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Newb needs help

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

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  1. ssorensonaz


    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.
    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

    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


    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

    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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