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I am currently making an automotive circuit that controls some automotive accessories (personal project not for production). In my circuit, I will be controlling some automotive relays and in any relay implementation I have seen so far there has always been some sort of flyback protection. I understand that you should always place a flyback protection but I am having doubts on what kind should be used. In the past my default setup would be a diode.

enter image description here

Recently, though, I have read that using a diode is actually detrimental to the relay as it will reduce the lifespan of the relay. When I try to research more about this topic, articles from google seems to be on the contrary, it could be very well be that these articles meant that a diode is better than nothing.

Another point of evidence that contradicts the claims is that automotive relays come in variants that have either a built in diode, or a built in resistor. Articles on google about flyback protection mostly mentions a diode as the go to component, very few of them mentions using a resistor or a resistor/diode combo.

Is it true that using a diode alone is not ideal as a flyback protection against inductive loads? Is it maybe a relay specific thing? I could not understand why a diode is bad for the relay.

I read several white papers and my two cents on the matter is: It would seem that if it were in an ideal world, the best flyback protection is basically a short circuit as the back EMF can be spent on the wires very very quickly, but that would make no sense as there would be no current passing through coil, which makes it meaningless. Having said that, the next best thing is using a resistor as a flyback protection, when the magnetic field of the coil from the relay collapses, the coil basically becomes a current source (voltage source also depending how you look at it) and the energy can be safely dissipated into the wire traces and the resistor (not back through the main power supply and the gate drivers of the coil potentially destroying them).

enter image description here

This has a major drawback, though, as the resistor will always have some power going through it while the relay is on. The diode would have not have this "leakage" problem but my theory where the problem is with a diode: As the back EMF voltage goes below the diode forward voltage, it would be the equivalent of having no flyback while still having a little bit of power left that needs to be dissipated, how this translates to the physical world is the relay would actually take longer to fully release, thus arcing for a longer period of time on the contacts occur, thus destroying them prematurely. This is just my theory though, feel free to correct it.

Then there is a TVS diode, specifically made to shunt excess current very quickly, but is it better than a resistor though? Don't TVS diodes just have a lower forward voltage?

Is the resistor the best flyback protection for a relay, and maybe other inductive loads? If so, how do I know what resistance I should choose?

Assuming this relay, this is a bit tricky to balance since lowering the resistance too much means I have to use bulky high power resistors, having it too high would also mean the back EMF voltage could be too high also and destroy my drivers. I could always use a relay that has a built in resistor but I think that would cheat me out of the learning opportunity I have now.

There are also hybrids like a TVS + resistor combo but I'm not really sure how much better they are.

EDIT:

Here are two application notes that support the claim of diodes reducing the relay lifespan:

  1. White Paper: Coil Suppression Can Reduce Relay Life (Link does not work all the time)
  2. Automotive Relay Application - Page 3 claims that a diode parallel to the coil reduces the lifespan down to 20% for motor loads.
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    \$\begingroup\$ The contacts are almost always what fail on a relay. Using a diode will cause the relay to open more slowly. Could that make a small difference to the longevity? Maybe. \$\endgroup\$
    – Drew
    Nov 2, 2023 at 19:19
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    \$\begingroup\$ Please link the article where you read that using a diode was potentially detrimental. \$\endgroup\$
    – Andy aka
    Nov 2, 2023 at 19:47
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    \$\begingroup\$ My limited knowledge of relay construction means I cannot understand why a micro-weld condition (sticking) can be alleviated by contact open momentum (as they say) when the contacts are already stuck and not moving hence zero momentum. They say this: The “stick” force is normally well within the ability of the net opening force, aided by the momentum of the moving armature, to break the stick and effect contact transfer <-- it can't move if it's stuck and, if it's moving it ain't stuck so, I don't understand their argument. This was the first link you posted. \$\endgroup\$
    – Andy aka
    Nov 2, 2023 at 20:33
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    \$\begingroup\$ I mean, all the relays that I've ever studied have a parameter called "must disengage" voltage and, if you choose one that has this parameter stated in the data sheet then (despite me finding the TE document flawed), I can't see that using a diode will be a problem. Also, if the data sheet does not tell you about this sort of stuff then it's not worth using. \$\endgroup\$
    – Andy aka
    Nov 2, 2023 at 20:37
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    \$\begingroup\$ I also don't see the problem with the diode but also what this diode is doing... \$\endgroup\$
    – Fredled
    Nov 2, 2023 at 20:49

4 Answers 4

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Like everywhere else, this is a compromise between two things, the transistor surviving the flyback, and the relay armature getting released quickly to break the contacts.

