0
\$\begingroup\$

I have a 12V 22A DC actuator that I turn on/off with an automotive 80A 12V relay. Could turning it on and off cause damage to relay in the long run?

I understand a diode is required on relay's control input because of induction and voltage spikes that the electromagnet produces. A motor is actually a big inductor, so it might generate even higher voltage spikes and damage the relay.

I cannot put a diode in output of relay too, because it uses differential voltage to move actuators up and down.

\$\endgroup\$
1
  • \$\begingroup\$ Please show your circuit. \$\endgroup\$
    – Andy aka
    Commented Mar 22, 2023 at 10:07

1 Answer 1

1
\$\begingroup\$

I have 12V 22A DC actuator and I turn it on/off with an automotive 80A 12V relay. Could turning it on and off cause damage to relay in the long run?

Yes.

However, you have a fairly large standoff between your motor current and relay rated current, so you might hope that the relay contacts would be fairly heavy, and so wear slowly.

I understand in relay's control input a diode is required because of induction and voltage spikes that electromagnet produces. But motor is actually a big inductor, so it might generate even higher voltage spikes and damage the relay?

Yes. The fact that your relay is only rated 12 V might be a problem.

I cannot put a diode in output of relay too, because it uses differential voltage to move actuators up and down.

There are other things you can do to limit the high voltage that damages relay contacts.

One is a snubber. That's a low value resistor and a capacitor in series.

The capacitor is large enough to absorb the energy stored in the motor's kickback without increasing in voltage to a point where the relay contacts can arc. It's large enough to slow the rate of voltage rise so that the slowly-opening relay contacts can 'get ahead' of the rising voltage.

The resistor is large enough to protect the closing relay contacts from the current pulse from the charged capacitor, and is small enough to not allow too much voltage drop from the motor current passing through it. A few ohms usually suffices.

The other is a bi-directional TVS over-voltage suppressor diode. Make sure it will operate with your 12 V, and then clamp a only a few times that voltage. Its energy handling must be sufficient for the energy stored in your motor windings.

\$\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.