Current booster for small solar charger

Discussion in 'General Electronics Chat' started by bisctboy, Nov 3, 2008.

  1. bisctboy

    Thread Starter Member

    Nov 3, 2008
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    I have 2 solar cells connected in series to give me an ouput of 7.2v (3.6v X 2) and a current of 70 ma. I am building a solar charger for various handheld devices like my PDA, ipod, and cel phone. These require 5v to my USB connector to charge them which I get by putting in a 5v regulator. Is there anyway I can increase the ma's without adding solar cells in parallel? In other words, can I convert the excess volts I have into increased current via circuitry?

    Thanks,
     
  2. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Yes, you can. Power equals V X I , so as long as expect power out is less than power in (free energy guys will tell you this is rubbish), you can convert your voltage or current to whatever you like.

    You have 7.2 X 0.07A = 0.5W of input energy. You may use a buck converter to create 5V @ (0.5W/5V) 0.1A if it were 100% efficient. You can expect about 80-90% efficiency from such a converter, so realistically about 0.08-0.09A.

    Please keep in mind that, for example, Ipod batteries have about (3.7 X 1.2Ah) 4.44W/h of energy. In order to replace this energy, you're looking at 4.44/(0.9X0.5W) = roughly 10h of charging. This is assuming a 100% efficient charging process, which is not possible because they are clearly stepping down the 5V to 3.7V through another buck converter.

    Steve
     
  3. bisctboy

    Thread Starter Member

    Nov 3, 2008
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    Is a buck converter something I buy or can I make it?

    Riddle me this....I built a solar charger that powered the ipod while connected but it didn't seem to charge it? The solar cells are connected to two 3.6v li-ion 800 ma batteries in series. The batteries are connected to a 5v regulator. The regulator is connected to a female type A USB. When I connect the ipod to the USB connector, the "almost dead" ipod indicates it is charging. However, when I unplug the ipod from the charger after a couple of hours, the battery indicator has not moved up at all? In fact, my kid used the ipod connected to the charger yesterday for a 2 1/2 hour car ride while plugged into the charger and it never went dead. But when he unplugged it, the battery indicator showed it was "almost dead"?

    Thanks,
     
  4. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
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    It is honestly barely worth all the trouble of building, since you will not be losing much energy if you used a linear regulator instead. Basically, the linear regulator will throw away the voltage that you aren't using. There are some issues, since your voltage difference is 7.2-5 = 2.2V. Linear regulators have a dropout voltage, which is by definition the minimum Vin minus the output voltage. You should search for a low-dropout regulator that puts out 5V and can supply at least 250mA.

    Steve
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    Did you measure the current now from your solar panels in winter?
    Unless you are in Cuba or Mexico then I bet the current is much less than the rating.

    I live in the south of Canada and my solar garden lights last only 1 hour now after charging all day in the sun but in summer they glow all night long.
     
  6. bisctboy

    Thread Starter Member

    Nov 3, 2008
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    The solar panel puts out about 140 ma of current in full sun light. This being said, I built the solar charger inside of a plastic craft case that is hinged. The lack of charging the ipod I wrote about earlier occured when the case was closed and the ipod was running off the batteries in the charger. Why will the ipod run for hours but not charge during that same time? I have also plugged the ipod in the charger and left it alone without playing it. I still get the same result of not charging after a couple of hours?

    Thanks,
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    You said that you tried to charge two 3.6V batteries in series (they charge to 8.4V) from your solar panels. They drive a 5V regulator that powered the ipod but didn't charge its battery. Then the solar panels/battery/regulator does not produce enough power to power the ipod plus charge its battery.
    I think your solar panels do not fully charge the battery to 8.4V so its power is too low to charge the battery in the ipod.
    Maybe the USB connector needs resistors on its data pins for the charging circuit in the ipod to function.

    Which 5V regulator? A little 78L05 has an output current of only 100mA.
     
  8. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
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    Or maybe the fuel gauge isn't activated some how?

    Steve
     
  9. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Several months ago, I did some experiments using a 2W solar cell using indoor lighting using several values of carbon composition resistors as a load:
    50, 100, 200, 500, 1K Ohms.

    I was somewhat surprised to discover that for a given amount of light, the current through the load did not change, independent of the resistance of the load!

    This lead me to conclude that a solar panel acts as a constant current source, the current depending upon the amount of solar energy that the cell is exposed to.

    Due to these observations, I suggest that attempting to use a voltage regulator in conjunction with a solar cell will be fruitless. The voltage regulator itself, such as an LM317, has fixed current load requirements in order to maintain guaranteed regulation. Some of this current is wasted in the resistor network that sets the output voltage. Additionally, linear voltage regulators have a fixed "dropout" voltage; ie: minimum differential between the input and output voltage. Fixed 7805 regulators have a dropout of 2v, LM317's have a 1.7v dropout.

    I suggest that there are two basic ways one might charge a battery using a solar cell;
    1) Use a voltage clamp circuit to limit the maximum battery voltage.
    2) Monitor the battery voltage; once the limit is reached, disconnect the battery from the charge source. The latter might be accomplished by using a voltage comparator and a MOSFET.
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    The solar panels charged a 2-cells lithium battery. If the voltage became more than 8.4V then the lithium battery might have caught on fire!

    A lithium battery cell is charged with constant current until its voltage reaches 4.2V when it is about 70% fully charged. Then the charger is supposed to switch to constant voltage and a comparator detects when the charging current is reduced to about 3% of the battery's rated output current then the charger shuts off.
     
  11. nyarlathotep

    New Member

    Jul 15, 2009
    3
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    from my research i have found that a LM2937 is a better regulator to use (only .5A though)
     
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