> Here's a (rough) worked example on a real product

>

> Panel here

>

>

>

>

http://www.mitsubishielectricsolar.com/images/uploads/documents/specs/MLU_spec_sheet_250W_255W.pdf
>

> Graphed result here

>

>

>

> dl.dropboxusercontent.com/u/30808964/PV%20panel%20mitsubishi%20V43.jpg

> 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

> leve.

> 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.

>

> Method.

> 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 (!)

>

>

>

> SO

>

> 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.

>

> QED.

>

> E&OE.

>

> ___________________

>

>

>

>

> On 28 April 2016 at 20:52, RussellMc <

spamapptechnzSTOPspamKILLspamgmail.com> wrote:

>

> > On 28 April 2016 at 19:33, Justin Richards <

justin.richardsKILLspamspamBeGonegmail.com>

> > wrote:

> >

> >> 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.

> >>

> >> â€‹Sort of, maybe.

> > Close to "yes in many cases"

> >

> > If part of a panel becomes shaded â€‹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

> > is

> > 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.

> >

> >

> > Russell

> >

> >

>

> --

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