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OK to connect parallel PV strings without fuses?


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5 hours ago, weber said:

Because of course, it's all about safety, so you have to charge people AU$254 to read the real thing!

Yup, I've also lamented that kind of nonsense before. ZAR900 for that kind of thing this side of the water, which is about 90AUD. For a third of that price I would not think twice, but they always price it just around the pain threshold.

Of course this is nothing compared to what it costs to get access to the NMEA2000 paperwork.

4 hours ago, weber said:

This forum allows way too short a time for edits.

Seems it is a new "feature" since a recent update.

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It has to do with earth faults (eg a short from the PV cable to roof material for example). An earth fault can cause one fuse to get bypassed, and the system will continue working normally as long as

See attached sketch showing what I think is happening in @Javi Martínez paulty panel.  A significant portion of the current generated by the good string is being shunted by the string with the shorted

It's getting crowded in here. @phil.g00, you seem to be missing the fact that in this case there are three strings in parallel. So if any group of cells gets shorted out, e.g. by a failed bypass

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The reason I am unconvinced about fuses, is not that they aren't a legal or safety requirement.

Yes, I'm convinced to install them, but I am unconvinced that they actually blow and notify me that I have a circulating current issue.

A PV panel will only conduct in reverse after Voc is exceeded, and at Voc the parallel strings provide zero current by definition.

Now a blown bypass diode could provide sufficient voltage mismatch, but I think you'd still be splitting hairs whether you'd get enough current to blow a fuse in say a 3 string array.

Maybe even a four string array would be pushing it too.

Granted, if there is 1.5X the rating the fuses will blow, I just think the likely-hood of this condition carrying on merrily under that current level is high.

A blocking diode would stop this, but I wouldn't use one because of power loss ( .7V x whatever current in each string ( say 8A) , about 5Watts/string).

So put fuses in, but can I suggest this as well.

This would cost no power during normal operation, use lightweight wiring, and indicate if you had a problem and where it was.

It uses two pin bidirectional two colour LED's, during normal operation they would be off. In a circulating current condition the combination of LED colors would indicate in which string the issue is. The resistors would be sensible values found by initially adjusting a pot.

 

image.png.bbeeee7ef674faa1d3a3e30b409f9375.png

 

 

 

 

Edited by phil.g00
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Legal requirements aside, 1 fuse is sufficient and two fuses may require a lot more wiring as all the -ve string legs have to be brought down to the fuse box instead of just being commoned on the roof. I 'd still have an MCB on the common negative leg though.

No difference electrically if fused on the negative leg and commoned on the positive leg, but the over way round is more the convention.

 

 

 

 

 

Edited by phil.g00
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20 hours ago, phil.g00 said:

The reason I am unconvinced about fuses, is not that they aren't a legal or safety requirement.

Yes, I'm convinced to install them, but I am unconvinced that they actually blow and notify me that I have a circulating current issue.

A PV panel will only conduct in reverse after Voc is exceeded, and at Voc the parallel strings provide zero current by definition.

Now a blown bypass diode could provide sufficient voltage mismatch, but I think you'd still be splitting hairs whether you'd get enough current to blow a fuse in say a 3 string array.

Maybe even a four string array would be pushing it too.

Granted, if there is 1.5X the rating the fuses will blow, I just think the likely-hood of this condition carrying on merrily under that current level is high.

A blocking diode would stop this, but I wouldn't use one because of power loss ( .7V x whatever current in each string ( say 8A) , about 5Watts/string).

So put fuses in, but can I suggest this as well.

This would cost no power during normal operation, use lightweight wiring, and indicate if you had a problem and where it was.

It uses two pin bidirectional two colour LED's, during normal operation they would be off. In a circulating current condition the combination of LED colors would indicate in which string the issue is. The resistors would be sensible values found by initially adjusting a pot.

 

image.png.bbeeee7ef674faa1d3a3e30b409f9375.png

 

 

 

 

Im not sure if it always would work. I undestand you analize the middle point and what would happen if the failure was above the middle?

