Calculations Needed

 I'm trying to set up a remote camera in my backyard using a solar
panel and a 12 volt battery.  The camera used 12 volts, 200mA.  The
battery that I was given is a 12 volt, rated at 10 amp hours. I also
have a 12 volt battery rated at 7 amp hours.  The area where I live
gets about 5 to 5.6 hours of sunshine per day.  I was thinking of
getting a 5 watt solar panel to keep the battery topped off.  Can
someone give me some idea as to whether this will work or if I need a
larger battery or larger solar panel.  Also, is there a web site that
someone can point me to where i can find the correct calculations used
ti determine how to size my solar power system?

Thanks in advance.

  Looks like the camera will use 2.4 watts x 24 hrs ~58 watt-hrs a day.
A 5W panel with 5hrs of sun would only get about half of that back, so
you need some sort of timer trick to only keep the camera on half the
time or get a bigger panel.

 robertbbm@tidni.com wrote:

 > I'm trying to set up a remote camera in my backyard using a
 > solar panel and a 12 volt battery.  The camera used 12 volts,
 > 200mA. The battery that I was given is a 12 volt, rated at
 > 10 amp hours. I also have a 12 volt battery rated at 7 amp
 > hours.

Do not skimp on the battery size. There may be quite a few
days of no sun. Plus it's kinder to the battery if you don't
discharge it to much.

 > The area where I live gets about 5 to 5.6 hours of sunshine
 > per day.  I was thinking of getting a 5 watt solar panel to
 > keep the battery topped off.

Also don't skimp on the panel. 10W or more is preferable to
get the battery charge up if there was a long dark spell.
And use a good charge controller. I would recommend ones
made by "Solar Converters". See:
http://www.solarconverters.com/
http://www.solarconverters.com/product_frame.html

 > Can someone give me some idea as to whether this will work

Yes, see two of my customer projects:
http://www.redrok.com/led3xassm.htm#jory
Jory has a very high end WIFI camera that consumes about 1A.

http://www.redrok.com/led3xassm.htm#dg240
And later I made this which is probably a better choice
than Jory's.

 > or if I need a larger battery or larger solar panel.

Probably.

 > Also, is there a web site that someone can point me
 > to where i can find the correct calculations used to
 > determine how to size my solar power system?
 > Thanks in advance.

Duane

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 On Mon, 18 Feb 2008 13:09:04 -0800 (PST), robertbbm@tidni.com wrote:


First you have to figure out exactly how much energy you will be using.
For example, is this a day-only camera, or will you be running it 24/7.

Is this for year round use?  If so, you want to size your system for the
worst month.

Let us say, for example, that you live in the Tampa Florida area, and that
you want to run this device 24/7/365

Your DC Load is 12*0.2 = 2.4 Watts * 24 hrs = 57.6 watt-hrs per day

The Tampa area averages 4.5 effective sun hours per day during the worst
month of the year.  And you need to figure about 20% losses in your system
due mostly to inefficiencies in battery charging.

I find it simpler to do the calculations using amp-hrs rather than watts.

So

57.6wH at 12V --> 4.80Ah (per day)
20% losses/safety factor --> 4.8*1.2--> 5.76Ah/day

divide that by the Sun hours to get your array current requirement:

5.76 / 4.5 --> 1.28 amperes

Now you can look at a vendor to find a panel that puts out at least 1.28A
at 12V.

The smallest and least expensive I see is a 30 watt panel from Global Solar
that puts out 1.7A at 17.5V.  ($172.29  http://store.altenergystore.com  )

You need to put out more than 12V in order to charge a 12V battery, and you
use a charge controller to keep from overcharging.

The next smallest panel I see on this particular web site is a 24W panel
that is just a bit undersized at 1.24A

So it looks as if you will need 1 30 watt module to run this camera
24/7/365 in the Tampa Fl area.

Battery capacity is calculated by looking at your load, allowable depth of
discharge, and required "days of autonomy" (i.e. how many days can you go
without sun and still have power).

5 days of autonomy, and a 50% depth of discharge is generally recommended.
Although with premium batteries, you can sometimes figure an 80% DOD.

So your daily consumption * 5 / 0.5 --> comes out to a battery requirement
of about 58AH at 12V

Obviously, if you don't run your camera 24/7/365, or if you live in a
different area, or ...  the inputs to these calculations change.


--ron

 
And you were doing so well too, Ron. So five days autonomy and 50%
DOD. 5 x 50 = 250. I think the batteries would be well and truly dead
long before you could get 250% out of them.

You really do need to learn that there are two types of "Depth Of
Discharge". They are very different and should not be confused.

The first is "Daily Depth of Discharge" - This is the amount that can
be used per day to meet X number of days of autonomy. For five days
autonomy this can not exceed approx 16% DDOD for a battery at C100 and
an allowed Maximum Depth Of Discharge of 80%.

The second is "Maximum Depth Of Discharge" - This is the amount that
can be used before severe and lasting damage to the battery occurs.
The most common MDOD is in the range of 70% - 80%. The high MDOD can
be tolerated because in a properly designed system this level of
discharge would a rare occurrence.

Always check with the manufacturer for the "Maximum Depth Of
Discharge, allowed for a battery.