led's

"Mr. Man-wai Chang" <toylet.toylet@gmail.com> wrote in message
news:kte3gb$12l$1@dont-email.me...
On 1/08/2013 7:24 AM, John Larkin wrote:
On Wed, 31 Jul 2013 21:14:04 +0800, "Mr. Man-wai Chang"
Avago Technologies
Red non-diffuse round LED,1mA 1.6V
http://uk.rs-online.com/web/p/visible-leds/0826521/
Is it the one you were talking about?

If red is the same color as green, yes.

Have you ever touched a real one? :)
Click to expand...
Someone as dim as you should probably steer clear of sarcasm!
 
On 8/1/13 6:59 AM, John Fields wrote:
On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.
You've got cause and effect backwards here. Forcing 20ma through it will
Click to expand...
cause ~3.2v drop across the LED. It is not necessarily the case that
3.2v will cause 20ma. Diodes have an exponential curve relating voltage
to current, a slight change in voltage can have a significant change in
current. Which is why you want to have some other device (eg, a
resistor) to help set the current.
Ergo, 3 volt LED.
Ergo, Nope.
Click to expand...
 
On Thu, 01 Aug 2013 11:10:08 -0700, Daniel Pitts
<newsgroup.nospam@virtualinfinity.net> wrote:

On 8/1/13 6:59 AM, John Fields wrote:
On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.
You've got cause and effect backwards here. Forcing 20ma through it will
cause ~3.2v drop across the LED. It is not necessarily the case that
3.2v will cause 20ma. Diodes have an exponential curve relating voltage
to current, a slight change in voltage can have a significant change in
current. Which is why you want to have some other device (eg, a
resistor) to help set the current.

Ergo, 3 volt LED.
Ergo, Nope.
Click to expand...
---
It appears that you've entirely missed the point, which is that with
20mA forced through the LED, the voltage across it will vary from a
minimum of 2.8 to a maximum of 3.8V.

Such being the case, there will be some number of LEDs which, when
20mA is forced through them, will drop 3.0V.

Those LEDs, then, because of their own internal resistance:

Vf 3V
Rs = ---- = ------- = 150 ohms,
If 0.02A

will pass 20mA when they're excited by a non-current-limited 3 volt
supply.

--
JF
 
On Wednesday, July 31, 2013 7:28:42 PM UTC-4, John Larkin wrote:
On Wed, 31 Jul 2013 08:00:56 -0700 (PDT), George Herold

gherold@teachspin.com> wrote:



On Tuesday, July 30, 2013 12:00:38 PM UTC-4, John Larkin wrote:

On Tue, 30 Jul 2013 07:15:57 -0700 (PDT), George Herold <gherold@teachspin.com



wrote:

On Monday, July 29, 2013 5:25:15 PM UTC-4, John Larkin wrote:



On Tue, 30 Jul 2013 01:28:23 +0800, "Mr. Man-wai Chang"

snip previus tuff

I was surprsded by the sub-nA light output. At roughly 100 mV/decade







current, I'd have guessed that the voltage drop would be so low that



it wouldn't have enough energy to make photons.



Ahh, but it's those electrons on the far end of the exponential 'Boltzmann tail' that have enough extra thermal energy to make it into the depletion region. (I'll have to try I-V curves for LED's at low temperatures someday, dunking into LN2 should be pretty easy.)







Well, now you're talking about physics and science and stuff like that..

Grin.. well lots of electronics looks like a branch of science to me.





Seems to me that you need X volts to make a photon of X electron-volts energy.





Can thermal energy bootstrap an electron to make a photon that's more EV than



the voltage across the junction?

There was some news 'splash' from (maybe) mit of someone getting more light out of an led than power in.. at some ridiculously low current.



(The good thing about electrons is there are a whole lot of them, so there are a few with many kT's of kinetic energy.)



There's also this 'common' physics demo where they look at the led wavlength and forward voltage drop for a bunch of different color leds and plot things up to get a measure of Planck's constant. (A bit of a bogus experiment when you look at the details.)







Modern LEDs are remarkable gadgets. It would be fun to explore the corner cases,



"someday" as you say.







We're currently experimenting with cheap (like, $12) semiconductor lasers to



find some that accidentally make clean picosecond light pulses when whacked by



one of our laser drivers. The manufacturers certainly don't know if they might.



Most laser data sheets are pitiful.



Just spent over a grand on Thorlabs hardware to hold a laser in place and get



some of its light into a fiber.

Ouch.. the only thing more expensive than optics is high vacuum.



Speaking of modulating diode lasers, Cliff Stoll (who's quite a lovable 'character'.) Was visiting the other day. He does a bunch of educational outreach and uses a modulated diode laser, beam splitter, and cheap corner cube reflector to measure the speed of light. But what he needs is a cheap (fast) photodiode detector. I put him on to Phil's book... but I've been thinking it might be a nice project.

