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I am using an electromechanical relay in a relatively medium-current project, in the range of 3-5 amps. I have designed it so the coil is controlled by a microcontroller through an optocoupler, with the 'primary'(low voltage) side at 5V and the 'secondary' (high voltage DC) at 24V. Normally I would use an SSR to reduce parts count, but SSRs with a current rating of more than 2 amps are prohibitively expensive compared to coil relays.

The optocoupler has a phototransistor output, with the collector at 24V. The IR diode is controlled with logic-level 5V, intended to safeguard the MCU in case of a failure mode.

How do I design a snubber to prevent the optocoupler from being destroyed during switching? I have seen circuits using diodes and RC circuits, but few outline formulas or application notes(surprisingly, the AoE talks little about this).

(The optocoupler is the SFH6156-2T from Vishay, the relay is a MCHRM1-S DC24V from Multicomp)

Diagram minus snubber sub circuit

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  • \$\begingroup\$ Many people just use a diode across coil if switching is not too offen. \$\endgroup\$ Feb 4 at 4:55
  • \$\begingroup\$ So you mean a snubber over the coil? What do you value in the snubber, cheapest possible solution, or something that prolongs the relay life? And you don't need another snubber over the contacts to prevent arcing and sparking, which could jam the MCU? \$\endgroup\$
    – Justme
    Feb 4 at 9:29

2 Answers 2

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The relay coil is nominally 800Ω so the 1Ω resistor is doing virtually nothing.

You should have a diode or a diode + zener diode or diode + resistor across the coil, otherwise when the phototransistor turns off the emitter voltage will go very far negative, likely causing the transistor to break down. The opto output is rated at 70V so you can do either. Using the two-component option may extend the electrical life of the relay by speeding drop-out.

schematic

simulate this circuit – Schematic created using CircuitLab

Your LED current is extremely high if the drive voltage is actually 5V, exceeding the absolute maximum limit on the datasheet, you should also check the CTR limits and allow for aging (which is accelerated by high LED current).

I would prefer a buffered drive circuit (with much lower LED current 5-10mA) to drive a 30mA relay at 24V, using a beefy output transistor (like a 600mA part), to be on the conservative side. Inductive loads like relay coils are hard on the transistor (a high Vce and 30mA will exist simultaneously for a short time during switching) and the opto is not really rated for that service- there is no SOA curve for the BJT. You will notice that the first circuit stresses the transistor the least, but the relay somewhat more, so there is a trade-off and some unknowns. Or use a relatively expensive photo-MOS SSR to drive the relay (you still need the diode or diode+).

If you go with direct drive I would test samples to destruction (if possible to observe that) to see how far you are from the SOA limits and stick with the one qualified supplier of optos. The consequence of failure would tend to be the relay stuck on.

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  • \$\begingroup\$ The 1 ohm resistor is per the datasheet. Prioritizing the optocoupler, wouldn't it then be easier to use the optocoupler as a low-side switch, returning the relay to ground with the flyback diode attached as described in Tweed's response? The optocoupler is driven by an MCU with 5V logic(no 3.3V sadly), and can only source 40mA of current per pin. \$\endgroup\$
    – lemon
    Feb 4 at 20:21
  • \$\begingroup\$ It doesn't matter, low vs. high on an isolated transistor as the two are in series. \$\endgroup\$ Feb 4 at 20:45
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The relay coil requires 30 mA at 24 V, and the opto can handle 50 mA continuously, so the numbers look fine. You'll just need the usual "freewheel" diode across the relay coil to limit the voltage on the opto when it switches off.

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  • \$\begingroup\$ with the anode connected to ground(no conduction in normal operation)? Wouldn't this damage the optocoupler(or is it reversed)? \$\endgroup\$
    – lemon
    Feb 4 at 7:42
  • \$\begingroup\$ Yes, anode to ground. No, no damage -- the idea is that when the opto cuts off, the relay current flows through the diode until it decays to zero, during which time the emitter of the opto is no more than one diode drop below ground, with a total of less than 25 V across it. \$\endgroup\$
    – Dave Tweed
    Feb 4 at 12:03

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