No there is current through the caps, just not through the diodes.
Wrong! It is the current in the diodes that prevent reverse currents (that damage the polarized layer inside the electrolytic capacitor) from flowing in the capacitor. To say this another way, the diode clamps the reverse voltage that would otherwise appear across the capacitor, depolarizing it.
Unable to post the whole circuit (legal) sorry.
What BS! You do not have to post the "whole" circuit; just enough to provide a realistic simulation of what the series capacitors are subjected to.
If an electrolytic capacitor is reversed-biased, it breaks down the polarized layer, punches through, and leaks. Spice does not model this behavior directly (a Spice Capacitor is a pure, idealized device). So follow along
Nigel Goodwin; you might learn something about what a simulator can show.
First with only "ideal" capacitors, that unlike real electrolytics, exhibit no real-world depolarization behavior:
V(x) is the +-20V applied AC voltage, and appears across C1 and C2 in-series.
V(y) is the voltage that appears across C2. Note that it reverses the polarity across C2 by -10V during every cycle, depolarizing it.
Y(x)-V(y) is the voltage that appears across C1. Note that it reverses the polarity across C1 by -10V during every cycle, depolarizing it.
To prevent C1 and C2 from reverse polarity, look what happens if we pre-charge both capacitors to 10Vdc before we apply the sine wave. In Spice, this is done with the .IC (Initial Conditions) directive:
Notice that the reverse polarity problem is solved.
V(y) is the voltage that appears across C2. Note that the polarity across C2 is always positive; no depolarization.
Y(x)-V(y) is the voltage that appears across C1. Note that the polarity across C1 is always positive; no depolarization.
So how do we "automatically" pre-charge the capacitors? That is where the diodes come in. Here is another simulation run with the diodes added. I also shortened the time frame, so we can see what happens during the first few cycles of the applied AC wave.
Notice that the reverse polarity problem is solved because the diodes "rectify" the applied AC and precharge the capacitors.
V(y) is the voltage that appears across C2. Note that the polarity across C2 is always positive within one diode drop; no depolarization if the electrolytic will tolerate a small reverse bias of 0.6V. I(D2) shows the current through D2.
Y(x)-V(y) is the voltage that appears across C1. Note that the polarity across C1 is always positive within one diode drop; no depolarization if the electrolytic will tolerate a small reverse bias of 0.6V. I(D1) shows the current through D1.
Note that I(D1) is much larger than I(D2). This is because V1 went positive on the first cycle, so D1 does mostly all the work of charging both capacitors on the first half-cycle of AC. There is an assymptotically decreasing bit of current in both diodes on subsequent half-cycles. After hundreds of cycles, those current pulses will continue, but they are tiny, only having to replace the charge lost due to a bit of leakage in the capacitors.
For those that think it is ok to make an AC capacitor without adding the diodes will have to convince me and themselves that the capacitor maker thinks it is ok to reverse bias the electrolytics repeatedly, cycle-by-cycle, allowing the capacitor to repeatedly break down.
Read page 13 of this CD Application guide...