Electronic – How to discharge a capacitor being charge by an op-amp

You had the right idea with a voltage divider, but as you say resistors will always dissipate power. The answer is therefore a capacitive voltage divider. It works just like a resistive voltage divider except that the impedance goes down with frequency. Your frequency is well known and controlled, so a capacitive voltage divider is quite appropriate. Adjust the capacitors so that the peak output voltage is the 200V you want to charge the output cap to, and put a diode to this output cap like you would from any other 200V AC source.

One thing to watch out for though. The power line will occasionally have large voltage spikes. These will make it thru the capacitive divider proportionally just like the intended voltage. This could possibly overcharge the output cap on rare occasions. It might be a good idea to put some kind of clamp circuit accross the cap that turns on when the voltage is a little above the normal value. This could save a few random mysterious field failures years after installation.

Added:

Here is what I was talking about in the comment:

You have 220V AC sine coming in, so the peak voltage is 220V * sqrt(2) = 311, and the peak to peak voltage is twice that or 622V. C1 and C2 must be sized to reduce that p-p to 200V, so C2 must have 2.11 times the capacitance of C1.

Think about the circuit with only C1 and C2. The node between these two now is at 200V p-p. Diode D1 pushes the DC bias on the capacitors so that the negative peak is 0V or more, and D2 pushes it so that the positive peak is 200V or less. When C3 is at less than 200V, then the circuit acts like a charge pump and each cycle will add a little charge onto C3. How much depends on the absolute value of the capacitors and the cycle frequency. Once C3 reaches 200V the AC peaks just hit the limits where the diodes conduct but don't actually transfer any current. In other words, once C3 is fully charged, there is no more power drain from the AC line.

Note that while there is no power drain when C3 is fully charged, there is current. However, this current is 90 degrees out of phase with the voltage, which is how no power is tranferred. Technically this presents a poor power factor. However, in general the AC power grid is more inductive than capacitive and power companies actually maintain banks for capacitors to try to even this out. In most cases you're actually helping the power company by offsetting some local inductive load. Also from what you say this is apparently quite low power. A few mA reactive current either way isn't a issue.

The one-terminal source is referred to ground, which has a specific symbol that you also used.

1. The 1-terminal supply is like a battery in which the negative terminal is automatically connected to ground. The 2-terminal is more like a battery: you can stack more or put it inside a circuit, so it simply creates a voltage. Then you have to reference it to ground if you want.

2. In the first circuit, the capacitor is connected between the low terminal of the voltage source and ground, so basically it's not supplied by anything. You have to put a ground symbol in the negative terminal of the voltage source, to make the circuits equivalent.
   As I said in n. 1, since the 2-terminals supply is not referenced, you are not creating any potential across the capacitor.

3. Especially in this case, yes, 0 V is ground.

P.S. It's not the best to put horizontal grounds, consider turning it.

This is the equivalent of your second circuit with a 2-terminal supply