Start with what you know
You know what the weather is like where you live. You know approximately how many hours of sun you get during the day. Combine this knowledge with the size of you battery bank and you will start to get a picture for what size your solar panels need to be.
We determined earlier that I need at least a 22 Ah battery to power the one radio I need powered over a 24 hour period. However, something I did not mention earlier is that this number is misleading. Keep in mind there are periods that I do need to sleep. During these periods I will have the radio turned off. It will not be drawing any current off the battery. When it is daylight, the battery itself will be charging.
Again, we really need to keep these things in mind. Sometimes we are drawing power from the battery, sometimes we are charging the battery, sometimes we are doing both.
We know that the radio I am using draws, on average, 750 mA/hour. Therefore, we need to put 750 mA/hour back into the battery to keep it at a full charge. Since there is a time that there will be no daylight, we have to adjust for that.
The easiest way to adjust for non charging times is to look at an almanac for your city and look at the amount of daylight available on the shortest day of the year. At my location, that is around December 21. I found a number of sunrise/sunset calculators via Google, and the one I happened to pick is located here.
For 2009, it shows December 20 as the shortest day, with sunrise at 7:33 AM and sunset at 4:58 PM. Doing some math that gives me about 9 1/2 hours of daylight for charging. Not too shabby! I do not really care about any other days of the year – I want to look at the worst case scenario, so that’s why I have picked the shortest day.
Now that I know this, I know what I am up against. I will consume 18 Ah of current. I have 9.5 hours to replenish. So, I need to charge the battery at a rate of 18/9.5 = 1.89 A/hour to keep a full charge on the battery.
Some basic Ohm’s Law
Way back when there was this German guy named Ohm. He came up with a law that showed how current, voltage, and resistance were all related. Along the way, power got thrown into the mix and it was found that power, in watts, is equal to the voltage times the current. Or, written out, it would be P = E * I. Some people like to change E and I around, so the formula appears as P = I * E. See? This formula is now as easy as PIE to remember!
By knowing any two of the values, we can quickly find the third value.
‘To find the current in Amps, or I, the formula becomes I = P / E.
To find the voltage, in Volts, or E, the formula becomes E = P / I.
And finally, as mentioned above, to find the power in watts, or P, the formula is P = I * E.
What do we do now?
Now we start getting into the fun part of the whole thing. We know from above that we need to put 1.89 Amps into the battery to keep it charged on the shortest day of the year. That’s all fine and good, but solar panels are not specified by their output in amps! Actually, they are, but the current in Amps is shown at the specific voltage output of the panel.
For instance, I found a BP Solar panel with the following ratings:
- 10W
- 16.8 V
- .59A
This means this particular panel will put out 16 Volts when connected to a load – in this case a battery – and is sized to be a 10 watt panel. By using the formula above, I = 10W / 16.8 V, we can see that we will get .59 Amps out of it, or 590 mA.
You may also see designations talking about Voc and Ioc. These are ‘O’pen ‘C’ircuit voltages and currents. For a common 12V gel cell, something with a Voc up to about 21 volts would be okay. This particular BP panel above would be something we would want to look at. However, it is a bit on the small side. This can be good or bad depending on your needs.
Remember: As long as the voltage going into a rechargeable battery is greater that the float voltage of the battery, the battery is going to charge. Just don’t overdo it by shoving 30V into a 1.5V battery! A few volts over and you are okay. Going hog wild will get you into trouble.