Electronic – Understanding the capacitors in an op amp circuit

Your theory is correct, but not complete.

When you start discharging capacitor - voltage will drop. When you draw 50% of energy from capacitor - capacitor voltage will drop by 50% by 30%. With power supply connected in parallel - capacitor voltage will not drop so deep, because current will just start flowing from supply, and you will be unable to get energy from capacitor.

Edit: image removed.

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Many power supplies have rating determined by temperature.

If you have power supply with transformer - you can overload it by 20% with no problems at all if you "let it rest" later.

If you have supply based on voltage converter (transformerless) this is more complicated - you may have problems with voltage stability, voltage drops, temperature etc. In general - you shall not overload transformerless power supply, even for a short time.

And one more thing about big capacitors - they are able to draw very large currents when they are charged. Supercapacitors should be charged from current limiter.

I'll do the circuit analysis.

This is an inverting amplifier with a gain of $$ |A_V| = \left| \frac{R_1 || -\frac{j}{\omega C}}{R_2}\right| = \left| \frac{R_1 / R_2 }{1 + j \omega R_1 C} \right| = \frac{R_1 / R_2 }{\sqrt{1 + (\omega R_1 C)^2}} $$ which gives you all of the information you need:

• At low frequencies (\$ \omega \approx 0\$), the gain is $$ |A_V| = \frac{R_1}{R_2} $$ so the DC gain of this amplifier is the same as it was without the capacitor.
• At high frequencies, the \$1/\omega\$ term makes the gain shrink, so high frequencies noises and sharp edges are filtered out.
• The cutoff frequency of the amplifier is at $$ \omega R_1 C = 1 \implies \omega = \frac{1}{R_1 C} $$ which is fairly high, since \$C\$ is small.

Finally (thanks to LvW), if your circuit is ringing, this capacitor adds an extra pole in the amplifier's frequency response, which can increase the phase margin and make the circuit more stable. This is a bit more complex and depends on the properties of the op-amp, so I won't go into detail.