I just recieved my VW solar panels and have a question.

[Airspeed]

I took them out in the sun and measured each of the four I bought ($15 ea. Ebay) they all measured 14.5 volts in partial shade.
Each one has a diode built inline on the wire.
I also bought a solar charge controller that has a blocking diode as well,Should I remove the diodes that are inline on each panel since my controller has one built in? It would be alot easier to manage the panels without the diodes hanging off each panel. Would it be benificial to leave them on?
These panels are supposed to give me 5 watts each,is there a way to measure that?
I am going to mount them on an aluminum frame and use an aluminum tripod off my old telescope so I can set them in the sun.
It's gonna be so nice to not have to worry dead batteries anymore. Between the little Honda E300 generator I just got and these solar panels I should be set. Thanks in advance for any suggestions on the diode questions! Aaron

[bdosborn]

Yup, you can cut them off if your controller has a blocking diode:
http://www.windsun.com/tech_tips.htm

Here's a *relatively* cheap watt meter. Its the cheapest I've found on the net but still more than I'd like to pay (but I probably will anyways).
http://www.powerwerx.com/product.asp?ProdID=3809

But, you could use a HF meter to measure voltage (V) and current (I).
Watts=V*I (you should get about 0.3A from one panel in full sun.)
http://www.harborfreight.com/cpi/ctaf/d ... mber=92020

Make sure to post pictures of your array after its done, it sounds like a great project. And at $3/watt, more than half of what a typical panel costs.

Bruce

[Chuck Craven]

Airspeed
With each panel giving out 14.5v unloaded. At 5watts each panel max current will be 350 milliamps. Total current from the two panels will supply 0.7 amps or 700 milliamps. If you leave the diodes in you do not need the controller. The panels will never over charge a 12v lead acid battery. To test the full load voltage you need to put a resistor across the panels in full sun. The value of the resistor would be 40 ohms at 5 watts. A smaller wattage resistor will work just watch it doesn’t get to hot. You should see about .35 amps or 350 milliamps through the resistor and measure the voltage across the resistor. This will be full load voltage for the panel. The voltage should be less than 14.5volts. This is the max charging voltage the panels can produce. Look at your battery voltage charts and the panels can’t produce a high enough voltage to over charge your battery no mater what its amp hour ratting of the battery is. If you sill want to use the controller than take out the diodes. You will gain .7volts more charging voltage to the controller.

Chuck

[Airspeed]

I have four 5 watt panels (20 watts) The panels max power is 18 volts ,they were putting out 14.5 volts in partial shade and each panel has a diode built in to the supply wire. I thought this could overcharge a battery? The charge controller also shuts off the batt if power gets too low and it has a built in meter and some other special features so I cant see it hurting anything? I will get rid of the diodes on the panels and just use the one in the controller if doing so will give me more power. Thanks! Aaron

[Airspeed]

This is the controller I bought, I think I paid $25 for it. Aaron

[Chuck Craven]

Ok! Then maximum current supplied to the battery would be about 1.4 amps. A car charging system is set for 13.8 volts when charging a car battery, which is high for a deep cycle battery. Rule of thump for a deep cycle battery is: not to charge more than 1/10 of the ratted current for half of the Amp hour ratting. So say 70 HR /2= 35 amps /10= 3.5 Amps per first hour. Then trickle charge the battery to full charge. A 70HR deep cycle battery trickle charge is about 1.5 amps or less. With 20 watts and your controller you will not have any problem over changing the battery. Now remember it only can charge when the sun is out!
Unless you are hooking up a wind generator too!

Chuck

[Airspeed]

Wind generator,hmm? That's a great idea! heck,maybe I will throw a water wheel in any flowing water at the campground. I could start selling power to all the other campers!
Thanks again! Aaron

[Chuck Craven]

Be careful the state will start taxing you as a monopoly. Go to jail do not collect $200.


Chuck

[Airspeed]

Yea,I'd probably have to do extensive environmental impact studies to see if my wind and water turbines were causing any damage to the mosquito population. I guess I better leave the power biz to PG&E.
Aaron

[Chuck Craven]

Owes is Wiener - gies (WE energies). We call mosquito’s birds of pray! In Minnesota and Wisconsin it’s our state bird.

Chuck

[bdosborn]

Here's a couple of points to consider:

A solar panel will drive a short circuit very nicely. The short circuit output of a solar panel is about 90-95% of its rated output so a fuse is very important thing to have with a panel. You can use Chucks method to determine the exact output of your panel but you can also just measure the short circuit amps if you want to see if its putting out about what it should (just measure the output with an amp-meter and it should be around 0.3A per panel). Panel output amps is determined by the amount of sunlight on the panel and voltage is determined by temperature. Manufacturers use an artificial light source to simulate full sunlight at 80F to determine panel output so its unlikely you'll ever see a total of 5W from each panel in actual usage anyways.

Most battery manufacturers recommend a float voltage of about 13.3V for a lead acid battery. A controller will ensure you don't overcharge your batteries if you leave the panels on them all the time since your panel will put out 14.5V or higher. Even a 1.5A trickle charger will shut itself off when the battery is fully charged (well, a good one will).

The charging characteristic Chuck discussed isn't quite right. Battery manufacturers typically recommend charging at 1/10 of the amp hour rating of the battery, not 1/10 of 50% of the amp hour rating. But that's just splitting hairs, it doesn't affect your setup.

emiller has a windmill setup for his trailer and I'm building one too.
http://tnttt.com/viewto ... highlight=

Bruce

[Chuck Craven]

Hi Bruce!

The reason I put the 50% in there is for people first starting out experimenting with battery charging. It’s a 50% safety factor, so the battery doesn’t get over heated and warp a plate. Then they get upset their system was damaged. Once one has some experience charging batteries, they will quickly find out what the ratted charging current is for the battery they have. With a 20-watt solar panel I don’t think he will damage the battery unless some thing get shorted. Then it will be quite obvious what went wrong. Good point on the fuse should have suggested that! Thanks!


Chuck

[bdosborn]

I just measured my single VW panel output and got the following:

14.05 volts
0.24 amps

And its a beautiful sunny day. Thats about 3.4 watts. Not as good as I would have hoped but good enough. Of course I'm using the cheapy HF meter so who knows what I'm really getting.
Bruce

[Airspeed]

Do you have the older or newer model? what type of connector did it have ? Thanks! Aaron

[bdosborn]

Here's the one that I have:



Bruce