Solar panel maximum charge rate
metron9
Posts: 1,100
The solar panel spark fun sells for 11 bucks, How much battery storage is it capable of charging on a typical sunny summer day?
At 4.5V and 100ma short circuit does this mean you could charge a 4.5V battery with 100ma in one hour? and fill a 600ma battery in 6 hours?
or is the math not that simple?
Now suppose we have a 1.5V battery pack, is this a bad combination because you will have to waste the excess voltage or can you use the extra power to charge a 1.5V battery faster?
I suspect you could charge 3 * 1.5V batteries in series and then use the power in parallel.
Must be some good links on the subject but I thought i would ask here as many of you have perhaps gone through these steps in your own experiments.
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Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!
At 4.5V and 100ma short circuit does this mean you could charge a 4.5V battery with 100ma in one hour? and fill a 600ma battery in 6 hours?
or is the math not that simple?
Now suppose we have a 1.5V battery pack, is this a bad combination because you will have to waste the excess voltage or can you use the extra power to charge a 1.5V battery faster?
I suspect you could charge 3 * 1.5V batteries in series and then use the power in parallel.
Must be some good links on the subject but I thought i would ask here as many of you have perhaps gone through these steps in your own experiments.
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Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!
Comments
Take the voltage produced by the PVA and multiply it by the maximum amperes or milli-amperes produced. That will give you an idea of the maximim VA (or wattage if you will) it can produce. I usually use 75-80% of that at first, just for calculation purposes and call that new figure predicted power. Cloudy days, low sun angle, dirty surfaces, and various other inefficiences WILL come into play, so the final answer may have to be obtained empirically. This is just a rough estimate.
Now take the total rated AH (ampere hours) of the battery in question and divide it by the "predicted power" figure from above. That will give you the TIME required to charge that particular battery. That proprtion of sunny, daylight hours is the percentage of charge as well.
Somewhat needless to say, make sure the voltage produced by the PVA is in EXCESS of that required for the battery. Also remember to place a reverse protection diode in the charging circuit, so you don't turn the PVA into a TOASTER at sundown!
Regards,
Bruce Bates
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So lets take one of the batteries on Spark Fun as an example.
Solar Cell:
www.sparkfun.com/commerce/product_info.php?products_id=7845
Battery:
www.sparkfun.com/commerce/product_info.php?products_id=335
Voltage times maximum amperes:
4.5 X .01 = 0.45
mAH divided by predicted power:
2700 / .045 = 60,000.
60, 000 what? Seconds? If all variables exist in a perfect world, then thats 16.6 hours.
Actually I think that should be .1 (one tenth) making a result of 60.
Again, 60 what?
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- - - PLJack - - -
Perfection in design is not achieved when there is nothing left to add.
It is achieved when there is nothing left to take away.
4.5 * .100 = .45 '4.5v * .1 amps (100ma)
.45 * .75 = .3375 'safety margin mentioned above
2.7 /.3375 = 8 '2.7amp hours (2700mah) which gives 8 HOUR charging time which seems about right
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When the going gets weird, the weird turn pro. -- HST
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Paul Baker
Propeller Applications Engineer
Parallax, Inc.
Post Edited (Paul Baker (Parallax)) : 12/8/2006 2:09:03 AM GMT
Bruce,
Are you saying that:
In stead of shining light on the PVA and charging a battery during the day, and getting light by discarching the battery into the PVA at night...
We would be able to shine light on the PVA to charge the batteries, but only be able to make s'mores from the heat at night?
(Light In => Electricity Out: should mean Electricity In => Light Out; not Electricity In => Heat Out)
Sorry - Couldn't resist.
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John R.
8 + 8 = 10
At 4.5V and 100ma short circuit does this mean you could charge a 4.5V battery with 100ma in one hour? and fill a 600ma battery in·6·hours?
This is more like 8 t0 10 hours at Full Sun Light The Hole Time
This depends on what mil-amp-hr battery that you are going to use the more the mil-amp-hrs the battery the more time you need
or put more that one panel together to get more mil-amps
One thing i can tell you is that if you are only going to charge one sell with solar panel i would use a LM317 reg set·the·to current 100 milamps
that way you drop the voltage so low it will take longer to charge
I know this because i have play with·them before when talk about [url=mailto:4.5@100ma]4.5@100ma[/url] short circuit you will only get about 1 volt· at the most maybe
And this is at Full Sun Light And a Clean Panel and New when they get older they do not put out as much power
I have a lot Solar Lights in my yard and some time i have to repair them
This is not a bad price for those panels
if you are going to have more than one cell i would tell to at least have 3.0 volts above the battery pack that you are going to use
I·would also use a diode after the LM317 reg so that the battery dose not drain back throught the reg and or the·solar panel
If you need any more help in matter please let me know
I hope this helps you
Sam
Post Edited (sam_sam_sam) : 12/8/2006 2:16:24 AM GMT
Batteries are chemical devices and excessive charge rates do physical harm to the useful life of the battery.
Both overheating and outgassing may occur.
Try to keep the battery near a full charge and merely trickle charge to top off.
The optimal balance is to work back from the the AH capacity of your battery group and factor it by the optimal charge rate to retain useful life. If you have too much sun, you can always add more batteries and have a deeper reserve. I suspect the best approach is a trickle charge of a Gel cell chemistry with large reserves compared to minimal use [noparse][[/noparse]even going into sleep modes or shut downs on a frequent basis].
Trying to make a solar powered vehicle or solar powered iPod is likely to struggle with intermitant functionality. The duty cycle is a big factor and may demand more panels with a lot more batteries.
