Prev: Low Power Satellite Based Laser / Imaging System Could EasilyTrack Activity Disturbing Cheap Reflecting Fibers
Next: Analysis- With an Introduction to Proof, 4-E-Instructors SM is available at affordable prices. Email me at instructors.team[at]gmail.com if you need to buy this. All emails will be answered ASAP.
From: Jon Kirwan on 29 Jan 2010 17:04
On Fri, 29 Jan 2010 12:37:49 +0100, Bernhard Kuemel
<bernhard(a)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
From: Don Klipstein on 30 Jan 2010 00:49
In <5cb25$4b61eed3$557f6e77$9089(a)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(a)misty.com)
From: default on 30 Jan 2010 12:02
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. 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)
--
From: John Larkin on 31 Jan 2010 19:16
On Sat, 30 Jan 2010 12:02:54 -0500, default <default(a)defaulter.net>
wrote:

>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. 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

From: Jon Kirwan on 31 Jan 2010 21:04
On Sun, 31 Jan 2010 16:16:04 -0800, John Larkin
<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote:

>On Sat, 30 Jan 2010 12:02:54 -0500, default <default(a)defaulter.net>
>wrote:
>
>>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. 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
First  |  Prev  | 
Pages: 1 2 3
Prev: Low Power Satellite Based Laser / Imaging System Could EasilyTrack Activity Disturbing Cheap Reflecting Fibers
Next: Analysis- With an Introduction to Proof, 4-E-Instructors SM is available at affordable prices. Email me at instructors.team[at]gmail.com if you need to buy this. All emails will be answered ASAP.