Key Concepts
- Module current increases nearly linearly with light intensity.
- Module operating voltage is relatively insensitive to the light intensity,
dropping about 5% in 10% of full sun.
Definitions
Power point-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).
Duty Cycle-Percentage of time that an application is expected to
operate.
AM1, AM1.5-For all intents and purposes - Full sun illumination
intensity on a clear day at noon.
Calculations for Systems Without Batteries
Voltage Considerations
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.
Current Calculations
Find the minimum current needed for the application: Imin
Determine the minimum light intensity (threshold intensity) under which the
application will run: Lmin (The table below gives a rough idea
of light intensity under various conditions. Intensity is rated as a percentage
of full sun intensity (also called AM1.5)
|
Energy Available at Various Light Conditions
Relative to Full Sun
|
|
Condition |
Intensity
(% of full sun) |
| Full
sun-panel square to 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 moduel square to
sun |
84% |
| Inside
window, single pane, double strength glass, window module at 45 degree
angle to sun |
64% |
| Indoor
office light - at desk top |
0.4% |
| Indoor
light - store lighting |
1.3% |
| Inoor light
- home |
0.2% |
|
Calculate the required full sun current specification for your module: Imod
Imod = Imin x 100% / Lmin
Chose a module that matches the voltage required and the current, Imod
calculated.
Note: 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 Calculations for Applications using Direct Power
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.
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
Calculate average current draw: Iavg. This is equal to the
current draw of the application times the duty cycle.
Estimate the average illumination on the module, Lavg (i.e. 4
hours of full sun per day averages to Lavg = 4/24 = 16.6% of
full sun average illumination over the day). See table above for help on this.
Calculate the module current requirement. Imod = Iavg
x 100% / Lavg.
Select the module that matches the voltage required and current Imod
calculated.
Example Calculations for Applications with Batteries
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.
Iavg = Iapp x duty cycle = 20 ma x
8hr/24hr = 6.67 ma
Lavg = 100%x 4/24 = 16.67%
Imod = 6.67 ma x 100 / 16.67 = 40 ma
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.
Iavg = Iapp x duty cycle
= [3mA x (23hr 50 min)/24hr] + 300mA x (10min/24hr)
= [3mA x .993] + [300mA x .0069] = 5.05mA
Lavg = 100% x 2/24 = 8.33%
Imod = 5.05mA x 100/8.33 = 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 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.
Iavg = Iapp x duty cycle
= 2A x (4 days x 6hr/30 days) = 2A x (4 days x 6hr/(30days x 24hr/day)
= .35A = 35mA
Lavg = 100% x 4.5/24 = 18.75%
Imod = 70mA x 100/18.75 = 373mA
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.
