Solar Light: supplementing charging to battery

Discussion in 'General Electronics Chat' started by Jan Luthe, Feb 1, 2015.

  1. Jan Luthe

    Thread Starter Member

    Jan 10, 2015
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    In continuation of my threads “LED: current, voltage questions” and “Solar Light Circuit Board”, let’s see if I learned something:

    I have 10 solar lights and would like to supply extra power to supplement the charging from the solar panel so that the lights stay on longer.

    1) My power supply (WallWart) : 120VAC, 19 watt, output:12VDC, 800ma, 16.5VDC no load

    2) Each light is 7 feet apart for a total run of 70 feet.

    3) I would like to supply maximum 50ma to each light at a max of 1.3v. Since the power supply may be running at night I want to ensure that the inductor doesn’t get too high a voltage and in turn blow the LED. 10 lights at 50ma= 500ma which a safe load for the power supply.

    4) I want to run 18 or 20 gauge stranded wire from the power supply to the middle light (go 35feet each way) and hook up the lights in parallel. Any significant concern about voltage drop over 35 feet?

    5) Since the batteries may be full at some time I will assume a voltage output of 16.5 v and battery voltage of 1.3 volts. 16.5 – 1.3= 15.2v, 15.2/50ma=304 ohms, 15.2 X .05= 0.76 watt. Use 300 ohm 1 watt resistor installed on the positive end of the battery of each light (easiest place to solder)?

    6) Do I require a diode at the power supply to prevent ‘feedback/draining battery’ when the power supply is disconnected from the 120AC?

    Any other suggestions/comments?
     
  2. wayneh

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    I think you'll find that the solar panels will contribute very little to this arrangement. You might as well just use 12V landscape lights. No batteries to worry about.
     
  3. Jan Luthe

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    Jan 10, 2015
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    My solar post lights are already there. With this thread I am trying to do 2 things:
    1) Keep my lights on longer (possible all night) when I am having a party outside and
    2) See if I am understanding correctly what I have learned in this forum the past few weeks.
     
  4. wayneh

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    Sounds OK. Be aware though that driving 50mA into the battery may require more than 1.3V at the power supply. You typically control the current and the voltage will seek its level.

    Using a much lower DC supply voltage would be far more efficient, with more of the energy ending up as light instead of heat in the resistor.

    No concern, it should be fine as long as you solve the weathering issues.

    You'd want a resistor rated to at least double the expected heat, to keep it from burning up. But anyway this alone is not a good arrangement since it will keep forcing 50mA into the battery forever, regardless of state-of-charge, and this may be too much for the battery to tolerate. Are these nicads? Nicads can tolerate a higher trickle current than other chemistries although others are catching up. You might consider a 10-20mA target instead. This might be enough to light your LEDs while not being a risk to your batteries. Some experimentation with this might be useful. 50mA seems like a lot.

    You said the supply is DC, so it probably has a bridge rectifier in it already. However you will need a diode at each light, since you are effectively putting all the batteries in parallel. You need a diode at each to isolate them from the others.
     
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  5. Jan Luthe

    Thread Starter Member

    Jan 10, 2015
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    1) I am using this power supply because that is one that I have that has a relatively high current (800ma) output.
    2) I am suggesting 50mA because that is the current draw from the battery when the LED is on. There is an inductor on the CB to boost the voltage.
    3) The battery in each light is 900mAh, 1.2V, Ni-Mh. Please educate me on "forcing 50mA into the battery". I was thinking that once the battery is at 1.3V it would no longer take any mA from the power supply because the voltages are equal and there is no 'driving force'. Please advise.
    4) Does it matter if I don't use a diode at each light? When the power supply is disconnected from the 120VAC then all the batteries would be a 'bank' of batteries and all the lights would stay on for the same length of time. This would be advantageous for the lights that were in the shade and didn't receive a full charge from the sun.
    5) Is the 300 ohm the correct value?
     
  6. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    Your battery is the fuel tank
    When the tank is full, no more fuel can be put in it.

    The battery is drained at night, by the LED's, so solar supplementation will not lengthen the time lights are on.

    If you want the lights to stay on longer you must increase the capacity of your batteries, not the size of the charger.
    If you doubled the capacity (900mAh) and each light had two rechargeable batteries in parallel (1800mAh) the lights would
    stay on longer and your increased charging rate via solar would be a good thing to have. :)
     
  7. spinnaker

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    Oct 29, 2009
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    Maybe I lost it in the thread but I did not see the purpose of the walwart.

    What is it's purpose? Why is a solar light project using 120vac? Do you have an inverter??
     
  8. wayneh

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    That's fine. It's just inefficient.
    These must be fairly bright lights?
    But the voltages are not equal. The resistor only drops voltage in proportion to the current flowing through it, and that current will not drop below 50mA even if the battery is fully charged. This might cause the battery to explode. You need a charge regulator of some kind so that the current drops to a safe level.
    Well, it will work to some degree, and better if all the batteries are the same age to start off. Over time though, they will become unbalanced. Diodes are very cheap, by the way, and drop ~0.7V across themselves. This takes a little heat off the resistor.
    That would limit the current to 50mA, but that's not a good enough solution, in my opinion. 50mA into a 900mAh battery is a charging rate of 0.056C. You may want to visit Battery University and look up the manufacturer's charging instructions for your batteries.
     
  9. spinnaker

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    VERY inefficient especially if the wall wart is being used for charging (still not sure what the OP is doing with a 120 vac charger in a VDC application).

    Not only are you going to loose lots of energy due to heat that could otherwise be used to charge the battery. The wallwart is simply not designed to charge batteries. Charging batteries is a very complex process to do it properly.

