Electrical – the purpose of a mains X2 Security Capacitor and how to determine its value

I may not be obvious on inspection, but this is a very clever circuit known as a "valley fill circuit". Here are a large number of examples of them - images are live linked to pages.

Importantly - the diodes are NOT just there for protection - they provide a current steering mechanism so that the capacitors charge in series from Vin and effectively discharge in parallel into the load. When Vin valls to Vin_max / 2 the capacitors provide current to the load and "fill in" the voltage valley.

To answer the specific questions:
- The capacitors are effectively in series when charged and the rules for series connected capacitors apply - so the effective capacitance is half what it would be if one capacitor was used - BUT the circuit needs two capacitors to work - see "Energy?" below.

Energy?:

Assign each capacitor capacitance = C Farad. Assign Vinmax the value V Volts.

If one cap of value C was placed across Vr then Vmax on cap = V so
enerrgy in cap = 0.5 x C x V^2 = C.V^2/2

Now place 2 caps of C each as shown in the diagram.
Each cap will charge to a max of V/2.
Current to charge them will flow from Vr through top cap, through hozizontal diode and then via bottom cap to ground.
Energy in one cap = 0.5 x C x (V/2)^2 = C.V^2/8
As there are 2 caps the stored energy will be double this =
C.V^2/4
which is half what we got before. If we used cap of C/2 in the original arrangement with a single cap energy max =
0.5 x (C/2) x V^2 = C.V^2/4 which is the same as for the 2 x C caps in series.

The REASON that energy is halved and effective capacitance is halved is that energy stored increases with V^2. Halving the voltage per cap reduces energy to 1/4 of original and havfing 2 caps brings the total back to 50% of before. Saying that this is equivalent to a single cap of half the value just means that, as we saw, that a single cap of C/2 capacity would have stored the same energy if it had V placed across it.

This circuit is a voltage halver.
I have seen circuits that have 3 caps in series with steering diodes to reduce Vout to Vin/3 and you could notionally extend this to N stages.

Here is a 3 stage valley fill circuit from here . This seem a tad mind boggling at 1st glance but is very straight forward. The 3 caps are series connected with joining diodes such that the diodes conduct when the capacitors are to charge from Vin. The tops of all capacitors are all connected via diodes to output (cap +ve to diode Anode) (top cap directly to output) and the bottom of all caps have diodes to ground (cap -ve to diode cathode) (bottom cap direct to ground). Extention to more stages would be trivial - with diminishing returns as losses increase.

1. As others have stated, both specifications are important, and there are other reasons for wanting capacitance on the power lines: preventing voltage droop from the battery if it can't supply current quickly (battery chemistry impacts this significantly), I*R losses if the battery is far away, path for motor transients (though I'd expect diodes and smaller ceramic caps, not electrolytics, across the bridge transistors for the faster transients).

2. This is questionable; you've left out tolerance. If the dual caps each have +/-20% tolerance, and you get one at +20% and one at -20%, they're not going to share current evenly because they have different impedances, which could lead to one failing prematurely, and the other shortly afterwards since it may not be able to handle the load by itself. Also keep in mind that higher tolerances tend to cluster near the outer limits, because values closer to nominal are typically sold with tighter tolerance specs for more money; that is, a +/-20% part is not likely to be within +/-10% tolerance, because those will be the +/-10% parts, etc.

3. The reduction in capacitance is not recommended. If 15 uF were sufficient for the intended load, the original designer likely would have gone with a 15 uF electrolytic instead of a 1500uF electrolytic; for the same voltage and current ratings, the 15 uF would be substantially cheaper and smaller. If you're using a different kind of motor than the original designer had in mind, you'd need to look into the suitability of the lower capacitance yourself.

If you're really dead set on getting rid of that particular capacitor, I'd look at increasing the PWM frequency (so you can then reduce the bulk capacitance required), if that will work with the motor you're using (the motor will have its own mechanical and electrical time constants that your circuit has to deal with).

Another thing to consider is the impact of motor inductance that's in parallel with this capacitance. The higher ESR of an electrolytic can sometimes be beneficial in dampening potential LC oscillations that might otherwise be possible as a result of pulsing power to the motor with the bridge, particularly if the motor winding resistance is low. You'd need to look into the impedance of the motors you want to drive to check for potential resonance.

As a general note for future reference, Analog Devices (no affiliation with myself at time of this posting) has an article I've found useful on the general parasitic effects of capacitors in their "Ask The Applications Engineer # 21 : CAPACITANCE AND CAPACITORS"