LEDs as photo diodes

[Bernhard Kuemel]

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

 

[Bernhard Kuemel]

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.

 

[Jon Kirwan]

On Thu, 28 Jan 2010 21:04:57 +0100, Bernhard Kuemel
<bernhard@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

 

[Jon Kirwan]

On Thu, 28 Jan 2010 19:51:34 -0800, Jon Kirwan
<jonk@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

 

[Jasen Betts]

On 2010-01-28, Bernhard Kuemel <bernhard@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.

--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[Bernhard Kuemel]

Jon Kirwan wrote:

On Thu, 28 Jan 2010 21:04:57 +0100, Bernhard Kuemel
bernhard@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?

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

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.

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?

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.

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

 

[Bernhard Kuemel]

Jasen Betts wrote:

On 2010-01-28, Bernhard Kuemel <bernhard@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.

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

 

[Bernhard Kuemel]

Bernhard Kuemel wrote:

Jasen Betts wrote:
On 2010-01-28, Bernhard Kuemel <bernhard@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.

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.

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

 

[George Herold]

On Jan 29, 5:20 am, Jasen Betts <ja...@xnet.co.nz> wrote:

On 2010-01-28, Bernhard Kuemel <bernh...@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.

--- news://freenews.netfront.net/ - complaints: n...@netfront.net ---

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.

 

[Jon Kirwan]

On Fri, 29 Jan 2010 12:37:49 +0100, Bernhard Kuemel
<bernhard@bksys.at> wrote:

snip
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.
snip

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:

: R2
: ,-------+------+------+------/\/\-----
: | | | |
: | | | |
: / \ --- \ ---
:current ^ \ / D1 / R1 --- C1
: source | --- \ |
: \ / | / |
: | | | |
: | | | |
: '-------+------+------+---------------

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:

: ,-------+------+------,
: | | | |
: | | | |
: / \ --- \ ---
:current ^ \ / D1 / R1 --- C1
: source | --- \ |
: \ / | / 10M |
: | | | |
: | | | |
: '-------+------+------'

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.

: ,-------+------+-------,
: | | | |
: | | | |
: / \ --- --- |
:current ^ \ / D2 --- C2 |
: source | --- | |
: 2 \ / | | |
: | | | \
: | | | / Rmeter
: +-------+------+ \
: | | | / 10M
: | | | \
: / \ --- --- |
:current ^ \ / D1 --- C1 |
: source | --- | |
: 1 \ / | | |
: | | | |
: | | | |
: '-------+------+-------'

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

 

[Don Klipstein]

In <5cb25$4b61eed3$557f6e77$9089@news.inode.at>, Bernhard Kuemel wrote:

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.

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 (don@misty.com)

 

[default]

On Thu, 28 Jan 2010 21:04:57 +0100, Bernhard Kuemel
<bernhard@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?

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

Thanks, Bernhard

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)
--

 

[John Larkin]

On Sat, 30 Jan 2010 12:02:54 -0500, default <default@defaulter.net>
wrote:

On Thu, 28 Jan 2010 21:04:57 +0100, Bernhard Kuemel
bernhard@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?

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

Thanks, Bernhard

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

 

[Jon Kirwan]

On Sun, 31 Jan 2010 16:16:04 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 30 Jan 2010 12:02:54 -0500, default <default@defaulter.net
wrote:

On Thu, 28 Jan 2010 21:04:57 +0100, Bernhard Kuemel
bernhard@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?

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

Thanks, Bernhard

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.

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