> 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 <
RemoveMEapptechnzTakeThisOuTgmail.com> wrote:
>
> > On 28 April 2016 at 19:33, Justin Richards <
spamBeGonejustin.richardsspamBeGonegmail.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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