The following are considerations and suggestions for choosing the size of solar module needed for your application. The discussion is divided into 2 main parts: (I) Direct use of power from the module and (II) applications including battery storage to insure the availability of power when the module is not generating power.
These calculation examples are for using PowerFilm's OEM modules, but could be used as a basis for further solar system development.
|Iusage = Iapp * hours used|
|The operating voltage and current which produce the maximum power from the module. (Forcing the module to operate at a higher or lower voltage results in a less efficient operation).|
|For all intents and purposes - Full sun illumination intensity on a clear day at noon.|
|Current consumption of application|
The voltage of the module should be selected so that the power point voltage is near the required operating voltage of the application. As a rough estimate, you can figure that the power point voltage is about 75% of the open circuit voltage.
|Condition||Intensity (% of full sun)|
|Full Sun- Panel square to the sun||100%|
|Full Sun - Panel at 45 degree angle to Sun||71%|
|Light overcast||60 - 80%|
|Heavy overcast||20 - 30%|
|Inside window, single pane, double strength glass, Window and Module square to sun||91%|
|Inside window, double pane, double strength glass, Window and Module square to sun||84%|
|Inside window, single pane, double strength glass, Window and Module at 45 degree angle to Sun||64%|
|Indoor office light - at desk top||0.4%|
|Indoor light - Store lighting||1.3%|
|Indoor light - Home||0.2%|
Imod= Imin x 100% / Lmin
Module performance is usually specified in terms of current @ a specific voltage (i.e. 50mA@3V) which gives performance at a specific operating point. This operating point is usually close to the power point. Some modules are specified at full sun and others at lower intensities such as 1/4 Sun. This is done to simplify selection. 1/4 sun is a more typical intensity used by portable electronics and is often chosen as the threshold intensity.
Example 1: A radio to be powered by the module requires 9 mA at 3 Volts to operate. You want the radio to operate with any illumination above 20% of full sun.
Imod= Imin x 100% / Lmin = 9 mAx100/20 = 45mA.
Thus you need a module which will produce 45mA at 3 V under full sun illumination.
Example 2: Same as Example 1, but the given operating light is office light. Lmin = 0.4%.
Imod= 9mAx100/0.4 = 2,250mA. This is a very large module for a radio. A better solution may be to use a smaller module coupled with a battery which recharges from the module when left in a window.
Example 3: You want a Flashing LED for a point of purchase display which works under store illumination. The flasher circuit uses an average of 0.1mA at 2.4 Volts to power 5 LEDs.
Imin = 0.1mA. Store lighting gives Lmin = 1.3% Then: Imod= 0.1mA x 100/1.3 = 7.7 ma @ 2.4V. Alternatively, you might look at the low-light specifications where performance is given at 0.4% of full sun (about 400 Lux). This can be normalized to the 1.3% level.
I_app = Current draw of application (or load)
I_usage = I_app * hours used = total current used in application
I_mod = I_usage / (avg solar hours * Light intensity)
Average Solar hours = see bottom of this page
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 13.5 to 14.5 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.
Example 4: 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_usage = I_app * hours used = 20mA * 8 hours = 160mAh
I_mod = I_usage / (avg solar hours * light intensity) = 160mAh / (4 hrs * 100%) = 40mA
Example 5: A mobile phone draws 3mA in the standby mode and 300mA in the 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_usage = I_app * hours used = (3mA * 23.83 hrs) + (300mA * (0.167)hrs)
I_usage = 71.5mA + 50mA = 121.5mA
I_mod = I_usage / (avg solar hours * light intensity) = 121.5mA / 2 * 100% = 60.75mA
If the charging voltage of the phone is 4.2V, you will need a 4.2V, 60mA module at the very least to supply all needed power from the module.
Example 6: 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_usage = I_app * hours used = (2A * 6 hrs * 4 days) = (2A * 24 hrs) = 48 Ah
I_mod = I_usage / (avg solar hrs * light intensity) = 48 Ah / (4.5 hrs * 30 days * 100%) = 356 mA
If the boat is used 4 days per month with the days separated by equal time intervals, a 15.4V 350mA 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.
If a battery is fully charged and a solar panel is connected, over-time the battery may become over-charged. This over-charge damages the battery and reduces its lifetime. See your battery manufacturers specifications for maximum charge voltage. (note: Lithium batteries in particular can catastrophically fail when over-charged, always use proper protection circuitry when charging lithium batteries).
If the battery is used as a "buffer" between the solar panel and the load then protection circuitry may not be needed. Protection circuitry may not be needed if the load will prevent the batteries from ever reaching 100% charged.
Different areas of the world and different times of the year can greatly affect the average full sun illumination.
Light intensity hitting the solar module can also be approximately calculated by knowing the solar panels exact angle in relation to the sun in clear skies. A solar panel angled exactly perpendicular to the sun would yield cosine(0 degrees) = 1 or 100% sun. A solar module angled 30 degrees away from the sun would yield cosine (30 degrees) = 0.866 = ~ 87% sun
If you have any questions about specifying solar for your project, feel free to contact us and our engineers can help you with any of your questions.