I was looking over the circuit design for OpenEVSE. I was specifically interested in the GFCI portion of the circuit: enter image description here

This circuit measures the current difference between Line and Neutral on a 120/240v 50-60Hz power source using a current transformer on the bottom two pins of the pin heade.

AFAIK this circuit triggers an interrupt on an MCU if the difference in current is larger than 20mA (50mA??).

I assume the 2 zeners are to avoid damaging the circuit from large current(voltage) spikes.

Both op amps are powered with +5v. It seems to me that this circuit will only trigger on the positive half of the AC wave. Since we are talking about 50-60Hz, this could mean there is a whole +/-25ms from the time where a short is created to when it is detected. Not to mention the time for the contactor to open...

I am wondering if I am overreacting and this is typical of GFCI circuitry. This circuit is usually hooked up to a 30+ amp breaker. Is my analysis correct? Is this an acceptable amount of delay to interrupt the power source in a ground fault condition?

My thought would be to have the op amps in an "absolute value" configuration and trigger on both halves of the ac waveform, saving up to 25ms.

  • \$\begingroup\$ But will it detect a DC fault to ground? That is, a fault after the rectifiers in a SMPS?... \$\endgroup\$
    – bobflux
    Dec 16, 2019 at 22:25
  • \$\begingroup\$ The first amplifier stage is inverting and then rectifying so it will detect on the negative half-cycle, I think. \$\endgroup\$
    – Transistor
    Dec 16, 2019 at 22:30
  • \$\begingroup\$ @peufeu No i don't think it will detect a fault in an SMPS but then the power supply itself should probably have it's own protection. I know in the case of this circuit that electric vehicles do have protection built-in on the DC charging side. \$\endgroup\$ Dec 16, 2019 at 22:35
  • \$\begingroup\$ @Transistor Yeah your right, same result tho. \$\endgroup\$ Dec 16, 2019 at 22:36
  • 1
    \$\begingroup\$ @james Absolutely. I came up with my own implementation requiring the MCU and a latching "GFCI" circuit that I came up with (essentially a SR latch with a variable trigger voltage) to provide a logic LOW. If either goes HIGH, then the contactor is de-energized. \$\endgroup\$ Dec 17, 2019 at 19:25

1 Answer 1


How long is that 10uF capacitor going to take to charge? a lot of half-cycles, that's how long. It doesn't make sense to load the output of the first stage that way. So yeah, unsafe.

  • \$\begingroup\$ The cap should charge instantaneously, there is no resistance between it and gnd. It should, on the other hand, discharge slowly as D3 prevents it from discharging through the op-amp's "pull" half of the output. \$\endgroup\$ Aug 6, 2020 at 17:39

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