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I briefly encountered a designer who wished to use an MOV instead of a flyback diode with a relay coil. I don't know much about the application, but why would you want to do that?

The only reason I can think of is that the coil is normally energized and seldom de-energized, and the MOV acts as a speed-up circuit for de-energization, burning the inductive energy faster than a flyback diode.

That said, why not use a zener or a resistor, especially given that MOVs have limited lifetimes? What other options are there?

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Using a varistor or a combination of zener diode and a normal diode instead of just a single diode has following advantage:

After turning off the coil of a relay the energy stored in the inductor should be dissipated (turned into heat) as fast as possible.
If this is not done the electromagnet lets go the contacts too slowly causing the contacts to wear much faster.
In addition the use case may require the relay release time to be as short as possible.

The energy of the coil is dissipated faster the higher the voltage across the varistor/diode.

For details see my answer to a similar question. It compares the simulation of a circuit with just a single fly-back diode and a circuit with a diode and an anti-serial Zener diode.

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  • \$\begingroup\$ One or two days ago somebody posted a link to a paper/app note about this topic. \$\endgroup\$ – Curd Apr 10 '18 at 14:10
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As has been pointed out in other answers, the basic diode snubber arrangement extends the relay drop-out time and the speed of contact opening.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Diode snubber. (b) Resistor snubber. The diode can be left out but doubles the power consumption of the circuit.

I checked a 24 V Finder relay in my stock and measured 800 Ω coil resistance and 2.8 H inductance de-energised and 9 H when the armature was pressed in.

At 24 V the energy stored in the relay would be \$ \frac {1}{2} LI^2 = \frac {1}{2} L{\frac {V}{R}}^2 = \frac {1}{2} \frac {24}{800}^2 = 4.5 \ \text {mJ} \$.

We know from the maximum power transfer theorem that maximum power transfer occurs when the load resistance (R1) is equal to the source resistance (the 800 Ω of the relay coil). The reader can prove to their own satisfaction by calcuus that the maximum time delay will occur when the relay is short-circuited at the instant of power-cut. (For simplicity assume that D1 is ideal.)

Running this through a simulation results in the graphs below.

enter image description here

Figure 2. The switches are opened at t = 1 ms. The time to drop to half-current for circuit 1a is shown by the blue line and occurs 7.3 ms after the switch opens. The time to drop to half-current for circuit 1b is shown in orange and is 3.8 ms.

The relay will drop out at twice the speed for the resistor snubber.

... MOV instead of a flyback diode with a relay coil. ... but why would you want to do that?

If the MOV can be rated to break down just above the relay voltage then it may be close to the optimum relay load to minimise drop-out time. It may behave faster than a diode and uses one component instead of two as shown in Figure 1b.

The only reason I can think of is that the coil is normally energized and seldom de-energized, and the MOV acts as a speed-up circuit for de-energization, burning the inductive energy faster than a flyback diode.

I agree with that.

That said, why not use a zener or a resistor, especially given that MOVs have limited lifetimes? What other options are there?

I think we've covered this.

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Reasons not to use only a MOV:

  • "Everyone" uses a diode.

  • I do not know any cases where a diode isn't fast enough to protect the transistor (which is the diode's main function).

  • I think MOVs are more expensive than a simple diode.

Reasons not to use zener diode + resistor:

  • a diode is cheaper

  • a zener diode isn't going to be faster than a diode

  • a diode is fast enough

You should ask that designer why (s)he wants to use a MOV while almost everyone else just uses a diode.

If the relay coil is fed with AC then maybe a MOV would be a solution but using AC to power a relay's coil is either rare or not allowed (for most relays as far as I know).

Next time, when someone suggests a weird/strange/non-standard solution, immediately answer: "Great idea, I am curious, why would you do it like that?" so that you can learn something new because who know, it actually might be a good idea. If it is, the person should be able to convince you with facts.

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    \$\begingroup\$ It's neither weird nor strange nor non-standard to use not only a single fly-back diode but a diode with an anti-serial Zener diode or Varistor if the inductance is the coil of a relay. \$\endgroup\$ – Curd Apr 10 '18 at 14:20
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    \$\begingroup\$ @Curd Read carefully, I never wrote that using a MOV or zener in addition to the diode is weird or non standard. In my opinion only using a MOV is non-standard. In 99% of commercial products only a diode is used. I am still not convinced that adding a MOV and/or zener is "better" or "needed". It might be "safer" though but perhaps only a diode is "good enough" already. \$\endgroup\$ – Bimpelrekkie Apr 10 '18 at 14:38
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    \$\begingroup\$ It has nothing to do with safety. The relay contacts will last longer (more cycles) and the relay can be used to do faster switching. \$\endgroup\$ – Curd Apr 10 '18 at 14:46
  • \$\begingroup\$ @Bimpelrekkie It was mentioned in passing by a colleague, and I didn't have access to the designer directly, and my colleague was at a loss as well. \$\endgroup\$ – schadjo Apr 10 '18 at 14:47
  • \$\begingroup\$ If a Varistor is used you don't need a diode. If you use a Zener diode you need another normal diode in anti-series because the Zener would be forward biased when relay is on. \$\endgroup\$ – Curd Apr 10 '18 at 14:48

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