I have been building boost converters. My current revision is working well at about 300 W, 55 V in to 86 V out. The main problem I now face is that the output capacitor is getting quite hot.

If I leave the boost converter running for 10 minutes or so the output capacitor (330 μf, 200 V, 105°C) becomes too hot to touch, at a guess I would say 80 to 90°C, where I think it claims 105°C. I don't really want to push it to its limits.

What is the reason radial electrolytic capacitors get hot? Does it have to do with the speed they are charged and discharged?

What would be an appropriate fix to the issue? Would it be to have more capacitors sharing the load?

  • \$\begingroup\$ Check also if the resonant frequency is well beyond the used "frequency boost". \$\endgroup\$
    – Antonio51
    Sep 13, 2022 at 10:00

1 Answer 1


This is due to the dissipation factor (DF) of the capacitor.

Capacitors are not ideal components in the real world. They have some resistance (impedance) and that manifests itself as ohmic power losses when a current is applied. The bigger your ripple current, the more power is dissipated in the capacitor, and the hotter it gets.

This can be problematic in aluminium electrolytic capacitors because heat reduces the operating lifetime of the component.

You can reduce the temperature by replacing the capacitor with one that has a smaller dissipation factor. Many parametric search systems on parts website (e.g. Mouser, DigiKey) don't list the DF as a parameter, but do list the rated ripple current, so you can look for parts with higher ripple current ratings as a way to find lower DF caps.

  • \$\begingroup\$ thanks, would using 2 capacitors actually share out the ripple load reducing the issue? these caps were actually salvaged from broken ebay Chinese boost converters and i have more of them. \$\endgroup\$
    – Jay Dee
    Sep 12, 2022 at 20:32
  • 2
    \$\begingroup\$ Using 2 caps halves the ripple current, and since losses are RI^2, dissipation in each cap will be one quarter. So they will run much cooler. Note the temperature inside the cap is higher than on its surface, because the cap is cooled by air. \$\endgroup\$
    – bobflux
    Sep 12, 2022 at 20:50
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    \$\begingroup\$ Yes, as bobflux said, two caps in parallel will halve the power dissipation in each cap (or close to it - there may be some variation in manufacturing) and increase the surface area for thermal dissipation, thus reducing the temperature. If your capacitor's datasheet provides thermal resistance specs for the package you can use those to estimate the internal temperature of the capacitor, but it's quite rare to find those numbers published for caps. As a rule of thumb, assume the temperature rise above ambient inside the cap is twice that of the temperature rise above ambient on the surface. \$\endgroup\$
    – Polynomial
    Sep 12, 2022 at 21:13
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    \$\begingroup\$ @JayDee I hate to say this, but broken Ebay Chinese boost converters would not be my first choice of components for new projects like this. I suspect it is likely they were made down to a price and consequently the components are very unlikely to be top quality. As the answer says, go and buy high quality capacitors. \$\endgroup\$ Sep 12, 2022 at 21:31
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    \$\begingroup\$ @JayDee That'd be the ripple voltage, since you're measuring the voltage across the capacitor. That doesn't tell you the ripple current. There are two ways to measure the ripple current: either use a current probe, or insert a small value shunt resistor in series with the capacitor and measure the voltage across it. The voltage drop across the resistor is proportional to the current (V=IR). So if you used a 50mΩ shunt resistor, you'd measure the voltage across it and calculate the current as I=V/0.05. The shunt resistance must be kept low as to minimise the affect on the measurement. \$\endgroup\$
    – Polynomial
    Sep 12, 2022 at 21:52

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