Calculations for Systems with Batteries
Voltage Considerations
For battery charging applications, the operating voltage of the module should be at least as high as the charging voltage of the battery. This is higher than the battery's output voltage. A single NiCd battery has a typical output voltage of 1.2 volts, but requires 1.4 Volts for charging purposes. A 12 Volt lead acid battery needs a charging voltage from 14 to 15 Volts. In cases where a blocking diode is required to prevent the battery from discharging through the solar module when the module is in the dark, an additional 0.6 V is required. As an example, a battery pack with 3 NiCd batteries, which operates at 3.6 Volts, needs a module with either 4.2 or 4.8 V depending on whether a blocking diode is used.
Is a blocking diode required?
When the solar module is in the dark and still connected to the battery, it is simply a forward biased diode and can drain current from the battery. This is less of a problem for amorphous silicon modules than single crystalline modules, but can still be a problem if the module is in the dark a large percentage of the time. The leakage rate also drops dramatically if the open circuit voltage of the module is significantly larger than the output voltage of the battery. For applications that get sun daily, diodes can probably be ignored if the module is sized correctly. If the application is going to spend extended time in a case or drawer, however, a blocking diode would be advisable. Each application should be evaluated individually for this choice.
Current Calculations
1. 
Calculate average current draw: I_{avg}. This is equal to the current draw of the application times the duty cycle.

2. 
Estimate the average illumination on the module, L_{avg} (i.e. 4 hours of full sun per day averages to L_{avg} = 4/24 = 16.6% of full sun average illumination over the day). See table above for help on this.

3. 
Calculate the module current requirement. I_{mod} = I_{avg} x 100% / L_{avg}.

4. 
Select the module that matches the voltage required and current I_{mod} calculated.

Example Calculations for Applications with Batteries
Example 1: A yard light draws 20mA and you want it to work for 8 hours per night. You estimate that you get the equivalent of 4 hours of full sun per day.
I_{avg} = I_{app} x duty cycle
I_{avg} = 20mA x 8hr / 24hr
I_{avg} = 6.67mA
L_{avg} = 100% x 4hr/24hr
L_{avg} = 16.67%
I_{mod} = I_{avg} x 100% / L_{avg}
I_{mod} = 6.67mA x 100% / 16.67%
I_{mod} = 40mA
Example 2: A mobile phone draws 3mA in standby mode and 300mA in talk mode. It is assumed that the phone is used in the talk mode for an average of 10 minutes per day, while in the standby mode for 23hrs and 50 minutes. The phone can get an equivalent of 2 hours of direct sunlight per day. Find the module size needed to keep the phone charged.
I_{avg} = I_{app} x duty cycle
I_{avg} = {3mA x [(23hr 50 min)/24hr]} + [300mA x (10min/24hr)]
I_{avg} = {3mA x [(23hr x 60min) + 50 min]/(24hr x 60min)] + {300mA x [10min / (24hr x 60min)]}
I_{avg} = [3mA x .993] + [300mA x .0069]
I_{avg} = 5.05mA
L_{avg} = 100% x 2hr/24hr
L_{avg} = 8.33%
I_{mod} = I_{avg} x 100% / L_{avg}
I_{mod} = 5.05mA x 100% / 8.33%
I_{mod} = 60mA
If the charging voltage of the phone is 6V, you will need a 6V, 60mA module at the very least to supply all needed power from the module.
Example 3: A fishing boat has a 12 volt battery system which powers a trolling motor and depth finding equipment. The boat is in use 4 days out of every month and requires an average of 2A for 6hrs of use per day. The boat will get an average of 4.5hrs of sunlight per day. Calculate the module size needed considering a monthly cycle.
I_{avg} = I_{app} x duty cycle
I_{avg} = 2A x [(4 x 6hr)/30 days]
I_{avg} = (2A x 1000mA/1A) x (24hr / 720hr)
I_{avg} = 2,000mA x 0.0315
I_{avg} = 63mA
L_{avg} = 100% x 4.5hr/24hr
L_{avg} = 18.75%
I_{mod} = I_{avg} x 100% / L_{avg}
I_{mod} = 63mA x 100% / 18.75%
I_{mod} = 336mA
If the boat is used 4 days per month with the days separated by equal time intervals, a 14V 400mA module should be sufficient to store enough energy to run the boat. However, if the boat were used 2 consecutive days, there would not be enough time to fully recharge the battery before the next day's use. If the capacity of the battery is sufficient, this will not be a problem, but if the capacity of the battery is such that only one day's energy can be stored in it, more charging capacity will be needed and the calculations will have to be redone on a daily cycle.
