- Calculate the required full
sun current specification for your module: I
_{mod}

*I*_{mod} = I_{min} *x
100% /*** L**_{min}_{
}_{
}
- Chose a module that matches
the voltage required and the current, I
_{mod} 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.

*I*_{mod} = I_{min}
*x 100% /*** L**_{min} =

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. **L**_{min} =
0.4%.

*I*_{mod} *=
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.

*I*_{min} = *0.1mA.
Store lighting gives*** L**_{min} = 1.3%

*Then:*** I**_{mod}
= 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.

**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:
**I**_{avg}. This is equal to the current draw of the application
times the duty cycle.

- 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.

- Calculate the module current
requirement.
**I**_{mod} = **I**_{avg} x 100% /
**L**_{avg}.

- Select the module that matches
the voltage required and current
**I**_{mod} 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.

*I*_{avg}*
= ***I**_{app} x duty cycle = 20 ma x 8hr/24hr = 6.67 ma

**L**_{avg} =
100%x 4/24 = 16.67%

**I**_{mod} =
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.

*I*_{avg}*
= ***I**_{app} x duty cycle

*= [3mA x (23hr 50 min)/24hr]
+ 300mA x (10min/24hr)*

*= [3mA x .993] + [300mA
x .0069] = 5.05mA*

**L**_{avg} =
100% x 2/24 = 8.33%

**I**_{mod} =
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.

*I*_{avg}*
= ***I**_{app} x duty cycle

*= 2A x (4 days x 6hr/30
days) = 2A x (4 days x 6hr/(30days x 24hr/day)*

*= .35A = 35mA*

**L**_{avg} =
100% x 4.5/24 = 18.75%

**I**_{mod} =
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.