Datasheet recommends driving relay in an uncommon configuration

Question

I am looking at the datasheet of G8G relay(s,) specifically the section where they recommend how to drive the relay and placing the reverse diode.

I wonder about their statement:

OMRON recommends coil driver circuit (b) and (c) for coil surge suppression. However the circuit (d) is not recommended because it may negatively affect the durability performance.

1. Why is d) - placing the reverse diode across the relay - not recommended? This is the most-recommended way I can find if I google how to drive relays. I have known this way of driving relays for years.

2. They say OMRON recommends this hookup but they don't mention their source and I was unable to find the source of their statement. In what way would this configuration "negatively affect" the durability performance *of the relay I suppose)?

3. Regarding circuit (b,) why does this make sense? After the low-side transistor turns off, the relay's coil will start generating a large voltage at its top-side, and it won't have a way to draw current from somewhere like (a), (c) and (d) which all connect the top side of the relay to its low side.

Answer 1

Option (d) clamps the voltage across the coil to a low value (forward drop of the diode) compared to the other options. When the voltage is low, the rate of decay of the coil current is slow, meaning it takes longer to decay to zero, and takes longer to reach the point where the contacts start to open.

This has two issues:

1. The force applied to open the contacts is lower than expected, causing the contacts to open more slowly than intended, which may cause an arc to occur between the contacts. Arcs can cause pitting (deterioration) of the conductive surfaces, reducing service life of the relay.

2. It may be possible that the next turn-on signal arrives while the relay coil current is still flowing, causing a "reverse recovery" event (refer Appendix below), which can lead to failure of the electronic coil driver.

Issue #1:
This is supported by an Omron publication that mentions that the diode-only solution (option d) may have "release time that is delayed too much":-

Link for image above:
https://omronfs.omron.com/en_US/ecb/products/pdf/auto_precautions.pdf
Under heading "(6) Contact Protection Circuit (Arc Reduction)".

Issue #2: UPDATE IN RESPONSE TO COMMENTS
Some commenters on this answer have expressed doubt that diode reverse recovery, a phenomenon on a time-scale measured in micro-seconds (or nano-seconds in the case of switch-mode diodes), would have any effect on a mechanical relay operating every few seconds.

Well, the answer is that it doesn't affect the relay itself - but it can affect the driver of the relay coil, leading to field failures.

My experience with this issue was in factory automation. The failed device was usually the transistor conducting the coil current, this typically failed short-circuit, so the relay was stuck energised. These failures occurred only a couple of times per month in a population of several hundred relays - not enough to severely affect productivity, but very annoying when called out to fix the problem at 2am.

Root-cause analysis revealed that this failure only occurred when slow (50Hz) diodes were used as the coil-current catch diode ("free-wheeling diode"); these have Trr measured in hundreds of micro-seconds, and Irr that can be 3 to 10 times their forward current, and these parameters are very temperature dependant (both get much worse with increasing temperature).

A slow diode (long Trr and large Irr) can hold the Vce (Vds) of the BJT (MOSFET) up at the supply voltage while the BJT (MOSFET) collector (drain) current can reach many multiples of the coil current. This can stress the transistor and exceed its SOA.

The key issue here: it only takes one event (where the time between the last turn-off, and the next turn-on, is too short), and it's game over for the poor transistor.

In my case, we simply replaced the slow diodes with fast diodes (Trr < 1us).
Result: no more 2-am call-outs.

Here is more complete list of possible solutions:

1. Use a fast diode (Trr < 1us under all operating conditions).
2. Use a more sophisticated transistor driver to ensure the SOA is never exceeded under any condition (even if the diode is slow).
3. Ensure the coil current has gone to zero before attempting to turn on the coil again. If that is done, you can use whatever diode, transistor, and transistor-driver circuit you like.

But these solutions are usually more complex than simply ensuring the de-energisation time of the coil is sufficiently short. This is usually done by having sufficient reverse voltage across the coil during de-energisation, and that voltage is usually much larger than diode Vf.

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3. Regarding circuit (b,) why does this make sense? After the low-side transistor turns off, the relay's coil will start generating a large voltage at its top-side, and it won't have a way to draw current from somewhere like (a), (c) and (d) which all connect the top side of the relay to its low side.

When the BJT turns off, the voltage at BJT collector will go very high until the zener breaks over, causing the coil current to flow into zener cathode, out the zener anode, and then through the power supply to the other terminal of the coil. Thus, the coil sees a negative voltage compared to when the BJT was on. The voltage will be the difference between the zener break-over voltage, and the supply voltage. So, if the supply is 12V (say), then the zener could be 18V, so the coil sees -6V when BJT turns off.

This option is better at protecting the transistor from over-voltage than (a) or (c), since the zener can be placed right at the transistor, allowing it to clamp the transistor voltage effectively without worrying about external parts of the circuit. However, with (a) or (c), the current path during de-energisation includes any external wiring to the nearest power supply de-coupling capacitor; this will play a role in determining the actual peak voltage stress seen by the transistor.

