<span class="highlight">Solar</span> <span class="highlight">panel</span> current measure help

Hi John,
Thanks for your reply!!
So the power output would be read in watts, and the reading would depend on the sun's output to some degree?
Thanks for your info - I appreciate your wisdom :)

Good advice John,

I might add that multiplying the short circuit current with 16.5V will be closer to the mark because Isc is around 5% higher than Imaxpower.

Of course, the sun needs to be perpendicular to the panel's surface for the measurement to be meaningful.
And in real life, the max wattage one can expect from the panel is at least 10~15% lower because of cell heating. And this is with a 'true' MPPT solar regulator.
If not 'true' MPPT, the max useable wattage will drop by another 10~15%.

Have a great weekend.
cheers, Peter

Hi Paul

We have a Tvan with a solar panel. We have a meter to read input, and so far the one panel on the roof of the Tvan has only generated 1 Amp/hr of power. It is not angled to the sun as it is flat on the roof. So far, we have been unimpressed with its capability.

Laurie

Laurie

The TVan has a reputable brand solar panel.
But it is small - only 50 W. Good enough to top the batteries not much more.
Plus one also try to set up in the shade which makes it even less useful. Which is why a portanble set connected parallel is a good option.

CJ

Paul,

Power = voltage multiplied by current. The voltage is not that which you measure with no load on the panels, but as we are interested in power, is the voltage at which maximum power occurs. That's about 17 - 18 volts, though as Peter (a Queenslander!) says it drops with increasing temperature, so for him, 16.5 is maybe better.

There are a lot factors in getting an absolutely right figure. The panel should be perpendicular to a line drawn to the sun. (A good way of establishing the correct angle is to simply aim the panel so as to cast the biggest possible shadow.) Panels are spec'd at a radiation level of 1kW per square metre; in southern Australia at this time of year you won't achieve that radiation level from direct sunlight, even at noon. Any cloud has an interesting effect - scattered cloud may increase total radiation by reflection, or reduce it if interposed between panel and sun.

Using the approach suggested you'll get near enough for your purposes, but I don't claim absolute accuracy!

Cheers

John