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21 hours ago, phil.g00 said:

The reason I am unconvinced about fuses, is not that they aren't a legal or safety requirement.

Yes, I'm convinced to install them, but I am unconvinced that they actually blow and notify me that I have a circulating current issue.

A PV panel will only conduct in reverse after Voc is exceeded, and at Voc the parallel strings provide zero current by definition.

Now a blown bypass diode could provide sufficient voltage mismatch, but I think you'd still be splitting hairs whether you'd get enough current to blow a fuse in say a 3 string array.

Maybe even a four string array would be pushing it too.

Granted, if there is 1.5X the rating the fuses will blow, I just think the likely-hood of this condition carrying on merrily under that current level is high.

A blocking diode would stop this, but I wouldn't use one because of power loss ( .7V x whatever current in each string ( say 8A) , about 5Watts/string).

So put fuses in, but can I suggest this as well.

This would cost no power during normal operation, use lightweight wiring, and indicate if you had a problem and where it was.

It uses two pin bidirectional two colour LED's, during normal operation they would be off. In a circulating current condition the combination of LED colors would indicate in which string the issue is. The resistors would be sensible values found by initially adjusting a pot.

 

image.png.bbeeee7ef674faa1d3a3e30b409f9375.png

 

 

 

 

Im sure It would blow. Imagine six string of two 60 c panels. There is a sharow in one of them, so in that panel current pass thtough bypass diode. If the panels were separated, Vo of that string would be 60 V and 72 V for the other five. wich woul be the equilibrium if you join all? Vo would be a Voltage from 60 yo 72, depending on radiation, passing current un reverse trough the string in shadow. That would be un float stage, or at the end of absortion stage, when no power is needed and PV Voltage go Up.

Edited by Javi Martínez
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The midpoint is at equipotential in normal operation, therefore no current flow.

A sufficient voltage difference, say caused by blown bypass diode in any half-string upstream or downstream will result in a voltage difference between either + & midpoint or - and midpoint.

This will upset this balance, probably insufficiently enough to blow a fuse if there were insufficient parallel strings ( my opinion), but easily sufficient enough to provide the 20mA (limited by the resistors) that the LEDS need to operate.

Javi,  I concede that the more strings in parallel the more strings can contribute  to blowing a fuse, but what you are forgetting is at 72Voc there is no current output from any string.

It is an open circuit voltage by definition, at that voltage all the healthy strings in parallel are not capable of delivering current, never mind 1.5x their sc current.

So the fuses will not blow at Voc voltage, they will only blow at a voltage that is capable of producing a combined current above sufficiently higher that the Voc of  60V of the faulty panel. 

Because at 60v Voc of the faulty panel no reverse current will flow again by definition, so to achieve 1.5X sc current in reverse you probably need a few more volts to drive the reverse current.

Now remember there is an MPPT input impedance in parallel with the faulty string as well which also providing an alternative current path, so I don't think it is a definite conclusion that a fuse will blow at all.

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25 minutes ago, phil.g00 said:

The midpoint is at equipotential in normal operation, therefore no current flow.

A sufficient voltage difference, say caused by blown bypass diode in any half-string upstream or downstream will result in a voltage difference between either + & midpoint or - and midpoint.

This will upset this balance, probably insufficiently enough to blow a fuse if there were insufficient parallel strings ( my opinion), but easily sufficient enough to provide the 20mA (limited by the resistors) that the LEDS need to operate.

Javi,  I concede that the more strings in parallel the more strings can contribute  to blowing a fuse, but what you are forgetting is at 72Voc there is no current output from any string.

It is an open circuit voltage by definition, at that voltage all the healthy strings in parallel are not capable of delivering current, never mind 1.5x their sc current.

So the fuses will not blow at Voc voltage, they will only blow at a voltage that is capable of producing a combined current above sufficiently higher that the Voc of  60V of the faulty panel. 

Because at 60v Voc of the faulty panel no reverse current will flow again by definition, so to achieve 1.5X sc current in reverse you probably need a few more volts to drive the reverse current.

Now remember there is an MPPT input impedance in parallel with the faulty string as well which also providing an alternative current path, so I don't think it is a definite conclusion that a fuse will blow at all.

