# GCM circuit design redux

I'm coming back to this design again after some time away from it. Previously I was trying to design the circuit to be completely passive, but that hasn't wound up being very reliable, so now I'm resigned to attempting it with an isolated power supply.

This little excerpt might be a little hard to read, but the "relay test" inputs each come from the load side of a contactor that switches North American split-phase 208/240 VAC (so each leg is 120 VAC from ground 180 or 120 degrees phase separated). The 150k resistor is a 1W flame-proof resistor, followed by a 1 mA current limit and a 1k resistor to ground. The voltage across the 1k resistor is compared to 1V. The output of the comparator isn't shown, but it lights an optoisolator that passes the status on to the logic circuitry.

The questions I have here are about the portion of the circuit shown.

1. Is the 1 mA regulator serving a purpose here or should I just remove it? My own paranoia led me to put it in to insure that if the path to ground was poor enough that any potential residual current would be limited to 1 mA, but the big series resistor is supposed to be guaranteed to fail-open, so there should be no way for its value to reduce by enough... right?

2. As designed, the comparator and its optoisolated output are completely isolated (one detail missing from the screen grab is that pin 8 of the comparator - its Vcc - is connected to Viso - the isolated +5 supply from DC1). Does this have any implications for the potential voltage the comparator could see if the ground impedance rose to infinity? Should I ground the isolated supply output? It's an isolated supply because I'm trying to insure that there's a good and proper "moat" completely around all of the HV circuitry. The one ground connection allowed is from the low side of the two 1k sense resistors in the GCM circuit. There are no other "exits" from the HV playpen other than components rated for many thousands of volts of isolation.

The goal here is to detect when the ground impedance rises above no more than 100Ω/V, or in this case, 24kΩ. The secondary side of the optoisolator has a peak-hold, so it's ok that only the peaks rise above the comparator threshold.

• There are many problems with this design, starting with the fact that you can't compare the voltage across R20/R22 with the voltage across R13 unless they're referenced to the same node, such asVGND. Beyond that, I can't figure out exactly what it is you're trying to measure -- you mention both residual current and ground impedance, but this circuit doesn't appear to measure either of those. – Dave Tweed Nov 13 '17 at 12:52
• Whether or not to connect the negative output of DC1 to ground is one of the questions I’ve asked. It sounds like your answer is “yes.” – nsayer Nov 13 '17 at 16:25
• If there is a high impedance to ground, then less current will flow through R20/R22, so the voltage across it will decrease. The ground reference for the input to the relay test terminals is separated from this circuit’s ground by an imaginary resistance we are trying to measure (or at least threshold). – nsayer Nov 13 '17 at 16:26
• So you're trying to confirm that you can drive at least 1mA through the ground connection, is that it? if so, it should be clear that limiting the current to no more than 1mA is completely counterproductive. If the component tolerances are off in the wrong direction, or if the line voltage is a little low, it may become impossible to trigger the comparator at all. Why did you give up on your first approach (from two years ago)? – Dave Tweed Nov 13 '17 at 16:51
• You’ve stated the problem correctly. I can adjust either the current regulator or the threshold to separate them. That’s easy. But one of the questions is whether or not to bother with the regulator at all. I gave up on the current design because it is temperature sensitive - false positives (that is, tripping) in Summer. My best guess is that the optoisolator secondary contributes to that, so the current design attempts to move the threshold test to the primary, which requires isolating the power supply for that (which I was previously trying to avoid). – nsayer Nov 13 '17 at 17:34