0
\$\begingroup\$

I have this beefy brushed motor (24VDC, 30A at full load). It's relay driven (not negotiable due to requirements of electromechanical safety) so start and stops give problem. The interruption rating of the relays is just at the limit, with substantially no derating.

At startup I limit the current surge with a power NTC. This motor runs for about 30s about 10 times a day so heating and recooling is not an issue. Need to experiment but it should be safe for inrush.

The other problem of course is at switch off. Relays are (as always) rated at resistive load and over 30V there is risk of arching.

Now, theorically, if I put, say, a varistor to suppress the overvoltage there shouldn't be problems. The thing I fear is the size of that varistor.

Estimating (from similar motors) and inductance of 500-700mH, the usual formula at 30A gives me about 200-300 J of energy in the windings. I guess that a lot of these won't reach the varistor but that's still a significant amount of energy. Also I only find ratings for short pulses (usually the 8/20µs or the 10/1000µs). With that inductance I guess the transient would be in the ms range.

So, what's the solution? A really big varistor? An RC or RCD snubber (like for triacs)? Just let the contacts suffer and die? (it's switching 10 times a day...)

EDIT: OH silly me. I just need a bridge of freewheeling diodes. For some reason I was fixated with varistors but I have no problem with a slower current decay on the motors.

However I'm still interested to know if the theory part was correct

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

The DC motor is an inductive load. A common solution against voltage spikes with all relay switched inductive loads when switching off is to connect a silicon diode across (prallel with) the inductive load. Cathode connected to plus, so normally non conducting. The diode must at least be able to withstand the motor's working voltage and the motor's current, which latter it will bleed for a short moment after switchoff.

\$\endgroup\$

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.