May be, It would be interesting study reverse effects. I did a test to a 72 c solar panel with a high level of radiation.

I conected  a Charge in parallel and the results were:

If we apply Vo+0.6 V It starts to flow current.

If we apply Vo+5V It flows 3.8 A against solar panels.

At Vo+8.6 V It happens something like and avalanche current and flows the 10 A that the charger can supply.

Edited by Javi Martínez
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So maybe in the region of Vo +1.5 V is sufficient to to cause 1.5X sc current to flow in reverse.

But that would be 1.5V +Vo for every panel in the string.

Going back to your example of 6 parallel strings of two panels, that would mean that at approx. 73.5 +61.5V = 135V is sufficient to blow a fuse.

 Now you have the MPPT presenting a constant changing input impedance to the panels, but lets say its runs about 80% of the healthy Voc of 144V thats around 115V. (Too low to force a reverse current)

Then when it is not accepting power that input impedance goes up and voltage will rise to 135V, but the question is at a 135V operating point is the combined currents sufficient to supply 1.5x Isc.  Remember contrary to the charger you used in your test, the healthy strings act like a charger that can supply less current the higher the voltage rises.

My guess is with 6 strings in parallel there is sufficient. 

But the more panels in a string and the less parallel strings the less likely that scenario becomes. 

Your choice of 6 strings of two series panels is different to the original poster's 3 x 3 configuration.

 

Edited by phil.g00
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Therefore the check to see if a configuration would actually blow a fuse would be:

If the operating curve of the panel could deliver  (1.5sc/Number of healthy strings in parallel)  at a

(voltage of sum of Voc +1.5V of healthy panels + Voc +1.5V of unhealthy panel)/(Sum of Voc's of healthy panels)

For the 6 by 2 scenario that means is your panel can deliver  1.5/5  (30% of Isc) at 135/144 (94%) of Voc the fuse will blow.

Fuse curves aside, if the PV panel operating curve can't do this the fuse won't blow.

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20 minutes ago, phil.g00 said:

So maybe in the region of Vo +1.5 V is sufficient to to cause 1.5X sc current to flow in reverse.

But that would be 1.5V +Vo for every panel in the string.

Going back to your example of 6 parallel strings of two panels, that would mean that at approx. 73.5 +61.5V = 135V is sufficient to blow a fuse.

 Now you have the MPPT presenting a constant changing input impedance to the panels, but lets say its runs about 80% of the healthy Voc of 144V thats around 115V. (Too low to force a reverse current)

Then when it is not accepting power that input impedance goes up and voltage will rise to 135V, but the question is at a 135V operating point is the combined currents sufficient to supply 1.5x Isc.  Remember contrary to the charger you used in your test, the healthy strings act like a charger that can supply less current the higher the voltage rises.

My guess is with 6 strings in parallel there is sufficient. 

But the more panels in a string and the less parallel strings the less likely that scenario becomes. 

Your choice of 6 strings of two series panels is different to the original poster's 3 x 3 configuration.

 

Yes, i think for 3x3 you are right. Always? I don't know. Remember cloud edge effect. We can have in that case a radiation of 1400 w/sqm

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This just a illustrative graph that I grabbed of google for increased irradiation, I'm not sure you can make this claim either remember in a 2 panel situation you are at 94% Voc, in a 3 panel situation the % Voc would be higher.

It would appear that approaching Voc  at any level of radiation the output current falls off a cliff.

image.png.f9ce3656044e1611f72e6ee096f66b3d.png

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Yes, I accept that, the question is will there be sufficient current to blow a fuse to prevent this happening.

I would love fuses to blow, I just think that the faith in them to actually blow is misplaced.

I think that there is plenty of wriggle room for a faulty condition to exist unnoticed.

That little circuit above provides a visual indication of the presence of circulating current and in what direction.

It would require manual intervention, ( so would a blown fuse), but It would be far more sensitive.

By the way a value for R in that circuit would be around 300 ohms 0.5Watt.

