Good Afternoon,

Excuse my ignorance as I'm not an electrical engineer, I'm just trying to build a device to help grade school students. In the simplest form, I'm trying to setup a solar powered LED light box. In searching, I realize that basic topic has been covered many times. However I couldn't find a schematic that would cover these specifics.

In a relatively small clear acrylic box (say 2" x 4"), I would like to mount a solar panel that would power a single red or green "Super Bright" LED bulb. It would be switchable as only the red or green LED needs to be illuminated at a time, not both at the same time.

Some of the challenges I foresee are as follows:

  • The LED must be able to stay illuminated all school day, every day.
  • As I'm sure you remember, often times the only light in the school room comes from the overhead florescent tubes.
  • In order for these to see sustained use, I greatly prefer not to involve batteries as I'm trying to keep build costs down and make sure they are useful long term and not shelved when the batteries die.

Excuse the crude design, but this is what I'm envisioning: enter image description here

To start playing with, I have already ordered the following:

Single Pole Double Throw, 3-Pin Terminals, On/Off/On Three Position Switch

LED Bulbs

Super Bright LED - Red 10mm COM-08862

  • 625-630nm Red
  • 2.1-2.3VDC Forward Voltage

  • 80mA Forward Current

  • 30 degree viewing angle

  • 10,000-12,000 MCD output

Super Bright LED - Green 10mm COM-08861

  • 515-520nm Green

  • 3.0-3.4VDC Forward Voltage

  • 80mA Forward Current

  • 30 degree viewing angle

  • 16,000-22,000 MCD output

With regards to the housing/box, I'm thinking of using something similar to this: enter image description here

Can you help point me in the right direction with ideas for the components and/or a schematic to make this work?

Unfortunately they can't be plugged in. They will need to sit on individual student desks (w/o power) and can't have electrical cords running all around the class (tripping hazard).


closed as too broad by Leon Heller, PeterJ, Brian Carlton, Daniel Grillo, Axeman Sep 30 '15 at 13:44

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ A schematic is required. Question should be closed. \$\endgroup\$ – Leon Heller Sep 24 '15 at 0:04
  • 1
    \$\begingroup\$ What is the goal of this device? 80mA LED's are pretty bright, typical signal LED's use 1-10mA. \$\endgroup\$ – helloworld922 Sep 24 '15 at 0:05
  • \$\begingroup\$ Have you considered plugging it into the wall? That would be a lot easier than a solar panel. \$\endgroup\$ – Houston Fortney Sep 24 '15 at 0:13
  • \$\begingroup\$ Your first step is to test the power available from your solar cell under indoor lighting conditions. That dictates everything else ... like, finding another power source, or blinking lights to save power. \$\endgroup\$ – Brian Drummond Sep 24 '15 at 11:00

I'm sorry, but you're on the wrong end of how things work. Let's start with sunlight.

Very bright sunlight (directly overhead) has a visible power density of about 500 watts/sq meter, and a brightness (illuminance) of about 100,000 lux. A common standard for indoor lighting is about 250 lux. As a rough number, this suggests that normal indoor lighting has a power density P of $$P = \frac {250}{100,000} \times {500} =1.25 \frac{watt}{m^2} $$ This isn't quite right, but it's close enough for a quick analysis.

With a box size (and therefor a solar cell size) of 2" x 4", this produces a power at the solar cells of $$P = .05 m \times {.1 m} \times {1.25} = 6.25 mW$$ The current efficiency champion for silicon PV cells is a bit less than 20%, so the maximum electrical power your box can produce under fluorescent lighting is on the order of $$P_e = 6.25 mW \times 0.2 = 1.25 mW$$ Assuming you can get a photocell whose maximum power point is at an output voltage of about 3.5 volts (to match your green LED), this suggests that the best current you'll get is $$i = \frac{P_e}{V_f} = \frac{1.25 mW}{3.4 volts} = ~0.37 mA$$ You'll note that this is about 0.5% of the maximum current, and while it will probably be visible I'm afraid it probably won't be very impressive.

As a sanity check, here is a solar cell about 40% bigger than your nominal box, which puts out a nominal 6 volts at 50 mA. Applying the brightness factor derived before suggests a classroom output of .125 mA. Agreement to a factor of 3 is pretty good for this kind of analysis.

Since a solar cell's output current is the important factor, Houston Fortney's circuit (with solar cell instead of power supply) will work as well as anything, and you can even get rid of the resistors.


I recommend you consider a solution that plugs into the wall. Maybe this one? If you must use a solar panel, it will need to be a big one.


simulate this circuit – Schematic created using CircuitLab

  • Use 1/2 watt resistors.
  • If you go with a solar panel and the voltage changes, recalculate the resistors with Ohm's Law
  • \$\begingroup\$ You could even use a USB port for power if that's any better. \$\endgroup\$ – Houston Fortney Sep 24 '15 at 0:25

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