- 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.