24v GFCI’s?

[winnie]

However as winnie pointed out, is is also possible that the ground fault returns on an intended path, but this is true for all GF common circuits.

When you have a grounded secondary, there are a huge number of paths available for current that do not involve the residual current detector (zero sequence current transformer), and thus a GFCI is very likely to trip on a fault.

When you have an ungrounded secondary, then you significantly reduce the number of available paths, which reduces the chance that a residual current detector will trip on a fault.

In the limiting case of an RCD on the output of an ungrounded transformer, most if not all of the fault current paths will return via the RCD and thus it is unlikely to detect the faults. For the OP, if the decision is made to put ground fault detection on the 24V lighting circuits, I'd suggest that the residual current detection hardware hit individual lighting circuits, not a single (essentially useless) RCD on the output of the 50A 24V transformer

-Jon

 

[LarryFine]

All it could do is detect an imbalance between the currents on the two nominally ungrounded conductors.

I'm suggesting that a supply-side conductor would need to be grounded for a load-side contact to cause that imbalance.

 

[LarryFine]

Touch it to your tongue

No, thank you; once was enough. Nothing like the shock I got from the 90v battery in my parents' "portable" AC/DC tube radio.



I will check 1.5v batteries with my tongue and a finger. That just gives a slightly bitter taste.

 

[LarryFine]

If there is a pathe for current to flow (i.e. a complete circuit), the GFCI will see current leaving the device, but not returning. This is true of any zero sequence CT sensor.

Again, without a grounded supply conductor, there is no pathway for that current difference, thus no shock to detect.

 

[synchro]

If a non-grounded 2-wire secondary feeds the line side of a 2-wire GFCI, then if there's any current flowing on one line input wire then there has to be an equal and opposite polarity current on the other line input wire (i.e., they'll have the same amount of RMS current). This is because there's essentially no where else for the current to go if there is not a third conductor tied to the secondary such as ground.

A portion of the line input current will be the current drawn by the electronics in the GFCI, but again this current from the secondary will be the same on each of the two line inputs. The remaining current will flow through the GFCI load terminals, but the current through each load terminal will be the same as the current through its corresponding line input terminal (minus the current to power the electronics). Therefore the current through both load terminals will be the same irrespective of whether the loads are placed across these terminals, to "ground", or to another potential (as long as the voltage and current ratings of the GFCI are not exceeded). Therefore there will be no current difference to cause the GFCI to trip. Ground has no particular significance in this situation because the secondary conductors have no connection to ground before entering the GFCI (except for maybe gigohms of leakage).

As mentioned in other posts there has to be two faults to ground (or to another common conductor), one on each GFCI load terminal, in order for them to draw current through the GFCI when the transformer secondary is ungrounded. However, this will not cause different currents to be drawn from the two load terminals when the secondary is ungrounded, and therefore the GFCI will not trip. To create a current difference that would make the GFCI trip there would have to be a path for current to flow around the GFCI between the line and load terminals (basically a bypass path around the GFCI). With a grounded secondary this path of course would be "ground'.

 

[jim dungar]

Again, without a grounded supply conductor, there is no pathway for that current difference, thus no shock to detect.

The first fault turns an ungrounded system into a grounded system, this is why ground detectors are now required by the NEC. On large systems the coupling capacitance can be enough to provide a path.

I know the OP is about 24V. But the physics do not change, it just means some issues may not be found to not come into play.

 

[morepower]

You cannot offend me and please don't put anybody above yourselves just because we are mods. I was incorrect. I totally understand your post and I agree with you. I looked at 680.51 referencing 680.23(A)(2) and I didn't look closely at it. I assume all of 680.23.

That section makes no sense anyway as winnie pointed out.

Back to the OP--- how do you put gfci on 24v---

I am surprised that this is 24v ac. If it were DC which I suspect it is then there wouldn't be an issue.

Nope, the lighting is definitely ac.

 

[winnie]

Again, without a grounded supply conductor, there is no pathway for that current difference, thus no shock to detect.

With an ungrounded secondary the path must be from one supply conductor to the other. As you note, a single gfci on the transformer secondary could not detect this.

But say the transformer feeds 100 separate lamps. You could have 2 gfcis each feeding 50 lamps, 10 gfcis each feeding 10 lamps, or 100 separate gfcis.

A single fault would not cause much current flow. 2 faults could cause current flow, and the more the circuit is divided up into separate gfci zones, the greater the chance that a fault would cause detectable current flow. This would need to be flow from one gfci to another. Any flow of current from one gfci, through two faults and back to itself we ould simply look like load.

Jon

 

[synchro]

I'm thinking the following would provide GFI protection for a 24VAC lighting circuit using a standard 120V GFCI.

The 24V circuit would not be ungrounded or grounded. Instead, it's connected by a jumper to the load side neutral of the GFCI. A ground fault current on the 24VAC side would then be seen as a common-mode current on the load side of the GFCI, and so the GFCI would trip if the fault current exceeds 6 mA.

