PV systems

JohnD · 27 Nov 2014

I have been reading the data sheets for a popular PV inverter
http://www.fronius.com/cps/rde/xbcr...rnational/42_0426_0074_EA_149621_snapshot.pdf
page 34

it says:

"Make sure that the grid neutral conductor is grounded. For IT networks (isolated networks without grounding), this is not the case, and operation of the inverter is not possible."

and also

"The inverter is equipped with a universal current-sensitive RCMU according to DIN VDE 0126-1-1. This monitors residual currents from the solar module up to the inverter grid connection and disconnects the inverter from the grid when an improper residual current is detected."

plugwash · 27 Nov 2014

JohnW2 said:
Let me try an example, then. Say there is a 10A load and, per westie's proposition, say that 5A of that is coming from the PV (connected to final circuit downstream of the house's RCD, hence 'avoiding' that RCD) and 5A from the grid, via the house's RCD. There will be 5A in both L and N of that RCD, hence balanced. Now add a 30mA L to E fault on the loaded circuit. The total 'load' increases to 10.030A, hence 5.015A from each source. 5.015A therefore goes through the RCD's L, but only the 5.000A due to the 'proper load' goes through the RCD's N. The L-N imbalance in that RCD is therefore only 15mA, although the fault itself is 30mA. With those figures, one would need a 60mA fault on the circuit to achieve a 30mA imbalance in the house's RCD. ... or am I still missing things or getting things wrong?

You are still failing at basic circuit analysis.

If you actually do the circuit analysis properly based on your assumptions you will find the current in the RCDs N is 4.985A

JohnW2 · 27 Nov 2014

DetlefSchmitz said:
The question then, is whether the generator (inverter) end RCD needs the neutral to be earthed on the inverter side. I am leaning towards the view that it does not. As it is at earth potential anyway, it would still be the case that live to earth current in the 'middle' would cause an imbalance.

I don't see that. The current flowing in the connections to the L and N output terminals of the inverter must be the same (per Mr Kirchoff). That inevitable truth does not change if one connects those terminals to an RCD (and to nothing else), so the L and N currents through the RCD have ten also got to be the same. The only way that there can be an L-N imbalance in the RCD is if some current is finding it's way back to one of the inverter's output terminals which bypasses the RCD. In practice, about the only credible way that can happen is when the inverter's N terminal is connected to earth and an L-E fault (or maybe N-E fault) arises on the other side of the RCD, such that some current gets back to the N terminal of the inverter without going through the RCD.

That is really just 'Physics 101' (courtesy of Mr Kirchoff), so I'm very confident that it's true

Kind Regards, John

plugwash · 27 Nov 2014

an RCD protects against "earth faults" where the N-E interconnection is on one side of the RCD and the fault is on the other side of the RCD. As I said before where the power is coming from is irrelevent, all that matters is where the fault is and where the N-E interconnection is.

JohnW2 · 27 Nov 2014

JohnD said:
I have been reading the data sheets for a popular PV inverter... it says: "Make sure that the grid neutral conductor is grounded.

What exactly does that mean? They presumably are not suggesting that one should install an explicit N-E link in one's grid installation, particularly not if it is TT or TN-S? Are they merely saying that the grid installation cannot be IT?

JohnD said:
"The inverter is equipped with a universal current-sensitive RCMU according to DIN VDE 0126-1-1. This monitors residual currents from the solar module up to the inverter grid connection and disconnects the inverter from the grid when an improper residual current is detected."

Fair enough - but, as I've been implying, it's very difficult to see what there would be for it to detect in the absence of a N-E link on the PV side, because, in the absence of such a link, I can think of no way that a 'residual current' (which I take to mean L-N imbalance) could arise.

