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I have come across these two app notes:

Coil Suppression Can Reduce Relay Life

Application of Coil Suppression with DC Relays

enter image description here

Both of them agree on the fact that a common diode + Zener diode in series is the best solution for spike suppression all the while maintaining a very fast energy drop-off.

However, the second app note shows a -24 V transient with such a schematic. That seems to me way out of a standard open-collector output. What could be done to keep the benefits of the Zener + diode without having the risk of frying an output?

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3 Answers 3

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If you need fast turn off you'll use a transistor that can withstand the voltage (and has adequate safe-operating-area for the switching itself). The voltage is the zener voltage at the coil current prior to switching plus a diode drop. Typically that's on top of the supply voltage so the transistor has to withstand the sum of the two voltage.

Aside from logic circuits, there's no "standard open-collector output", each design will be different.

Here is a simulation with a 100mA coil and a diode vs. a 24V zener.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

Magnetic circuits such as relays typically have a much lower 'hold' current than the operating current, so if your current has to drop to, say, 20mA typically before the contacts start to open the zener circuit will be about 5x faster in this example.

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  • \$\begingroup\$ I suppose a Schottky diode would have an even longer drop-out time with a 0.3V transient. And a push-pull that effectively short the coil would be longest. But would it be possible to inject a reverse current to collapse the field quickly? \$\endgroup\$
    – PStechPaul
    Jan 11, 2023 at 23:04
  • \$\begingroup\$ Yes, a Schottky would be slower but only very marginally unless the coil operating voltage is extremely low. Most of the voltage as the field dies down is dropped across the internal resistance when there is just a diode- and it starts out as the coil voltage ignoring the diode drop. To force a current through an inductance very quickly you need to apply a very high voltage, it's just physics. \$\endgroup\$ Jan 11, 2023 at 23:17
  • \$\begingroup\$ A better solution might be to implement an economizer circuit so that the coil is held in with just minimum current. Then there would be much less energy to dissipate and release would be quicker. \$\endgroup\$
    – PStechPaul
    Jan 12, 2023 at 5:02
  • \$\begingroup\$ I am a bit intrigued by the "hold current" point that you make. Why would the contacts being open, or closed, impact the kickback in the relay coil? If this is about a change in the magnetic circuit as the actuator moves, I would expect this to be wildly dependent on the specific relay. On the other hand, if you mean that after energization we can reduce the current down to a safe holding current, and then opening the relay is faster - this is much clearer and easier on my brain. Can you elaborate a bit? Thanks! \$\endgroup\$ Jan 12, 2023 at 8:35
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    \$\begingroup\$ @VladimirCravero Sure you could do that- reduce the current (but it's more complex and a mechanical shock could open the contacts) so it's not often done. Typically minimum pull-in vs. max drop-out is a large ratio (eg. 7:1 for JS1a, 5:1 for G7L AC or DC coil) but you also have to account for PSU tolerance and (the big one) temperature. So 10% of rated voltage is enough to guarantee a JS1a will drop out at 20°C but if the temperature is colder it could be too high. \$\endgroup\$ Jan 12, 2023 at 10:04
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This is simply due to the zener (+ diode) voltage. If -25 V is too much, you could use a lower voltage zener that fits your application.

Keep in mind this will increase the dropout time. The entire point of this setup is to keep the voltage across the coil higher so the field dies faster.

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Depending on your application, you could connect your relay in a 2 switch forward configuration:

schematic

simulate this circuit – Schematic created using CircuitLab

To activate the relay, both switches need to activate. The relay then sees the full voltage of the power supply.

If either switch is opened, you get into the "single diode" scenario, which causes the relay to open slowly. But if both switches are opened at the same time, the voltage on the relay becomes the reverse of the operation voltage, which helps with the magnetic field collapsing. The voltage as seen at the switches is never higher that the power supply with a diode drop on top.

Note that one important thing with this setup, is that you need your main voltage rail capacitors to be big enough to handle the sudden inrush from the relay, without the voltage climbing too high for the other components

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