 

Edited by phil.g00
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You can see IV curve data in this capacitive swept i do in a 36 cells 20 wat panel:

4,561290323 1,083870968
4,669892473 1,083870968
4,669892473 1,083870968
4,778494624 1,083870968
4,887096774 1,083870968
4,995698925 1,083870968
5,212903226 1,083870968
5,538709677 1,083870968
5,97311828 1,079569892
6,407526882 1,079569892
6,841935484 1,079569892
7,167741935 1,079569892
7,602150538 1,075268817
8,03655914 1,075268817
8,579569892 1,075268817
9,013978495 1,075268817
9,339784946 1,075268817
9,774193548 1,070967742
10,3172043 1,070967742
10,7516129 1,070967742
11,18602151 1,070967742
11,51182796 1,070967742
12,05483871 1,066666667
12,48924731 1,066666667
12,81505376 1,058064516
13,24946237 1,049462366
13,57526882 1,032258065
14,00967742 1,019354839
14,55268817 1,002150538
14,87849462 0,980645161
15,20430108 0,95483871
15,53010753 0,929032258
15,85591398 0,898924731
16,29032258 0,868817204
16,61612903 0,834408602
16,83333333 0,795698925
17,15913978 0,756989247
17,15913978 0,756989247
17,59354839 0,675268817
17,81075269 0,632258065
18,13655914 0,589247312
18,35376344 0,546236559
18,46236559 0,503225806
18,67956989 0,464516129
18,78817204 0,421505376
18,89677419 0,387096774
19,00537634 0,356989247
19,11397849 0,322580645
19,22258065 0,292473118
19,3311828 0,266666667
19,3311828 0,240860215
19,43978495 0,219354839
19,43978495 0,197849462
19,5483871 0,184946237
19,5483871 0,167741935
19,65698925 0,150537634
19,65698925 0,141935484
19,65698925 0,129032258
19,7655914 0,120430108
19,7655914 0,111827957

First data is voltage and second current

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So lets take 20V as Voc and Isc at 1Amp ( near as dammit to make the maths easier).

And say we need 1.5Amps by the combined healthy strings to blow the fuse.

 then 94% (4  panel string) = 18.8V       -----     +/- 400mA available from healthy string,  1.5A/.4 = 3.75, so 4 healthy strings would be needed to blow a fuse in a 5 string array.

92% = ( 3 panel string) =18.4V               -----     +/- 520 mA available from a healthy string, 1.5A/.52 = 2.88, so 3 healthy strings would be needed to blow a fuse in a 4 string array.

85.5% = (2 panel string) = 17.1V           -----      +/- 770mA available from a healthy string  1.5A/.77  = 1.94, so 2 healthy strings would be needed to blow a fuse in a 3 string array. (The parallel MPPT would also be taking current at this voltage so maybe 2 healthy strings is insufficient.)

So fuses certainly wouldn't help in a 3X3 configuration of this particular panel.

I just realized that the actual % Voc are based on a 36V panel not a 20V panel so I'll redo this again, I wont be allowed to edit this shortly

 

Edited by phil.g00
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I am going assume 3 bypass diodes  ( I dont know if they do that for a 20W panel, but because they do it at higher wattage I think it suits our purpose.

Assuming Voc of 20V ans Isc of 1 amp again.

So for 20 Voc -1/3 20Voc +1.5V = Voc of faulty panel = 13.2

A healthy panel needs 20Voc +1.5V to pass 1.5 x Isc.

So  a healthy 4 panel string has a VOC of 80V, and 4 panel string with a blown bypass diode needs (3x21.5 + 13.2V = )77.7 V to pass 1.5A in reverse.

Each healthy string at 77.7/4 = 19.425V is capable of around 0.22Amps . 1.5A/.22A = 6.8 . Therefore healthy 7 strings in a 8 string array.

3 panel string  has a voc of 60V, and a 3 panel string with a blown bypass diode needs (2x21.5V + 13.2V = ) 56.2V to pass 1.5A in reverse.

Each healthy string at 56.2/3 = 18.73 is capable of delivering 0.43Amps. 1.5/.43 = 3.48 . Therefore four healthy strings in a 5 string array.

