RCD wiki + Common Misconceptions about RCDs

[sparkticus]

It is something which I thought might occur but had never investigated. On a sketch I have just done my statement is not true. A N-E fault will not affect the operation - in fact it will make the RCD more sensitive.

Most 30mA RCDs operate around the 25mA mark. Say we have a circuit with 1A flowing through the RCD L conductor. 15mA is flowing N-E so we have 0.985A flowing through the RCD N conductor.
Now someone receives a 10mA shock from the L conductor, only 0.975A flows through the RCD N conductor thus causing operation (25mA imbalance) with only a 10mA shock current.
I am in no way advocating N-E faults btw

That is a really interesting thought. For my own understanding if there is a "background" N-E leakage of say 20mA the RCD should sit there and not trip. If a L-E leakage develops, say a 20mA then the differential is zero. So the N-E leakage has essentially raised the RCD trip threshold somewhat? That is, a further 25mA (total 20mA + 25mA = 45mA) may be required to trip the RCD (assuming the RCD will trip with a 25mA differential)? Is my thinking right?

 

[westie101]

The issue was that of taking the combined neutral and earth of a PME supply through the meter before deriving the installation's 'earth' from it (i.e. before invoking the 'S' of TN-C-S). This was BAS's suggestion as a possibly way of implementing bernardgreen's idea about automatically disconnecting both N&E without (perhaps!) technically violating the prohibition of having a disconnecting device in the protective earth. I suspected that this delay in spilting the PEN conductor into N & E until after the meter would not be allowed. Do you have an answer to that question?

Interesting one, in my view if it was immediately after the meter it would probably comply with ESQCR as the "customer's" installation begins at this point. But then again a device exists in the form of the main isolator(s) on the CU ............. But at the wrong end of the tails!

Unfortunately I have no figures for voltages as we get the customer to switch the supply off before we attend and tend to repair the fault before switching on again.

I do know that the shocks can be from the metalwork to walls or floors so would expect that there could be some fairly high impedances in place

 

[ricicle]

That is a really interesting thought. For my own understanding if there is a "background" N-E leakage of say 20mA the RCD should sit there and not trip. If a L-E leakage develops, say a 20mA then the differential is zero. So the N-E leakage has essentially raised the RCD trip threshold somewhat? That is, a further 25mA (total 20mA + 25mA = 45mA) may be required to trip the RCD (assuming the RCD will trip with a 25mA differential)? Is my thinking right?

No - if the N current through the RCD is already less than the L then any extra current flowing 'out' from the L to an earth fault or shock will only make the difference at the RCD greater because that too is not returning through the N.

 

[JohnW2]

That is a really interesting thought. For my own understanding if there is a "background" N-E leakage of say 20mA the RCD should sit there and not trip. If a L-E leakage develops, say a 20mA then the differential is zero. So the N-E leakage has essentially raised the RCD trip threshold somewhat? That is, a further 25mA (total 20mA + 25mA = 45mA) may be required to trip the RCD (assuming the RCD will trip with a 25mA differential)? Is my thinking right?

My head is getting all twisted up with this one. What you say was clearly ricicle's original thought (in his post which he subsequently deleted, but which I quoted in my initial response to him), and I had no difficulty in not questionning what he had suggested, since it seemed intuitively correct. However, he then questioned himself, and his second argument (which you have just quoted) also seemed quite convincing.

Thinking aloud about this further ... I think we can all agree that the direction of the RCD imbalance due to a L-E fault is 'L current > N current'. The question which therefore aises relates to the effect of a N-E fault on the current through the N side of the RCD, and hence the direction of the L-N imbalance which results (i.e.whether it adds to or subtracts from the L>N imbalance due to the background N-E fault). In his second argument (which you are challenging) ricicle suggested that a N-E leak would reduce the flow through the N side of the RCD, thereby causing an imbalance in the same direction (L>N) as a L-E fault. I initially accepted that. Indeed, if you think of the simplest situation of a TN-C-S setup, the N and E are actually joined just a little way upstream of the RCD. A N-E fault in the installation therefore actually bypasses the N side of the RCD, therefore clearly reducing the RCD's N current and supporting ricicle's second argument - leading to the counter-intuitive (for you and I) bottom line that an N-E fault actually decreases the threshold for tripping in response to an L-E fault!

I initially thought this was straightforward, but my brain now disagrees. What do you think?

Kind Regards, John.

 

[JohnW2]

No - if the N current through the RCD is already less than the L then any extra current flowing 'out' from the L to an earth fault or shock will only make the difference at the RCD greater because that too is not returning through the N.

Yes, as I've just written in response to Sparkticus, having happily accepted your initial suggestion (since it sounded intuitively correct), I think I have now convinced myself that your opposite, second, argument is actually the right one!

I think the intuitive problem arises from the fact that we (at least I) subconsciously think of the N-E and L-E faults as being mirror images, but that's not the case. In fact, as you say, in both cases the RCD's N current is less than it's L current, in both cases because some of the current flowing through L has found a return path to the supply transformer which doesn't involve going through the N side of the RCD.

Thanks for bringing this up, and going publically through your 'thought reversal' - since it's certainly taught me something that I would never have previously dreamed was true!

Kind Regards, John.

 

[sparkticus]

I initially thought this was straightforward, but my brain now disagrees. What do you think?

I am struggling with it. I think I need to draw it out but just a thought experiment for a moment. Imagine a simple toroid with N current wrapped around one side and L current wrapped around the other.

Now imagine the output of N is connected to E with a current of 20mA.

Now imagine the output of L is connected to E with a current of 20mA.


The magnetic flux at the middle of the toroid is zero. So a magnetic trip at that point will not trip.



