What happens when you flip the inputs of an op amp?

[jim hardy]

I just mean that Vout = -(R2/R1)*Vin given by steady-state analysis doesn't mean we can realize a stable linear amplifier with that relation.

Maybe. Then again. Maybe this pencil is in 0g and it's already upright. We don't know because those details have not been provided so why assume them. And that's my point and why I answer it depends. If one puts a 741 in place of the ideal opamp will it be stable? No definitely not. So maybe don't put a 741 there. What circuits are legal to place there? Unknown. The problem does not specify that.

Fair enough.

I think we all know how the circuit operates
and we realize that

That formula is true only at two infinitesimal points
It does not represent describe_{(is a better verb-jh)} the circuit's behavior at any other point.
So it cannot be used to evaluate the circuit for stability, linearity, or prediction of behavior.
In short it does not represent the circuit under discussion.

In other words, you can't use a single linear equation to describe a discontinuous function even if a computer simulation tells you it's okay.
You've got to either bound Vo as @NascentOxygen did in post #45
or use an inequality instead.

And as teachers we should encourage beginners to recognize that.

No more semantics and hair splitting on this one for me.

Over and out.

old jim

and @ still won't autocomplete when i click it, until after about six tries and a preview.

 

[sophiecentaur]

In other words, you can't use a single linear equation to describe a discontinuous function even if a computer simulation tells you it's okay.

Precisely and precisely. A good computer simulation could, however, recognise what's happening and spot the situation. The sort of simulation you usually come across is just like a student whose good at Maths but who never built anything.

 

[jim hardy]

I'm on an iphone so I can't mark up the image (or at least I don't know how to) but in second line from the top A*Vo/Vc is reduced to A*A in the line below. This is not a valid substitution because vo=A(Vb-Va)=A*Vb

thanks, now i see it

shoulda been $A=\frac{Vo}{Vb}$ instead.old jim

ps
@alan123hk Great graphic, it broke down the communication barrier.
Nicely Done !

 

[Abdullah Almosalami]

The relation V_out = A(v⁺ - v^-) is incomplete, it's a simplification. There are strong conditionals attached to it restricting it to only those instances where V_out is not pinned near either rail.

A more complete representation would be V_out = A(v⁺ - v^-) iff (V_{o max}> V_out > V_{o min})

So before applying that simplified rule, you must first establish that V_out is going to lie within the allowable range, and if it doesn't then that rule doesn't apply.

-edited-

Well sure. The saturation points are there, but it doesn't change my question.

 

[jim hardy]

but it doesn't change my question.

So my question then is what is the difference?

Are you asking "When you flip the inputs ?"

Meaning that the linear and continuous function inside the red rectangle no longer describes the behavior of your circuit
so any conclusions drawn from it are irrelevant.

You now need to describe the circuit with a nonlinear discontinuous function.

The plots of the two functions just happen to cross at two points as shown .

That's the difference.

 

[LvW]

Well sure. The saturation points are there, but it doesn't change my question.

Abdullah A. - does that mean that you have the feeling that your question was not yet answered? (In spite of more than 70 contributions?)

 

[Abdullah Almosalami]

Abdullah A. - does that mean that you have the feeling that your question was not yet answered? (In spite of more than 70 contributions?)

Oh my bad I was just working my way down the responses and I'm not even all the way through yet. But certainly my question has been answered a while back at this point! And with much gratitude for the amount of knowledge put in here beyond my expectations! I might even revisit this post after I've gone through more coursework and into the details of the internal circuitry of amplifiers and such to look at the responses from a new perspective once more.

 

[Averagesupernova]

I really don't understand why this thread has the momentum that it seems to. Switching the inputs of an op-amp from inverting negative feedback to positive feedback as shown in the very first post of this thread does in no way imply that their analysis should be the same. If one cannot accept this from the beginning then one needs to be prepared for a difficult road ahead. That is not to say that I have not been fooled by some strange op-amp configurations. I certainly have.

 

[jim hardy]

I really don't understand why this thread has the momentum that it seems to.

For me it has been a months long search for that good sentence.
But it took @alan123 's picture in post 63 to unlock my alleged brain.

I think perhaps many of us people who are drawn to science struggle for words ? I know i do.
I struggle for math, too
best i can do for this circuit is

$Vout = 12 X \frac{(\frac{Vout}{2} + {Vin}) } {abs(\frac{Vout}{2} + {Vin}) }$

......

but i digress... what counts is we got there. It took all of us, though.

Good thread !

old jim