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From: Bernhard Kuemel on 28 Jan 2010 15:04
Hi seb!

I used LEDs experimentally as photo diodes, yielding up to 1,4V with
green LEDs when near a 60W incandescent light bulb. Right now I put 10
LEDs in series to get higher voltages, but the voltages don't add?! All
I get is a voltage similar to the individual voltages - up to about 400
mV with yellow LEDs.

How can that be?

No current is measurable when I close the circuit with the multimeter
even at the 200 uA range.

Thanks, Bernhard
From: Bernhard Kuemel on 28 Jan 2010 15:08
Bernhard Kuemel wrote:
> Hi seb!
>
> I used LEDs experimentally as photo diodes, yielding up to 1,4V with
> green LEDs when near a 60W incandescent light bulb. Right now I put 10
> LEDs in series to get higher voltages, but the voltages don't add?! All
> I get is a voltage similar to the individual voltages - up to about 400
> mV with yellow LEDs.
>
> How can that be?

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.
From: Jon Kirwan on 28 Jan 2010 22:51
On Thu, 28 Jan 2010 21:04:57 +0100, Bernhard Kuemel
<bernhard(a)bksys.at> wrote:

>Hi seb!
>
>I used LEDs experimentally as photo diodes, yielding up to 1,4V with
>green LEDs when near a 60W incandescent light bulb.

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

>Right now I put 10
>LEDs in series to get higher voltages, but the voltages don't add?! All
>I get is a voltage similar to the individual voltages - up to about 400
>mV with yellow LEDs.
>
>How can that be?

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

>No current is measurable when I close the circuit with the multimeter
>even at the 200 uA range.

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.

>Thanks, Bernhard

Jon
From: Jon Kirwan on 28 Jan 2010 22:54
On Thu, 28 Jan 2010 19:51:34 -0800, Jon Kirwan
<jonk(a)infinitefactors.org> wrote:

>... 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.

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
From: Jasen Betts on 29 Jan 2010 05:20
On 2010-01-28, Bernhard Kuemel <bernhard(a)bksys.at> wrote:
> Hi seb!
>
> I used LEDs experimentally as photo diodes, yielding up to 1,4V with
> green LEDs when near a 60W incandescent light bulb. Right now I put 10
> LEDs in series to get higher voltages, but the voltages don't add?! All
> I get is a voltage similar to the individual voltages - up to about 400
> mV with yellow LEDs.
>
> How can that be?

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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