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Wannabe Geek


Topic # 172012 8-May-2015 12:04
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Hi guys

I have installed 9 solar led lights down my drive after a few days i realized that they were not staying on long enough so that i have done is installed a timer and transformer and wired them up in series the transformer is 12vdc so 12/9=1.333 volts witch should be fine to run the led lights. this has worked fine but has stuffed the batterys so i installed a Capacitor-Electrolytic-1000uF-16V-105C-L-ESR into every light to replace the battery. doing this has not worked so i am now at a loss where to go from here. I was wanting to disconnect the solar panels and just use the timer to turn the leds on and off. any help would be great     

here is a pic of the led light:
Click to see full size


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  Reply # 1300874 8-May-2015 17:06
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You need a resistor across each light's battery terminals to form an equal voltage divider for each light.  Without a DC resistance path, there is no way to even hope equal voltage across each.  Not sure about a value right now, it's been a long week, but suspect around 100 Ohm or maybe even less.  It will waste a small amount of power, but nothing to worry about.

I would rather run a 1.2V voltage regulator to power them all in parallel.




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  Reply # 1300892 8-May-2015 17:44
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Are you wanting to run the lights off the transformer or to charge the batteries from it and let the solar cell still turn them off and on overnight?




Richard rich.ms

 
 
 
 




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Wannabe Geek


  Reply # 1301060 9-May-2015 09:06
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Hi Richard 

Yes i want to run the lights off the transformer and not use the battery's or the solar panels at all.

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  Reply # 1301190 9-May-2015 15:35
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I would clip all the components off the board except the LEDs and then just put an appropriate resistor from the 12v supply for how much current you want thru the LEDs and then parallel all the lamps up to the 12v cable.

Not sure that you would be able to keep the inbuilt 1.2 to whatever stepup circtuit working on a series supply easily.




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  Reply # 1301193 9-May-2015 15:51
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Those solar garden lights actually have a small switchmode step up power supply in them. (that green component that looks like a resistor will actually be a coil) As even elcheapo LEDs need at least 2V before they start working. And white ones need around 3V for them to work well. (Just from looking at the LEDs that Jaycar sells) Which means that switchmode supply could be producing as much as 18V assuming the LEDs inside that light are wired in series.


Either you need to remove those switchmode supplies. See how those LEDs in each light are wired. (Series , Parallel, Or a combination of series and parallel) So you can see how much voltage is needed for each light. Then you can calculate what value dropping resistor will be needed to run the LEDs in each light.

But Niels idea will be easier for you to do. Just leave the super capacitors in each light (so the switchmode supplies will work properly) Then cut the wires that go between the solar panel and the curcuit board in each light. And to make a 1.2V voltage regulator, just buy an LM317 voltage regulator (Jaycar sell them) And wire the adjust pin to ground. The other pins are self explanatory. Only problem is that regulator will get very hot running it off 12V. So if you have access to a 5V power supply use that instead. And you will still need to attach the regulator to a heatsink.






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  Reply # 1301224 9-May-2015 17:55
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Nah, they are not in series. I have some of a similar lamp and they have pads for 5 LEDs in it but mine came with only 2 populated. All paralled up.




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  Reply # 1301247 9-May-2015 18:50
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White LEDs are rated approximately 3V each (thereabouts) at rated current (just like red is 1.6V and green is 1.8V).  They will work from about 2V as someone said.

A 1.25V regulator will not get very hot even at 12V, but 5V is better.  The dissipation will be the voltage drop (e.g. 12V - 1.25V = 10.75V) multiplied by the current draw (probably around 25mA, guess) equals less than 300mW.  Multiply this by the number of fittings, and it can add up quickly and then require a heatsink.  Running off 5V will dissipate about 100mW per fitting (based on the guessed 25mA current draw), so a third the heat compared to running off 12V.

Using an LM317T voltage regulator is very easy, and it is hard to damage.




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  Reply # 1301248 9-May-2015 18:52
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I think I've found the datasheet for that solar light controller:
www.mikrocontroller.net/attachment/158139/QX5252.pdf




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  Reply # 1301267 9-May-2015 20:02
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yes that is the chip 5252F

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  Reply # 1301302 9-May-2015 22:03
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Looks like your inductor is 47uH so the total LED current is 75mA (from datasheet) which is 12.5mA per LED.  With a half decent current like that, the forward voltage will be about 3.3V so about 1/4W.  Adding losses, it will use about 300mW per fitting which is 240mA at 1.25V.  Given that the batteries are 1Ah, if they are fully charged (i.e. full sunlight most of the day), then it will last 4h.

So with 240mA, if you use a 5V power supply and LM317T voltage regulator, then you will probably need a small heatsink for the regulator as it would dissipate 0.9W per fitting.

If you use an LM317T as a current regulator (IC plus 2 resistors, set to 75mA) off a 5V power supply, connect the output to the the 6x parallel LEDs (disconnect inductor), then the (current) regulator will dissipate only 125mW and need no heatsink at all.  I would do this solution, a 5V power supply with a current regulator inside each fitting.  This will run 13 fittings off a 1A supply.




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