Kind Regards, John

mikeyd · 27 Nov 2014

JohnW2 said:
Let me try an example, then. Say there is a 10A load and, per westie's proposition, say that 5A of that is coming from the PV (connected to final circuit downstream of the house's RCD, hence 'avoiding' that RCD) and 5A from the grid, via the house's RCD. There will be 5A in both L and N of that RCD, hence balanced. Now add a 30mA L to E fault on the loaded circuit. The total 'load' increases to 10.030A, hence 5.015A from each source. 5.015A therefore goes through the RCD's L, but only the 5.000A due to the 'proper load' goes through the RCD's N. The L-N imbalance in that RCD is therefore only 15mA, although the fault itself is 30mA. With those figures, one would need a 60mA fault on the circuit to achieve a 30mA imbalance in the house's RCD. ... or am I still missing things or getting things wrong?

Kind Regards, John

I am rubbish at explaining. Would it help if you thought of, and treated the PV as a a simple light bulb (or lamp before anyone gets picky), just one which provided power instead of used it? Just as with a lamp, in terms of line and neutral currents, what goes in must come out. Any imbalance therefore must be seen (in full) by the RCD.

DetlefSchmitz · 27 Nov 2014

Yes, you are both right. Just thinking out load really - always dangerous.

The upshot is probably that most PV installations have a shortcoming. (Not mine I hasten to add, as I don't use an RCD.)

It would be nice to get an official opinion. Whatever that means.

JohnW2 · 27 Nov 2014

plugwash said:
The key thing to remember is that current flows in loops. So when a "fault to earth" happens current flows from the supply, "LINE" through the fault, through the "earth"*, through the connection between neutral and earth and back to the neutral of the supply.

No argument with that.

plugwash said:
If the connection between neutral and earth is on the upstream side of the RCD and the fault is on the downstream side of the RCD then the fault will create a corresponding imbalance in the RCD regardless of which side of the RCD the supply is on.

I don't really understand your point here. 'Upstream' and 'downstream' are surely defined in terms of what side the supply is on?

For an RCD to experience an L-N balance, all that is necessary is that there is an N-E link on one side of the RCD and a L-E (or maybe N-E) fault on the other side - as you say, regardless of which side is the 'supply'.

Kind Regards, John

JohnW2 · 27 Nov 2014

plugwash said:
JohnW2 said:

Let me try an example, then. Say there is a 10A load and, per westie's proposition, say that 5A of that is coming from the PV (connected to final circuit downstream of the house's RCD, hence 'avoiding' that RCD) and 5A from the grid, via the house's RCD. There will be 5A in both L and N of that RCD, hence balanced. Now add a 30mA L to E fault on the loaded circuit. The total 'load' increases to 10.030A, hence 5.015A from each source. 5.015A therefore goes through the RCD's L, but only the 5.000A due to the 'proper load' goes through the RCD's N. The L-N imbalance in that RCD is therefore only 15mA, although the fault itself is 30mA. With those figures, one would need a 60mA fault on the circuit to achieve a 30mA imbalance in the house's RCD. ... or am I still missing things or getting things wrong?

Click to expand...

You are still failing at basic circuit analysis. If you actually do the circuit analysis properly based on your assumptions you will find the current in the RCDs N is 4.985A

I need some help, then, if you could. The voltage and impedance of the 'proper load' being unchanged, the total (obviously both L and N) current through that 'proper load' will still be 10A. You are saying that, rather than the 50:50 split between grid and PV supplies that existed before the fault arose, with the fault present less than 50% will go through the grid neutral, and hence more than 50% (5.015A) through the PV neutral. I guess I'm going to have to sit down and draw this out because, trying to do it in my head, I don't see why that should be the case!

Kind Regards, John

mikeyd · 27 Nov 2014

plugwash said:
If you actually do the circuit analysis properly based on your assumptions you will find the current in the RCDs N is 4.985A

Please don't anybody point out that the assumptions are based on someone turning up the dimmer switch next to the sun at the exact time of the fault

mikeyd · 27 Nov 2014

JohnW2 said:
I need some help, then, if you could. The voltage and impedance of the 'proper load' being unchanged, the total (obviously both L and N) current through that 'proper load' will still be 10A. You are saying that, rather than the 50:50 split between grid and PV supplies that existed before the fault arose, with the fault present less than 50% will go through the grid neutral, and hence more than 50% (5.015A) through the PV neutral. I guess I'm going to have to sit down and draw this out because, trying to do it in my head, I don't see why that should be the case!

