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Op-amp basic circuits questions

Q

qwerty

Jan 1, 1970
0
I've been reading about some basic op-amp circuits. The book has the
circuit and then it gives the analysis. The analysis doesn't make
much sense to me. For example, it says that when the op-amp is
working in it's linear area (sorry if the translation from the book's
lanugage is sloppy) the two inputs have the same voltage.

But how do we know that the op-amp is working in its linear area? I'm
trying the circuits on SPICE and it seems that if I use a circuit
with feedback (http://img58.imageshack.us/img58/4422/opampgm9.png)
the op-amp will work in its linear area, but if I remove the feedback
then it won't. Why?

Should I worry too much about those details or should I just learn
the circuits and move on?
 
J

John Popelish

Jan 1, 1970
0
qwerty said:
I've been reading about some basic op-amp circuits. The book has the
circuit and then it gives the analysis. The analysis doesn't make
much sense to me. For example, it says that when the op-amp is
working in it's linear area (sorry if the translation from the book's
lanugage is sloppy) the two inputs have the same voltage.

Yes. This is another way to say that an ideal opamp has a very high
differential gain.
But how do we know that the op-amp is working in its linear area?

Its output is not saturated against either of the supply rails, but is
producing a live, controlled (by the feedback network) output signal.
I'm
trying the circuits on SPICE and it seems that if I use a circuit
with feedback (http://img58.imageshack.us/img58/4422/opampgm9.png)
the op-amp will work in its linear area, but if I remove the feedback
then it won't. Why?

If you remove the feedback, it is almost impossible to keep the input
signal small enough and well enough matched to keep the two inputs
almost exactly (but not quite) exactly the same. This is what
negative feedback does. If the two inputs are not almost exactly the
same, the output voltage changes in the direction that makes the two
input voltages more alike. When they are almost perfectly matched,
there is only enough difference between them (when multiplied by the
high differential gain of the amplifier) to produce the output voltage
that got them to that state of match.
 
T

Tim Auton

Jan 1, 1970
0
qwerty said:
I've been reading about some basic op-amp circuits. The book has the
circuit and then it gives the analysis. The analysis doesn't make
much sense to me. For example, it says that when the op-amp is
working in it's linear area (sorry if the translation from the book's
lanugage is sloppy) the two inputs have the same voltage.

But how do we know that the op-amp is working in its linear area? I'm
trying the circuits on SPICE and it seems that if I use a circuit
with feedback (http://img58.imageshack.us/img58/4422/opampgm9.png)
the op-amp will work in its linear area, but if I remove the feedback
then it won't. Why?

The bit about the inputs being at the same potential is not a property
of op-amps. It's a property of op-amp circuits using negative feedback
(pretty much all of the ones which work well). Without feedback there is
no way for the op-amp to alter the potentials on its inputs. With
feedback the output is driven one way or the other (we design the
circuit so that it's the right way) till there is zero voltage
differential at the inputs.


Tim
 
A

Alan B

Jan 1, 1970
0
But how do we know that the op-amp is working in its linear area? I'm
trying the circuits on SPICE and it seems that if I use a circuit
with feedback (http://img58.imageshack.us/img58/4422/opampgm9.png)
the op-amp will work in its linear area, but if I remove the feedback
then it won't. Why?

John summed it up quite well. The large open loop gain of the device makes
even the tiniest difference in potential at the input appear as an enormous
swing at the output. That's why op amp design (as differentiated from
comparators) uses degenerative feedback - the gain is thus made
controllable, and easily determined through mathematical calculation.
Should I worry too much about those details or should I just learn
the circuits and move on?

That depends entirely on how well you wish to understand the fundamentals
of the craft.
 
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