com> wrote:
> 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:
>
> > >http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/opampvar2.html
>
> > >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.
> 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
> above.
> And so this is the importance of the "infinite open-loop gain" that
> the perfect op amp has.
>
The oscillations you described are a byproduct of your discrete, step-
by-step characterization of the op-amp. Real op-amps do not make
instantaneous changes. Rather, the voltage on the output rises or
falls until equilibrium is established through the feedback to the (-)
input. This all happens very quickly of course, but it is still a
continuous process. Oscillations occur when the delay and overall gain
of the feedback loop "reinforce" particular frequencies. It is
possible to get an op-amp to oscillate between discrete states, but
this requires special techniques, such as non-linear circuit elements
or topologies.
--
Joe
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