Moonlight solar panels

metron9metron9 Posts: 1,100
Any one know if a solar panel puts out any voltage under moonlight?

Reason I ask is I just finished a on off switch using an ADC to measure the voltage on the panel. I have to wait for the next full moon to see if that will do the trick unless anyone has any experience with moonlight and solar panels.

I am using the small panel from spark fun. I use a 10k and 2.5 k resistor voltage divider and measure the voltage from a 1.1V internal reference on a mega48 chip. I just dont want the unit to stay off under strong moonlight.

Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!


  • 7 Comments sorted by Date Added Votes
  • Tracy AllenTracy Allen Posts: 6,180
    edited January 2007 Vote Up0Vote Down
    Hi Metron,

    Are you wanting to harvest energy to power a circuit, or just trying to measure the light level? The light level in full moonlight is about 0.05 lux max, around 0.0005 watt per square meter. Not much.

    Solar panels designed for power generation in full sunlight are "leaky" and at low light levels most of the energy is lost to internal leakage currents. There are some panels (on "solar calculators" for example) that are designed for use in normal indoor lighting, but I wouldn't count on them for output below 10 lux. It is a different matter if you only need to measure the moonlight. A solar panel is not the best option for that.

    Tracy Allen
  • metron9metron9 Posts: 1,100
    edited January 2007 Vote Up0Vote Down
    No not trying to harvest any moonlight energy. But I am amazed at the voltage produced on the 2.5 x 3.5 solar panel from spark fun. I am using it in a solar collector night light project where I read the voltage on the solar panel, when it gets low enough I turn on the LED's and start drawing from the battery.

    I have a small AA penlight and at 2 feet it still produces over 3V on the panel. Right now I use a 10K from the panel to a 2.2K to ground and measure the voltage from the center of those resistors, at 3V it measures .75V The ADC measures the voltage and when it goes under .75V I turn on the LED's, Tonight was a bit cloudy and it turned on the system at about 10 minutes to actual sunset for my area. I was thinking even though the 3V from a flashlight is not much in the way of current it still triggers the ADC to turn the system off.

    So I am wondering how much voltage they might produce in full moonlight. If the moonlight generates over 3V I may have to drop the resistor values to 5K and 1.25K or less to eat up the low current condition to use this method as the on off switch. The voltage divider would be always on so I don't want to bleed too much current from the panel as it charges a 4V SLA battery, that is why I am looking for the power this panel can generate by full moonlight. As you know an ADC can measure voltage at very very low current. I just recalculated and I was in error before, I guess with a 10k-2.2k divider I am only using .0003A (1/3 of a milliamp) at 4.5V (full sunlight) I thought I was at 3ma. So I can drop those resistors to a 2K and 500 ohm divider that will draw only 2ma max. I don't think the moonlight could drive the voltage up to 3V at that current.

    I will know on Feb 17th's full moon if it's clear, although this product will be used in the summer as well. I could use smaller resistors and another fet and just turn on the current to measure the panel voltage but I try to design with the fewest parts possible.

    It's a little tougher than I thought it would be to find twilight using a solar panel as the input because of the external factors. I don't want the light to go on too soon, and I don't want it to go off if there is a full moon or a bonfire reflecting light to the panels.

    Of course before I post I usually have new ideas and try them out mid sentence. I just tried a 2.2k/500ohm divider and I think that will work without being to sensitive to low current voltages.

    Think Inside the box first and if that doesn't work..
    Re-arrange what's inside the box then...
    Think outside the BOX!

    Post Edited (metron9) : 1/29/2007 4:25:05 AM GMT
  • latigerlillylatigerlilly Posts: 114
    edited January 2007 Vote Up0Vote Down
    At night, even with moonlight, solar panels drain (not produce) electricity. That's why they put diodes between the solar panels and the batteries that they charge. It is to prevent draining the batteries at night. To activate a function at night with your basic stamp, a photoresistor circuit may be better.
  • JordenLouisJordenLouis Posts: 1
    edited July 2013 Vote Up0Vote Down
    Thats really amazing I heard here. Moonlight can also be a source of solar energy..Great!! I am excited to learn about the solar energy factors more and more..
  • Duane C. JohnsonDuane C. Johnson Posts: 955
    edited July 2013 Vote Up0Vote Down
    Hi metron9;

    I think you are thinking about this a bit incorrectly.

    PV panels are a collection of cells generally strung in series. In your case 3V sounds like a 5 or 6 cell panel.

    With perfect cells, ones with no leakage current, the open circuit voltage is independent of light intensity. So the output voltage would be the same whether the light is from sunlight or the greatly dimmer moonlight.

    Ok, practical cells do have leakage current so the output voltage is somewhat dependent on the light intensity. However, the leakage current is highly dependent on other factors such as temperature and age.

    Most design PV based light sensors based on the output current the cells produce rather than the voltage.
    Here is an example:

    Note, this circuit amplifies the current from the cell while maintaining zero volts across the cell. The advantage is there is no leakage because there is only a few mV across the cell. This is more complicated than you require.

    As a compromise, a load could be applied to the cells so reduce the output voltage. This resistive load, generally consumes much more current so the leakage current is insignificant which results in a quite linear response and independent of temperature.

    In your case using a 3V panel as a sensor you could apply an appropriate resistor as a load to bring the voltage down to the pin switching point of about 1.65V.

    You could use a much smaller sensor such as a BLUE LEDs as a PV sensor. This has an open circuit voltage greater than 1.65V.

    And some other LED based circuits used with BEAM robotics .

    Duane J
    262 x 163 - 2K
  • TubularTubular Posts: 2,882
    edited July 2013 Vote Up0Vote Down
    Yes you absolutely do get power from moonbeams.

    We made some telemetry devices that had a nice low quiescent current. And on full-ish moon nights we'd get readings through the night
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