Kind Regards, John

Did you see my reply above John (at 20:26)? You are thinking too hard. The PV is just another load / appliance. It can't have a different current through it's line compared to it's neutral.

plugwash · 27 Nov 2014

JohnW2 said:
plugwash said:

JohnW2 said:

Let me try an example, then. Say there is a 10A load and, per westie's proposition, say that 5A of that is coming from the PV (connected to final circuit downstream of the house's RCD, hence 'avoiding' that RCD) and 5A from the grid, via the house's RCD. There will be 5A in both L and N of that RCD, hence balanced. Now add a 30mA L to E fault on the loaded circuit. The total 'load' increases to 10.030A, hence 5.015A from each source. 5.015A therefore goes through the RCD's L, but only the 5.000A due to the 'proper load' goes through the RCD's N. The L-N imbalance in that RCD is therefore only 15mA, although the fault itself is 30mA. With those figures, one would need a 60mA fault on the circuit to achieve a 30mA imbalance in the house's RCD. ... or am I still missing things or getting things wrong?

Click to expand...

You are still failing at basic circuit analysis. If you actually do the circuit analysis properly based on your assumptions you will find the current in the RCDs N is 4.985A

Click to expand...

I need some help, then, if you could. The voltage and impedance of the 'proper load' being unchanged, the total (obviously both L and N) current through that 'proper load' will still be 10A. You are saying that, rather than the 50:50 split between grid and PV supplies that existed before the fault arose, with the fault present less than 50% will go through the grid neutral, and hence more than 50% (5.015A) through the PV neutral. I guess I'm going to have to sit down and draw this out because, trying to do it in my head, I don't see why that should be the case!

Current in the PV neutral must equal current in the PV live.

The only place the current from the fault can go after passing through the fault is the NE link which is on the grid side of the RCD.

Therefore (assuming only a single fault*) for the currents in the system to balance the imbalance in the RCD must be exactly equal to the current in the fault.

JohnW2 · 27 Nov 2014

mikeyd said:
I am rubbish at explaining. Would it help if you thought of, and treated the PV as a a simple light bulb (or lamp before anyone gets picky), just one which provided power instead of used it? Just as with a lamp, in terms of line and neutral currents, what goes in must come out. Any imbalance therefore must be seen (in full) by the RCD.

Indeed, I've already invoked the "what goes in must come out" concept, in the name of Mr Kirchoff.

However (to both you and plugwash), I think you've probably made me understand at least one of the flaws in my argument (and my attempt at mental circuit analysis!). ... Whereas the normal load current (without a fault) was shared 50:50 between PV and grid supplies, when one introduces the, say, 30mA, L-E leak, that can only affect supplies that have a N-E link. Hence, with the grid side neutral earthed and the inverter neutral not earthed, all 30mA would have to be 'lost' from the grid side's neutral.

However, going back to where this all started, if the inverter neutral, as well as the grid neutral, were earthed, then I would think that the current in the neutrals of both supplies would reduce as a consequence of the fault, thereby making the imbalance on the grid side less than the full size (30mA of whatever) of the L-E leak.

Kind Regards, John

JohnW2 · 27 Nov 2014

plugwash said:
Current in the PV neutral must equal current in the PV live. ... The only place the current from the fault can go after passing through the fault is the NE link which is on the grid side of the RCD. ... Therefore (assuming only a single fault*) for the currents in the system to balance the imbalance in the RCD must be exactly equal to the current in the fault.

Thanks. Yes, as you'll see from what I just posted, I think I eventually twigged that.

However, as I said, I think this is only true so long as the grid neutral, but NOT the inverter neutral, are earthed. If both are (separately) earth-referenced, per the comments that started this discussion a few pages back, I think that things change - more in the direction of what I was incorrectly saying before!

Kind Regards, John

plugwash · 27 Nov 2014

JohnW2 said:
However, as I said, I think this is only true so long as the grid neutral, but NOT the inverter neutral, are earthed. If both are (separately) earth-referenced

If there is a N-E interconnection on both sides of the RCD then most likely the RCD will trip even in the absence of any fault.

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