Solar cable

[mitiempo]

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

A panel of a given size with too small a wire will give less output where it counts if wire too small is used. It is not much different than a large flow of water through a small tube - it is a restriction.

Why would anyone pay for a good solar panel and then restrict its output by saving a few dollars on the wire?

 

[lw395]

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.

Assuming you are just wiring the panel straight to the batteries, not using a converting regulator, that's probably true.
Where it will go wrong is if part of the panel is in shade. Then the volts drop as well as the panel impedance rising.
It depends on how inconvenient/costly fat cable is.
It might be better to have a better panel position at the expense of cable loss for instance.
It might be that the critical thing is getting a charge at low illumination angles, rather than getting max charge at max illumination. That may just deliver extra charge that the batteries cannot absorb.

The question is, is the resistance enough to drop the voltage below 14.4 or so volts, in the circumstances that matter to you, with the way you use your charging system. That will vary in individual cases.

The minimum cable size is governed by the temperature rise in the cable at max output, please don't set fire to your boat!

Hope that helps?

 

[mitiempo]

Am about to install a new solar panel rated at 80w so I guess I'm looking at 4-5 amps at best.
Colin

A panel of 80 watts should never be wired direct to the batteries, a controller is necessary.

 

[VicMallows]

a controller is necessary.

Agree. But that controller could be a competent and diligent human keeping a close eye on a voltmeter.

 

[VicS]

Agree. But that controller could be a competent and diligent human keeping a close eye on a voltmeter.

Use thin enough wire to connect the panel and you wont even need to keep an eye on the voltmeter

 

[mitiempo]

Agree. But that controller could be a competent and diligent human keeping a close eye on a voltmeter.

They would need a good battery monitor so they know the status of the battery as voltage of a charging battery isn't effective. Wired direct in bright sun the batteries will see 4 or 5 amps @ 17 volts.

It would also help if the competent and diligent human has no life.

 

[Como]

Wired direct in bright sun the batteries will see 4 or 5 amps @ 17 volts.
D

Wired direct, the voltage across the battery would be held at the battery voltage of say 14.5 volts if it was fully charged already. If not fully charged it will be lower.

 

[VicMallows]

Wired direct, the voltage across the battery would be held at the battery voltage of say 14.5 volts if it was fully charged already. If not fully charged it will be lower.

No. I can assure you that it will readily rise to 16 or 17v. Useful if you deliberately want to equalise the battery for a suitable period, but punishing otherwise.

 

[lw395]

No. I can assure you that it will readily rise to 16 or 17v. Useful if you deliberately want to equalise the battery for a suitable period, but punishing otherwise.

Agreed.
There is a popular myth that lead acid batteries will absorb lots of current and limit the voltage to their gassing voltage. Even when it's not fully charged, it does not take an immense current to get the volts up to 16 or so. That's if it is purely DC under consideration.
A pulser circuit can get peak voltages of over 50V with not much power.

The more measurements you do on batteries, the more complex you realise they are!