RCD wiki + Common Misconceptions about RCDs

[sparkticus]

Just as I said before, you again seem to be assuming that the 20mA current flowing through your N/E resistor will flow through the N side of teh RCD - but it won't.

But are you sure it won't? because that is the source of my confusion, it's just taken me a while to get there

If there is a 5V PD between N and E from the supply (TN-S system) as I described why won't the current flow between N and E? Again, I'm not saying that I am right but it seems to me that it should cause a N - E current. Now I am the first to say that with all of the RCD testing I have done (probably tested thousands of RCDs over many years) I have not seen a scenario where the RCD fails to trip for such a reason I describe but in theory this scenario (in my mind) should exist.

 

[echoes]

I think the terms L-E current and N-E current are distracting. e.g. if there was a path to earth half-way along a heating element, where we could measure a voltage of 120V, would that be classed as L-E fault or N-E fault? The answer: is does not matter. The only difference is that the PD giving rise to fault current is greater at the 'live' end of a load than the 'neutral' end. And this is true whether the load is 1 ohm or an open circuit.

What we need to consider is that at the point of the fault, there is a current flowing, that does NOT flow through the RCD. Kirchoff's current law tells us that the sum of currents into a node is zero, which is why the fault currents add up.

 

[JohnW2]

Just as I said before, you again seem to be assuming that the 20mA current flowing through your N/E resistor will flow through the N side of teh RCD - but it won't.

But are you sure it won't? because that is the source of my confusion, it's just taken me a while to get there
If there is a 5V PD between N and E from the supply (TN-S system) as I described why won't the current flow between N and E? Again, I'm not saying that I am right but it seems to me that it should cause a N - E current.

Right. I think that long last realised why we have been talking at cross-purposes! I'm not sure if you said this explicitly before, but I see you now write:

If there is a 5V PD between N and E from the supply

If that is the case, then I have to agree with you that a current due to introducing a N-E path (fault) within the installation will flow through the N side of the RCD and therefore will (***in the absence of other factors - see below), as you have been persistently saying, have the effect of reducing the L-N imbalance in the RCD in response to an L-E fault - i.e. as you have been saying all along, the N-E fault would (subject to ***) have the effect of increasing the L-E fault threshold of the RCD.

However, I've been talking/thinking about the situation (as with TN-C-S) in which there is essentially no potential difference between the supply neutral and 'earth' (which in this context refers to the installations CPCs and bonded metal etc.). As I said (and which probably confused you, if you were thinking of a pd between neutral and earth), in the absence of any loads on the installation, introducing a N-E fault then has no effect. However, if there are loads resulting in current through the RCD, introducing a N-E fault will provide a partial return path for the load current which bypasses the N side of the RCD, so that the current in the N side of the RCD will reduce as a result of the N-E fault - with all the consequences I've stated which have been confusing you!

*** So, with a pd between incoming neutral and 'earth' (as quite possible with TN-S and TT), but no loads on circuits supplied by the RCD, the effect of a N-E fault will be to increase current through the N side of the RCD. If, on the other hand (as with TN-C-S), there is no pd between incoming N & 'E', but there are loads being supplied by the RCD, then the effect of an N-E fault will be to reduce current through the N side of the RCD. If (probably the real-word situation with TN-S or TT), we have both of those things going on (pd between incoming N&E AND loads on the RCD), then whether an N-E fault increases current through N side of RCD (hence increasing threshold for L-E faults trips) or decreases current through N side of RCD (hence decreasing threshold for L-E fault trips) will depend upon which of those two opposing processes wins - the greater the load on RCD-supplied circuits, the more likely that 'my', rather than 'your', situation will prevail. With TN-C-S, I think that I probably 'win' all the time, provided there is at least some load being supplied by the RCD.

So ... as with so many things in life, it seems like 'it depends'!

Have we perhaps (almost?) got there?

Kind Regards, John.

 

[sparkticus]

I think the terms L-E current and N-E current are distracting. e.g. if there was a path to earth half-way along a heating element, where we could measure a voltage of 120V, would that be classed as L-E fault or N-E fault? The answer: is does not matter. The only difference is that the PD giving rise to fault current is greater at the 'live' end of a load than the 'neutral' end. And this is true whether the load is 1 ohm or an open

We are only using the terms L, E, N etc for the sake of clarity. I see your point in terms of what is essentially a voltage divider analogy but my scenario is different (though as I accept I may be wrong)

What I described is a current actually flowing "out of" the Neutral output of an RCD to earth. The current I described was 20mA. Please keep in mind that the current I describe is derived from the PD between N and E of a TN-S system (from the supply Neutral not from Phase) We can agree (I hope) that such a current is possible right? If you agree then that current (20mA) flows in the Neutral side of the RCD toroid. Then if a 30mA current flows from the L output of the RCD then the RCD toroid is "seeing" and net magnetic flux derived from the current differential of 10mA.