NiCads seem to prefer a charge rate of about 1/30 of the AH capacity, but can get by with much lower. Ocassional quick charges at 1/10 are okay. Take a look at The Art of Electronics for discussion.
Almost all batteries don't like to go below 50% charge and quick charging from such lows may destroy the battery. Lithium batteries are certainly this way and likely to be the worst for solar energy [noparse][[/noparse]though that can charge faster].
Also, the solar panel will require a blocking diode to avoid discharging the battery when not charging. So, some overvoltage is needed for the diode and batteries generally are tolerant of some excessive voltage.
Putting in a regulator can reduce your output capactity by 30% - a huge waste. So I would try to put in another diode on the supply side to have the battery provide operational output in roughly 5.5v to 3.0v range from a solar cell that may be puting out 7.0v to charge a 6.0v battery. The additonal diode will also have the micoprocessor do a low voltage shutdown while the battery is above the 50% discharge. So the whole thing will self-correct when the sun comes out again. Batteries are not noisy, so that isn't a regulation concern and the swings are somewhat mild, even with solar charging.
Try to understand what is happening electro-chemically if you don't want to be changing batteries often. After all, that defeats the whole purpose of the solar panel as a charger.
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"If you want more fiber, eat the package.· Not enough?· Eat the manual."········
I had a guy ask me how he could get more light out of one of these garden lights like Menards sells for 45 bucks aof 10 units. I pulled one apart, it uses a charge pump and one 1.5V 600mah battery. Open circuit on the charge pump is about 6V, pulse width is 30% or so at (if I remember ) 150HZ or so. The LED draws about 7.5 mA when on.
What I did is replace the LED with a 2n7000 Mosfet. I useed a 5V battery source and no resistors with a luxeon 1W led. The pulse width modulated signal from the circuit drives the fet and the led draws about 35mA. The light output (attached) is very nice. The solar panel and charging of the battery as well as turning on and off by daylight is still don with the original circuit.
I ordered one of those car battery solar panels, 400ma at 12V charge rate, and a small lead acid battery , 3inches x 2 inches size.
The person who asked me this has a larger yard device that he is using this garden lamp in, and has room for additional solar charg panels and batteries. I thought a complete package that has the charging circuit in it would be a better way to go instead of building a charge circuit.
I ordered the 850260 solar battery charger from http://batterycountry.helpdeskconnect.com
Trying 12V supply at the PWM through the fet I get about 100ma average through the LED. (the luxeons I am using can go 250mA)
See BIG pictures at www.metron9.com/led/led.html
1. Standard garden light led at 7.5 mA
2. Luxeon at 35mA
3 and 4 from a greater distance but same as 1, and 2.
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Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!
You have good point about this
Try to keep the battery near a full charge and merely trickle charge to top off.
The optimal balance is to work back from the the AH capacity of your battery group and factor it by the optimal charge rate to retain useful life. If you have too much sun, you can always add more batteries and have a deeper reserve.
This something to keep in mind
I use the regulation when i have used larger solar panels than that
Thanks for reminding about his part
Putting in a regulator can reduce your output capactity by 30% - a huge waste. So I would try to put in another diode on the supply side to have the battery provide operational output in roughly 5.5v to 3.0v range from a solar cell that may be puting out 7.0v to charge a 6.0v battery. The additonal diode will also have the micoprocessor do a low voltage shutdown while the battery is above the 50% discharge. So the whole thing will self-correct when the sun comes out again. Batteries are not noisy, so that isn't a regulation concern and the swings are somewhat mild, even with solar charging.
Sam
Suppose you use 6 - 1.2V rechargables in series for a 7.2 capacity
If you used a 7.5V solar panel, -.6 diode drop (6.9V charge max) would you get automatic trickel charging at the 6.9V level and not have to worry about overcharging?
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Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!
I thinik the issue you're missing here is charge rate. manufacturers specify for each rechargable battery they produce one or two rates of charge that they recommend. One is for fast charging and the other is for normal charging. Both are generally based on a fixed voltage.
The charge rateis specfied in mA hours per period, such as:
Fast charge = 2000 mA for 5
Slow charge = 500 mA for 20 hours
There are generally heat concerns with fast charging where there are generally noewith slow charging.
I suspect by "trickle charge" you mean tapered charging or top-up charging. From my own point of view, since you have not limited the charging current in any way, you do not have a proper taper or top-up methodology.
Regards,
Bruce Bates
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Assume the solar panel charger can only deliver a slow charge of X mA.
Typically to get a full charge the charger puts out a larger voltage than the battery can hold and circuit is needed to stop charging as well as the rate of charge circuit.
The question really is, if the battery full charge capacity is greater (7.2V) and the charger can only deliver 6.9V, when the battery gets to 6.9V current stops flowing and there is no need for a circuit to disconnect the charger. correct?
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Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!
I connected a 1 Farad Capacitor and plotted the charge time (attached) About 3 seconds to a full Farad at 2V
A Farad as I recall can deliver 1 AMP for 1 Second.
Trying some math here
If the capacitor can hold 1000mA and charges in 3 seconds, is that not 333mA per second charge rate?
So 1 Farad is 1000mA draw that down to 0 over 1 hour (3600 seconds) you have 0.277 mA draw per second
A 600mA battery can deliver 600mA for 3600 seconds at 600mA per second so 600/.277 so the battery is 2,166 Farads
2,166 farads times 3 seconds = 6498 seconds is 1.8 hours
I have to go to a concert so I will be back to ponder these numbers. Have a go at it if you wish but I am trying to understand how 60ma measured current is perhaps 333mA given the right circuit unless my math is wrong ...
Picure is 1 Farad capacitor charge at 59mA measured, what does the scope calculation for this plot come too? 59mA or more?
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Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!