This is my first time using dual-coil latching relays (Kemet EC2-5TNU), and I'm running into some strangeness when breadboarding out the below circuit: enter image description here

The coils switch without issue when sending a pulse to either transistor, and the LEDs simultaneously pulse without issue as well. The thing I'm seeing is when I release the voltage driving the base of either transistor...this causes the opposing LED to do a brief and dim pulse. As an example, I'll pulse the SET coil...on the trailing edge of that pulse, instantly the RESET LED will pulse dimly. And vice-versa.

I know this is probably very 101, but I just can't seem to find an explanation in my searches. I have flyback diodes in place as you can see. Is it possible (using the example above) that when the SET coil's magnetic field collapses on the trailing edge of my trigger, it's inducing something in the opposite coil somehow, such that the opposite LED is getting forward-biased for a moment? Yeah, I'm spit-balling here obviously.

I did find an extensive Omron app note that recommended putting an extra diode in series with the Set or Reset coil when doing certain configurations (Page 23 here). Those configurations didn't seem to apply, but I figured I'd at least try...and lo and behold, it did indeed fix the issue. But I'm curious as to the why? Here's the updated schematic for that: enter image description here

I'm scratching my head, since besides the shared VCC and ground, they're independent circuits...right? Wondering if somebody might be willing to give me a hint as to what is happening here. Is it really necessary that I have those extra diodes to use a latching relay like this, or are they just a band-aid for something I'm overlooking here? Thanks for any input!

Edit 1: After Jasen’s comments I decided to open up one of my latching relays to get a visual on it. For the record I had a whiskey and made a burnt offering to the relay gods first. Picture should be attached if anybody is curious.

enter image description here

Edit 2: OK, so I scoped it with my original circuit above (without those extra 2 diodes D3 and D4)...I put the probe at the Collector of Q1, which is part of the Reset circuit...then I pulsed Q2 to trigger the Set coil. On the trailing edge of that pulse, the scope shows a dip of about -4V from the Reset coil before coming back:

enter image description here

So I'm thinking the presence of this momentary negative voltage is forward-biasing the LED. And the reason the Reset LED no longer pulses with D3 in the circuit is because during that negative voltage, D3 is reverse-biased...thus the LED never sees the negative voltage and stays in its inactive state. Does that make sense?

  • \$\begingroup\$ I think that @Jasen observations are on the right track. You could connect the LEDs to ground rather than +5V and drive them with the same signals that drive the transistors. If your controller has enough high side drive current. \$\endgroup\$ – Dan1138 Feb 2 '20 at 2:56
  • \$\begingroup\$ Thanks @Dan1138, my goal was to possibly trigger this from different sources at my leisure, so was trying to avoid directly driving the LED from the controller. \$\endgroup\$ – boochyan Feb 2 '20 at 21:20
  • \$\begingroup\$ Yes that makes sense, but the reason OMRON suggests this circuit is to speed up the relay since Tau=L/R and when the diode is high R the time constant Tau is small. The crosstalk from Set to Reset is high impedance and is not relevant. \$\endgroup\$ – Tony Stewart EE75 Feb 3 '20 at 20:26

the two windings are wrapped on the same former in opposite directions, this forms a type of transformer so you can put energy into one coil and get a little out of the other coil.

  • \$\begingroup\$ Thanks for the response, interesting! You're saying that the 2 coils are actually wrapped around the same physical object? Are dual-coil latching relays generally constructed in this way? I'm feeling a little confused by the datasheets that show the 2 coils on opposite ends of the physical relay...maybe I should just crack one open and have a look. Either way, sounds like this is expected behavior...is adding the 2 extra diodes as I did a legit way to deal with this issue would you say? A consequence of trying to get 2 for the price of one by switching the coil and LED together... \$\endgroup\$ – boochyan Feb 2 '20 at 1:18
  • \$\begingroup\$ most latchin relays have a permanent magnet that holds the switch in each position, because of that you can drive them from a single winding if you can drive it with opposite polarities. but that's not convenient electonically, so the provide two windings so that you cah have unipolar drive. \$\endgroup\$ – Jasen Feb 2 '20 at 3:26
  • \$\begingroup\$ Cool, yeah I'd looked into using single coil latching relays, but did seem to generate more complexity in other ways, so I went with the dual. Just for kicks, i did actually bust one open (I put a pic in the post). \$\endgroup\$ – boochyan Feb 2 '20 at 21:09
  • \$\begingroup\$ I guess I'm still not understanding the "why" of what I'm seeing. I know when the magnetic field of a coil collapses, the polarity of the coil inverts and makes a spike...this makes the flyback diode forward-biased allowing that energy to dissipate. And as you say, during this moment the opposite coil is also getting a little bit of action. Does this induced pulse of energy create a similar collapse on the opposite coil, but that voltage quickly dissipates to below the power rail, so it's forwarding-biasing that LED? \$\endgroup\$ – boochyan Feb 2 '20 at 21:46

Since VL=LdI/dt increasing the diode voltage allows a faster rate of recovery. VL is the diode + zener voltage drop.

  • \$\begingroup\$ Thank you Tony. So adding in that 2nd series diode makes a greater total voltage drop, and thus makes for fewer cycles to get that spike dissipated. \$\endgroup\$ – boochyan Feb 2 '20 at 21:11
  • \$\begingroup\$ I posted an "Edit2" above. I think I might have an understanding of this now, but if you'd be so kind as to have a look....I'm curious if my conclusion makes sense to you? Thanks! \$\endgroup\$ – boochyan Feb 2 '20 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.