LEDs as photo diodes

The LEDs work otherwise. Each has about 1.6V forward voltage as determined with the multimeter and block in reverse direction. They all shine when powered with 18.8V.

Not disagreeing, but how did you measure the 1,4V?

Probably would help to know more about what you are doing to measure things as well as the general setup.

The LED die is often tiny and the currents are likely fairly small. You describe a 60W incandescent bulb as the source. The black-body radiation temperature is said to be about

2500-2600C. Planck's law then describes the distribution curve over wavelength.

Assume everything from about 600nm and shorter _may_ produce an electron. Out of a radiance of 76.3 W/cm^2-steradian at

2550C, I get about 1.12 W/cm^2-steradian at and shorter than 600nm. This suggests that about 1.47% of emitted light is emitted in wavelengths short enough to yield electrons in your LED. You can compare this with stated luminous efficiencies for 60W bulbs that are a little over 2% -- but they include some longer wavelengths in that figure, I think. The LED itself will have a small quantum efficiency, as well. Over all wavelengths shorter than 600nm, this might average to about 10% of the 1.47% mentioned above, I'd guess. So maybe 0.15% or a factor of 1.5e-3.

An incandescent bulb emits power over a sphere that is roughly uniform in all directions. The area of the sphere is

4*pi*r^2. So if you place your LED at a distance of half a meter away, the sphere has about 3.1 m^2 surface area. Now the LED itself may be (.7mm)^2 or 490e-9 m^2. This means that if you get the LED nicely lined up facing the light adn the LED encapsulation doesn't reflect away any of the light nor absorb it on the way to the die, that the LED die may intercept about 1.5e-3 * 60 watts * (490e-9/3.1), or about 1.4e-8 watts converting to current. Assuming a work function voltage near your 1.4V figure (which is stretching it, I think, as I believe it should be higher than that), that would be about 10nA or so. That's not going to tweak a 200uA scale much.

If you placed it a great deal closer, say 10cm from the center of the bulb, this would work out to 25 times more (5 times closer, squared) or about 250nA. Still 1/4th of 1uA.

However, I'm out of my element here. Don K. might step in and fix up my words here with some better analysis for you. But that's how I see what you are facing, for now.

Jon

Should read more like "... about 1.5e-3 * 60 watts * (490e-9/3.1), or about 1.4e-8 watts. Converting to current and assuming a work function voltage near your 1.4V figure (which is stretching it, I think, as I believe it should be higher than that), that would be about 10nA or so."

Jon

not enough current to produce a higher voltage into the load your meter presents

try a series-parallel arrangement.

FWIW, I've seen weak photo-diode behaviour from glass-cased 1N914 diodes too.

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I made some new measurements. DMM is Metex M-3800, DC 2V range, negative probe connected to LED cathode via alligator clip test leads, positive probe to anode. I held the LED at the insulation of the alligator clips to avoid creepage current through my fingers. LEDs were held horizontally at the 60 W incandescent lightbulb (brand Lightway), touching it just opposite to the socket. The leds top is a flat, frosted arrow shape. I tried to find the position of maximum voltage and sometimes it appeared the voltage rised with time (within 20s). Maybe there was an input capacitance being charged up.

green LED (V):

1.288

yellow LEDs (some have an orange tint; mV). The first 2 or 3 are not part of the 10 series LEDs.

270 530 550 680 590 690 560 590 580

The 10 yellow LEDs are put in a breadboard. Same DMM, alligator test leads to the anode of the first LED and cathode of the last LED. The 60 W light bulb may not have been as close as with the measurements above.

To exclude any effects of the bread board I soldered 2 yellow LEDs in series. Just soldered a cathode and an anode terminal together, no PCB etc. The individual LEDs produced 490 mV and 650 mV respectively. Both in series measured 500 mV.

Anyone care to reproduce this experiment. Connect 2 LEDs in series and measure the individual and series photovoltages close to (touching) an incandescent light bulb?

Thanks for your effort of explaining the unmeasurably small short curcuit current. This is no big problem for me, since I have no reason to believe it should be otherwise. But what puzzles me is that the photovoltages don't add up.

Bernhard

5 pairs of parallel LEDs connected in series give 1065 mV. Clearly above individual photovoltage, although not 5 times as much. A 3x3 configuration gave me 1700 mV, pretty much 3 times the individual voltage.

About the 2 diodes that individually give 490 mV and 650 mV individually and 500 mV in series ... could it be that the diodes only allow their individual photocurrent so the total current would be limited by the weaker diode and my DMM would measure the voltage across a sense resistor and so rather measures current than voltag? I guess a FET might measure voltage more correctly.

Bernhard

The 3x3 config measures 0.1 uA (the minimum on my DMM). 10 LEDs parallel measure 0.5 uA. Photovoltage of 10 paralell yellow LEDs is 1300 mV.

Bernhard

This sounds like the answer. The 10Meg (or lower) input impedance of your meter is shorting out all the current. If you've got any large value resistors (100 Meg or 1 G ohm) you could try putting one in series with the meter. The meter is then the bottom of a voltage divider and you can do the math to figure out the voltage from the LEDS... course you're still drawing some current from them..... otherwise perhaps a FET opamp as buffer.

George H.

