Solar powered USB help?
September 8, 2012 8:52 PM   Subscribe

Calling all electrical engineers! Need some clarity on how to figure out solar charging rates and consumption of a USB device...

Ok, hivemind...I'm having a hard time keeping milliwatt hours and milliamp hours straight here, and I could really use someone with some electrical knowledge to keep me from going crazy.

I've got a USB powered device that the specifications say uses 5VDC/1.0A for power. I have been using a 6600 mWh rechargable Lithium ion battery to power the device when I need it to be portable (which is often) and it's been dandy...I regularly get 15+ hours of uptime from it. But now I'm looking at the possibility of a permanent outdoor home for this device, and thinking about how to make it run from solar cells.

Given the above, can you help me figure out how large a solar panel I would need to power the thing during the day and build up enough charge to last it through the night? For argument's sake, assume average sunlight levels for the US...not Arizona, but not overcast all the time either. Are we talking square inches or square feet here?

Any suggestions for how to put something like this together?
posted by griffey to Science & Nature (3 answers total) 5 users marked this as a favorite
 
Best answer: Okay, so I work directly with solar power for off-grid telecommunications equipment, so I hope I can help with your very small application. The first thing you want to do is calculate the cumulative monthly watt-hour (Wh) load of your device:

5V 1A = 5.0W.

5.0W x 24 hours x 31 days = 3720 Wh, or the same as 3.72 kWh. We'll use kWh for these calculations.

The next thing you want to know is your location, climate and latitude. One good basic calculator is the US DoE NREL pvwatts application. Go there and choose San Diego, leave it at the defaults (1.0 kW panel size system), leave all the other defaults as they are, and look at the monthly kWh a system will generate in San Diego. Now go back and do the same thing with the same size of solar system in Minnesota or Maine, a place that has poor sunlight hours. Notice now the kWh monthly production sucks in December and January?

For off grid solar things that need to be completely independent you will want to calculate based on the worth months of the year, which will be from November to end of February. December/January will be your worst months.

It will help that the system you are trying to create will run entirely from DC so you will avoid the losses involved with an AC to DC inverter.

If you are in a northern state, I recommend one 180W size solar panel, a small charge controller such as the SS10, and a couple of 12V 12Ah size AGM lead acid batteries. You can get the batteries for $23 each with free shipping from eBay. I also recommend a DC circuit breaker on the positive wire from the solar panel to the charge controller so that you can flip a switch for disconnect if necessary.

On the load side, buy and install a very basic DC-DC converter to turn 12V into 5V, such as the Meanwell SD-15A-05. This will take a 9V to 18VDC input and output a flat 5V.

So, basically: Solar panel sufficient for your needs, DC disconnect switch, charge controller, batteries, and DC-DC converter. It's up to you to figure out or buy some kind of weatherproof enclosure....
posted by thewalrus at 9:26 PM on September 8, 2012 [3 favorites]


If you get 15 hours on a 6.6Wh battery, then that means your average draw is closer to 0.4W, not 5W (which is the typical peak/max draw for USB). That gives you a daily load of about 10Wh.

You can buy solar panels, charge controllers, and batteries on Amazon. You could probably do some sophisticated calculations but I would just buy a panel that's nice and big for the task. Multiply the solar panel's power rating in watts by the hours of full sunlight and you get a rough idea of how many watt-hours per day you can expect from your panel. This is a wiggly calculation and the correct way to do it is per thewalrus above, but for a small system you can just overengineer it. A 5W panel should do it - two hours of sunlight per day and you're good (or 4 hours of 1/2 sunlight, etc). You can buy these on Amazon for $40, they're about one square foot. Get a charge controller and a battery too. You can get a small 12V -- even a 5 amp-hour battery gives you 60 Wh which is six days of power.

Maybe I'm fudging it too much. Would a 5W panel be okay? I really think it would be. If this is super mission critical, maybe run the numbers more carefully (or else just go with a 10W).
posted by PercussivePaul at 9:43 PM on September 8, 2012


your average draw is closer to 0.4W

Yeah, 5V/1A is just max ratings for a bus-powered USB device. Think of your solar panel + battery + gadget as a gas pump, gas tank, and engine. The engine uses about 10gal/day, and you need a big enough pump to put that much back into the tank (on average, even with short / cold / cloudy days) before the tank runs dry.

You can have a bigger tank (like a 5Ah battery -- big enough for 6 days) and a marginal pump (5W) and maybe be OK. Or you can have a smaller tank and a bigger pump, but remember: you can't store up more energy than the battery can hold -- when it's full, it's full.

And finally, as thewalrus notes, you'll need a DC-DC converter, to change 12V from your battery to 5V for your device. (You could wire up a 5V battery pack, but all the cheap/standard equipment is designed for 12V lead-acid batteries, so you'll end up saving money by running the pump (solar panel) and tank (battery) at 12V, and just converting down to 5V to give the engine what it wants.)
posted by spacewrench at 10:59 PM on September 8, 2012


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