Phil is fond of ebay APDs, which are apparently surplus from some

expensive projects. How fast do you need? Pulse or sine wave?
Click to expand...
Yeah, all good questions. I'll have to send an email to Cliff. I'm guessing it'll be easier to modulate with a sine wave. So at 100MHz I'd get a full 2*pi phase shift with a path lenght of 3 meters. That seems reasonable. But I guess a bit slower would work too ~50 MHz. (Sometimes I make something, measure it, and then define the spec.)

I'm not sure we can use ebay as a source, unless I can buy a few build it and then get several hundred more. And do I need APD's? I've been reverse biasing all sorts of diodes lately. The optoelectronics PD's I'm using list a maximum reverse bias of 30V, I've had 'em up to 60V and no problem. (I ran out of voltage.) I was wondering if I could make garden variety PD's avalanche.

George H.
--



John Larkin Highland Technology, Inc



jlarkin at highlandtechnology dot com

http://www.highlandtechnology.com



Precision electronic instrumentation

Picosecond-resolution Digital Delay and Pulse generators

Custom laser drivers and controllers

Photonics and fiberoptic TTL data links

VME thermocouple, LVDT, synchro acquisition and simulation
Click to expand...
 
On Thu, 01 Aug 2013 08:59:26 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Tue, 30 Jul 2013 18:33:40 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 08:49:21 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 30 Jul 2013 21:11:06 +0800, "Mr. Man-wai Chang"
toylet.toylet@gmail.com> wrote:


1. Would a 3V LED short itself when connected to a 3V DC input?

There is no "3V LED"

---
I don't believe that's true, since some LEDs will drop 3V at their
rated current.

OK, show me a data sheet for a "3 volt LED."

---
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.

Ergo, 3 volt LED.
Click to expand...
Cool, Digikey can quit listing resistors by ohms and start listing
them all by voltage. That will save a lot of people doing hard math.




--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation
 
On Thu, 01 Aug 2013 14:13:26 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 11:10:08 -0700, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 8/1/13 6:59 AM, John Fields wrote:
On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.
You've got cause and effect backwards here. Forcing 20ma through it will
cause ~3.2v drop across the LED. It is not necessarily the case that
3.2v will cause 20ma. Diodes have an exponential curve relating voltage
to current, a slight change in voltage can have a significant change in
current. Which is why you want to have some other device (eg, a
resistor) to help set the current.

Ergo, 3 volt LED.
Ergo, Nope.

---
It appears that you've entirely missed the point, which is that with
20mA forced through the LED, the voltage across it will vary from a
minimum of 2.8 to a maximum of 3.8V.

Such being the case, there will be some number of LEDs which, when
20mA is forced through them, will drop 3.0V.

Those LEDs, then, because of their own internal resistance:

Vf 3V
Rs = ---- = ------- = 150 ohms,
If 0.02A

will pass 20mA when they're excited by a non-current-limited 3 volt
supply.
Click to expand...
It's meaningless and circular (and not predictive) to force "Ohm's
Law" onto a nonlinear device. It is a good way to blow up LEDs.

Ohm's Law isn't even a law. It's a statement that some devices more or
less sometimes have a linear relation between voltage and current.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation
 
On Thu, 1 Aug 2013 18:11:59 +0100, "Ian Field"
<gangprobing.alien@ntlworld.com> wrote:

"Mr. Man-wai Chang" <toylet.toylet@gmail.com> wrote in message
news:kte3gb$12l$1@dont-email.me...
On 1/08/2013 7:24 AM, John Larkin wrote:
On Wed, 31 Jul 2013 21:14:04 +0800, "Mr. Man-wai Chang"
Avago Technologies
Red non-diffuse round LED,1mA 1.6V
http://uk.rs-online.com/web/p/visible-leds/0826521/
Is it the one you were talking about?

If red is the same color as green, yes.

Have you ever touched a real one? :)

Someone as dim as you should probably steer clear of sarcasm!
Click to expand...
And innocent LEDs.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation
 
On Thu, 1 Aug 2013 10:30:01 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Wednesday, July 31, 2013 7:28:42 PM UTC-4, John Larkin wrote:
On Wed, 31 Jul 2013 08:00:56 -0700 (PDT), George Herold

gherold@teachspin.com> wrote:



On Tuesday, July 30, 2013 12:00:38 PM UTC-4, John Larkin wrote:

On Tue, 30 Jul 2013 07:15:57 -0700 (PDT), George Herold <gherold@teachspin.com



wrote:

On Monday, July 29, 2013 5:25:15 PM UTC-4, John Larkin wrote:



On Tue, 30 Jul 2013 01:28:23 +0800, "Mr. Man-wai Chang"

snip previus tuff

I was surprsded by the sub-nA light output. At roughly 100 mV/decade







current, I'd have guessed that the voltage drop would be so low that



it wouldn't have enough energy to make photons.