However, it may be that an RCD toroid is really wired as a transformer. ----- and I need to think about that for a moment.

 

[ricicle]

It's simples really. Each coil on the Toroid creates a magnetic flux but will oppose not compliment each other thus cancelling each other out when currents are equal.
Any difference between currents will make one flux bigger than the other thus leaving a net flux which, if big enough, will induce current in the trip coil.

 

[JohnW2]

I initially thought this was straightforward, but my brain now disagrees. What do you think?

I am struggling with it.

I was, too - but I think I'm now at peace with msyelf!

I think I need to draw it out but just a thought experiment for a moment. Imagine a simple toroid with N current wrapped around one side and L current wrapped around the other.
Now imagine the output of N is connected to E with a current of 20mA.
Now imagine the output of L is connected to E with a current of 20mA.
The magnetic flux at the middle of the toroid is zero. So a magnetic trip at that point will not trip.

No argument about that - which is obviously the basis of the 'intuition' which ricicle, you and myself initially applied. However, the flaw in that thought process is that is NOT what happens with N-E and L-E faults. Try this thought experiment with your toroid....

Start with 50 mA flowing through both coils. No net flux. No trip. Partially bypass the 'N' coil with a resistor (simulating a 15mA N-E fault) such that the current through coil reduces to 35 mA. 15mA imbalance (in favour of L), not enough net flux for trip. Now increase the current flowing through the L coil to 65 mA, without any change in N current (simulating a 15 mA L-E fault). Current imbalance now 30mA. Now trips.

Does that help?

Kind Regards, John.

 

[sparkticus]

It's simples really. Each coil on the Toroid creates a magnetic flux but will oppose not compliment each other thus cancelling each other out when currents are equal.
Any difference between currents will make one flux bigger than the other thus leaving a net flux which, if big enough, will induce current in the trip coil.

I agree but in my post above I imagined L>E @ 20mA and N>E @ 20mA so the toroid is balanced with no net flux differential. I can't get my head around this. If what I say is true (I'm sure I am wrong) then an RCD handling two phase (two lives) accommodate any amount of leakage to earth provided both phases leaked the same amount. This is warping my mind at the moment.

 

[sparkticus]

It's simples really. Each coil on the Toroid creates a magnetic flux but will oppose not compliment each other thus cancelling each other out when currents are equal.
Any difference between currents will make one flux bigger than the other thus leaving a net flux which, if big enough, will induce current in the trip coil.

I agree but in my post above I imagined L>E @ 20mA and N>E @ 20mA so the toroid is balanced with no net flux differential. I can't get my head around this. If what I say is true (I'm sure I am wrong) then an RCD handling two phase (two lives) accommodate any amount of leakage to earth provided both phases leaked the same amount. This is warping my mind at the moment.

EDIT, dam - they would be 180 deg out of phase which I had not initially considered. Argh

 

[ricicle]

I agree but in my post above I imagined L>E @ 20mA and N>E @ 20mA so the toroid is balanced with no net flux differential. I can't get my head around this. If what I say is true (I'm sure I am wrong) then an RCD handling two phase (two lives) accommodate any amount of leakage to earth provided both phases leaked the same amount. This is warping my mind at the moment.

I dont want to get too deep here ( and my laptop battery is getting low).

A load using 2 phases would act in the same way as 1 phase and N. As there can only be one 'phase current'.

BUT - if the 2 phases were connected through the RCD to 2 seperate circuits and returning via a N which wasnt through the RCD then if the load currents (or leakage currents were equal) then there would be no trip. But getting 2 loads to draw the same current within a few milliamps would be quite a feat !

 

[sparkticus]

Start with 50 mA flowing through both coils. No net flux. No trip. Partially bypass the 'N' coil with a resistor (simulating a 15mA N-E fault) such that the current through coil reduces to 35 mA. 15mA imbalance (in favour of L), not enough net flux for trip. Now increase the current flowing through the L coil to 65 mA, without any change in N current (simulating a 15 mA L-E fault). Current imbalance now 30mA. Now trips.

Does that help?

It does and thank you but there is something about the no load situation I described which is confusing me to the point where I can't quite think through the logic clearly. One of those situations where I feel sure that I am wrong but I'm wrestling with wrong and it's got me in a Boston crab

 

[JohnW2]

Does that help?

It does and thank you but there is something about the no load situation I described which is confusing me to the point where I can't quite think through the logic clearly. One of those situations where I feel sure that I am wrong but I'm wrestling with wrong and it's got me in a Boston crab

Are you perhaps overlooking the fact that, in the absence of an underlying/background load (i.e. a pre-existing N current which can be partially diverted by a fault), a N-E fault will have no effect on an RCD - that being the reason why people sometimes get confused by the inconsistency of whether or not an RCD trips when they touch N and E together?

Kind Regards, John.

 

[sparkticus]

as I hit
A load using 2 phases would act in the same way as 1 phase and N. As there can only be one 'phase current'.

You are right. I am literally committing my confused thoughts to writing as I think of them. As soon I had hit submit I realised I was wrong on that.

But I am still thinking about this N-E current. It is still not that clear to me.

 

[sparkticus]

Are you perhaps overlooking the fact that, in the absence of an underlying/background load (i.e. a pre-existing N current which can be partially diverted by a fault), a N-E fault will have no effect on an RCD - that being the reason why people sometimes get confused by the inconsistency of whether or not an RCD trips when they touch N and E together?

No, I'm familiar with those scenarios. I think I need to sleep on this one which usually works. I often wake up in the middle of the night in a moment of complete clarity. At this particular moment "wrong" has got me in a headlock