I have a relay as an additional safe torque-off switch for a motor driver, that supplies the power to the MOSFETs. Now the main capacitive load is behind the relay, otherwise it will be destroyed after the first use, but to reduce some ringing when switching the high-side, a 10 - 22 μF capacitor (ceramic) should be added and this will be seen as a capacitive load by the relay.

The relay is the following: https://docs.rs-online.com/5098/0900766b80fa75dd.pdf

How can I decide, what the capacitive load limit of the relay is? The supply voltage is 48 VDC, the energy by the capacitors is 0.5CU2 = 25.3 mJ, which seems not a big problem on a first sight, but I'm no sure. As a final solution I see to use a resistor of around 22 kΩ next to the relay contacts, which should load the relay to around 36 V in 0.5 s.

What do you think about this capacitive load size?

  • \$\begingroup\$ Ic=Vc/ESR of cap which means the contacts burn with energy dissipated in contact resistance and Cap total ESR. You must show all variables in a schematic with specs for desired limits of 1A resistive load absolute max. At 48V. Then derate for longer life. \$\endgroup\$ May 29, 2021 at 15:22
  • \$\begingroup\$ What I don‘t see is how the energy plays a roll. I mean when I use a 1nF cap with 10mOhm ESR I don‘t expect the relay to stick, since the energy is too small. So somehow not only the maximum current must play a roll, but also the energy from the inrush current, and that threshold should somehow be refered to the capacitive load and the switching voltage \$\endgroup\$ May 29, 2021 at 20:25

1 Answer 1


Your problem was already considered by the SE Community: Relay contact sticking while driving a capacitive load

This TE relay is a good device (already used, but I prefer Omron or Finder because they characterize their relays better), and as Tony Stewart said can be used reliably up to 1 A at your 48 V. Between us, if you relax the expected life in terms of number of operations (e.g. you operate the relay rarely), then you can increase a bit the operating current.

As you say, the inrush drained by the capacitive load could stick your contacts for excessive heat. What matters is the dissipation in the relay contacts, for which we do not have the contact resistance value.

Usual solution is a series NTC or -- better -- a series inductor. We don't know the time of closure of the electric contact (not the mechanical operation), but we can assume something in the order of 10-100 us; if shorter the inductor will have a larger reactance, solving the problem completely. At 100 us the 22 uF has values of reactance XC in the order of few ohm: the inductor must provide enough reactance XL to limit the current. Let's say that XL is slightly larger than XC; then you end up with about L=100 uH. At 10 us XC will almost disappear increasing the theoretical inrush current value, but XL will be ten times larger.

[Update after comment on possible resonances] If an inductor is not a viable solution, then the NTC is, at the cost of some dissipation. For bulky slower inrush an auxiliary relay can be used to short circuit the dampening resistor, so putting together a "mechanical NTC". -- Otherwise a solid-state relay may be used. -- I prefer mechanical relays with double contacts (so 2-wy relays) and I put the contacts in parallel: in case of asymmetry (or pristine wearing of one contact), the one heating up sees its current going into the other and this ensures some protection.

  • \$\begingroup\$ An inductor does not work here, since it is a motor controller and any inductance in the high side source will increase rining, that is also the reason I want additional capacity after the relay, to counteract the parasitic inductance from the relay contacts \$\endgroup\$ May 29, 2021 at 20:28
  • \$\begingroup\$ You have then two problems: the inrush current and resonances. If you put a NTC you have some dissipation. When the inrush is slower, you can even cut the damping resistor with a relay after some tens ms up to seconds; here the problem is in the order of a hundred us or so. Otherwise, as they suggest in the other link, use a solid-state device: they say "triac", but in general a solid-state relay can work. \$\endgroup\$
    – andrea
    May 29, 2021 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.