Insulation resistance for devices

[Richard100762]

I was checking the IR on a socket circuit and found 0.4M between neutral/earth. The reading came from a garage door motor that was plugged in. once unplugged the IR was good. Does the circuit need to be cleared of all things plugged in when testing or does this constitute as a fault. I was called to do this due to spurious RCD tripping.

 

[bill01803]

Anything plugged in should not be included in test results but if it is causing the tripping it needs repairing.

 

[MJPD29]

You are only testing the fixed wiring unless you are also being contracted to carry out portable appliance testing. Another reason to unplug everything apart from the obvious false test results is you could damage some electrical components while IR testing.

 

[Pretty Mouth]

I was checking the IR on a socket circuit and found 0.4M between neutral/earth. The reading came from a garage door motor that was plugged in. once unplugged the IR was good. Does the circuit need to be cleared of all things plugged in when testing or does this constitute as a fault. I was called to do this due to spurious RCD tripping.

When I'm fault-finding a tripping RCD, I leave all equipment in circuit for the initial test, and remove piece by piece to see the effect it has on the IR.

0.4M ohm is a typical IR reading for SPDs. If you get 0.4M ohm at 500V but much higher at 250V, there's a good chance that it's due to surge protection. Was IR L-E acceptable?

 

[Lucien Nunes]

0.4M between neutral/earth. . I was called to do this due to spurious RCD tripping.

Obviously that in itself won't cause tripping, the IR would need to be 1000 times lower to leak a reasonable number of milliamps from neutral to earth under typical conditions. Even 0.4 megohms L-E is still 30 times higher than would cause a trip.

When testing current-using equipment, the test really ought to be between both L+N paralleled to earth, just as a PAT tester does it, regardless of whether the equipment is portable or hard-wired. As mentioned above that is quite a common result for VDRs tested at 500V and there might be a VDR in the door motor.

 

[Richard100762]

Obviously that in itself won't cause tripping, the IR would need to be 1000 times lower to leak a reasonable number of milliamps from neutral to earth under typical conditions. Even 0.4 megohms L-E is still 30 times higher than would cause a trip.

Being thick here too much Xmas cheer can you give me the maths on that please.

 

[Lucien Nunes]

There's very little voltage between N & E under normal conditions. Assume the circuit is running at something near maximum voltage drop of say 10V (4.3% of 230V) therefore there is 5V drop along the line conductor and 5V along the neutral. If we also assume that whatever is causing the IR is located in the worst place so that the 5V appears across that 0.4MΩ, then it's just ohms law and the leakage would be 5/0.4M = 12.5μA. I.e. so tiny that it's completely swamped by normal functional leakage. Even with a resistance 1000 times lower, at just 400 ohms, the worst-case leakage would still only be 12.5mA and less than the tripping threshold, and that's with 5V between N & E. Near the board on a TN-C-S supply, the drop might only be a fraction of a volt, in which case it is going to take a lower resistance still, perhaps just ten ohms or less, before it trips. That is how even solid neutral-earth faults can lie low until there is enough load on the system to drive up the N-E voltage.

Obviously if the low IR is located L-E instead then the leakage is 230/0.4M = 0.6mA. With no other leakage sources, a typical RCD that trips at 22mA would need a resistance below 230 / 0.022 = 10.5kΩ to trip it. This can be disconcerting when reading the IR on digital display that shows nnn.nn MΩ. The display can legitimately show 0.00 MΩ and yet the insulation is still high enough to not trip. Only by going to a higher resolution do you see the difference between say 12kΩ which probably won't trip, and 2kΩ which will.

 

[MJPD29]

Being thick here too much Xmas cheer can you give me the maths on that please.

Cheer or beer?

 

[Richard100762]

There's very little voltage between N & E under normal conditions. Assume the circuit is running at something near maximum voltage drop of say 10V (4.3% of 230V) therefore there is 5V drop along the line conductor and 5V along the neutral. If we also assume that whatever is causing the IR is located in the worst place so that the 5V appears across that 0.4MΩ, then it's just ohms law and the leakage would be 5/0.4M = 12.5μA. I.e. so tiny that it's completely swamped by normal functional leakage. Even with a resistance 1000 times lower, at just 400 ohms, the worst-case leakage would still only be 12.5mA and less than the tripping threshold, and that's with 5V between N & E. Near the board on a TN-C-S supply, the drop might only be a fraction of a volt, in which case it is going to take a lower resistance still, perhaps just ten ohms or less, before it trips. That is how even solid neutral-earth faults can lie low until there is enough load on the system to drive up the N-E voltage.

Obviously if the low IR is located L-E instead then the leakage is 230/0.4M = 0.6mA. With no other leakage sources, a typical RCD that trips at 22mA would need a resistance below 230 / 0.022 = 10.5kΩ to trip it. This can be disconcerting when reading the IR on digital display that shows nnn.nn MΩ. The display can legitimately show 0.00 MΩ and yet the insulation is still high enough to not trip. Only by going to a higher resolution do you see the difference between say 12kΩ which probably won't trip, and 2kΩ which will.

Thanks I’ll pour over that Boxing Day. Merry Christmas to you all.