Ahh, but it's those electrons on the far end of the exponential 'Boltzmann tail' that have enough extra thermal energy to make it into the depletion region. (I'll have to try I-V curves for LED's at low temperatures someday, dunking into LN2 should be pretty easy.)







Well, now you're talking about physics and science and stuff like that.

Grin.. well lots of electronics looks like a branch of science to me.





Seems to me that you need X volts to make a photon of X electron-volts energy.





Can thermal energy bootstrap an electron to make a photon that's more EV than



the voltage across the junction?

There was some news 'splash' from (maybe) mit of someone getting more light out of an led than power in.. at some ridiculously low current.



(The good thing about electrons is there are a whole lot of them, so there are a few with many kT's of kinetic energy.)



There's also this 'common' physics demo where they look at the led wavlength and forward voltage drop for a bunch of different color leds and plot things up to get a measure of Planck's constant. (A bit of a bogus experiment when you look at the details.)







Modern LEDs are remarkable gadgets. It would be fun to explore the corner cases,



"someday" as you say.







We're currently experimenting with cheap (like, $12) semiconductor lasers to



find some that accidentally make clean picosecond light pulses when whacked by



one of our laser drivers. The manufacturers certainly don't know if they might.



Most laser data sheets are pitiful.



Just spent over a grand on Thorlabs hardware to hold a laser in place and get



some of its light into a fiber.

Ouch.. the only thing more expensive than optics is high vacuum.



Speaking of modulating diode lasers, Cliff Stoll (who's quite a lovable 'character'.) Was visiting the other day. He does a bunch of educational outreach and uses a modulated diode laser, beam splitter, and cheap corner cube reflector to measure the speed of light. But what he needs is a cheap (fast) photodiode detector. I put him on to Phil's book... but I've been thinking it might be a nice project.

Phil is fond of ebay APDs, which are apparently surplus from some

expensive projects. How fast do you need? Pulse or sine wave?

Yeah, all good questions. I'll have to send an email to Cliff. I'm guessing it'll be easier to modulate with a sine wave. So at 100MHz I'd get a full 2*pi phase shift with a path lenght of 3 meters. That seems reasonable. But I guess a bit slower would work too ~50 MHz. (Sometimes I make something, measure it, and then define the spec.)

I'm not sure we can use ebay as a source, unless I can buy a few build it and then get several hundred more. And do I need APD's? I've been reverse biasing all sorts of diodes lately. The optoelectronics PD's I'm using list a maximum reverse bias of 30V, I've had 'em up to 60V and no problem. (I ran out of voltage.) I was wondering if I could make garden variety PD's avalanche.
Click to expand...
A modulated laser, ballpark 1 milliwatt, will make gobs of signal into
an ordinary photodiode, no need to avalanche. If you're using sine
waves, and can use a tuned amp, even better.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation
 
On Thu, 01 Aug 2013 12:42:51 -0700, John Larkin
<jlarkin@highlandtechnology.com> wrote:

On Thu, 01 Aug 2013 08:59:26 -0500, John Fields
jfields@austininstruments.com> wrote:

On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Tue, 30 Jul 2013 18:33:40 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 08:49:21 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 30 Jul 2013 21:11:06 +0800, "Mr. Man-wai Chang"
toylet.toylet@gmail.com> wrote:


1. Would a 3V LED short itself when connected to a 3V DC input?

There is no "3V LED"

---
I don't believe that's true, since some LEDs will drop 3V at their
rated current.

OK, show me a data sheet for a "3 volt LED."

---
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.

Ergo, 3 volt LED.

Cool, Digikey can quit listing resistors by ohms and start listing
them all by voltage. That will save a lot of people doing hard math.
Click to expand...
---
In your shop, anyway.

--
JF
 
"John Larkin" <jlarkin@highlandtechnology.com> wrote in message
news:5selv8th0moipoq5sce4pbh60lgmoo3kjs@4ax.com...
On Thu, 1 Aug 2013 18:11:59 +0100, "Ian Field"
gangprobing.alien@ntlworld.com> wrote:



"Mr. Man-wai Chang" <toylet.toylet@gmail.com> wrote in message
news:kte3gb$12l$1@dont-email.me...
On 1/08/2013 7:24 AM, John Larkin wrote:
On Wed, 31 Jul 2013 21:14:04 +0800, "Mr. Man-wai Chang"
Avago Technologies
Red non-diffuse round LED,1mA 1.6V
http://uk.rs-online.com/web/p/visible-leds/0826521/
Is it the one you were talking about?

If red is the same color as green, yes.

Have you ever touched a real one? :)

Someone as dim as you should probably steer clear of sarcasm!

And innocent LEDs.
Click to expand...
TBF - I blew a few up back in the days when the Candela rating was a
challenge to get some visible light out of the damn things.