The diode is just the most simple device to clamp the flyback and protect the transistor, but it will dissipate the flyback energy the slowest so the relay current drops slowest and the magnetic field collapses slowly and the force that keeps the contacts closed weakens slowly until the contact starts slowly moving.

The slowly opening contact may bounce and arc until the contact has opened far enough so it does not arc over any more.

So yes, the flyback diode over the coil is poor for the contacts.

A resistor could be used over the coil. It will just pass current and has to dissipate power when relay is on. You would select the power handling based on how much power it needs to dissipate and how much temperature rise is acceptable. You can select the resistance by estimating from the relay current how much voltage will be generated over the resistance when coil current decays. And the voltage must be less than what the transistor can handle before it damages.

Other options exist, such as zener diode combined with the regular diode, to limit the flyback voltage higher regardless of current, unlike with the resistor.

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  • \$\begingroup\$ I see so something has take the brunt of the coil, either the driver or the contacts. I generally would lean on the relay getting destroyed compared to the driver as the relay is an easier fix. On that note what option would you choose? On the second reference link it mentioned a diode parallel to the driver, i wonder how that would affect the drivers lifespan \$\endgroup\$
    – DrakeJest
    Nov 2, 2023 at 21:07
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    \$\begingroup\$ @DrakeJest A normal diode clamps at 0.7 V; many drivers can take more voltage than that and be entirely within their specifications. With a resistor or zener, you can choose the voltage freely. The higher the flyback voltage, the faster the relay switches. \$\endgroup\$
    – jpa
    Nov 3, 2023 at 14:27
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The TE connectivity paper explains the reasoning very well.
I will quote the core part here, but reading the paper is recommended.
TE Connectivity are leaders in their field.

TL/DR:

  • A diode across a relay coil slows release by providing an electromagnetically produced force which tends to hold the armature "operated" and in opposition to restoring spring forces. The reduction in restoring force results in slower speed breaking, which allows increased surface arcing and which can produce "microwelds" between the contacts. These welds are small enough that they are able to be proken by the restoring spring forces once the stored energy reduces BUT cause contact surface damage which is cumulative and leads to shortedned contact life.

  • Using a zener diode plus a diode across the coil instead of just a diode still provides a limit to the 'flyback' voltages but reduces hold in forces and limits damage.
    (See paper for a better explanation of this).


This diode shunt provides maximum protection to the solid state switch, but may have very adverse effects on the switching capability of the relay. It is important to realize that the net force available to cause the armature to open is the difference between the magnetic restraining forces and the spring opening forces, that each of these is varying in a manner to cause the net force to vary both with time and armature position. It is this net force which gives rise to the armature system velocity and energy of momentum as it attempts to effect armature and contact spring transfer.

A slowly decaying magnetic flux (the slowest is experienced with a simple diode shunt across the coil) means the least net force integral available to accelerate the armature open. In fact, rapid loss of the opening forces supplied by stiff NO contact springs, coupled with slowly decaying magnetic forces, can actually cause a period of net force reversal where the armature velocity is slowed, stopped, or even momentarily reversed until the flux further decays, finally permitting available spring “return” forces to cause transfer to continue.

It is equally important to realize that when the contacts of a typical power relay make, connecting very fast rising (e.g., resistive) medium or high current loads to the voltage source, a minute molten interface occurs between the mating contacts, giving rise to a microweld or stick condition that must be separated at the next opening transfer.

The “stick” force is normally well within the ability of the net opening force, aided by the momentum of the moving armature, to break the stick and effect contact transfer. However, the loss or even reversal of armature velocity (under conditions of simple diode shunting as described above), and accompanying loss of armature momentum needed to help break the contact stick, can result in failure to break the stick, and a contact “weld” is experienced.

The more rapidly the coil current decays, the less the magnetic hold back, and thus the greater the armature momentum and contact stick “break-ability.”

Obviously, this is optimized when no suppression is used. However, near optimum decay rate can be obtained by using a Zener diode in series with a general purpose diode. When the coil source is interrupted, the coil current is shunted through this series arrangement, maintaining a voltage equal to the Zener voltage (plus forward diode drop) until the coil energy is dissipated.

