Olin Lathrop wrote:

> Gerhard Fiedler wrote:
>>>   Out = Gain(PosIn - NegIn)
>>
>> Well, no, it doesn't. If it did, several common circuits wouldn't work
>> as intended (including the Schmitt-Trigger you used in your example),
> 
> No, the above equation shows what the opamp does in all cases, even
> when used in a circuit with hysteresis.  I did leave out clipping at
> the power rails, but thought it was understood a opamp can't generate
> a voltage outside the range it is given.  

You say "it was understood". If everything that is understood is really
understood, what was your point with your criticism in your previous
message? :)

> So to be more correct (but more cluttered and a little less easy to
> see the forest for the trees), here is what a ideal opamp does: 
> 
> Out = min(PosPwr, max(NegPwr, Gain(PosIn - NegIn))) 
> 
> In other words, it's the simple formula from earier with the result
> clipped to the supply rails.  This applies to Schmitt trigger
> circuits just as well as linear amplification circuits.

Again, it applies of course, but just as my mental image applies, with
the one additional condition that the gain must be high enough to not be
relevant. Which is given for the Schmitt-Trigger circuit.

Your point was that what I called my mental image has flaws that your
equation doesn't. You still have failed to provide one (except for an
anecdote of a totally incompetent EE for which both probably were not
suited).

And for the record, this is still far away from a real transfer function
of a real-world opamp. It should be obvious to you; you know enough
about opamps to be able to imagine a /real/ transfer function of an
opamp. I gave enough clues in my last email (which you'd have to look up
if you want them :)

Gerhard
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