Back then there were only red ones, and if you could get one to be visible
in sunlight - it didn't do it for long!
 
On Thu, 01 Aug 2013 12:47:49 -0700, John Larkin
<jlarkin@highlandtechnology.com> wrote:

On Thu, 01 Aug 2013 14:13:26 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 11:10:08 -0700, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 8/1/13 6:59 AM, John Fields wrote:
On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.
You've got cause and effect backwards here. Forcing 20ma through it will
cause ~3.2v drop across the LED. It is not necessarily the case that
3.2v will cause 20ma. Diodes have an exponential curve relating voltage
to current, a slight change in voltage can have a significant change in
current. Which is why you want to have some other device (eg, a
resistor) to help set the current.

Ergo, 3 volt LED.
Ergo, Nope.

---
It appears that you've entirely missed the point, which is that with
20mA forced through the LED, the voltage across it will vary from a
minimum of 2.8 to a maximum of 3.8V.

Such being the case, there will be some number of LEDs which, when
20mA is forced through them, will drop 3.0V.

Those LEDs, then, because of their own internal resistance:

Vf 3V
Rs = ---- = ------- = 150 ohms,
If 0.02A

will pass 20mA when they're excited by a non-current-limited 3 volt
supply.

It's meaningless and circular (and not predictive) to force "Ohm's
Law" onto a nonlinear device.
Click to expand...
---
Complete and utter nonsense, since it's a very good way to determine
the linearity of a device.

For example, quite some time ago I measured - at various voltages -
the voltage across, and the current through several incandescent lamp
filaments and, by using Ohm's law was able to plot the resistance of
the filament as a function of the voltage across the filament.

Was delta R/delta V a constant?

I think you know the answer to that one.

In much the same way, I used a diode instead of an incandescent lamp,
and was able to determine that a diode doesn't go ohmic until the
current through it causes the temperature of the diode to go high
enough to almost destroy it. I ran into measurement problems up
there, so the data's not all that reliable, but if there's any
interest I can do the experiment over again and tighten up the
environment.
---

It is a good way to blow up LEDs.
Click to expand...
---
Ya think so???

OK, try this:

1. Get an LED and run its rated current through it.
2. Measure the voltage dropped across the LED.
3. Disconnect the supply and set it to the voltage you measured.
4. Leave the ammeter in place and connect the voltage supply across
the unconnected ends of the ammeter and the LED.
5. Post the currents measured in step 1 and step 4.
---

Ohm's Law isn't even a law. It's a statement that some devices more or
less sometimes have a linear relation between voltage and current.
Click to expand...
---
You're confused.

Since Ohm's law is used to determine the relationship between voltage,
current, and resistance - AT A SINGLE POINT - it, alone, can say
nothing about the linearity of the relationship between voltage and
current.

For _that_ relationship to be determined, a sufficient number of
measurements must be made in order to determine the trend of dI/dV.

--
JF
 
On Thu, 01 Aug 2013 15:58:54 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 12:47:49 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Thu, 01 Aug 2013 14:13:26 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 11:10:08 -0700, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 8/1/13 6:59 AM, John Fields wrote:
On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.
You've got cause and effect backwards here. Forcing 20ma through it will
cause ~3.2v drop across the LED. It is not necessarily the case that
3.2v will cause 20ma. Diodes have an exponential curve relating voltage
to current, a slight change in voltage can have a significant change in
current. Which is why you want to have some other device (eg, a
resistor) to help set the current.

Ergo, 3 volt LED.
Ergo, Nope.

---
It appears that you've entirely missed the point, which is that with
20mA forced through the LED, the voltage across it will vary from a
minimum of 2.8 to a maximum of 3.8V.

Such being the case, there will be some number of LEDs which, when
20mA is forced through them, will drop 3.0V.

Those LEDs, then, because of their own internal resistance:

Vf 3V
Rs = ---- = ------- = 150 ohms,
If 0.02A

will pass 20mA when they're excited by a non-current-limited 3 volt
supply.

It's meaningless and circular (and not predictive) to force "Ohm's
Law" onto a nonlinear device.

---
Complete and utter nonsense, since it's a very good way to determine
the linearity of a device.

For example, quite some time ago I measured - at various voltages -
the voltage across, and the current through several incandescent lamp
filaments and, by using Ohm's law was able to plot the resistance of
the filament as a function of the voltage across the filament.

Was delta R/delta V a constant?

I think you know the answer to that one.

In much the same way, I used a diode instead of an incandescent lamp,
and was able to determine that a diode doesn't go ohmic until the
current through it causes the temperature of the diode to go high
enough to almost destroy it. I ran into measurement problems up
there, so the data's not all that reliable, but if there's any
interest I can do the experiment over again and tighten up the
environment.
---

It is a good way to blow up LEDs.

---
Ya think so???

