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checking a solar panel. - Page 2

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Posted by Eeyore on November 2, 2008, 10:28 am
 


fabtraninbox@gmail.com wrote:


Try it with a load of ~ 120-150 ohms. You'll need a resistor of several watts
dissipation. say a 120 ohm resistor of serval watts and 33 ohms 1/2 W to put
in series.



Will have to calculate.



There is no minimum amps. What is the battery capacity in Ah ? And what
technology e.g. lead-acid ?

Graham


Posted by Ron Rosenfeld on November 2, 2008, 1:23 pm
 
On Sat, 1 Nov 2008 20:28:24 -0700 (PDT), fabtraninbox@gmail.com wrote:


We have to make a number of assumptions to see if the panel is in working
condition; and you need to provide more information to assess whether it
can charge your battery.

Assuming you exposed the panel to the sun at noon, and the angle of the
panel was perpendicular to the "sun angle", at a time when there should
have been an irradiance of 1kW/M^2, it seems to be underperforming.

If my calculations are correct, 36 4" diam cells occupy an area of about
0.29 M^2.

I don't know what efficiencies were being obtained in 1993.  But with a 5%
efficiency, the power output should be about 14W.

You have provided Voc and Isc  (oc=open circuit; sc = short circuit).

On the few lower powered panels I just looked at, these values would be
higher than Vmp and Imp (mp=max power which is how panels are usually
rated). And the calculated output using Vmp and Imp is less than that
calculated using Voc and Isc.

Therefore, I believe we can assume that, for your panel, it's output will
be less than your measured Voc * Isc.

In your case:  19V x 0.2A = 3.8W.  So, under load, this panel might put out
3W or less.  And it will be even less efficient in your relatively hot
climate, assuming your temperatures provided are in the Celsius scale.

Clearly the panel is underperforming.  Whether this is something as simple
as connections that can be repaired, or internal degradation of the actual
PV cells, is something you might be able to determine by inspection, and by
measuring the output of the individual cells.

As to whether this will be sufficient to charge a battery, that depends
critically on the charging parameters for the battery under discussion. For
some batteries, that output would be insufficient to even make up for the
normal self-discharge; and for others it might be excessive.

You'll need to provide more data.

It would be helpful to assess the output of your panel under a load.
--ron

Posted by Eeyore on November 2, 2008, 6:50 pm
 

Ron Rosenfeld wrote:


I agree totally.

Graham


Posted by spaco on November 2, 2008, 5:23 pm
 Why did your friend sell the panel to you?  Had he already determined
that it didn't work well?  What was he using it for?

Pete Stanaitis
--------------------

fabtraninbox@gmail.com wrote:



Posted by Bill Kaszeta / Photovoltaic Re on November 3, 2008, 4:35 am
 On Sat, 1 Nov 2008 11:00:33 -0700 (PDT), fabtraninbox@gmail.com wrote:


I suggest you connect it to a 12-volt battery of any sort and measure the
resulting charging current with full sun on the PV module.

A good 64-watt module would charge at about 3.5 to 3.8 amps.


Bill Kaszeta
Photovoltaic Resources Int'l
Tempe  Arizona  USA
bill@pvri-removethis.biz

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