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I'm trying to design a ground impedance monitor. The idea is that the equipment in question needs to disallow operation if the ground wiring is disconnected or the connection is poor quality.

This is what I have now:

schematic

simulate this circuit – Schematic created using CircuitLab

The LED is actually the emitter of an LTV-844S opto-isolator. The blocking diodes are actually S1M (but I figured 1N4004 were similar enough for the purpose).

In testing right now, it takes a ground impedance of ~120 k-ohms to trip the system. That seems like it's way, way too high. My first instinct is to crank up the series resistance, but I'm hard pressed to see how I'm going to wind up with something that's suitably sensitive AND accurate (the spec calls for a trip level of around 12k - 100 ohms per volt), because the impedance level to trip is a tiny fraction of the series resistance, meaning that the series resistors' own tolerances will swamp the desired measurement.

Adding a huge amount of active circuitry is something I would not be happy about doing. At the moment, there is an isolated power supply powering the logic systems, but that power is on the "safe" side of the dividing line - where the detector of the opto-isolator lives. It's not really reasonable to bring that power across the line, because then the two sides won't be galvanically isolated.

EDIT: I've tried to work some more on this circuit, but I'm still not sure I have a good answer yet. There are competing goals at work that frustrate an easy solution. On the one hand, nominal AC voltage may vary anywhere from 100 to 130 VAC (this may need to operate internationally or on poorly regulated power) and the resistors have a 5% tolerance.

My latest attempt changes the series resistance to 240k and adds a 10 volt zener diode reverse biased in series with the opto-isolator. The hope there is that the zener tolerance of 3% being 3% of 10 volts will allow it to possibly "discipline" the relatively sloppy tolerance of the resistors. The idea is that until the voltage across the zener exceeds 10 volts, it won't conduct at all, and the higher the impedance to ground, the lower the voltage across the zener will be.

On the bench, this design appears to work well, but I'd like to hear from folks about this. My worry is that this design works with one set of components but will not be sufficiently reliable when copied.

Note that in the intervening time since I posted this question the actual supply voltage for this test circuit has been moved on the load side of a contactor. If the test fails, the contactor will open, meaning that there will be minimal exposure to dangerous voltages on the chassis when the ground is open.

Anybody have any ideas?

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  • \$\begingroup\$ How feasible would it be to use some sort of Point of Load isolated power solution? (ADI makes the isoPower chips, and there are also isolated point of load modules or Photovoltaic Isolators available as more traditional solutions, depending on the circuit's voltage/current needs.) \$\endgroup\$ Jan 15, 2015 at 4:35
  • \$\begingroup\$ I hate to add another power supply to this circuit, as there already is a 12 volt supply for the logic circuitry. From what I can figure, the big problem is setting a precise "cliff" for the current flow. Like, "conduct at least a minimum of exactly 2.5 mA or none at all." \$\endgroup\$
    – nsayer
    Jan 15, 2015 at 5:07
  • \$\begingroup\$ For a textbook-level overview of ground-check monitoring circuits see for example this. There's also a Wikipedia article on the topic but it's a rather low quality wiki entry. \$\endgroup\$ Jan 15, 2015 at 6:07
  • \$\begingroup\$ @RespawnedFluff The article you reference seems to be talking about situations where the monitoring and the monitored circuit are separated by some distance with a cable connecting them. That's not the case here. The only way to test the ground is to "leak" a small amount of current from one (or both - but not at the same time) hot lines and indicate a fault if that current does not flow. The problem is that the current must be quite small - 2.5 mA here - but the measuring tolerances are way, way tight. \$\endgroup\$
    – nsayer
    Jan 15, 2015 at 6:22
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    \$\begingroup\$ The incoming lines into the device are two single-phase hot lines (north american 208/240 VAC) and ground. There is no neutral. The amount of current I'm willing to "leak" to ground is far too little to close a relay, and I believe a relay would also not be suitably sensitive - the spec talks about detection thresholds of 100 ohms. I really don't want to try and find room for a power transformer in the chassis. If that's really the only way, then so be it, but that's why I'm asking. \$\endgroup\$
    – nsayer
    Jan 15, 2015 at 6:57

3 Answers 3

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Not only is the example circuit insensitive, it is mildly unsafe. For example if the supply side ground were to open the other ground (possibly connected to the equipment) is put at an AC "hot" potential. In this situation, even with the 47k resistors, someone touching the left side ground could receive a "mild" tingle, (a light AC shock).

A better idea would be to use a very small low voltage (safety agency rated) transformer across the Hot1 and HOT2 points, then take the secondary winding and use this to create a low voltage DC supply for the opto-isolator LED. Now use the two AC ground points to short out the LED. That way if either of the ground points ever becomes open you get a light pulse from the LED. Select the series resistor going to the LED low enough so that an extra 100 ohms across the LED (in a defective GND to GND connection) allows a voltage greater than the LED's turn on voltage. So with this setup any signal coming from the opto-isolator indicates a bad ground.

