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Here is my attempt at making a LED-based solar panel:

I learned that if an LED is reverse biased, it can yield a considerable voltage up to 2.5 volts. I also did this experiment at home and, using my multimeter, I found that the voltage produced by an illuminated white LED is a maximum of 2.5 volts; yet the current is too small, about 0.7µA. I mounted 25 White LEDs in series on the breadboard, and found that if the array of LEDs is exposed to bright sunshine, the output voltage is 29.7 Volts. The "H" on the meter's display is there because I pressed the data retention button, labeled "Hold", in order for me to take this picture. The value in the picture is the voltage yielded by the LEDs when exposed to afternoon sunshine, and is close to the max voltage, which is 32 Volts

The output current, however, is still too small, about 0.07µA. I connected a 10 Ohm resistor in series to the positive output of the panel, hoping that, based on Ohm's law, a considerable current would be forced to flow across the resistor, if we presume that the voltage is about 20V. This did not work, for when I set up the meter on the Ampere measuring function, and connected it in series with the circuit, I still got a reading of the aforementioned value, even with the resistor in series.

I presumed that the internal resistance of the LEDs in series could account for this failure, for it appears to be infinite.

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    \$\begingroup\$ So what is your question? And what do you expect? Everything you measured seems OK to me. Of course you will only get a tiny amount of current from these white LED. The current stays the same when you put them in series. Only if you put them in parallels, you might get more current, but then the voltage stays the same. \$\endgroup\$ – Stefan Wyss May 9 '18 at 10:47
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    \$\begingroup\$ Who is saying that an LED makes a good solar panel? \$\endgroup\$ – Andy aka May 9 '18 at 10:55
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    \$\begingroup\$ If you want a more usable amount of power from sunlight use a proper solar panel. There's a reason you see those on top of houses rather than thousands of LEDs. \$\endgroup\$ – Finbarr May 9 '18 at 10:56
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    \$\begingroup\$ Semiconductor diodes are generally not hurt by the reverse current caused by the photoelectric effect. \$\endgroup\$ – Olin Lathrop May 9 '18 at 11:59
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    \$\begingroup\$ Compare the area of each of those tiny LED chips to the area of a typical solar cell. The solar cell is thousands of times larger. It should be no surprise if the solar cell can source thousands of times more current. \$\endgroup\$ – Solomon Slow May 9 '18 at 13:17
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You can boost the current with this kind of setup, however you will need to parallel your strings like this:

schematic

simulate this circuit – Schematic created using CircuitLab

For each string you add you will get an additional 0.7uA, the downside of this is you'll need a lot of LED's.

You can however do useful things with a string of LED's like this, your generating roughly 20uW of energy. Like run a microprocessor

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Any semiconductor diode can be used as a photocell. However, most of them are very inefficient unless designed specifically for that purpose. White LEDs are particularly bad since they produce white light by re-emission of blue or near UV by phosphors. Those phosphors absorb some of the light hitting the LED from outside.

In any case, LEDs don't make good photocells. The total power they can provide will be quite small.

Your basic problem is that you seem to be confusing open circuit voltage with power. Any diode produces about its band gap voltage in reverse when you shine the right light on it and it is left open circuit. However, as you saw, if it can't produce much power then the voltage will rapidly collapse when you try to draw any meaningful current from it.

You can harvest the little power from these LEDs with a well-designed circuit. The first thing to do is to put a capacitor across the LEDs. That will slowly get charged up, asymptotically approaching the open circuit voltage. At that point you have a fixed amount of energy available, regardless of how little the LEDs can provide over time. What happens then depends on how you want to use the energy.

Keep in mind that just sensing the cap voltage and powering the circuit to do so has to take less current than the LEDs require, else the circuit will be a net negative. This can get tricky.

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  • \$\begingroup\$ Given the tiny amount of current the LEDs could provide, I wouldn't be surprised if a cheap capacitor loses more than it holds. \$\endgroup\$ – Mast May 9 '18 at 17:41
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You are missing two major issues.

First, LEDs are tiny, so an LED of your type will only intercept a teensy bit of power. Granted, the shape of the LED will tend to concentrate the incoming light, but the broader the LED beam (and whites are generally very broad) the less effective it is as a concentrator. And the diffusion caused by the phosphor is not remotely helpful, either. As you've noticed, you can get very nice voltages, but no current to speak of. Since power is voltage times current, this should not be a surprise.

Second, LEDs are not good at power generation. To begin with, unlike photodiodes or solar cells (and solar cells are basically very large photodiodes), the same quantum effects which make LEDs good at emitting a narrow band of wavelengths mean that an LED will only absorb light over the same narrow band of wavelengths. White LEDs are actually a blue LED with a phosphor coating, and phosphors don't work backwards to convert broad-spectrum light to a narrow blue one.

Put the two together and you've got a device which doesn't absorb a whole lot of light in the first place, and which ignores most of what it does see.

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  • \$\begingroup\$ LEDs actually absorb colors to the left of the base color on a chart: UV LEDs hardly anything, IR LEDs the most. Waterclear red is the most cost effective. \$\endgroup\$ – dandavis May 10 '18 at 1:43

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