What we need to consider is that at the point of the fault, there is a current flowing, that does NOT flow through the RCD. Kirchoff's current law tells us that the sum of currents into a node is zero, which is why the fault currents add up.

And Kirchoff is not (was not) wrong but please take a look at my RCD toroid scenario.

 

[JohnW2]

I think the terms L-E current and N-E current are distracting. e.g. if there was a path to earth half-way along a heating element, where we could measure a voltage of 120V, would that be classed as L-E fault or N-E fault? The answer: is does not matter. The only difference is that the PD giving rise to fault current is greater at the 'live' end of a load than the 'neutral' end. And this is true whether the load is 1 ohm or an open circuit.
What we need to consider is that at the point of the fault, there is a current flowing, that does NOT flow through the RCD. Kirchoff's current law tells us that the sum of currents into a node is zero, which is why the fault currents add up.

I think that's all true if the situation is as simple as you describe, with only one load (your heating element) served by the RCD and (unlike sparkticus's assumptions) no pd between incoming N & E. However, if either or both of those complications arise, then the whole thing gets much less straightforward - see my recent response to sparkticus.

Kind Regards,John.

 

[sparkticus]

Right. I think that long last realised why we have been talking at cross-purposes! I'm not sure if you said this explicitly before, but I see you now write:

If there is a 5V PD between N and E from the supply

If that is the case, then I have to agree with you that a current due to introducing a N-E path (fault) within the installation will flow through the N side of the RCD and therefore will (***in the absence of other factors - see below), as you have been persistently saying, have the effect of reducing the L-N imbalance in the RCD in response to an L-E fault - i.e. as you have been saying all along, the N-E fault would (subject to ***) have the effect of increasing the L-E fault threshold of the RCD.

 

[sparkticus]

*** So, with a pd between incoming neutral and 'earth' (as quite possible with TN-S and TT), but no loads on circuits supplied by the RCD, the effect of a N-E fault will be to increase current through the N side of the RCD. If, on the other hand (as with TN-C-S), there is no pd between incoming N & 'E', but there are loads being supplied by the RCD, then the effect of an N-E fault will be to reduce current through the N side of the RCD. If (probably the real-word situation with TN-S or TT), we have both of those things going on (pd between incoming N&E AND loads on the RCD), then whether an N-E fault increases current through N side of RCD (hence increasing threshold for L-E faults trips) or decreases current through N side of RCD (hence decreasing threshold for L-E fault trips) will depend upon which of those two opposing processes wins - the greater the load on RCD-supplied circuits, the more likely that 'my', rather than 'your', situation will prevail. With TN-C-S, I think that I probably 'win' all the time, provided there is at least some load being supplied by the RCD.

So ... as with so many things in life, it seems like 'it depends'!

Exactly, it will depend upon the circumstances you describe above. But what is interesting is that I have only started thinking about this in the past two days since this thread. I have never actually seen the scenario I described in the field (or perhaps not noticed it) I think I'm going to try to reproduce it on my bench on Saturday. Then I'll be at peace

 

[sparkticus]

In the meantime, may I refer you to my original post in this thread and ask if you have any comments about my wiki draft? Even though this is a 'DIY forum' I wouldn't be comfortable uploading (and wouldn't upload) anything to the wiki without getting the blessing of at least a few electricians, since I really don't think that would be appropriate.

I think it is an excellent idea John. As we have discussed in the past there is great confusion/disinformation (amongst electricians as well) about what RCDs do and do not do. I think what you have written already plus the discussions that have taken place in this thread just about covers it as far as I can tell.

 

[echoes]

I think my scenario is still valid with a PD between N & E.
I can't see how the fault currents flowing through the coils of the toroid would ever cancel and hold the RCD if the L-E and N-E are both faults to the same earth.
I suppose if one fault was somehow engineered to result in a fault current 180 deg out of phase with the other fault current, then they would cancel, but it's contrived.

 

[echoes]

If that is the case, then I have to agree with you that a current due to introducing a N-E path (fault) within the installation will flow through the N side of the RCD and therefore will (***in the absence of other factors - see below), as you have been persistently saying, have the effect of reducing the L-N imbalance in the RCD in response to an L-E fault - i.e. as you have been saying all along, the N-E fault would (subject to ***) have the effect of increasing the L-E fault threshold of the RCD.

