Electronic – How do capacitors affect Operational Amplifiers

It's like the "paradox" of the immovable object meeting the irresistible force. In reality, neither can exist.

A real power source will have some impedance. A real capacitor will have some impedance. Real wires connecting them have resistance and inductance.

So in reality when you slap a fairly 'stiff' power source across a fairly good real capacitor there's a spark and the capacitor charges through those series resistances with some ringing and stuff due to the inductances. Ignoring the inductances, the voltage difference would simply divide in ratio to the internal resistance of the capacitor and the internal resistance of the power supply and the wire resistance.

To answer your specific question: If the capacitor and voltage source and wires are ideal, you have a mathematical problem, like division by zero. It's of no consequence in the real world- it just illustrates that the ideal models of the power source and the capacitor and wires are insufficiently accurate to describe their real-world behavior. Modelling any one of those as a real part with real resistance (and inductance) will make the mathematical problem go away, but it won't likely give you an accurate indication of what is actually happening.

For example, if the wire (or the capacitor or the power supply) had 10m\$\Omega\$ resistance, and there is a 10V difference, you could predict you'd see 1000A (which is very high, but not infinity) and the capacitor would charge very quickly. In reality that isn't likely going to happen because of other non-ideal factors. If the 10m\$\Omega\$ was modeled as in the power supply, the power supply voltage would drop. If the 10m\$\Omega\$ was modeled as in the capacitor, the voltage would suddenly appear across the capacitor terminals. If the 10m\$\Omega\$ was modeled as in the wire, the voltage would appear across the wire. But none of those is very realistic.

If you modeled as a circuit with no resistance at all and a tiny bit of inductance (even superconducting wires of any length have inductance) then a simple mathematical model would predict ringing that would persist forever, energy sloshing back and forth between the inductance and capacitance at an angular frequency of \$\omega_0 = {1 \over \sqrt{LC}}\$.

As TEMLIB pointed out, a comparator is not an op-amp, even if it happens to be drawn as a triangle with with 2 inputs and one output.

A typical design for a comparator is an open drain comparator, which can be modeled like this:

^{simulate this circuit – Schematic created using CircuitLab}

Here, OA1 is an ideal op-amp. If you connect Vout to V-, you won't get a voltage follower.

Additionally, even if you stick to the world of "true op-amps", there are some restrictions on what op-amps you can use and how you use it. For example, some op-amps are not unity gain stable. Since a voltage follower has by definition a gain of one, trying to build a voltage follower with one of these will always be unstable, and you won't get the desired voltage follower behavior.

You could start with an advertised unity gain stable op-amp and drive it unstable by what you connect the output to and what input waveform you give to the op-amp. It is also possible to do the reverse and add components to stabilize an unstable configuration. There's a wide variety of literature available online on op amp stability.