ssorensonaz
- Feb 12, 2013
- 2
- Joined
- Feb 12, 2013
- Messages
- 2
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.
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.