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Suggestion:

schematic

simulate this circuit – Schematic created using CircuitLab

There should be a diode across the MOSFET (D1). There is already a diode included with the MOSFET but it's good to double it with an external diode. To avoid flyback surge from the coil of the relay, from the switch of the relay and from the load, place varistors or strong TVS diodes as on the schematic (the schematic editor doesn't have varistor symbol so I put this). Varistors dissipate more power but TVS diodes are more accurate. IMO varistors are better in this case.

And add a solid capacitor (C1).

A diode between ground and drain would protect only against negative current coming from the coil. I don't know if it's necessary.

Resitors are non-sens since they either impede the current needed to actuate the relay or reduce the voltage. Resistor parallel to the coil will either waste energy when the MOSFET conducts or will be ineffective in dissipating the energy.

EDITUM:

Commenting on your comment on the application note: They show a typical circuit which uses a N-MOSFET. Whereas you use a P-MOSFET. I recommend using a N-MOSFET. See second schematic. Here again, TVS diode can be replaced by varistors.

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  • \$\begingroup\$ I have not really used a varistor before, how do you spec varistors for this application? Digikey filters it , by max DC voltage, MIN,TYP,MAX, ENERGY, Capacitance. Can you please share a quick rundown on how i would choose my varistor. I have no space for big varistors only smd ones \$\endgroup\$
    – DrakeJest
    Nov 2, 2023 at 21:01
  • \$\begingroup\$ I used a PMOS just because i already had a bunch of them from a previous project. I really dont mind switcing to an NMOS. does it really matter that much though? \$\endgroup\$
    – DrakeJest
    Nov 2, 2023 at 21:09
  • \$\begingroup\$ In your case, the only important parameter is working voltage. It should be less than what the MOSFET drain can support and equal or slightly more than 12V. The larger the better. Radial provide higher energy dissipation (varistors act as energy dissipators) but if you want small size components, you can also use SMD varistors. capacitance is small enough not to interfere with the coil. Maybe very large varistors could. 10mm radial should be ok. \$\endgroup\$
    – Fredled
    Nov 2, 2023 at 21:12
  • \$\begingroup\$ Your bottom circuit works , I have applied this in the 90s and had no hassles in Auto electric use . \$\endgroup\$
    – Autistic
    Nov 2, 2023 at 21:14
  • \$\begingroup\$ There is less dissipation through NMOS but here, it's not an issue. The coil is not using that much of power. PMOS are just a little bit more difficult to control in my opinion. If you know what you are doing it's ok. \$\endgroup\$
    – Fredled
    Nov 2, 2023 at 21:15
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Let me collate everything I have gathered so far:

To answer the main question of does adding a flyback diode to a relay coil reduce its longevity.

YES IT DOES

This is coming from articles from industry leading companies:

  1. TE Connectivity: (TE Connectivity links sometimes say access denied keep trying eventually it will open up)

    White Paper: The Application of Relay Coil Suppression with DC Relays

    White Paper: Coil Suppression Can Reduce Relay Life

    White Paper: The Application of Relay Coil Suppression with DC Relays (found by @periblepsis)

    Application Notes: Automotive Relays

  2. Panasonic

    Application Notes: Automotive Relays User Guide

For detailed summary/explanation you will have to read the answers of other users as they can/have explain it better

So what flyback protection do you use?

At the end of the day it is up to the designer to evaluate which solution is best for you. The application notes From Panasonic and TE Connectivity offer great tips in making your relay circuit as effective as it can be such as load placement, dynamic behaviors, and of course coil suppression methods.

According to TE Connectivity application note, the life span of a relay can be reduced to as little as 20% of its lifespan depending on the coil suppression method used

enter image description here

But if you do choose a resistor, diode, zener as your method of choice Panasonic application note recommend these values (do check your relays manufacturer datasheet/application note for your values)

enter image description here

What is the best compromise?

A single diode parallel to coil is the best protection to your relay driver but it also comes at the cost of the worst influence on relay lifespan. In my opinion the best compromise is a zener diode parallel to the driver. As it offers the least influence on relay lifespan and does not have prolong the release time of the relay contacts which might be important on high current application. I think @Fredled made a good example circuit.

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  • \$\begingroup\$ "... to up to 20% ..." is ambiguous. Possibly better to put either to as little as 20% or by up to 80% boost up. \$\endgroup\$
    – Russell McMahon
    Nov 8, 2023 at 13:11
  • \$\begingroup\$ @RussellMcMahon thanks for pointing that out, i followed your suggestion and updated the answer. \$\endgroup\$
    – DrakeJest
    Nov 12, 2023 at 17:28

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