(Ideally the one ground side should be a separate known good ground - such as an Earth ground).

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    \$\begingroup\$ It's possible that the diagram is misleading you. There is only one ground. I'm a bit confused as to how the isolated supply would be "shorted out" by being grounded. I get the idea you're proposing - you power the LED in an isolated way but connect the anode to ground. Why would an isolated supply "care" that you did that? \$\endgroup\$
    – nsayer
    Jan 15, 2015 at 7:10
  • \$\begingroup\$ You need a constant current source. \$\endgroup\$
    – GR Tech
    Jan 15, 2015 at 12:17
  • \$\begingroup\$ Would that be as simple as, perhaps, an LM134 between the common 1N4004 cathodes and the opto emitter anode? That would certainly eliminate the series resistor tolerance as a source of error. \$\endgroup\$
    – nsayer
    Jan 15, 2015 at 15:03
  • \$\begingroup\$ Adding an LM334 and a set resistor of 130 ohms yields a current flow of ~500 µA. With that in the circuit, the sensitivity is now somewhere between 47 k and 91 k. Better, but still too high. It may be necessary to start tuning the secondary side, though. \$\endgroup\$
    – nsayer
    Jan 15, 2015 at 18:12
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Does your incoming AC supply consist of a Hot, Neutral, Ground or is it two Hot conductors and a ground but no neutral?

I have an older Hi-Pot tester that uses a circuit that monitors the voltage between the Ground and Neutral terminals. It seems very effective and obviously CSA thought that it was okay because the device does have CSA certification.

There is no easy way to do this reliably without either some decent amount of current flowing down the ground conductor or monitoring the voltage difference between two points that both connect to ground somewhere. Of course, the Neutral is bonded to the grounding conductor back at the power distribution panel (breaker box) most of the time.

I do have another suggestion: you could dump a significant amount of current down the ground line as a short pulse. These pulses would be of sufficiently short duration and sufficiently far apart so as to not cause electrocution should someone be in the path of an open ground connection.

I don't know what the safety / regulatory authorities would say about that, though.

Bottom line: this is relatively easy if you have a Neutral connection that is bonded to Earth Ground somewhere in your distribution system. It is significantly more difficult if you do NOT have an available Neutral conductor.

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  • \$\begingroup\$ Alas, this is a hot-hot-ground device. No neutral. One of the design goals - if possible - is to have the leak current be 5 mA or less so that you can use the device on a household GFI protected outlet if necessary. UL compliance is the idea, so clearly they're on board with the whole leaking-stuff-to ground thing. The fact that the source is on the load side of the contactor ameliorates the safety concerns, as if the test fails, the contactor will open and the safety risk will disappear. \$\endgroup\$
    – nsayer
    May 26, 2015 at 4:52
  • \$\begingroup\$ I have a couple of ideas that I hope to try over the next little while. Question: is this two phases of a 3-phase circuit (208 Vac) or is this a 120 / 240 split-phase supply (240 Vac)? *** And - does physical size matter greatly? \$\endgroup\$ May 28, 2015 at 2:29
  • \$\begingroup\$ It can be both 208 and 240 VAC. It can even be 120VAC hot-neutral. It can also be european 220 VAC hot-neutral, though I am content with changing the series resistance for that case. Physical size is not unimportant. (Lack of) complexity is paramount. PCBs have a square-inch cost to them. I cringe at introducing active circuitry here, as that means coming up with a way to power it. \$\endgroup\$
    – nsayer
    May 28, 2015 at 6:29
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If I understand your question correctly, check out GFCI - Ground Fault Circuit Interrupter. You can get these in hardware stores (at least in Canada).

Depending on the manufacturer, they come as either a wall outlet or a panel breaker (a special one).

Basically, it monitors the amount of current that flows in/out of your live and identified conductors ("identified" is commonly referred as "neutral") and the moment they differ by a very small amount (less than a few mA), trips the circuit. I believe it also monitors the bonding conductor ("bonding" is commonly called "grounding") and trips if the latter gets disconnected.

This maybe a good starting point for you, try to find how they work and replicate if out-of-the-box solution is not for you.

Good luck and stay safe.

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  • \$\begingroup\$ GFCI's trip if the sum of the current on Hot and Neutral are't the same -- meaning that there is a ground fault. It doesn't assess the quality of the connection to ground. \$\endgroup\$ Jun 1, 2015 at 20:42
  • \$\begingroup\$ What @ScottSeidman said. The device has a separate GFI elsewhere. This question is about a GCM circuit. \$\endgroup\$
    – nsayer
    Jun 1, 2015 at 21:32

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