OK, try this:

1. Get an LED and run its rated current through it.
2. Measure the voltage dropped across the LED.
3. Disconnect the supply and set it to the voltage you measured.
4. Leave the ammeter in place and connect the voltage supply across
the unconnected ends of the ammeter and the LED.
5. Post the currents measured in step 1 and step 4.
---

Ohm's Law isn't even a law. It's a statement that some devices more or
less sometimes have a linear relation between voltage and current.

---
You're confused.

Since Ohm's law is used to determine the relationship between voltage,
current, and resistance - AT A SINGLE POINT - it, alone, can say
nothing about the linearity of the relationship between voltage and
current.

For _that_ relationship to be determined, a sufficient number of
measurements must be made in order to determine the trend of dI/dV.
Click to expand...
---
Over the range of interest of either current or voltage.

--
JF
 
On Thu, 01 Aug 2013 15:58:54 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 12:47:49 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Thu, 01 Aug 2013 14:13:26 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 11:10:08 -0700, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 8/1/13 6:59 AM, John Fields wrote:
On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.
You've got cause and effect backwards here. Forcing 20ma through it will
cause ~3.2v drop across the LED. It is not necessarily the case that
3.2v will cause 20ma. Diodes have an exponential curve relating voltage
to current, a slight change in voltage can have a significant change in
current. Which is why you want to have some other device (eg, a
resistor) to help set the current.

Ergo, 3 volt LED.
Ergo, Nope.

---
It appears that you've entirely missed the point, which is that with
20mA forced through the LED, the voltage across it will vary from a
minimum of 2.8 to a maximum of 3.8V.

Such being the case, there will be some number of LEDs which, when
20mA is forced through them, will drop 3.0V.

Those LEDs, then, because of their own internal resistance:

Vf 3V
Rs = ---- = ------- = 150 ohms,
If 0.02A

will pass 20mA when they're excited by a non-current-limited 3 volt
supply.

It's meaningless and circular (and not predictive) to force "Ohm's
Law" onto a nonlinear device.

---
Complete and utter nonsense, since it's a very good way to determine
the linearity of a device.

For example, quite some time ago I measured - at various voltages -
the voltage across, and the current through several incandescent lamp
filaments and, by using Ohm's law was able to plot the resistance of
the filament as a function of the voltage across the filament.
Click to expand...
What makes sense is to plot current vs voltage.

When you plotted resistance, was it E/I or dE/dI?

Was delta R/delta V a constant?

I think you know the answer to that one.

In much the same way, I used a diode instead of an incandescent lamp,
and was able to determine that a diode doesn't go ohmic until the
current through it causes the temperature of the diode to go high
enough to almost destroy it. I ran into measurement problems up
there, so the data's not all that reliable, but if there's any
interest I can do the experiment over again and tighten up the
environment.
---

It is a good way to blow up LEDs.

---
Ya think so???

OK, try this:

1. Get an LED and run its rated current through it.
2. Measure the voltage dropped across the LED.
3. Disconnect the supply and set it to the voltage you measured.
4. Leave the ammeter in place and connect the voltage supply across
the unconnected ends of the ammeter and the LED.
5. Post the currents measured in step 1 and step 4.
---

Ohm's Law isn't even a law. It's a statement that some devices more or
less sometimes have a linear relation between voltage and current.

---
You're confused.

Since Ohm's law is used to determine the relationship between voltage,
current, and resistance - AT A SINGLE POINT - it, alone, can say
nothing about the linearity of the relationship between voltage and
current.
Click to expand...
That's goofy. It's not a law, it's just two numbers that you enjoy
dividing because you think it means something.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation
 
On Thu, 1 Aug 2013 21:23:20 +0100, "Ian Field"
<gangprobing.alien@ntlworld.com> wrote:

"John Larkin" <jlarkin@highlandtechnology.com> wrote in message
news:5selv8th0moipoq5sce4pbh60lgmoo3kjs@4ax.com...
On Thu, 1 Aug 2013 18:11:59 +0100, "Ian Field"
gangprobing.alien@ntlworld.com> wrote:



"Mr. Man-wai Chang" <toylet.toylet@gmail.com> wrote in message
news:kte3gb$12l$1@dont-email.me...
On 1/08/2013 7:24 AM, John Larkin wrote:
On Wed, 31 Jul 2013 21:14:04 +0800, "Mr. Man-wai Chang"
Avago Technologies
Red non-diffuse round LED,1mA 1.6V
http://uk.rs-online.com/web/p/visible-leds/0826521/
Is it the one you were talking about?

If red is the same color as green, yes.

Have you ever touched a real one? :)

Someone as dim as you should probably steer clear of sarcasm!

And innocent LEDs.

TBF - I blew a few up back in the days when the Candela rating was a
challenge to get some visible light out of the damn things.

