: Paralleling PV panels with differing illumination levels.
RussellMc email (remove spam text)
part 1 3536 bytes content-type:text/plain; charset="utf-8" (decoded base64) <CACZQxzMAtf4NRR5=Nq4ARMph7pd0_ukS0h9KGojO0uRPE764QA@mail.gmail.com>
Here's a (rough) worked example on a real product
Graphed result here
1362 x 1544 resolution.
200 x 227 version
[image: Inline images 1]
This is only at 100 90 80 70% full sun but shows what happens.
Green circles show optimum and paralleled current at the selected light
V is set to Vmp at full power as before.
IF I did it right then results are "interesting".
900 W/m^2 loses little
800 loses rather more - about say 0.2/6.3 or about a minimal 3 %
BUT 700 W/m^2 loses LESS than 800 W/m^2.
Their lines or mine may be wrong.
Drop vertical from peak power point on power-V curve to relevant V-I curve.
This is mpp for that % insolation.
Draw line horizontal left to show optimum I loaded.
For 100% curve draw line (red) vertically downward to x axis.
This is Vmp at 100% light.
>From intersection of red line and white VI lines draw horizontal lines
(thin red) to Y axis to get Ixx at Vmp100.
Compare differnces of related black and red lines.
HOWEVER - just realised - just looking where the vertical red line
intersects the CYANish power-V curves shows how much loss you get - you can
see peak power at x% insolation and off-peak power when paralleled.
Clear and easy.
AND you can see that the 800 W/m^2 curve loses more power than the 700
W/m^2 one does (!)
Simple method (Agh!)
Draw line vertical from power-V curve for 100% sun to X axis.
Intercepts with other power-V curves show power loss in this case.
On 28 April 2016 at 20:52, RussellMc <gmail.com> wrote: apptechnz
> On 28 April 2016 at 19:33, Justin Richards <gmail.com> justin.richards
>> Does this imply that two separate arrays, one facing East the other West
>> (due to limited roof realestate) could be connected in parrallel without
>> the need for a dual tracking inverter with only a small performance hit.
>> Sort of, maybe.
> Close to "yes in many cases"
> If part of a panel becomes shaded then either
> - the max current for all cells in the same series string is the current
> that the shaded cell generates
> - or if the shaded cell has protection diodes then for N cells in series
> and 1 shaded cells thyen
> current max is as before but
> Vpanel_now = Vpanel x n/(n-1) - 1_diode_drop
> For panels illuminated evenly but at 2 different levels.
> Working through my stack exchange answer, for 100% and xx% illuminations,
> down to about xx >= 50% it looks fairly benign.
> For ery low xx it can still be remarkably good.
> In my 2nd examples, for 20% insolation the 20% panel makes 79% of the
> current it would at optimum but at aboyt 43/39ths the voltage so power drop
> 79% x 43/39 = 87% of the power it would otherwise make.
> This is if the 100% panel still works at the old MPP.
> Odds are the combination has a different Vmp and the end result will be
> BETTER than calculated above.
> As xx insolation falls there comes a point that Voc is <= the operating
> voltage of the 1st panel and you get nothing.
> In my SE answer that occurs at about 5% insolation (bottom line shown is
> 10%) so you don't lose much.
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