 

[synchro]

With the above circuit, though, you'd need to consider the amount of current a large ground fault on the 24VAC side could put through through the neutral of the GFCI since the transformer will deliver 5X more current at its 24VAC output than is supplied to its input.

A simple thing would be to have a resistor limit the current rather than having a solid jumper between primary and secondary, since very little current is needed just to trigger the GFCI.
Or separate runs off of the 24V secondary could each have its own OCPD with a rating no more than the 120V GFCI breaker, so that a single fault would not be a problem.
There are probably other alternatives like having a 2-pole breaker interrupt both 120V line and neutral as in the 404.2(B) Exception if the neutral current gets too high.
The resistor makes more sense although I suspect you'd need one that is UL approved for the purpose.

 

[winnie]

I'm thinking the following would provide GFI protection for a 24VAC lighting circuit using a standard 120V GFCI.

The 24V circuit would not be ungrounded or grounded. Instead, it's connected by a jumper to the load side neutral of the GFCI. A ground fault current on the 24VAC side would then be seen as a common-mode current on the load side of the GFCI, and so the GFCI would trip if the fault current exceeds 6 mA.

View attachment 2553491

But that is grounding the 24V secondary, via the primary neutral rather than an EGC/GEC. It would have the benefits and problems of a grounded secondary.

-Jon

 

[morepower]

I'm thinking the following would provide GFI protection for a 24VAC lighting circuit using a standard 120V GFCI.

The 24V circuit would not be ungrounded or grounded. Instead, it's connected by a jumper to the load side neutral of the GFCI. A ground fault current on the 24VAC side would then be seen as a common-mode current on the load side of the GFCI, and so the GFCI would trip if the fault current exceeds 6 mA.

View attachment 2553491

No, this wouldn’t work because the code requires an isolation transformer with total separation between the primary and secondary.

 

[LarryFine]

No, this wouldn’t work because the code requires an isolation transformer with total separation between the primary and secondary.

However, a 24v GFCI would require that one secondary conductor tie to an electrode (system) to function, as with any circuit.

 

[winnie]

However, a 24v GFCI would require that one secondary conductor tie to an electrode (system) to function, as with any circuit.

This is not quite correct. A gfci is a residual current detector. To detect a ground fault you need current flow between the fault and ground, and thus would need a grounded circuit.

But if the goal is detection of current flowing outside of the intended path, you don't need that ground electrode.

 

[LarryFine]

This is not quite correct. A gfci is a residual current detector.

Come again?

To detect a ground fault you need current flow between the fault and ground, and thus would need a grounded circuit.

Isn't that what i said?
"However, a 24v GFCI would require that one secondary conductor tie to an electrode (system) to function, as with any circuit. "
A secondary conductor tied to an electrode system defines a grounded circuit.

But if the goal is detection of current flowing outside of the intended path, you don't need that ground electrode.

How else can the leakage current be caused and thus detected?

 

[winnie]

Additional, one could have a current limited source (say at a different frequency eg 400hz) placed between the 24V secondary and a ground electrode. This source would not itself be a shock hazard, and thus would not defeat the intent of leaving the secondary ungrounded, but would create a signal that could be used to detect ground faults.

Jon

 

[winnie]

How else can the leakage current be caused and thus detected?

For someone to get shocked, current would need to flow out of the intended circuit through one fault, through the person, and back into the intended circuit through another fault.

If the circuit segment feeding the first fault, and only the first fault, is measured by a residual current device, then the fault current will be detected. Similarly for the second fault.

If a single rcd measures the circuit leading to both faults, then the unintentional current path will look like load current.

So I absolutely agree that detection of a ground fault requires some sort of connection to ground to create a detectable signal to ground.

But do detect unintended current flow between two circuit points does not require a ground connection.

 

[synchro]

A GFCI measures and responds to the common-mode component of the currents flowing in the conductors that pass through its toroidal CT. The common-mode current is the sum of the instantaneous (or equivalently vector) currents in each of these conductors, divided by the number of conductors. Unless there's another closed circuit present between the line and load sides of the GFCI which bypasses around the GFCI toroid, then the common-mode current through the toroid must be zero according to Kirchhoff's current law.
There is some current drawn by the electronics within the GFCI but that current flows between the conductors and therefore has no common-mode component.

Now if the line side of the GFCI is fed by the secondary of an isolation transformer with no other conductors attached to it, then there's no possibility of a closed circuit bypass path being formed around the GFCI. Therefore it should not trip regardless of the load even if the GFCI output has a fault path to ground or to the primary circuit, as long as the maximum ratings of the GFCI device are not exceeded. In this case the common-mode current is always zero because the isolated secondary winding of the transformer cannot source a common-mode current (just as a delta winding cannot source zero sequence currents).