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I know that to measure the actual power of the solar panel I need to measure the voltage & current at maximum power (Imp), i.e. connected to the load. However I can't connect the panel to any load at the moment, so I wonder if I can at least estimate the power by measuring simple parameters such as open-circuit voltage and short-circuit current (Isc). To at least understand if I need to order a bigger panel.

My panel is rated at 17V and with no load I'm measuring around 19V in direct sunlight. Short circuit current is 1.15A. For my application I need around 20W of power.

Is there any typical relation between Isc and Imp? Will it be higher than Isc or lower? (My solar panel seems to be a monocrystalline type if that matters)

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  • \$\begingroup\$ Please link the data sheet for the panel and explain which part of the world you are in when you measured 19 volts in direct sunlight. \$\endgroup\$
    – Andy aka
    Oct 28, 2023 at 11:39
  • \$\begingroup\$ @Andy - Belarus probably. See chart in my answer. \$\endgroup\$
    – Russell McMahon
    Oct 28, 2023 at 11:45
  • \$\begingroup\$ @floppydisk Where are you located? I gavesome stats for Minsk. || How much power you need may not be your primary concern - how much energy/day may be. || . How much energy do you need in a day. Those figures suggest a ~= 20W panel in not ideal conditions. But (see my answer) if you get 20W at peak you'd get maybe 20 Watt hours daily now (late October) and 10 Watt hours per day on average in December. \$\endgroup\$
    – Russell McMahon
    Oct 28, 2023 at 11:51
  • \$\begingroup\$ @Andyaka there's no datasheet, it's a Chinese panel "rated" at 600W \$\endgroup\$
    – floppydisk
    Oct 28, 2023 at 14:14
  • \$\begingroup\$ 600 W - wow. Output in noonday sun on a good clear day pointed directly at sun should be about 200W x Area-in-square-metres. Almostr every PV panel that I've seen* has had approximately genuine ratings marked on them. A 600 Watt panel is about 3 square meters - that's VERY large. 17V is a rather unusual Vmp voltage rating. What is specified Imp. || Please answer my questions above re desired Watt-hours per day. || Can you supply a photo of panel and rating plate, and dimensions|||| * I've been involved with about 350,000 PV panels BUT most were very small compared to yours. \$\endgroup\$
    – Russell McMahon
    Oct 29, 2023 at 0:12

1 Answer 1

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You can obtain a reasonable estimate using Voc and Isc.

Vmp will typically be in the Voc x 0.80 to 0.90 range.
Higher efficiency panels will have a higher Vmp/Voc ratio.
You panel's claomed 17V Vmp is 17/19 = 0.895, which is higher than I'd expect from most panels in the market. Typical module efficiencies are around (very roughly) 20% for monocrystaline cells. You can get higher but they are less usual.

Isc is slightly higher than Imp, but not vastly so. From not recently renewed memory - about 5% to 10% higher.

You say "full sunlight".

Standard ratings are at 1000 W/m^2.
Actual insolation at true noon with the panel pointinmg directly at the sun on a smogless cloudless day can be somewhat above that.
Panel heating will decrease output noticeably, but that's unlikely to be an issue for brief testing unless the panel has been heating in the sun previously.

Watch for very thin high cloud which is nearly invisible but which can noticeably affect insolation.

IF the panel is rated at 17.0 Vmp and you are seeing 1.15A then IF the rating is correct I'd expect somewhat less than 20W Wmp.
However, most panels are reasonably accurately marked (for most values of most) so I'd expect that better pointing, better sun and a cool panel would come closer to 20 Watts.

A panel is pointing AT the sum when its shadow is parallel with its major axis AND of maximum length. As you alter the azumith angle from 0 degrees (panel lying flat on ground) the shadow will grow in length, reach a maximumwhen pointing at the sun and then decrease as the angle continues to increase towards panel-vertical.


Here is solar and more inmformation for Belarus and for Minsk from the fabulous www.gaisma.com site.

The kWh/m^2/day line in the "Minsk, Belarus - Solar energy and surface meteorology" chart gives you the mean monthly daily insolation. I term this "Sunshine Hours" (SSH). in equivalent "sunshine hours"

Currently at 1.44 SSH/day it's not looking marvellous, and is getting worse :-( . 0.5 SSh/day in December :-( .
Your 20W panel well pointed throughout the day will make about 10 Watt hours in December - enough to mostly charge a single 3300 mAh 18650 LiIoncell with a suitable buck converter and charger. That's an average day.

enter image description here


Added:

Where are you located?
I gave some stats above for Minsk.

How much power you need may not be your primary concern - how much energy/day may be.
How much energy do you need in a day.
Your output figures suggest a ~= 20W panel in not ideal conditions.
But (see my answer) if you get 20W at peak you'd get maybe 20 Watt-hours daily now (late October) and 10 Watt-hours per day on average in December. If you want 20 Watts for one hour in December you'd need 2+ panels oif the type you have now. If you want 20 Watts for N hours you'd mneed 2N panels - or more. These need to be optimally pointed as the sun moves, or you need to accept even less output.

Hopefully you are somewhere sunnier than Minsk :-) :-( .

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  • \$\begingroup\$ Thanks for your detailed answer :) I am indeed somewhere sunnier than Minsk - currently in Georgia (lat 41) \$\endgroup\$
    – floppydisk
    Oct 28, 2023 at 17:58
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    \$\begingroup\$ @floppydisk Gaisma for Tbilisi in Georgia here. || \$\endgroup\$
    – Russell McMahon
    Oct 29, 2023 at 0:05

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