Solar Panel Package
How to calculate power consumption?
I would like to setup a solar + Battery solution for time lapse photography.
How do I tell what battery i would need to run my camera during night time hours and Solar Panel to charge it during daylight hours.
I was looking at this package http://store.sundancesolar.com/20wadosoenst.html However, how will I know (or approximate) how much the battery will drain while no light is available and how long it will take to recharge to full capacity?
Any help or starting point is appreciated.
Thanks very much for the info!!! this is awesome.
So am I right to calculate that the maximum draw in this system could be ~3amps before draining off more than can be replenished??
Which would be….4W?
You need to know the power consumption of your apparatus to begin with, but I think that will be very modest compared to the type of things that page you linked to says it can do. This is even more true if your camera supports time-lapse shots since then it puts itself in stand-by mode while it waits to take the shots. If it doesn’t, then the power consumption may go up fast…
What you have to know to evaluate that set-up is two things:
- for electric stuff power is voltage times current,
- batteries are measured by the amount of charge it can hold, and charge is current times time.
So, if the battery can hold 32amp-hour it means it can sustain a 1Amp current for 32 hours, or 2Amp for 16 hours, etc, you get the idea. To find the current you’ll be draining from the battery you divide the power consumption (the 1W above) by the battery voltage (let’s assume 12V, the site is not clear on this) which gives aproximately 0.083Amp. So that battery can last for 32/0.083 = 384 hours, which seems to be enough.
Of course you need to recharge the battery, and if your power consumption were greater than what you can replenish it, it would be no good. But from the site it says the solar panel can provide 6amp-hour a day, which for the 1W power consumption gives 6/0.083 = 72 hours > 24 hours, so you’d be safe.
Now, if the power consumption were 5W then you’d be in trouble…
Assuming you want to use only the power you can replenish (taht is, you want to run the whole thing for days), the calculations are as following:
According to the site, the solar panel can provide 6 amp-hour a day (working with 5 hours of sunlight, I guess, which can be a pretty good estimate since you may have cloudy days, the position of the sun may have to be more or less straight up, etc). That means if you’re leaving the camera on continuously (i.e., 24 hours a day) you can only use 6 amp-hour / 24 hours = 0.25 amp, which works out to a power consumption of 0.25 amp * 12 V = 3W (remember power is voltage times current, and the battery voltage is 12V). Thus, 3W is the most you can use *continuously*. If your camera can put itself in a saving mode (or a stand-by mode) and wake-up only when it needs to get the shot, the power consumption should go down a lot. So, assuming in stand-by mode the consumption is small and you need to shoot only a few seconds, say 5 seconds, each minute, then you’d be operating in total only 1/12 (5sec/60 sec/min) of a day, which means two hours. in this case, you have 6 amp-hour / 2 hours = 3 amp, and the total power available would be 3amp * 12V = 36W.
Things a re a little more complicated due to several factors: the power consumption in stand-by mode may be small but not negligible, and it may add up, you probably will have some power drain depending on how you connect your camera to the solar panel, etc, but if do the calculations leaving some allowance for these things you should be ok.