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I have a small control with a linear power supply, bridge rectifier, and smoothing capacitor. It turns on a motor for just 10 seconds or less and in the worst case only every 3 minutes, but maybe much longer.

When the motor is on, the ripple measured across the capacitor is around 600 mV or around 0.42 Vrms.

The capacitor is 1000 μF, 35 V. In the datasheet ESR is not shown but tan = 0.12 @ 120 Hz

So I used the calculation:

2 × 3.14 × 100 (100 Hz for me) × 0.001 = 0.628

then 0.12 / 0.628 = 0.191r

and then 0.42 / 0.191 ≈ 2.2 A

All the capacitors I have seen for this value appear to be in the same range 0.11-0.14.

This figure seems to be very high, but I am not basing that on anything, just surprised.

So the capacitor fitted has a ripple of 1945 mA at 100 kHz (derated by 80% for 120 Hz), I guess this is OK because of the low duty.

My question is:-

  1. Have I done the calculations correctly?
  2. If I have, am I using the correct method to select a capacitor, i.e. use the calculated ESR to arrive at the ripple current, then compare this with the ripple handling rating for that capacitor until one is found with the right combination. Interestingly, I found a couple of H-ripple capacitors, but they ended up with low ESR and therefore the current would be higher anyway.

I know the best answer would be to reduce the ripple voltage. But at the moment, I am just looking at an existing board and using it to understand what is going on.

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  • \$\begingroup\$ Why don't you put several capacitors (maybe smaller capacitance each if you don't need the resulting larger overall capacitance) in parallel? This would share the ripple current to manageable levels, and would also reduce the ESR \$\endgroup\$
    – anrieff
    Commented Mar 2, 2023 at 9:24
  • \$\begingroup\$ (at 100 kHz the k looks extraneous - ?) \$\endgroup\$
    – greybeard
    Commented Mar 2, 2023 at 10:16
  • \$\begingroup\$ Yes - this is a good idea, but is the workings i used correct? \$\endgroup\$
    – Gordon
    Commented Mar 2, 2023 at 15:54

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