connecting the 2 inputs of an opamp

[boog]

I was tinkering with a quad amp the other day, ended up with hella oscillations.  Started trying different value caps in parallel with different components on the board (very interesting in helping to understand/hear what happens where).  Well, I accidentally dropped one on my perfboard while tinkering and the oscillation quit.  I looked at where the cap fell and what actually happened was that the lead for the cap had connected the inverting and noninverting input of the second stage.  My question is, does this mean that the second stage of the opamp was bypassed, thus lowering the overall noise/gain/electronic bru-haha?  Or, if that question is not answerable, what happens when you connect the two inputs of an opamp?

[R.G.]

Opamps are "differential amplifiers". They amplify the difference in voltage between the + and - inputs by the open loop gain of the amplifier. By shorting the pins, you forced the difference between them to be zero (note to purists: OK, OK, *very close to zero*) and that reduced anything the outside of the opamp was doing down to the point where it could not oscillate.

You can't "bypass the second stage" of an opamp by shorting the inputs. Many opamps have either full or simplified schematics of the insides. It's just transistors and the like, connected together on a very tiny PCB. In older opamps and a few current ones with external compensation pins, you could mis-use the compensation pins by pulling the inputs inactive and forcing a signal into the compensation pins. That turns out not to be very useful in general.

[boog]

Ah-ha!  More stuff that's over my little head.  When you say "anything outside the opamp" would that mean anything before the inputs only or anything as in ANYTHING?  And am I correct then if I were to say that the only way for an opamp to work "correctly" is to have different levels of voltage entering at the two inputs?

[R.G.]

Ah-ha!  More stuff that's over my little head.  When you say "anything outside the opamp" would that mean anything before the inputs only or anything as in ANYTHING? 

Right - anything on the outside of the chip.

And am I correct then if I were to say that the only way for an opamp to work "correctly" is to have different levels of voltage entering at the two inputs?

Yes. All "normal" opamps as we use them today have differential inputs - that means that the amplifier only amplifies the difference in voltage between the two inputs. If you short the two inputs together, that eliminates any difference in voltage between them, and that means that no signal gets through.

The amplifier part amplifies the difference between the inputs by a big number, maybe 100,000 to 1,000,000. Feedback is arranged to use that gain from the output to make the - input follow the + input around. But you need at least some difference between the two inputs to make the output move around. In effect, the - input follows the + input around, the difference between the two with feedback being reduced to the input signal divided by the gain of the opamp. That's tiny. So for all normal applications, the two inputs are within a gnat's eyelash of one another all the time. The are at the same voltage, to within a very small amount, but they are not at *exactly* the same voltage. That tiny difference is what makes the amp work. If you short them, the difference is zero, and the amp has nothing to amplify.

[boog]

Ok; so that means that what I didn't actually have a loss of oscillation, I had a loss of amplification?  The signal left the first stage of the amp, travelled to the second stage (the shorted stage), was not amplified; travelled through the gizmos and whatsits in the second feedback loop (connected to the - input) and out the second output into the third stage.  So rather than amplify the signal 4 times, its only amplied 3 times.

And...R.G. this is at least the second of my posts in which you've come back w/ some crazysmart information.  Thank you very much for your patience and knowledge.  I really appreciate it!