MRW said:
Hi Jon:
Ideally, just using the opamp golden rules, if I were to calculate the
close loop gain for an inverting opamp configuration, I would get Vo/
Vi = - Rf / Ri. I've been told that the "minus sign is due to the fact
that the op-amp is inverting." But if my inputs are flipped for the
same inverting configuration (+ becomes -, - becomes +), then my
calculation steps would still be using the same methods and
assumptions, so I would still get Vo/Vi = -Rf / Ri (?), right?
Would it be valid to say that since the feedback network is still the
same as the correct inverting configuration, but the opamp inputs are
just flipped, we can just flip the sign on the close loop gain?
There's no mathematical step for this?
Heres a very simple analysis:
A feedback of the output voltage Vo into the positive or negative input can
be represented as x*Vo where x is the fractional amount of Vo that we are
feeding back. 0 <= x <= 1. We cannot stick any less unless we have some
other means of amplification(such as another op amp).
Since you know that for open loop op amp one has
Vo = A*(V2 - V1)
where V2 is the non-inverting input and V1 is the inverting input.
If we add a feedback path to the non-inverting then we have V2 = V2 + x*Vo
or
Vo = A*(V2 + x*Vo - V1)
solving for Vo gives
Vo = A/(1 - x*A)*(V2 - V1)
The same on the inverting pin gives
Vo = A/(1 + x*A)*(V2 - V1)
So here is the main difference, for the negative feedback we have the
amplification factor
A/(1 + x*A) = 1/(1/A + x)
Notice that this factor is always between A/(1 + A) and A.
For the positive feedback we have
A/(1 - x*A) = -A/(x*A - 1)
With positive feedback we invert the output for certain x but more
importantly we get infinite gain when x = 1/A and its very unstable around
this point. If we are slightly below 1/A then we are non-inverting but
slightly above we are inverting. Also there is no way to reduce the below a
factor of 1. So here it is always amplification(which is key).
The equations themselfs don't look much different but the - sign is a big
deal.
Op amps are made with large A and not small A. For a real op amp we have
1/(1/A + x) ~= 1/x
or
1/(1/A - x) ~= -1/x
but in the second case we have unstability because x might be around 1/A
(1/A is close to 0).
Now if you understand all that its not hard to see that a real op amp
doesn't like positive feedback. We could use it on an ideal opamp and get
gain for anything above 1. But in the real world the op amps will surely it
1/A when fed back into. You cannot make x so large to feed back all of the
input and because it actually does this in a continuous way(the op amp has
to "gradually" turn on) it will end up "hitting" 1/A and just loose control.
So while its true that they look like they are inverses of each other. i.e.
1/x vs -1/x, In reality the second case is very unstable and is only a
theoretical possibility.
In any case the negative feedback does everything the positive feedback can
and more. (we can get gains from 1 to infinity. (very similar to positive
feedback case)
Hope this is more clear,
Jon