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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.

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  • \$\begingroup\$ But will it detect a DC fault to ground? That is, a fault after the rectifiers in a SMPS?... \$\endgroup\$ – bobflux Dec 16 '19 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 '19 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\$ – Matthew Goulart Dec 16 '19 at 22:35
  • \$\begingroup\$ @Transistor Yeah your right, same result tho. \$\endgroup\$ – Matthew Goulart Dec 16 '19 at 22:36
  • \$\begingroup\$ A half wave gap @ 50Hz = 1/100Hz =10 ms not 25 ms. Your assumption is correct trip level. 20mA peak * 412 Ohms x gain=100 = 812mV with a comparator threshold of 5V/6= 833 mV Sure FW would be better \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 17 '19 at 0:52

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