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USB Solar Charger Current Question

Discussion in 'General Electronics Discussion' started by ajgiampa, Jul 25, 2011.

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  1. ajgiampa

    ajgiampa

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    Jul 19, 2011
    Hi everyone,

    I recently built a modified version of the USB solar charger shown on the DigiKey website. Here's the schematic:

    [​IMG]

    Since I'm using (2) 6V 450mA solar panels in parallel, I modified the circuit to use a 5V DC/DC converter rated to 1000mA instead of the 500ma DC/DC converter shown in the parts list and circuit diagram. The intent was to build a charger that would charge at a faster rate, much like the USB chargers that plug into an AC outlet from which I measure up to 900mA when charging my HTC-brand cell phone.

    The problem I'm running into is that I only get around 420mA from the solar charger in full sun. To test the circuit, I removed the solar panels and attached a 6V 1000mA power supply, expecting to draw near 1000mA during charging. But instead, I again only drew around 420mA from the circuit. This leads me to believe there's something in the circuit that is preventing more current from passing.

    Any ideas about what could be causing this would be greatly appreciated.

    Thanks...
     
  2. MagicMatt

    MagicMatt

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    Jun 15, 2011
    Presumably you have checked the obvious, and tested how much current the unit you are trying to charge will draw if connected directly to a power supply?
     
  3. Laplace

    Laplace

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    Apr 4, 2010
    Looking at the datasheet for the V7805-1000 it shows a specified input range of 6.5~32 Vdc and output current of 1000 mA. However, that 1000 mA is not a hard limit since the performance charts are shown for output current over 1000 mA. The more likely source of your problem is the minimum input voltage of 6.5 Vdc. What is the output voltage of your 6 volt solar array, especially under full current load?
     
    Last edited: Jul 26, 2011
  4. ajgiampa

    ajgiampa

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    Jul 19, 2011
    Yes, when connected directly to a source such as a USB port or AC wall adapter, the phone draws around 900mA.
     
  5. ajgiampa

    ajgiampa

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    Jul 19, 2011
    I'll check the full load voltage of the arrays on the next sunny day I get. But to simulate a greater than 6.5V source, I re-tested the circuit with a 12V 1250mA power supply and got the exact same current draw through the circuit when charging the cell phone....around 420mA.
     
  6. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Another possibility is related to the MBR150 diode. USB power voltage is specified as between 4.75 and 5.25 volts. However, the forward voltage drop of the diode would be about 0.5 volt @ 500 mA. So the solar charger may only be providing 4.5 volts while the AC adapter may be providing 5.25 volts. That difference in output voltage could account for the difference in charging current. Something to check.
     
  7. MagicMatt

    MagicMatt

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    Jun 15, 2011
    Surely that proves the issue is on the output side. Given there's only two components after output, it's got to be something simple (assuming the DC/DC itself is ok).

    Check voltages at each point. I'm betting Laplace is right - my MP3 player charges VERY slowly at <5V input, and at <4.5V input it stops and refuses to charge altogether (presumably to protect the Li-Ion cell).
     
  8. ajgiampa

    ajgiampa

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    Jul 19, 2011
    I checked the voltages at each point in the circuit and you are correct that the voltage drop from the diode is about 0.4 volts - thus the device is only getting ~4.6V through the USB cable.

    I also confirmed that the voltage output from the AC chargers I've been testing with is 5V through the USB cable to the device.

    Could a solution to this be getting a DC/DC converter that outputs ~5.5V at 1000mA? Do these even exist?

    Thanks again for all your help...
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
    2,820
    Jan 21, 2010
    At least part of the solution will be a MUCH larger input capacitor, bypassed by several smaller ones to reduce the ESR.

    The problem with the solar panel is that it is almost a constant current supply under load, and because the DC-DC converter won't be MPPT, it may well be operating the panel at a point where it cannot supply the full rated output.

    edit: This answer relates to the suggestion about using a DC-DC converter, not the circuit in the top post.
     
    Last edited: Aug 6, 2011
  10. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    The 7805 is a linear series regulator. Have you thought about using a shunt regulator instead? If the open circuit voltage of the solar array is 6.0 volts, try using a silicon power diode in series to drop 0.7 volt followed by a 5.1 volt zener power diode to shunt the current to ground. Then the load will be able to pull off all the current at a 5.0 volt load voltage. Sure, you will be "wasting" the current if it is not going to the load, but what were you going to do with that unused current anyway? But you will need a good I-V curve for the solar array in order to make good decisions.
     
