Solar cable

noelex said:
its simple, look at the cable you are connecting to, and use the same to the battery, or just ask the supplier for the recommended cable which he will happily supply.

Lots of people do this. I have even had professional installers tell me there is no point going to a bigger wire because "no more electricity can get out" :)
It is much better to size the cable properly with the sort of calculations that have already been posted. Even if the wires exiting the panel are small the voltage drop over a short distance will be small, but if the same sized cable is extended a long way, the voltage loss can be considerable.
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Thanks for that. You answered a question that I hadn't (yet) asked.
 
VicS said:
True the panel resistance is high but you dont want to loose 1.2 volts in the wiring.
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You won't lose anything like that much.

If I have my figures correct, the resistance of 20m of AWG14 (2 sq mm) is less than 0.2 ohms. I have a feeling that the internal resistance of a solar cell is of the order of 10 ohms or more, you really won't notice the additional resistance of the wire.

Yes you will still get a small voltage drop across it - but that just be volts that would otherwise have been lost in the cell itself, the impact on charging current will be only a couple of percent.
 
To give up 5w is annoying. A further problem is that although the Voc is generally about 20v this is with a cell temperature of 20 C. Solar cells are usuall operating at about 40C, which significantly drops the voltage, if you combine slight shadowing with voltage loss in the wiring you can drop below the battery voltage at least for the Vmpp.
Try and keep the voltage loss to a max of 5%, 3% is better.
 
Trop Cher said:
Does this mean that you can use an ordinary electrial block connector to join the cables from two solar panels (20watt each) together and then to a single cable to the regulator?
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Well you could inside, proper waterproof connectors, e.g. Bulgin would be infinitely better & crimped connections with heatshrink are much better than blocks...
 
Yes you are right Nigel, in fact that is the point I was trying to make.
The start of the post was a quote from a previous post, sorry for the confusion.
I am struggling with my new iPad which doesn't seem to allow quotes.
 
noelex said:
Yes you are right Nigel, in fact that is the point I was trying to make.
The start of the post was a quote from a previous post, sorry for the confusion.
I am struggling with my new iPad which doesn't seem to allow quotes.
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I see what you mean, I didn't read your entire post the error way mine, sorry.

Reference the iPad, you mean quotes this: "" ? They are on the .?123 button on the iPad, but you can always use the italic button in the toolbar above. I also recommend the apps Unicode Character Map Free or Unicode Maps, they both give you easy acess to symbols like Ω etc.

Written on my iPad ...
 
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bedouin said:
To be honest the internal resistance of a solar panel is so high that the resistance of the cable connecting it to the battery should not be an overwhelming consideration. So I don't think you need to stick to the 3% drop rule - you should be fine going up to nearer 10%. I think 2 sq mm would be fine for the task.

It is more normal to quote in sq mm than diameter - and you should also be able to tell by the price
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VicS said:
True the panel resistance is high but you dont want to loose 1.2 volts in the wiring.
Click to expand...

bedouin said:
You won't lose anything like that much.

If I have my figures correct, the resistance of 20m of AWG14 (2 sq mm) is less than 0.2 ohms. I have a feeling that the internal resistance of a solar cell is of the order of 10 ohms or more, you really won't notice the additional resistance of the wire.

Yes you will still get a small voltage drop across it - but that just be volts that would otherwise have been lost in the cell itself, the impact on charging current will be only a couple of percent.
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You said, "So I don't think you need to stick to the 3% drop rule - you should be fine going up to nearer 10%."
According to the arithmetic I learnt 10% of 12 volts is 1.2 volts! My argument is that you dont want to loose 1.2 volt in the wiring .


All the advice on sizing wiring for solar panels and the tables and online calculators is based on a maximum of 3% volts drop. Its not my figure ... its the figure you will find used throughout.
 
I cannot think of any electrical connection where you want to lose 3% or more to voltage drop.

10% of the voltage of the panel at max of about 17.5 volts is 1.75 volts - not a big deal when the panel is at full output. But when the panel is not at full output it may have an output of 15 volts or less. 10% of that drops it below an effective charging voltage.

I just installed a 90 watt panel (about 5 amps) and the run from the panel was about 26' there and back. I used 10 awg for less than a 2% voltage drop.
 
VicS said:
You said, "So I don't think you need to stick to the 3% drop rule - you should be fine going up to nearer 10%."
According to the arithmetic I learnt 10% of 12 volts is 1.2 volts! My argument is that you dont want to loose 1.2 volt in the wiring .


All the advice on sizing wiring for solar panels and the tables and online calculators is based on a maximum of 3% volts drop. Its not my figure ... its the figure you will find used throughout.
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When charging a battery what you really care about are Amps, not Volts. The battery will not obey ohms law. The voltage at the battery terminal will be determined primarily by the state of charge of the battery (on the assumption an active regulator is not being used).

If my guesses are correct and the internal resistance of the panel is say 10 ohms then a 0.2 ohm resistance is going to reduce the charging rate by 2% - not really enough to be noticeable
 
The internal resistance of the panel makes no difference.
The panel outputs a specific voltage which will drop if not in full sun. What matters is the voltage drop from the panel to the controller and from the controller to the batteries.
 
mitiempo said:
The internal resistance of the panel makes no difference.
The panel outputs a specific voltage which will drop if not in full sun. What matters is the voltage drop from the panel to the controller and from the controller to the batteries.
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Only if the panel is infinitely large, or attached to an effectively infinite resistance - neither of which really apply. If you connect a 50W panel across a 1ohm resistor you will not get 50W dissipated in the resistor however bright the sunlight
 
When next on the boat, and when we have some decent sunshine, I will test this out empirically. (say adding 5m of thin bell wire?). I have 2x40W panels in parallel. I will report back, but my hunch is that there will be negligible difference in the charge current.
 
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