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Newb needs help

Discussion in 'General Electronics Discussion' started by ssorensonaz, Feb 12, 2013.

  1. ssorensonaz

    ssorensonaz

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    0
    Feb 12, 2013
    I'm trying to setup a solar powered wifi camera. That is, I can find the camera, and it's powered by a USB port. Does anybody know if those things might use anything like the full power from a USB port? I need to know so I can properly size a battery, solar panel, and charge controller.

    Being new to this, i don't know if i've done this right, but it seems like i'd need a bigger panel and battery than i wanted (size is a problem) for this project.

    If anybody is willing to partake, here's my notes so far:

    Any camera which can run off USB power, will at maximum, consume the following:

    500mA * 5 Volts = 2.5 Watts,

    2.5 Watts * 24 Hours = 60 Watts.
    And
    500mA * 24 hours = 12,000 mAh of power if run from a battery all day.
    Multiply by 2 to allow for never drawing the battery below 50% charge, we need a 24 Ah battery.

    So, if we say we will get some sun everyday, but have enough battery for 24 hours, we should be in good shape.

    We'll just need a 24 AmpHour battery
    A Solar panel which can provide 60 Watts over the sunny period. 60 Watts div. by 5 hours = 12 Watts panel, is the minimum size, up to double that would be fine..

    A a charge controller for the Solar panel to feed the battery,


    If it will fit in a bird house sized box, we're good. Just need a sunny location with a southern sky exposure.

    Panels: 70 miliwatts per square inch., x 5 hours, = 350 Milliwatts hours per square inch per day.

    1 square foot is 144 square inches, or 50.4 watts per "day."

    Charge controller calcs:
    12 Watt Panel / 12 Volts = 1 Amp, add safety factor makes int 1.25 Amps Array Short Circuit Current (minimum controller input current)
    Max DC load current = 2.5 Watts / 5 volts = .5 Amps. (Minimum Controller output current)

    If we double the panel we need a 24 watt / 12 volts * 1.25 = 2.5 Amps for the array short circuit current and
    The max DC load doesn't change. So, total is No larger than 2.5 Amp controller.

    Of course, this all may be moot - the camera probably doesn't draw anywhere near the full power of the USB port.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,220
    2,695
    Jan 21, 2010
    That's 60Wh

    That assumes you use a linear regulator. and 24AH at 12V is 288Wh.

    if you have an 85% efficient switchmode regulator then you need 2 * 60 / 0.85 = 102Wh

    And that is approximately 8.5Ah

    Assuming you want to provide up to 120Wh of charge, and you have 6 hours of equivalent sunlight, then you need a 20W panel if you use a swichmode regulator.

    For your original linear regulator, you would need a 288/6 = 48W panel

    You can't get watts per day. Watts are already "per second"

    The rest of your calculations will be in error due to the issues above...
     
  3. ssorensonaz

    ssorensonaz

    2
    0
    Feb 12, 2013
    Yep, In over my head

    Thank you. I think i'll need to do some more studying. Looking at charge controllers, which is what I think you are referring to by "regulator," I never saw anything about linear vs. switch mode, though I did see MPPT.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,220
    2,695
    Jan 21, 2010
    No, the regulator is the thing that gives you 5V for USB from the 12V from your battery.
     
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