DescentBB

i have two solar panels that are 3v each and 100mA,

im building a solar powered airplane and i want to use just those two panels to fully charge this 8.4v 600mA battery that powers everything on the plane.

so im wondering if anyone knows how to solve this circuit problem im having. its not going to be charging while in flight of course.

or is this setup impossible with only those two panels

You need to figure out how many milliamps the plane will draw while flying. It's possible the thing will work with those panels, though you need to try it and see. Just because the batteries are 800ma doesn't mean it draws that much constantly.

To charge the battery, you need to supply current. This can be a low amount or a high amount but there needs to be current flowing into the battery.

This requires the charger (solar panels + additional circuitry) to apply more voltage to the battery than the batteries voltage.

See, the battery has a voltage that it is charged to. This acts very loosely like a reverse biased diode. You need to overcome this drop to allow current to flow into and charge the battery.

After you can supply enough voltage to overcome the batteries voltage, it's a matter of keeping the current under the max charging current but higher than a trickle which would simply keep the battery at it's current voltage.

Solar cells rating is typically in bright sunlight (ideal conditions) and their power output will not be as good in overcast or cloudy skys.

Not knowing the specifics of solar cells, I'd guess that the voltage output will remain near 3V in all but the lowest light conditions, but that the current will diminish rapidly. In that case, I'd recommmend using 3 or 4 panels to provide the required voltage.

Those 2 panels will only put out 6V if wired in series which is not enough to put current back into the battery. Using 3 would be cutting it really close and probably only a trickle charge at best.

Now, you could build a voltage multiplier (charge pump) and use just one cell but those do have efficiencies less that 100%, so your current of 100mA would be cut by 10% or more.

Also, if the battery is NiCd then you need to completely discharge it or keep it very near full charge all the time or you will experience memory effects where each cell remembers it's previous charge and thereby reduces it's capacity.

A 600mA/Hr battery should take 6 hours to charge at 100mA rate however due to internal resistance losses it will more than likely take 8 hours or more at that rate.

Hope this helps.

SS is 100% correct; you would either need 4 panels or a voltage multiplier. You need to get that charge voltage higher then the batteries unloaded voltage to push the current in.
Just like how ur 12v car battery charges at 13-15v.

What type of battery is it tho? NiMH? LiPolly?

with a voltage multiplier, the current would sink proportionally, because otherwise you would get more energy out of the system than you put in.
if this was possible, everyone would use a 0,5V mini solar cell and up the voltage for his 5m wingspan hotliner

Good point Floyd, forgot to put that in my post

If you have 1 cell providing 100mA and you quadruple the voltage, you would cut the current to 25mA (given 100% efficiency). More typical would be about 20mA or less.

Basically... Power = Voltage X Current.

So your solar cells are rated 3V at 100mA which is .3W

Substitute .3W for Power and you get...

.3W = Voltage X Current

So if your voltage is quadrupled to 12V, you can see...

.3W = 12V X Current

Current = .3W/12V = 25mA

I would stick with a slow charge rate of 1/10 C or;
if your battery was 1 amp hour 100 ma.

So a 600 Ma battery (6/10 th of an amp) should be slow charged at 60 ma. 100 ma. will produce extra heat which kills the batteries.

A quick charge at .. ONLY IF THE BATTERY IS CAPABLE, and for 1- 1.5 hours max is at .5C - 1C

Also don't forget the solar cell system needs a diode or two to block current going through them... deduct 1.5 V per diode from your supply voltage.

http://www.angelfire.com/electronic/hayles/charge1.html Awsome info.

its a NiMH battery.

im not willing to burn 50 bucks on another solar panel nor add more weight.

so how what kind of voltage amp would i need? (I dont really care how long it takes to charge)

and one more thing, should i put in a switch so that the juice in the battery doesnt flow back into the solar panels when the battery's in use?

That \"switch\" is a diode. And my crystal ball was at the shop so I assumed you had Nicad.

That charger thingy thing is what you need but you need 22volts dc to run that particlar do-dad...
That means doubling the amount of solar cells and no soup for you.

Kromedit: Not sure why the URL doesn't work here either, probably doesn't like commas or something, so for now:
http://tinyurl.com/eepjc

Umm I'm not sure why BBCode didn't work there so just copy and paste the link and take a boo.

