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Author Topic: Using a pv cell as a toaster  (Read 405 times)
ovonrein
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« on: January 09, 2012, 11:49:41 AM »

This is a quick question to address a gap in my PV knowledge that has shown up in another thread which, I fear, has grown so long by now that it may put a lot of people off who know the answer to the following simple question:

How does a single solar cell behave WHEN POWERED, ie when fed energy?

This situation can arise in any cell-string when only one cell is blacked out and the negative voltage forced on to that cell isn't so negative as to trip the diode on that circuit.

Thanks.
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JamesE
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« Reply #1 on: January 09, 2012, 11:57:17 AM »

If it won't work the diode then it's irrelevant - the forward voltage drop in a diode is a nominal 0.6 volt, rising a little (say 0.7 volt) with rising current. Thus the reverse voltage on the cell section is trivial. If it's less than 0.6 volt then it's night time!
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ovonrein
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« Reply #2 on: January 09, 2012, 12:39:27 PM »

I see the error of my ways.  Thanks, James.
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ovonrein
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« Reply #3 on: January 09, 2012, 01:30:43 PM »

Sorry, I thought I had it.  Turns out, I don't.  Help me again, please...

The Sanyo HIT250W (3x20) when irradiated with 500W/m2 has this IV curve, roughly:

33V/4.5A
40V/0.5A

Divide the V by 60 and we find to the IV curve per cell:

0.55V/4.5A
0.67V/0.5A

Before the shade arrives over one cell, all 60 cells are running at 33V and 4.5A.  The moment a shade arrives over one cell, that cell pulls the entire circuit down to 0.5A (say).   To get there, the sunny strings move to 26.67V leaving 6.33V for the affected string - which won't trip the diode.  In this string, 19 cells are running at 12.67V and the shaded cell must therefore run at -6.33V, ie it is consuming power.

Where am I going wrong?
« Last Edit: January 09, 2012, 01:46:46 PM by ovonrein » Logged
HPSauce
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« Reply #4 on: January 09, 2012, 04:58:20 PM »

Hi ovonrein, see that you have started a new thread on this in order to try and understand some more detail.

I made a diagram of the situation and I think you are correct in your analysis, the shaded cell will start to absorb some energy in the event of the MPPT not responding to the changed conditions. In this case 3.16Watts, not too much.

I found a couple of papers which cover the subject in some detail.
The first is
http://www.f09.fh-koeln.de/imperia/md/content/personen/wiesner_wolfgang/veroeffentlichungen/26ieee.pdf
and relatively easy to understand but appears to be fairly old.

A more modern paper is
http://www.mpi-halle.mpg.de/mpi/publi/pdf/9077_09.pdf
but is very detailed.

In summary, hot spots can develop within the cell - even emitting some light - and could cause damage. The panel diodes are added to prevent this.
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ovonrein
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« Reply #5 on: January 09, 2012, 05:09:31 PM »

I love you, HP.  What a gem you have dug out here http://www.f09.fh-koeln.de/imperia/md/content/personen/wiesner_wolfgang/veroeffentlichungen/26ieee.pdf.  It almost seems to me that you have failed to take in its significance.  Let me attach the beautiful IV chart.  I had no idea (though some suspicion) how the amps lift off below -10V.  That's just awesome.  Already at -6.33V your translation into 3.16W is conservative - the little fellow is happy to burn more.  Now we only need to think what the MPPT will do.  Oh, but we know.  It will push up the circuit voltage and so the little fellow gets a little warmer yet.  I wonder... 

Good enough for this thread.  Back to the main feature.


* IV.PNG (26.59 KB, 585x384 - viewed 130 times.)
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HPSauce
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« Reply #6 on: January 09, 2012, 07:19:11 PM »

I should just say that the exact voltage/power of the shaded cell will depend on the load characteristic of the inverter which I don't know. This will then update as the MPPT tries to respond to the change.
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ovonrein
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« Reply #7 on: January 09, 2012, 07:35:20 PM »

Yeah - I was wondering about that, too.  It looks to me as though by the MPPT increasing V, it will push the shaded cell more and more negative, whereby allowing that cell to (reverse) pass through more I, which, if I understand well, will allow I across the entire circuit to come off the floor.  At which point V in the sunny cells will fall a little, thus allowing the shaded cell to go further -V.  At some point the diode will trip but before that happens, there is some interesting games playing out with the MPPT.

On the other hand, one cell's consumption increase should be swamped by the production in crease from 59 cells, so perhaps the MPPT will rather quickly push the string voltage up to the point where the shaded string diode trips?
« Last Edit: January 09, 2012, 07:39:46 PM by ovonrein » Logged
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