[Neuroscience] Re: Op Amps - The Voltage Follower Circuit
(by r_s_norman from _comcast.net)
Sun Feb 15 10:32:31 EST 2009
On Sun, 15 Feb 2009 00:48:56 -0800 (PST), Bill
<connelly.bill from gmail.com> wrote:
>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.
You got it!
>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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