Back then there were only red ones, and if you could get one to be visible
in sunlight - it didn't do it for long!
Click to expand...
When we started using the original Cree SiC LEDs, we ran 50 mA through
them. I used two 74F38 sections in parallel, and something like 39
ohms to +5. Nowadays, 1 mA is about all you want for a blue panel
indicator.

Lots of people find blue, especially bright blue, to be annoying.

Trick: if all you have is 3.3 volt logic,


+5--------BLUE_LED------resistor-------0/3.3_logic


because when the logic is 3.3, the available LED voltage is only 1.7,
and that won't turn on a blue LED.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation
 
On Thu, 01 Aug 2013 15:39:01 -0700, John Larkin
<jlarkin@highlandtechnology.com> wrote:

On Thu, 01 Aug 2013 15:58:54 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 12:47:49 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Thu, 01 Aug 2013 14:13:26 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 11:10:08 -0700, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 8/1/13 6:59 AM, John Fields wrote:
On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.
You've got cause and effect backwards here. Forcing 20ma through it will
cause ~3.2v drop across the LED. It is not necessarily the case that
3.2v will cause 20ma. Diodes have an exponential curve relating voltage
to current, a slight change in voltage can have a significant change in
current. Which is why you want to have some other device (eg, a
resistor) to help set the current.

Ergo, 3 volt LED.
Ergo, Nope.

---
It appears that you've entirely missed the point, which is that with
20mA forced through the LED, the voltage across it will vary from a
minimum of 2.8 to a maximum of 3.8V.

Such being the case, there will be some number of LEDs which, when
20mA is forced through them, will drop 3.0V.

Those LEDs, then, because of their own internal resistance:

Vf 3V
Rs = ---- = ------- = 150 ohms,
If 0.02A

will pass 20mA when they're excited by a non-current-limited 3 volt
supply.

It's meaningless and circular (and not predictive) to force "Ohm's
Law" onto a nonlinear device.

---
Complete and utter nonsense, since it's a very good way to determine
the linearity of a device.

For example, quite some time ago I measured - at various voltages -
the voltage across, and the current through several incandescent lamp
filaments and, by using Ohm's law was able to plot the resistance of
the filament as a function of the voltage across the filament.

What makes sense is to plot current vs voltage.
Click to expand...
---
Of course, since that's all that's available, but it doesn't really
need to be plotted, - except for esthetics - it just needs to be
acquired, the E/I division done, and then the quotient plotted against
voltage.

Would you like to see the plots?
---

When you plotted resistance, was it E/I or dE/dI?
Click to expand...
---
E/I
---

Was delta R/delta V a constant?

I think you know the answer to that one.

In much the same way, I used a diode instead of an incandescent lamp,
and was able to determine that a diode doesn't go ohmic until the
current through it causes the temperature of the diode to go high
enough to almost destroy it. I ran into measurement problems up
there, so the data's not all that reliable, but if there's any
interest I can do the experiment over again and tighten up the
environment.
---

It is a good way to blow up LEDs.

---
Ya think so???

OK, try this:

1. Get an LED and run its rated current through it.
2. Measure the voltage dropped across the LED.
3. Disconnect the supply and set it to the voltage you measured.
4. Leave the ammeter in place and connect the voltage supply across
the unconnected ends of the ammeter and the LED.
5. Post the currents measured in step 1 and step 4.
---

Ohm's Law isn't even a law. It's a statement that some devices more or
less sometimes have a linear relation between voltage and current.

---
You're confused.

Since Ohm's law is used to determine the relationship between voltage,
current, and resistance - AT A SINGLE POINT - it, alone, can say
nothing about the linearity of the relationship between voltage and
current.

That's goofy.
Click to expand...
---
Not at all, it's just true.

If you choose to contest the veracity of my claim, then I expect you
to prove, mathematically, that I'm wrong.

Can you?
---

It's not a law, it's just two numbers that you enjoy
dividing because you think it means something.
Click to expand...
---
Wrong.

It is a law, and it defines the instantaneous relationship between
voltage, current, and resistance in a circuit.

--
JF
 
On Thu, 01 Aug 2013 19:35:47 -0500, John Fields <jfields@austininstruments.com>
wrote:

On Thu, 01 Aug 2013 15:39:01 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Thu, 01 Aug 2013 15:58:54 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 12:47:49 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Thu, 01 Aug 2013 14:13:26 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 01 Aug 2013 11:10:08 -0700, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 8/1/13 6:59 AM, John Fields wrote:
On Wed, 31 Jul 2013 16:23:44 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Wed, 31 Jul 2013 08:58:12 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 30 Jul 2013 17:46:34 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:
http://www.avagotech.com/docs/AV02-1029EN

Those are not "3 volt leds." Their typ forward drop is 3.2, but they
are still not shown as being voltage operable; they are spec'd at 20
mA.

---
Well, with no voltage to drive current through the junction they
wouldn't work at all, so they are certainly "voltage operable".