John, are you sure about that?
Suppose PD N-E is 5V, and PD L-N is 250V, and these voltages are in phase. Therefore the fault currents introduced by say a 1k N-E resistance (5mA), and a 50k L-E resistance (5mA) will still imbalance the RCD to same effect as a single 10mA L-E current.

 

[sparkticus]

I think my scenario is still valid with a PD between N & E.

Yes I think it is. But that is not what I was trying to describe. Admittedly I could have described it better but my thoughts on it were evolving at the same time.

What I am saying is that on a TN-S or TT supply the small N-E current that could exist can slightly modify the trip threshold of the RCD. However, there are many other variables that could negate that effect.

I can't see how the fault currents flowing through the coils of the toroid would ever cancel and hold the RCD if the L-E and N-E are both faults to the same earth.
I suppose if one fault was somehow engineered to result in a fault current 180 deg out of phase with the other fault current, then they would cancel, but it's contrived.

The toroid will only produce a "trip magnitude flux" when there is a significant current difference between what flows in the N coil and the L coil, in this case approximately 30mA differential either way, even if both currents flow to a common node (earth) after the RCD. The phase (or phase relationship) at 50Hz should make no difference (though I thought it would for a few minutes yesterday but corrected myself) since the toroid probably would not respond that fast. Due to line dynamics the phase relationship between phase current and any neutral-earth current would probably shift offer time anyway adding another variable. If the frequency was much lower, say 10Hz then I think it would play as a variable.

I will try my scenario on Saturday and will provide an honest report of what I find either way.

 

[sparkticus]

John, are you sure about that?
Suppose PD N-E is 5V, and PD L-N is 250V, and these voltages are in phase. Therefore the fault currents introduced by say a 1k N-E resistance (5mA), and a 50k L-E resistance (5mA) will still imbalance the RCD to same effect as a single 10mA L-E current.

I don't think so. I think they will balance. keep in mind that the N-E 5Volts (and the subsequent 5mA that you mention) is not from the phase.
It is from the Neutral via the sub-transformer so it passes through the Neutral side of the toroid (to earth) just as the phase current (5mA that you mention) passes through the phase side of the toroid (to earth) Both sides of the toroid see the same current.

 

[echoes]

I look forward to your findings hot from the bench, and thanks (all) for distracting me from the TV to have a good old think!

 

[echoes]

John, are you sure about that?
Suppose PD N-E is 5V, and PD L-N is 250V, and these voltages are in phase. Therefore the fault currents introduced by say a 1k N-E resistance (5mA), and a 50k L-E resistance (5mA) will still imbalance the RCD to same effect as a single 10mA L-E current.

I don't think so. I think they will balance. keep in mind that the N-E 5Volts (and the subsequent 5mA that you mention) is not from the phase.
It is from the Neutral via the sub-transformer so it passes through the Neutral side of the toroid (to earth) just as the phase current (5mA that you mention) passes through the phase side of the toroid (to earth) Both sides of the toroid see the same current.

But for an RCD in balance, the current through the phase coil is equal and opposite to the current in the N coil. If one coil was connected up the 'wrong' way, the RCD would trip (vigourously!) - it would be as if the 2 coils were connected as one.
In this 5V N-E example, we have our 5mA N-E current essentially flowing the 'wrong' way (180 deg out of 'correct' phase) through the N coil.
And I think that if we have our 5V N-E PD such that the L-N was 255V and L-E was 250V, the same applies.

 

[JohnW2]

John, are you sure about that?
Suppose PD N-E is 5V, and PD L-N is 250V, and these voltages are in phase. Therefore the fault currents introduced by say a 1k N-E resistance (5mA), and a 50k L-E resistance (5mA) will still imbalance the RCD to same effect as a single 10mA L-E current.

I'm coming to believe that you are correct - as you've pointed out to sparkticus in a subsequent post, I've been forgetting about 'phase' (direction of current flow through RCD). Before reading your message, I drew the following, which appeared to indicate that the L-E and N-E fault currents were 'cancelling' in this situation...

Untitled

• JohnW2
• 9 Sep 2011
• 1

However, as one can see, the N-E fault current is actually going through the N side of the RCD in the opposite direction from usual (since neutral is the 'line', not the return, for this current. So, if the voltages on L & N are in phase, the fault currents will, indeed, add in the RCD. In fact, I think that everything about my diagram is correct, but, given the direction of flow through the N side of the RCD, the value of my IfN is negative, making the imbalance IfL + IfN, in trems of the unsigned values of the currents.

Is that right?

Kind Regards,John.