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I'm trying to add an indicator to a relay circuit; specifically, I want to know if it's open or closed in a way that can detect contact welding. (I considered a DPDT relay, but it looks like these could fail with one contact welded while the other is open.)

Is there any way to detect continuity in a circuit that may or may not have a live (110 VAC) voltage, without affecting any load connected to that circuit?

Consider this diagram:

circuit

Basically, is there any way to implement the "magic box" here to detect whether the circled switch has continuity across its terminals, without circumventing the function of the switch, and that will work with the other switches in any position? (Assume the "magic box" is allowed to have an independent power supply. The "output" of the magic box should be illuminating one of two LEDs to indicate either "open" or "closed".)

To clarify, the goal is for the "tester" to be built into a larger device; the less steps the user needs to perform, the better. Ideally, imagine the relay replaced with a wall switch. The switch would have two lights, one indicating the switch is in the "off" position, and one indicating it is in the "on" position. (But the "on" light needs to light up whether or not there is voltage across the switch.)

I could wrangle something by hooking up a DPDT switch across the switch being tested that shunts it between a test circuit and "live" circuit, but that's not ideal. (In "test" mode, obviously the mains circuit can't be completed, and in "live" mode, the switch will always appear to be open.)

Note: it's fairly important that any failure in the "magic box" either does not short the switch, or is reliably detectable via the LEDs. (The former is more preferable.)

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    \$\begingroup\$ The circuit you show is going to burn the moment all switches are activated so, what are you really trying to accomplish (rather than show a faulty looking circuit)? \$\endgroup\$
    – Andy aka
    Dec 22, 2022 at 18:18
  • \$\begingroup\$ The circuit on the schematic does not make sense. Depending on the real circuit there might be ways. \$\endgroup\$
    – Eugene Sh.
    Dec 22, 2022 at 18:25
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    \$\begingroup\$ @Andyaka, see edit(s). I replaced one of the switches with a load. (There are actually four switches, all needing the "magic box", but I don't think that's relevant. What matters is having at least a second, so it's obvious the highlighted switch might not be the only interruption.) \$\endgroup\$
    – Matthew
    Dec 22, 2022 at 19:43
  • \$\begingroup\$ @EugeneSh., can you elaborate? What "doesn't make sense"? Is it just that I elided the load, which I have now fixed? \$\endgroup\$
    – Matthew
    Dec 22, 2022 at 19:54
  • \$\begingroup\$ With the load it makes more sense. In such a case you can perform your test in steps. First determine if the "main" switch is closed (that is the circuit is "live"). Can do by measuring the voltage between the "lower" pole of the power supply and the "upper" pole of your tested switch. If it is disconnected, a simple continuity test will do. If it is connected, measuring the voltage on the tested switch will tell you if it is closed or open. \$\endgroup\$
    – Eugene Sh.
    Dec 22, 2022 at 20:13

2 Answers 2

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Something like this should work:
Updated to match your recent edit

schematic

simulate this circuit – Schematic created using CircuitLab

Ignore the Rsim resistors - they're only there to make the simulator happy.
The "Ground" nodes are not intended to represent "Earth Ground" - they're just the circuits common 0V reference. This entire circuit should be regarded as Live at all times and treated with the necessary respect. Visolated could be a battery supply, or an isolated DC supply powered by the incoming mains.
The first switch (SW1a) is just duplicating your diagram & doesn't serve any extra function in the circuit. It's simulated as a "Time Controlled Switch" so that the simulator can switch it on half-way through the simulation.
Note that there's nothing here which will differentiate between welded contacts and a normal "relay closed" state.

With SW1 off, the circuit relies on its "Visolated" supply to determine whether the relay contacts are open or closed. When they're open, Visolated supplies current through R2, D1, D2 & R1 to charge up C1 and switch on the NMOS M1. This then allows current to flow through Dgreen, turning it on, and effectively shorts out Dred, turning it off (not the most power-efficient method, but functional).
With the relay contacts closed, current from Visolated through R2 & D1 is 'shunted' to ground instead of flowing through D2 & R1, so C1 discharges through R3 and M1 turns off.
In this case, Visolated only supplies about 0.5mA (in addition to the LED drive current).

With SW1 on, we now have mains AC applied to the relay contacts.
When the contacts are open, current during the AC positive half-cycle flows through D2 & R1, again charging up C1 and turning M1 on. The voltage which C1 is allowed to charge up to is limited by the Zener D3, and R1's primary purpose is evident here because it's what limits the current flowing through the D3 to a sensible level.
During the negative half-cycle, no current flows through D2 & R1, and C1 will start slowly discharging through R3, but will not discharge enough during one half-cycle to cause M1 to turn off.
Similarly to the initial case with SW1 open, when the relay contacts are closed, current is shunted by the relay contacts instead of flowing through D2 & R1, so C1 discharges and M1 turns off.
In this case, Visolated is supplying more current than above since it's now effectively in series with the main supply. Assuming 110V mains then Visolated will be supplying about 10mA over the duration of the negative half-cycle, and nothing over the duration of the positive half cycle (due to D1 being reverse-biased).

