[Neuroscience] Re: Op Amps - The Voltage Follower Circuit

r norman via neur-sci%40net.bio.net (by r_s_norman from _comcast.net)
Fri Feb 13 22:02:18 EST 2009


On Fri, 13 Feb 2009 17:16:27 -0800 (PST), Bill
<connelly.bill 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.




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