Also, notice that the data sheet shows a range of from 2.8 to 3.8V
across the junction with 20 mA through it, so 3.0V dropped across the
junction with 20mA through it is certainly within the range of
possibilities.

Consequently, for that LED, connecting 3.0V to it directly will force
20mA through it.
You've got cause and effect backwards here. Forcing 20ma through it will
cause ~3.2v drop across the LED. It is not necessarily the case that
3.2v will cause 20ma. Diodes have an exponential curve relating voltage
to current, a slight change in voltage can have a significant change in
current. Which is why you want to have some other device (eg, a
resistor) to help set the current.

Ergo, 3 volt LED.
Ergo, Nope.

---
It appears that you've entirely missed the point, which is that with
20mA forced through the LED, the voltage across it will vary from a
minimum of 2.8 to a maximum of 3.8V.

Such being the case, there will be some number of LEDs which, when
20mA is forced through them, will drop 3.0V.

Those LEDs, then, because of their own internal resistance:

Vf 3V
Rs = ---- = ------- = 150 ohms,
If 0.02A

will pass 20mA when they're excited by a non-current-limited 3 volt
supply.

It's meaningless and circular (and not predictive) to force "Ohm's
Law" onto a nonlinear device.

---
Complete and utter nonsense, since it's a very good way to determine
the linearity of a device.

For example, quite some time ago I measured - at various voltages -
the voltage across, and the current through several incandescent lamp
filaments and, by using Ohm's law was able to plot the resistance of
the filament as a function of the voltage across the filament.

What makes sense is to plot current vs voltage.

---
Of course, since that's all that's available, but it doesn't really
need to be plotted, - except for esthetics - it just needs to be
acquired, the E/I division done, and then the quotient plotted against
voltage.

Would you like to see the plots?
---

When you plotted resistance, was it E/I or dE/dI?

---
E/I
---

Was delta R/delta V a constant?

I think you know the answer to that one.

In much the same way, I used a diode instead of an incandescent lamp,
and was able to determine that a diode doesn't go ohmic until the
current through it causes the temperature of the diode to go high
enough to almost destroy it. I ran into measurement problems up
there, so the data's not all that reliable, but if there's any
interest I can do the experiment over again and tighten up the
environment.
---

It is a good way to blow up LEDs.

---
Ya think so???

OK, try this:

1. Get an LED and run its rated current through it.
2. Measure the voltage dropped across the LED.
3. Disconnect the supply and set it to the voltage you measured.
4. Leave the ammeter in place and connect the voltage supply across
the unconnected ends of the ammeter and the LED.
5. Post the currents measured in step 1 and step 4.
---

Ohm's Law isn't even a law. It's a statement that some devices more or
less sometimes have a linear relation between voltage and current.

---
You're confused.

Since Ohm's law is used to determine the relationship between voltage,
current, and resistance - AT A SINGLE POINT - it, alone, can say
nothing about the linearity of the relationship between voltage and
current.

That's goofy.

---
Not at all, it's just true.

If you choose to contest the veracity of my claim, then I expect you
to prove, mathematically, that I'm wrong.

Can you?
---

It's not a law, it's just two numbers that you enjoy
dividing because you think it means something.

---
Wrong.

It is a law, and it defines the instantaneous relationship between
voltage, current, and resistance in a circuit.
Click to expand...
The way you use it, it's not a law, it's a definition.




--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
 
On 2/08/2013 1:11 AM, Ian Field wrote:
Have you ever touched a real one? :)

Someone as dim as you should probably steer clear of sarcasm!
Click to expand...
Is this 1mA LED military grade stuff?

--
@~@ Remain silent. Nothing from soldiers and magicians is real!
/ v \ Simplicity is Beauty!
/( _ )\ May the Force and farces be with you!
^ ^ (x86_64 Ubuntu 9.10) Linux 2.6.39.3
不借貸! 不詐騙! 不援交! 不打交! 不打劫! 不自殺! 請考慮綜援 (CSSA):
http://www.swd.gov.hk/tc/index/site_pubsvc/page_socsecu/sub_addressesa
 
On Thursday, August 1, 2013 3:51:40 PM UTC-4, John Larkin wrote:
On Thu, 1 Aug 2013 10:30:01 -0700 (PDT), George Herold

big snip other stuff
Click to expand...


Speaking of modulating diode lasers, Cliff Stoll (who's quite a lovable 'character'.) Was visiting the other day. He does a bunch of educational outreach and uses a modulated diode laser, beam splitter, and cheap corner cube reflector to measure the speed of light. But what he needs is a cheap (fast) photodiode detector. I put him on to Phil's book... but I've been thinking it might be a nice project.



Phil is fond of ebay APDs, which are apparently surplus from some



expensive projects. How fast do you need? Pulse or sine wave?