2 panel string has a V0c of 40V, and a two panel string with a blown bypass diode needs ( 21.5V + 13.2V = ) 34.7V to pass 1.5A in reverse.

Each healthy string at 34.7V/2 = 17.35V is capable of delivering 0.7 Amps. 1.5/.7  = 2.14. Therefore 3 healthy strings in a four string array, the caveat being that at this voltage the MPPT is probably also drawing current so 3 healthy strings maybe insufficient.

 

 

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 "a 60 cell 8 A ISC panel can be damaged with a reverse current of 2 or 3 A."

So what you are saying is if a fuse was to blow it would be too late to save the panels in the string anyway.

Ok,  by far the simplest solution is to put a blocking diode in place at a constant cost at full production of say .9 *ISC( 8A)  =7.2 * 0.7V (V loss  across the diode) = 5 Watts per string. 

You'll never get reverse current but you'll also never really get alerted to the issue.

I think it is fair to assume that a blown bypass diode will cause the output voltage of a cell to drop by a third. ( 3 bypass diodes/panel).

So taking your configuration into account there will be a voltage differential in my circuit before the faulty panel begins to conduct in reverse.(lets say 9V)

I am using that voltage differential to drive LED's, but the LED's could be replaced by say:

image.png.eead77acfa1067d7539025417cc4eff6.png

Then logical combinations of the relay's auxiliary contacts could identify the faulty string and open a heavier latching relay on the faulty string. This would draw no significant power in operation. The relay would have to latch or it would chatter, as opening the string would effectively switch itself off. All sorts of buzzers or lights could be triggered by this final relay.

  

Edited by phil.g00
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4 minutes ago, phil.g00 said:

 "a 60 cell 8 A ISC panel can be damaged with a reverse current of 2 or 3 A."

So what you are saying is if a fuse was to blow it would be too late to save the panels in the string anyway.

Ok,  by far the simplest solution is to put a blocking diode in place at a constant cost at full production of say .9 *ISC( 8A)  =7.2 * 0.7V (V loss  across the diode) = 5 Watts per string. 

You'll never get reverse current but you'll also never really alerted to the issue.

I think it is fair to assume that a blown bypass diode will cause the output voltage of a cell to drop by a third. ( 3 bypass diodes/panel).

So taking your configuration into account there will be a voltage differential in my circuit before the faulty panel begins to conduct in reverse.

I am using that voltage differential to drive LED's, but the LED's could be replaced by say:

image.png.eead77acfa1067d7539025417cc4eff6.png

Then logical combinations of the relay's auxiliary contacts could identify the faulty string and open a heavier latching relay on the faulty string. This would draw no significant power in operation. The relay would have to latch or it would chatter, as opening the string would effectively switch itself off. All sorts of buzzers or lights could be triggered by this final relay.

  

I'm not sure if you can do with a relay. We have round 80-100 V DC and we had to consider the open arc. Maybe a SSR would be better.

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I would rate final relay to break Isc and Voc anyway, but remember in operation this relay should operate for a voltage imbalance before there is any reverse current in the string.

Also wiring several contacts in series is a farmers way of improving the breaking capacity.

It would have to latch until it is manually reset.

I don't have a lot of experience with SSR's. 

But don't they need to be turned on and stay on until a current zero, I  don't think they'd fit this scenario, but maybe there is a logical configuration that would suit.

Edited by phil.g00
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22 hours ago, maxomill said:

there seems to be that 2 fuses are mentioned  a lot 

why 

It has to do with earth faults (eg a short from the PV cable to roof material for example). An earth fault can cause one fuse to get bypassed, and the system will continue working normally as long as no second fault develops. If you have no second fuse, a second fault potentially causes disaster.

If no second earth fault is reasonably expected, for example on the cabling from busbars to inverter it is rather unlikely as it all runs in PVC trunking with thick insulated cabling, then you don't need the second fuse.

Edit: Granted, if both fuses are on the ground, an earth fault developing at any point before that is not going to blow any fuses :-)

 

Edited by plonkster
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