Appendix: Diode Reverse Recovery
Diode reverse recovery is the process by which a diode changes from ON to OFF, ie: from conducting current in the forward direction, to blocking voltage in the reverse direction. It does not occur instantly; during this process, the diode may conduct a significant current in the reverse direction, which has to flow in the circuit formed by the power supply, the diode, and the BJT (not the relay coil). During much of this process, the voltage across the diode remains low, which causes large power loss in the BJT.

Link below: a good introduction to diode reverse recovery:
https://www.mikrocontroller.net/attachment/351267/Understanding-Diode-Reverse-Recovery-and-Its-Effect-on-Switching-Losses.pdf

UPDATE 2024-06-10
For those wondering if diode forward recovery could explain the failures I mentioned. Diode forward recovery would cause transistor collector voltage to be higher than expected (0.7V above supply voltage) at the moment of turn off. Yes, this is indeed a known phenomenon, and can cause problems, particularly in low-voltage circuits as explained by the legendary Jim Williams in this app note:

https://www.analog.com/en/resources/technical-articles/diode-turn-on-time-induced-failures-in-switching-regulators.html

However, in this application, diode forward recovery is very unlikely to have been the root cause because:

1. At the time, a 20MHz scope (the fastest I had) did not reveal any significant voltage over-shoot on the transistor.
2. The transistor was rated ~70Vcbo (40Vceo), the supply voltage was 24VDC.
3. In this application, this event (forward recovery) occurs each switching cycle; whereas reverse recovery occurs rarely. If forward recovery was the root cause, then the failure rate would have been much higher.

In this application, the transistor switching stress is always worse for turn-on (diode reverse recovery) than turn-off (diode forward recovery). The amplitude of the over-stress, its duration, and total energy dissipated, are all far worse for reverse recovery than forward recovery. As proof, I present the following images from a Toshiba publication, and the following statements:

Forward Recovery:
Link for image above:
https://toshiba.semicon-storage.com/info/application_note_en_20210831_AKX00995.pdf?did=141363

Reverse Recovery:
Link for image above:
https://toshiba.semicon-storage.com/info/application_note_en_20210831_AKX00995.pdf?did=141363

Amplitude

1. Forward recovery causes a voltage over-shoot on the transistor at transistor turn off. The magnitude is of the order of, say, 5 to 20V in very slow diodes. For this application, this may at worst double the voltage stress on the transistor.

2. Reverse recovery, however, can easily increase current stress by a factor of 5 or more, and gets worse as temperatures increases.

Duration

1. Trr is usually far greater than Tfr for any diode, refer:
   https://electronics.stackexchange.com/a/647827/341959

2. During forward-recovery, the transistor current starts at load current, and starts reducing immediately the diode becomes forward biased. The duration of the process is typically shorter than reverse recovery. Peak power in the transistor, as a factor of Po, is less than 2Po.

3. During reverse-recovery, the transistor voltage remains high while its current ramps up from zero to load (Io) plus diode Irrm. Only after the peak diode reverse current has occurred does the transistor voltage start to fall. Peak instantaneous power in the transistor is therefore:
   PQ(pk) = V x (Io + Irrm).
   Since Irrm can be ~5 times Io, PQ(pk)= V x 6.Io.

Power & Energy dissipated in the transistor
From inspection of the images, and the statements above, it is clear that the energy dissipated in the transistor at turn-on significantly exceeds that at turn-off.

Turn-on energy loss is typically 5 to 10 times turn-off energy loss. Therefore reverse recovery, rather than forward recovery, is more likely to push a transistor out of its SOA.

Answer 2

As a preface, the different methods are from the perspective of reliability and longevity of the relay contacts.

1&2) Regarding the method (d), it is the most common way you see because it is the most simplest and easiest way that generally provides an acceptable result. It is likely the most straightforward method of clamping the inductive kickback from the coil to prevent the transistor from damaging, but it is also a method which makes the coil current to decay the slowest. The magnetic force that pulls the contacts together decays slowly, and this is by far the worst for the relay contacts as they open slowly which allows for the contacts to slide longer against each other while passing current, heat up due to increasing contact resistance until the contat separate, and as the contacts move slowly, they arc and spark longer until the distance is large enough.

All the other methods dissipate the energy from the coil faster, so the magnetic field decays faster, and the contacts break open faster.

3. The (b) makes sense and it has a current path when transistor turns off. The point is that when transistor is on, current flows from battery through coil and transistor back to battery in a loop. When transistor turns off, the coil current starts to flow through the zener, so the loop is still from battery through coil and zener back to battery. The thing that is changed is that the zener clamps the theoretically infinite voltage to some level like 24V over the zener and that is something that the transistor can survive without damage. If the battery is 12V, it means that the coil has 12V in reverse over it, and the coil needs to push out the required current with 12V voltage drop, so energy dissipates faster with 12V drop.

With method (d), all the coil needs to do is to push the current out with only 0.7V drop of the normal diode until coil energy is dissipated.