I am not sure what's happening. But I can wander my mind around a little. A photon may produce an electron that isn't recombined and otherwise becomes an element of a net current. Without anything attached to the LED leads, these electrons gradually accumulate on one side and yield a potential difference, much like charging a capacitor. However, as the voltage difference builds up across the junction, a counter current develops and acts to limit the maximum voltage. I don't know about LEDs, but silicon diode detectors will limit out at around 0.5V-0.6V. So it does NOT surprise me to hear you measuring in that area for individual LEDs. I don't think you were surprised, either. And the key here is that the LED cannot be "seen" as a voltage source... it's not.

Another way to view the LED is as a current source that depends upon the light to set the current. That also doesn't work in the end, though. It won't produce high voltages, for example, if you attach a large resistor value to it, because internal discharging will extinguish that almost immediately.

The volt meter requires some current draw and places a load on all this -- a separate pathway for the current to go. This _also_ acts to discharge the capacitance.

Imagine the diode model looking something like this:

And your meter as a 10Meg resistor. In the above model, R1 is probably pretty large. For an LED, I've read once or twice from John Larkin (if I read him right, of course) that it is remarkably high. Since I already know that photodiodes (silicon) can readily be in the few GigOhm area and since I know that John knows this much, I have to guess that John was suggesting much higher than this, even. R2 is probably not very large. Perhaps tenths of Ohms? C1 is the diode junction capacitance and for a photodiode depends on area, in part, and on the thickness of the depletion region (which I think means it is a variable value because as C1 charges up the voltage across D1 rises and the depletion region thickness changes.) I think it varies by some root power of the thickness, but the basic idea is that C1 isn't fixed but probably varies depending upon the voltage you measured.

The current source there is basically what I talked about before -- the current generated by the incident light striking your LED.

Now, you place your volt meter with a 10Meg load at the end and measure 500mV to 600mV, let's say. You are basically measuring the voltage across C1, but note that your meter is (ignoring R2 for a moment) basically in parallel with R1 and since R1 is so large, your meter _becomes_ the new R1. So something like this:

You measure, let's say, 500mV. Into 10Meg, this would be

50nA. So 50nA is flowing in the meter itself and it is given that your current source must be producing that much for you to get the reading. Some more will be flowing via D1. Assume for now that it is 10% of the 50nA or 5nA. For a typical silicon diode (not an LED) it might change by say 100mV per 10-fold change in current. So if there were 50nA flowing through D1, another 100mV higher or 600mV total. Which would then suggest 60nA via your meter. So here, to read 600mV instead of 500mV, the photocurrent would need to rise from 55nA (50nA via meter plus 5nA via diode) to 110nA (60nA via meter and 50nA via diode.) To read 700mV would require a photocurrent of 570nA, 800mV a photocurrent of 5.08uA, and so on.

Now stack up two of these.

Let's look at the earlier case, with a 55nA photodiode current. Assume for now that both current sources are yielding this same value. What happens?

Well, you've got current source 1 and current source 2 stacked up. The current from current source 1 goes through current source 2 and then back via your meter. The two diodes are indeed stacked, so you might at first imagine that there would be 500mV + 500mV there. But wait. That would mean 1V across your meter. But your meter is 10Meg Ohm. That would mean 100nA. But you don't _have_ 100nA! You only have

55nA! So what does this mean?

Well, assume for a moment that the diodes adjust by that

100mV per decade of current I mentioned before. The current we have is 55nA, so: 55nA = (V/10M) + Id

But also,

V = 2*(500mV + 100mV*LOG10(Id/5nA))

Before, your reading was 500mV. Now, the value for the current bypassing via D1 and D2 (Id) will be about 28pA, instead of the prior 5nA we took before. MUCH LESS. And almost _all_ of the 55nA will go through the meter now. What happens then? The meter reading goes to about 550mV.

Not much different.

Now explore what happens when the photocurrent is _much_ higher. If you do, you will see that the voltages _do_ stack up nicely. Because the meter itself isn't the lowest impedance pathway anymore. D1 and D2 become much lower. The basic idea here is that the instantaneous slope of the effective resistance curve for D1 and D2 isn't the same all the time and varies depending on how much current is passing via them. Sometimes, they are very high compared to your meter; sometimes very low compared to it; sometimes about the same.

At least, it seems like that to me.

Jon

They should produce less voltage as solar cells than as LEDs - I would expect the measured 1.4V each, for total of 14V.

They may only be producing sufficient current (few microamps ???) to get your voltmeter to read a little more than with only one LED being used as a solar cell.

Some of the LEDs may not be aimed well at the light source for that matter.

- Don Klipstein ( snipped-for-privacy@misty.com)

The amount of current a LED produces as a light sensor is miniscule. The light must be "on axis" shining right into the die to get the voltage you measured - ALL leds must have the light shining on the dies to get the voltages to add. It takes a lot of care to get it right, but when you do they work just like you think they should. I know from empirical experimentation - with sunlight.

Seems axiomatic that in any solar system the voltage/current output is only as good as the weakest one(s).

You may not see any current. Your (digital) meter has a 10 meg ohm input impedance? That is about 1/8 th of a micro amp. 200 ua full scale, is not enough resolution most days (especially when you factor in the spec that says plus or minus one digit as most digital meters are spec'd)

The stack of LEDs may well be current limiting into the voltmeter input resistance, so adding more LEDs in a series stack makes no more apparent voltage.

John

I think I went into much detail already on this subject. The OP being Austrian (or at least, with an email suggesting it), I tend to imagine the OP wants the kind of deeper detail most Germans I know seem to prefer. So I provided some of it after double checking with the models shown in the Hamamatsu detector manuals I have laying about. (Using that idea as an analog for LEDs.) Hopefully, it was accepted okay.

Jon