- 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.