Solar Power supplies for the hobbyist.

Here's some simple steps for choosing parts to assemble a solar power supply for a project.
We will use a telemetry project as an example.
In this project, the Arduino sleeps most of the time.
Every 5 minutes it wakes up, takes some measurements, and then transmits its readings over a radio link.

Determine the average current consumption, I_a
A project will draw varying amounts of current, depending on what it is doing at the moment.
It is best to measure the current with an amp meter, but if that is not available we can estimate the current by examining the data sheets of all equipment involved.

In our example, we draw 3 mA most of the time, 10 mA while taking readings, and 200 mA while transmitting.
It takes 1 second to take the readings, and 1 second to transmit them.
We draw 3 mA for 298 seconds, thats 894 mA seconds.
We draw 10 mA for 1 second, thats another 10 mA seconds.
We draw 200 mA for 1 second for another 200 mA seconds.
Total:1104 mA seconds.
The average current draw over 5 minutes (300 seconds) is 1104/300, or 3.68 mA.
Lets round that up to 4 as a safety margin.

Determine the daily current consumption, I_d
I_d = 24 * I_a
24 hours times 4 mA = 96 mA Hours.

Determine the charging efficiency, E_c
This is affected by a number of things, including battery chemistry.
We never get as much energy out of a battery as we put in.
A typical sealed lead acid battery, which we will be using for our project, has a charge efficiency of 0.7, meaning we get 70 mA hours out for every 100 mA hours we put in.

Power dissipation in the charging circuitry also lowers the efficiency.
For our example project, E_c is 0.65.

Determine the equivalent hours of daylight, H_e
This is equal to the efficiency times a figure called Insolation.
The advertised current ratings of solar panels are given when the panel is in direct sunlight, and the sun is directly overhead.
Of course, the sun is rarely overhead, and clouds happen.

Insolation varies with time of year, latitude, and climatic factors.
We need to use the lowest insolation figure.

Insolation figures are available from a variety of sources.
In the USA the figures may be obtained here.

Our example project will be in Southwest Florida, where the lowest insolation figure is 3.

So in our example H_e =  0.65 * 3, or 1.95 hours, we'll round up to 2 hours.

Determine the minimum Solar Panel Current rating, C_r
C_r =  I_d /  H_e      
Our panel must supply a full day of power in a short amount of time.
In our example C_r = 96 / 1.95, or 49.23 mA.
Lets round that up, and set C_r equal to 50.

Determine the minimum battery capacity, C_m
C_m = (24 - H_e) * I_a
Our battery must supply current during the night hours, which for all practical purposes is 24 - H_e.
In our example, the battery must supply 4 mA for 22 hours, so we need a battery with 88 mA Hours capacity.

Review and revise
Now we need to sit back and think.
Our chosen devices will work fine on a typical day, but how many days are actually 'typical'?

In reality, there may be rainy, cloudy days, sometimes for days on end.
And what about other factors. What if a squirrel decided one day that your panel would be a good place to sun itself?
And consider the site. Are there trees or buildings which would shade the panel? Increase the current rating accordingly.

Having determined the minimal ratings, we need to decide how much of a safety margin we need, or want.
For our example project in SouthWest Florida, cloudy conditions rarely persist for more than a few days.
We may decide we want the system to carry through for 5 days of cloudy weather (or squirrel sitting), with a full recharge within 1 day.

So now we wind up with a battery capacity of 420 ma Hours, and a solar panel with a current rating of 250 mA.