It would be extremely unlikely for a failure in this circuit to result in a 'relay-closed' condition. The parts which would be most likely to fail almost invariably fail to an open-circuit condition. Choosing 'fusible resistors' for R1 & R2 would be a good idea.
But for a guaranteed fail-safe solution you could add a fuse at the D1/D2/Relay junction to separate D1 & D2 from the relay in event of any of the components failing to a short circuit.
The tricky part would be selecting a suitable fuse. It would need to have a relatively low fuse current rating - probably 100mA or less (definitely less than whatever the normal operating current of your load is) - but also have a breaking capacity which meets or exceeds the current rating of the circuit it's plugged into.

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  • \$\begingroup\$ As to "there's nothing here which will differentiate between welded contacts and a normal 'relay closed' state"... I think I see where you're going with that. I could probably figure out how to wire 'and' logic to that indicator with the relay's coil state, but for my purpose I don't need that; the indicator only needs to indicate if the relay is closed. In usage, it's enough for that to light when not expected to indicate a fault. Anyway, I don't think it's possible to tell if the contact has welded when the relay coil is energized. \$\endgroup\$
    – Matthew
    Dec 22, 2022 at 19:58
  • \$\begingroup\$ Two other questions. What modifications if any are needed to run this on 12V? Can it use "packaged" LEDs (i.e. ones with internal resistors that expect a straight 12V across the wires)? I'd prefer to use Alpinetech PL12Ms as the indicators. (I could use 5V indicators, though I'd have to go with a different brand, but panel-mount is sort of a requirement, and I believe those tend to be the "don't add a resistor" kind.) \$\endgroup\$
    – Matthew
    Dec 23, 2022 at 16:28
  • \$\begingroup\$ BTW, "Visolated is supplying more current than above since it's now effectively in series with the main supply"... is that going to be a problem for the PSU, especially a) if it's a cheap LED driver and/or b) if it's also powering some other stuff (fans)? (...or a problem for the fans, for that matter...) \$\endgroup\$
    – Matthew
    Dec 24, 2022 at 18:42
  • \$\begingroup\$ If I replace the two grounds on the detector side with connectivity to each other (as I would naïvely expect to actually wire this thing), it seems (according to the sim) I wind up with 110VAC in some... undesirable places? Does one of those need to be AC ground/neutral? \$\endgroup\$
    – Matthew
    Dec 30, 2022 at 16:12
  • \$\begingroup\$ Those "grounds" are only the 0V reference for that side of the circuit. Yes they should be connected together, but there's no connection from there to AC mains neutral or earth-ground. \$\endgroup\$
    – brhans
    Dec 30, 2022 at 16:47
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You ruled out DPDT relays because of the failure mode where one contact welds shut and the other opens. This is a common problem with a very simple solution: Use a safety relay with force-guided contacts, e.g. from the TE SR2M series.

Safety relays are designed in a way so that the failure mode described is avoided mechanically, i.e. the contacts are linked as such that if one welds, the other will not be able to open.

This relay does not seem fulfil the other requirements of your application. I provide it only to let you know those kinds of relays exist.

For higher contact ratings, there also are safety contactors with force guided contacts. A quick search revealed the Allen Bradley 100S-C with 23A rating as an example of a safety contactor. I'm sure there are lots more.

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  • \$\begingroup\$ Oof. Contact rating, 6A. The relays I currently have "penciled in" (G5RL-1A-E-TV8) are 16A plus inrush. I'm trying to build this to handle "anything" that could be plugged into an electrical outlet. I already consider not handling 20A a concession, albeit one I can live with as 15A limit is still pretty common. \$\endgroup\$
    – Matthew
    Jan 7, 2023 at 15:32
  • \$\begingroup\$ I did not say that this type of relay is the type to use in your application. I just added the part number from the top of my head to make it easier to find a datasheet of one kind of safety relay. Once you know they exist, you can probably find one that is suitable for your application. \$\endgroup\$
    – Crazor
    Jan 7, 2023 at 18:09
  • \$\begingroup\$ I updated my answer to mention safety contactors. \$\endgroup\$
    – Crazor
    Jan 7, 2023 at 18:18
  • \$\begingroup\$ Sure. But $240 each is also problematic. 🙂 \$\endgroup\$
    – Matthew
    Jan 7, 2023 at 19:05
  • \$\begingroup\$ Basically, I am not aware of any such relays that have comparable contact ratings and aren't cost-prohibitive. If the options are building the circuit as described for $50 and using a $10 relay vs. using a $250 relay... \$\endgroup\$
    – Matthew
    Jan 7, 2023 at 19:19

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