Electronic – Op-amp latch-up sanity check

Positive feedback and negative feedback can be combined to create oscillation (astability). Aside from that, I think you're correct. In the case of the integrator, there's never a step change in the output, so there's never a discontinuity in the capacitor voltage. The only limit is the output voltage range of the op amp.

Assuming an infinite gain with negative feedback implies that the voltages at the two inputs will be equal. Assuming an infinite gain with positive feedback implies that the output will be saturated (unless V+ exactly equals V-).

Even in a real op amp, there's very little current into the input pins (microamps at most). Likewise, the open-loop gain is usually very large at low frequencies (on the order of 10^5 or 10^6 V/V). So the ideal op amp rules are often a pretty good approximation. But all real op amps are limited by their supply voltages, so sometimes that limit is included in an ideal model as well.

This may help:

• Remember that when current flows into the RC junction of your op-amp that the voltage at that point will tend to rise.
• If the inverting input voltage rises the slightest bit above the non-inverting input voltage then the op-amp output will start to swing negative.
• The output swinging negative will, through the capacitor¹, tend to pull the inverting input down towards zero again where it stabilise (for the moment).

The result is that feeding current into the RC node causes the op-amp output to go negative.

Out of interest I also redrew the op-amp circuit next to the RC integrator shown below which gives the suggestion that the op-amp is amplifying the small voltage across C (assuming high R1) while having high impedance from the resistor/capacitor node. Not sure if that is a legitimate way to look at it.

That's correct. It might be better than you think. The simple RC circuit has the advantage that it's non-inverting but the disadvantage that it's non-linear. With a constant input voltage the output will be an exponential charge curve.

Putting the op-amp in as you have shown still allows the capacitor to charge up but maintains the top terminal at virtual ground. The advantage is a linear change in output. The disadvantage is that there is a minus sign on the integral obtained.

¹ You can think of a capacitor as holding the voltage across it as a constant in the short term. That means that if the voltage on one side is changed the voltage on the other side will try to change by the same amount.

From the comments:

One question. what is the orientation of the capacitor in terms of conventional current? i.e. if vin goes positive the capacitor is I assume negative on its right-hand side (nearest vout). Now vout goes negative and therefore reduces the voltage across the capacitor until the potential at X is zero?

I think your understanding is correct.

If V_in goes positive then current flows into the X node charging up C. (Remember the op-amp's voltage hasn't changed yet.) This tends to increase the voltage on the inverting input and that causes the output voltage to decrease. This draws some charge from the right hand side of C. Now the inverting input is pulled back down to zero volts but there is charge on C so there is a voltage across it. Since the conventional current flowed to the right there is a negative voltage remaining on the capacitor.