Yeah, all good questions. I'll have to send an email to Cliff. I'm guessing it'll be easier to modulate with a sine wave. So at 100MHz I'd get a full 2*pi phase shift with a path lenght of 3 meters. That seems reasonable. But I guess a bit slower would work too ~50 MHz. (Sometimes I make something, measure it, and then define the spec.)



I'm not sure we can use ebay as a source, unless I can buy a few build it and then get several hundred more. And do I need APD's? I've been reverse biasing all sorts of diodes lately. The optoelectronics PD's I'm using list a maximum reverse bias of 30V, I've had 'em up to 60V and no problem. (I ran out of voltage.) I was wondering if I could make garden variety PD's avalanche.





A modulated laser, ballpark 1 milliwatt, will make gobs of signal into

an ordinary photodiode, no need to avalanche. If you're using sine

waves, and can use a tuned amp, even better.
Click to expand...
Hey! That's interesting. Could I resonate the PD capacitance with some inductor? I could even tune it a bit with the PD reverse bias. (Or were you thinking of a tuned stage after the PD?) (I had this 'crazy' idea in the past about using a T-coil* as part of a PD front end... only to find that Phil H. had already done it.)

George H.

*this was soon after reading about T-coils in one of the Jim Williams' books.
--



John Larkin Highland Technology, Inc



jlarkin at highlandtechnology dot com

http://www.highlandtechnology.com



Precision electronic instrumentation

Picosecond-resolution Digital Delay and Pulse generators

Custom laser drivers and controllers

Photonics and fiberoptic TTL data links

VME thermocouple, LVDT, synchro acquisition and simulation
Click to expand...
 
On 08/02/2013 09:53 AM, George Herold wrote:
On Thursday, August 1, 2013 3:51:40 PM UTC-4, John Larkin wrote:
On Thu, 1 Aug 2013 10:30:01 -0700 (PDT), George Herold

big snip other stuff


snip

A modulated laser, ballpark 1 milliwatt, will make gobs of signal
into an ordinary photodiode, no need to avalanche. If you're using
sine waves, and can use a tuned amp, even better.

Hey! That's interesting. Could I resonate the PD capacitance with
some inductor? I could even tune it a bit with the PD reverse bias.
(Or were you thinking of a tuned stage after the PD?) (I had this
'crazy' idea in the past about using a T-coil* as part of a PD front
end... only to find that Phil H. had already done it.)

George H.

*this was soon after reading about T-coils in one of the Jim
Williams' books.
Click to expand...
Yeah, photodiode front end design is often an exercise in creative
desperation. Not a lot of stones remain unturned.

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
 
On Fri, 2 Aug 2013 07:13:30 -0700 (PDT), George Herold <gherold@teachspin.com>
wrote:

On Thursday, August 1, 2013 11:23:52 PM UTC-4, John Larkin wrote:
On Thu, 01 Aug 2013 19:35:47 -0500, John Fields <jfields@austininstruments.com

snip

You're confused.



Since Ohm's law is used to determine the relationship between voltage,

current, and resistance - AT A SINGLE POINT - it, alone, can say

nothing about the linearity of the relationship between voltage and

current.



That's goofy.



---

Not at all, it's just true.



If you choose to contest the veracity of my claim, then I expect you

to prove, mathematically, that I'm wrong.



Can you?

---



It's not a law, it's just two numbers that you enjoy

dividing because you think it means something.



---

Wrong.



It is a law, and it defines the instantaneous relationship between

voltage, current, and resistance in a circuit.



The way you use it, it's not a law, it's a definition.

Hey this is kinda interesting. (But let's not have a big John vs John confrontation.)

So last week I was running this workshop on noise. I knew I'd have some spare time while the attendees were doing stuff. So I took along a setup to measure the Johnson noise of a light bulb with a DC current going through it. (The measruements were a bit of a pain, I had to abandon the inductor I was using as a bias element and go with a simple resistor...anyway that's not important.)
So at some voltage across the light bulb I measured the current. And I took that ratio to be the resistance of the bulb. And then I assumed that the bulb would be making Johnson noise given by v^2 = 4kTR*BW. Where I'd see more noise because of increased temperature of the bulb.
(The idea was to try and measure the temperature.)
Do you think there is something wrong with this 'theory'?
Does the light bulb have resistance?
Does it have Johnson noise?
What's the 'correct' relation between them?

George H.
Click to expand...
Sure, any resistor has Johnson noise.

If you were to drive the filament through an inductor or a noiseless current
source, and couple a small AC signal into it with a capacitor, at higher
frequencies it would look like a resistor of some ohms, and it would have the
corresponding Johnson noise. If the filament is incandescent, the Johnson noise
will be high.

That equivalent Johnson resistance would not be the DC E/I and will (probably)
be close to the slope, dE/dI at the operating point. Thermal mass makes dE/dI
frequency dependant. The exact math is beyond my pay grade.

I bet a filament has mountains of low frequency noise. Might be interesting to
measure.




--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top