how does the filter capacitor work in a rectifier circuit?
We commonly call these capacitors "reservoir" capacitors. They act as a reserve of power during periods of no voltage in much the same way that a water reservoir can supply water during dry spells.

Continuing with the water analogy, we notice that during the dry spells the water level will fall as water is consumed and this will result in falling pressure on the water mains. Similarly if current is drawn from the capacitor reservoir as in your "With Load" graph you will see the voltage droop between cycles.

When the water supply is restored pressure builds up in the line. Since the reservoir needs to be refilled water will flow into it. Meanwhile the load is demanding water and, since water can't flow in and out of the reservoir simultaneously, it should be clear that the mains pressure feeds both the reservoir and the load.

Note that during the "With Load" period of your graph that the supply voltage exceeds the reservoir voltage for about 1/3 of the time. That means that it has to supply the average current in 1/3 of the time so the pulsed current will be about three times that of the load current.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. A semiconductor version of the rectifier. Note that R2 and D3 are there to enable visualisation of the un-smoothed full-wave rectified waveform on the simulator.




The bottom trace shows current into the capacitor as positive.
(1) Note that there is a 10 A current pulse on the first half-cycle to charge the reservoir capacitor.
(3) Note that 80% of the time that C is discharging at -0.4 A.
(2) Note that subsequent pulses are much shorter - about 20% of the cycle and peaking at almost 2.5 A.


Figure 2. Simulator results. (Click to enlarge.)