  11. ajgiampa

    ajgiampa

    6
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    Jul 19, 2011
    Hi everyone,

    I'm pretty sure I got this working. The device now charges at ~5V and draws nearly 840mA from both a solar array and a 6V power supply when used as inputs during testing. Here's what I did:

    I replaced the V7805-1000 voltage regulator with an adjustable LM317T regulator. Using a 470 ohm and a 1500 ohm resistor with this regulator, I'm able to get an output voltage of ~5.4 Volts (measured). This increases the USB output voltage (after the diode) to slightly more than the 5V needed for the device to draw the maximum amount of current available.

    Thanks for pointing out that the forward voltage drop of the diode was reducing the voltage seen by the device - I wouldn't have realized that. For some reason the device being charged didn't want to draw full current without the full 5V through the USB cable. But now that I'm getting the full 5V, the device charges at my desired rate.

    I'll do more thorough testing in the coming (sunny) days to see if anything gets fried, but so far everything seems to work.

    Thanks again!
     
  12. whilburn

    whilburn

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    Oct 25, 2011
    Using Ajgiampa's situation, i have a question (directed to anyone) regarding his equipment.
    He has two solar arrays that, in parallel, should output aprox. 6 volts and around 800 milliamps. That equals about 4.8 Watts on the input side. Lets figure about 50% efficiency to the output. That should be 2.4 watts to the output.
    Output 5 volts, at 2.4 Watts gives 480 milliamps on the output.

    Assuming efficiency is truly 50%, Is my math and expectations for his circuit correct?
    I started another thread with different questions but still concern a solar power supplying power to a DC-DC converter.
    FYI, i have been doing some research and, as far as switching regulators go, i saw a 7805 drop in replacement on a Y0utube video with the name "usb Solar Power". The author posted the following link
    Mouser Part #:580-OKI78SR5/1.5W36C - $3.93
    Datasheet for OKI-78SR
    http://www.murata-ps.com/data/power/oki-78sr.pdf


    Or

    maybe the following as a solution.
    http://www.linear.com/product/LT1302#overview
    I saw a video online demonstrating the low 2volts dropout with a continous 5 volt output. The thing that concerns me is that with a Solar Power source, the voltage may be there, but the power available might be a limiting factor.
    thanks
     
  13. MagicMatt

    MagicMatt

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    Jun 15, 2011
    Your maths and expectations would be correct, but I do not understand where you get the 50% from.

    In good strong sunlight, the output power of the solar panels is (pretty close to) 6V at 450mA.

    The output will drop as the light level decreases.

    Your 50% figure is probably accurate for a day on which there's a bit of cloud cover, or the unit isn't in direct sunlight, or it's mid-evening and the light level has dropped.
     
  14. whilburn

    whilburn

    6
    0
    Oct 25, 2011
    :)) mostly i made it up.....
    However, to be honest, i did a quick google of "lm7805 effeciency", for a quick %.
    Also, my % effeciency is focused on the regulator not the solar panel/cells.

    I bought two 1.5 watt solar panels from Harbor Freight, which has a male car-power plug on each one. They each produce about 12 volts at 125 ma. So I went to Dollar General and purchased (2) Male to two Female Car-Power Y combination cables so i could daisy chain and use a standard car-charge plug. I havent had time to do my own board yet, but needless to say, in full sunlight yesterday, my Iphone 4 indicated "this is not a suitable charging device".
    I realize that the Car charger is set up for much higher current output and it has a MC34063 converter in it. It also has the Data line resistors that sets the amp draw on the power supply.
    I dont know all i need to know to make it work ,but I was looking for a solution as a step down converter at this point where i could use one or both to charge an iphone.
    Thats why i was asking about my math.
    1 panel produces 1.5 watts. 1.5 = 12 volts x 125 ma
    so if i can get a 85% efficient DC-Dc convertor to work, then
    1 panel + converter produce 1.275 watts = 5 volts x 255 ma

    Then if I paralleled the inputs, I could use a dongle with the two inputs to charge an Iphone at 5vdc & 500 ma.

    If my logic proves incorrect, dont hesitate to offer advice. I dont have an oscope, although I am going to try and find a cheap one at christmas or build a usb 'scope.
    so, im operating blind other than my meter and using shunts to measure current.

    I used to work in embedded hardware/software, but I do system type testing now and i have only my own tools to rely on for my projects. :)
     
  15. ajgiampa

    ajgiampa

    6
    0
    Jul 19, 2011
    Wilburn,

    I can say that I've measured the current draw in bright sunlight through the USB cable of my circuit at nearly 840mA. This output is despite any inefficiency in the circuit, so the circuit is much more than 50% efficient.
     
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