    Get your a battery charger for the job. Something like on of these:
    http://www.solar-electric.com/sg-4.html
     
  10. spinnaker

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    You don't put a diode on the lights. You lights should already be diodes. You put it on the panel to keep the battery from trying to power the panel.
     
  11. Jan Luthe

    Thread Starter Member

    Jan 10, 2015
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    Spinnaker: The solar lights are 'complete' with circuit board with diode, inductor, rechargeable battery,panel, etc. The diode I was referring to was may be putting one between the battery and the wallwart DC power supply. Because the wallwart is converting AC to DC there is most likely a bridge rectifier that will prevent 'backfeeding' so I don't require the diode.
     
  12. Jan Luthe

    Thread Starter Member

    Jan 10, 2015
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    Spinnaker: What does "OP" stand for? It is a 120 VAC input with a 16.5 VDC output. I am trying to keep things simple and use what I have lying around. My 1.2 volt AA batteries have gone thru about 1000 cycles already so I am not excessively concerned about getting the charging cycle down pat. All I am trying to do is keep the lights on longer at night when I am outside (or turn them on at night when the battery(s) are 'empty). My main concern is that I don't 'blow/burn' all my 10 solar circuit boards. For the number of times I am going to use this and the mA consumed, efficiency is not high on my priority. Simplicity is a higher priority. Is my 'design' going to work and not blow all my circuit boards?
     
  13. spinnaker

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    OP is original poster.

    I am still confused why you are using an AC walwart in a solar application. You have solar panels. If you have AC available then why not not power the light direct off the walwart and safe the cost and complexity with batteries???
     
  14. Jan Luthe

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    Jan 10, 2015
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    WayneH: Looks like I am not interpreting something right and that is why I am here. To learn.
    "But the voltages are not equal. The resistor only drops voltage in proportion to the current flowing through it, and that current will not drop below 50mA even if the battery is fully charged."

    Are you saying that if I 'disconnect' right after the resistor so that no current flows thru, then the voltage on the end of the resistor will go up to the voltage that is at the beginning of the resistor?
    Spinnaker:

    Attached is picture of what I have. Details of circuit board and solar light are on my previous thread "Solar Light Circuit Board". Perhaps you can put a link on this thread for me as I don't know how to do that. What I am looking for is the optimum resistor/design for supplying DC to my lights to keep them on longer, particularly when the batteries were not fully charged because it was cloudy or the individual solar panels were covered with snow. I am suggesting supplying 50mA @ 1.3VDC because the solar light, when on at night, draws 50mA@ 1.25VDC (inductor converts this to about 20mA@3.2VDC for the LED). My main concern is that I want to minimize the chance that I blow my circuit boards and or LEDs.
     
  15. spinnaker

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    Sorry I am not going to look up your old thread for you. Why do you have two? Just copy and paste the URL. It is not that hard.

    I can't understand why you refuse to explain why you are using AC in a solar project.

    Adding a resistor is not going to keep your lights own longer. The only things that will do that area:

    1. Larger battery

    2. More efficient charging

    3. More efficient current limiting to your lights as in using a switching regulator.

    4. Dimming your lights using PWM.
     
  16. wayneh

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    Sep 9, 2010
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    That's exactly right. Try it with any resistor and your voltmeter. (If you don't have a multimeter, they are very cheap and sometimes free at Harbor Freight, if you have one of those nearby.) Put the resistor between a battery and your meter. With all resistors less than 1 meg-ohm or so, the meter will show nearly the same voltage as touching the battery itself. A small amount of current flows in the meter when you measure voltage, so at very high resistor value, that small current causes a voltage drop you can see. But with small ohms values, the voltage drop is very small. This is all about Ohm's law, by the way; V = I•R or ∆V = I•R where the voltage is measured across a resistor.

    One thing you might consider is adding more LEDs to each lamp. These would perform the same function as the resistor - dropping voltage - but they would produce light at the same time. The resistor produces only heat. You could add even 4-5 LEDs because each will drop about 3V. You'll still need a resistor, but it will be a lower value and will make much less heat. The LEDs would fill the role of blocking current between the lights, so you would not need a blocking diode anymore.
     
    Last edited: Feb 2, 2015
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  17. Jan Luthe

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    Jan 10, 2015
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    Spinnaker: Previous thread: http://forum.allaboutcircuits.com/threads/solar-light-circuit-board.106073/

    The only thing AC is what is powering the wallWart. What is going to the solar lights is DC from the wallwart. I was thinking that when the lights go out at night because the batteries are dead/empty I would plug in the wallwart into the wall. Then the output of the wallwart (VDC) would power the lights and they would turn on again. The reason I started a new thread because I thought the prevoius one was getting a bit long.
     
  18. wayneh

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    You could use a light-activated outlet to turn the wall wart on only at night. They sell them for Christmas lights. I think some will even turn on at dusk for a preset length of time, say 5 hours instead of all night, so that the lights are off during the middle of the night.
     
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  19. spinnaker

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    I understand that DC is powering the lights. That is a given. Why are you using a wallwart when you have solar panels??????? There is no need for a solar panel at all then. And no need for a battery unless you wanted lights during a power failure and with those lights it should be no big deal if they were not lite during infrequent power failures.

    If you need both wallwart and solar panels to make this all work then this is an awfully poor design.
     
  20. spinnaker

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    IMG_1818.JPG IMG_1821.JPG IMG_1824.JPG IMG_1825.JPG Here is a solar light project I designed. No walwart and it has been working fine for a couple of years now.,
     
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