Oh and you'd want about 10 volts for charging your pack so 12 Volts - 1.5 for the diode thingy is close enough to the max input for charging your pack.

And before you say anything, you won't have to worry about backflow with that diode in place. Just make sure you install it in the right direction. They're called one-way for a reason. Put them in backwards and you get zilch. (Well, theres one that you CAN plug in backwards, but thats another topic.) Oh, you'll need atleast two diodes to get close to that 1.5V drop. Silicon diodes only drop 0.6V - 0.7V on average. (Germanium diodes only drop 0.3V on average.)

http://encyclobeamia.solarbotics.net/ar ... diode.html
http://en.wikipedia.org/wiki/Diode

In electronics, a diode is a component that restricts the direction of movement of charge carriers. It allows an electric current to flow in one direction, but essentially blocks it in the opposite direction. Thus the diode can be thought of as an electronic version of a check valve. Circuits that require current flowing in only one direction will typically consist of one or more diodes in the circuit design.

Diodes are awesome critters to have handy. Radio Shack sells them all day long. Connect two of them in series (one after the other) and you should get approximately a 1.4V drop. Maybe a little more, maybe a little less. It all depends on how well the diode is made. A pair of 1N4002 should do you fine.

Found a circuit that just might work for your application.

http://www.maxim-ic.com/images/appnotes ... 7Fig1C.gif
(*Note...there is an error in the schematic that rates the inductors as 330uF and 22uF, this should be 330uH and 22uH)

This circuit generates 5V, however, the MAX1771 is capable of producing up to 12V by changing the feedback voltage (the pin labeled FB). Data sheet for the MAX1771...

http://pdfserv.maxim-ic.com/en/ds/MAX1771.pdf

The webpage for the chip is here...

http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1030

The support chip, the MAX866 is here...

http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1210

You can probably (as in 99.99% chance) get sample chips for free from Maxim, just make sure you get through hole instead of surface mount unless you really know your stuff electronically and can etch PCB boards.

The inductors shown in the schematic can be made by you with some magnet wire available from Radio Shack...

Kromedit: Massive radioshack URL fixed

To calculate how many turns and how large, this is a good start...

http://www.daycounter.com/Calculators/A ... ator.phtml

The other parts should be easy to find and could be substituted with equivalents.

Good luck and have fun!

Forgot to mention...

The circuit I showed you is just to get the output voltage to where you need it, you'll still need to limit the current. I'll dig something up or you can search for battery charging circuits or current limiting on google.

thanks ss, i appreaciate it

Ahh this is the best one yet

http://www.rowan.sensation.net.au/elect ... tepup.html

Actually screw the electronics double your solar cell input from 6V 100 ma. to 12V 100 ma.,

and yes its ~ .5-.8 or so volts per diode drop not 1.5 so doubling your cells, (diode per bank) gives you a drop of 1.5V on average, 12V - 1.5V = 10.5V peak, but thats only in full sunlight.

On average in actuall use I'd bet it is more like 10V or so. Oh and technically it should be a charge current of 60 ma. for that battery pack for an 8 hour charge, so if you want the worlds simplest regulator buy a 12V 60 ma. light bulb and place it in series in the charge circuit.

That simple setup should work tickity boo unless you live in a cave or the mines

Here's a very simple current regulator using a easily attainable voltage regulator device...
http://www.techlib.com/electronics/regulators.html

Thanks for fixing the Massive Radio Shack Link Krom. What was wrong with it? Seemed to work fine for me.

It was breaking the margins so everyone had to scroll left right to see the thread.

i gave up on the step up thing. instead i got a 15.4v solar panel at 100mA. just wondering if this is ok to charge the 8.4v 600mA battery i have.

i appreciate all the help yall gave

Your charge Voltage should be around 10 volts, and at 60 ma for a trickle charge.

You can use a three terminal regulator to supply a steady 10 V to your circuit.

http://www.fairchildsemi.com/pf/LM/LM7810.html

As for the extra 40 ma, if you want you can make a constant current regulator, or use a 12v 60 ma light bulb in series with your load.

Personally I'd use the bulb... it will show when current is high through the circuit by lighting brightly. And if there is a dead short... the bulb will light up and absorb the current or plain blow out if its a real funky power short.