How do you derive opamp equations if you flip the opamp in an inverting configuration?

I hope you aren't saying you connected the feedback resistor to the non-inverting input. Because that won't work.

Inverting has a series resistor from the signal input into the inverting input, with a feedback resistor from the output to the inverting input. The non-inverting input goes to ground, either directly or through a resistor that exists to keep things balanced DC wise.

Non-inverting has the same resistor from the output to the inverting input. Then a resistor from the inverting input to ground. Then the signal is fed into the non-inverting input.

There always has to be feedback from the output to the invering input.

Michael

Hello Michael:

I'm actually creating a what-if scenario. What if in an inverting opamp configuration the opamp inputs are flipped? How would I prove via calculations that it will not work? That was just my curiosity.

An ideal opamp is just a comparator and does not depend on which which input pin is used but only on the difference in the signal.

When you add a feeback loop then you need to make sure that you feed back to the right input else you increase your amplifciation differential instead of decreasing it(which an be useful but( isn't what an op amps about).

Your basic ideal op amp is

----- I1 ---|\\ | \\ R >-- O = G*(I1 - I2) | /

----- I2 ---+/

I1,I2 are your input's, O is your output and R is infinite resistance.

In this case I1 and I2 are identical. You flip them and it doesn't matter except on your gain(you will flip the O in sign).

If you use some feedback then fliping which input it connects too flips it in the equation but now your feedback gain differential changes in sign and now adds instead of subtracts. So you get infinite output. That is the reason why the feeback is connected to the inverting pin. If not then its kinda useless to use feedback(Maybe there are some uses for it though).

Jon

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?

No. The feedback becomes positive, the gain becomes infinite (or more!) and the opamp slams to one of the power rails.

Interestingly, with an ideal opamp model, some simulators will show a zero-input, zero-output state here, which doesn't happen in real life.

John

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

Your so called mathematical model is only valid for a linear circuit, and a comparator with hysteresis(this is what you get) is a *non-linear* circuit element, which nevertheless can also be descibed with a set of equations, but completely different ones. You can calculate the threshold values going from negative to positive and vice versa. just assume one of the stable stages and calculate which input is needed to change the output sign.

--
ciao Ban
Apricale, Italy

--
Very nice, but it seems you\'ve left out the effect of the delay
through the opamp.

Thanks, Jon! :-) Much clearer.

You should get the datasheet for the device. I covers all the configurations.

Don

No, the signals int the equations are arbitrary. One can include the delay into V2 - V1 but the effect of x is still the same in any case. Here the delay is a real factor but even with no delay we still have the issues.

I guess, ask yourself why the delay exists. Then ask yourself how x works. In a real op amp we do not instantaneously get a feedback gain of x. It ramps up to x(charging of capacitances or whatever that cause this). This ramping up is the problem. x starts at 0 and then goes to whatever value it does. So the gain goes from A/(1 - 0*A) = A to something like A/(1 - A/A) = oo. (since A is so large 1/A is close to zero to x doens't have to travel far to cause infinite gain).

This because of finite slew rate which I guess includes all the real world inner workings of the op amp that make it not be able to just jump to x and above having to traverse from [0, 1/A] (which a negative feedback op amp will do but it doesn' thave the issue at the point 1/A). If we could have the opamp jump to, say, x = (A+1)/A then we have a gain of (-)1 with no problem and the thing will work just fine(excluding thermal noises and stuff that might make it very unstable at that point). But since the true x is dependent on t and will rise to, say, (A+1)/A it will surely pass through

1/A which causes infinite gain.

It doesn't really matter about delay except that the same thing that causes delay causes x to have to "ramp" up. So in actuallity they are the same thing and I did mention that these are the reasons why there are problem with the feedback.

In a sense though I think that "delay" is a somewhat ideal word because in actuality there is no delay(except for the finite propagation of the EM fied). What happens is that the slew rate(which is still a kinda generalized effect) is very small and so we do not see the signal coming out until the slew rate "charges" up.

On some sense what we see out by puting in VI is Vo = Vi(t)*Ramp(t)

where Ramp is just the ramp function. What we call delay is how long the ramp function takes to reach 1. actually probably not 1 as it would depend on Vi(t) too and some other factors. But at t ~ = 0 we do have an output. It just such a small fraction of Vi that we don't really consider it as a signal and we kinda then just approximate Vo as Vo = Vi(t)*Step(t-d) and say that Vo is a delayed version of Vi(t).

Anyways. I think they are essentially the same factors(or atleast stem from the same effect). You can call it delay but then its harder to see equations that depend on x. Just think of x as having to step through from

0 to the value determined by the feedback path and you can see why it doesn't work(even in theory if we just assume that x is continuous, x(0) = 0, x(T) = 'x')

Hopefully that make some sense. I think your issue with delay is that you see it as sorta being turned on instantaneously after a certain delayed time(well, better that it is a shifted signal). This is just an ideal approximation and if it were true in reality then we could set a specific x and have positive feedback that worked(probably not). If we think of delay more as a ramp then and ask why it ramps then that same answer tells us that x has to ramp too. x ramping is the issue.

One way to think about positive feed back is to think of turning a bowl upside down. Lets suppose its perfectly sphereical. Now try and put a marble on the bowl that where it will be stable(not roll off the bowl). Its impossible except at one point. Turn the bowl right side up and it too is impossible but at one point. These two cases are distinctly different though. In the first the marble will roll off the bowl with any slight change and it is almost impossible to tell how. In the second the marble will stay at that point and any slight change will have it move back to that point. In the first case we have positive feedback and in the second negative.

An analogy with the op amp is that we have a "bowl" that as a hole half way up that stretchs all around the bowl. Are requirement is that we have to push the marble up, along the bowl, from the ground. As we do this we will cross the hole and the marble will fall out the bowl. If we could just pick it up and stick it at the top point then we could get stability... but this stability is still highly unstable. For the negative feedback we do not have the hole and we can actually release the marble anywhere in the bowl and it will go to the stable point.

Jon