If you need assistance, please send an email to forum at 4hv dot org. To ensure your email is not marked as spam, please include the phrase "4hv help" in the subject line. You can also find assistance via IRC, at irc.shadowworld.net, room #hvcomm.
Support 4hv.org!
Donate:
4hv.org is hosted on a dedicated server. Unfortunately, this server costs and we rely on the help of site members to keep 4hv.org running. Please consider donating. We will place your name on the thanks list and you'll be helping to keep 4hv.org alive and free for everyone. Members whose names appear in red bold have donated recently. Green bold denotes those who have recently donated to keep the server carbon neutral.
Special Thanks To:
Aaron Holmes
Aaron Wheeler
Adam Horden
Alan Scrimgeour
Andre
Andrew Haynes
Anonymous000
asabase
Austin Weil
barney
Barry
Bert Hickman
Bill Kukowski
Blitzorn
Brandon Paradelas
Bruce Bowling
BubeeMike
Byong Park
Cesiumsponge
Chris F.
Chris Hooper
Corey Worthington
Derek Woodroffe
Dalus
Dan Strother
Daniel Davis
Daniel Uhrenholt
datasheetarchive
Dave Billington
Dave Marshall
David F.
Dennis Rogers
drelectrix
Dr. John Gudenas
Dr. Spark
E.TexasTesla
eastvoltresearch
Eirik Taylor
Erik Dyakov
Erlend^SE
Finn Hammer
Firebug24k
GalliumMan
Gary Peterson
George Slade
GhostNull
Gordon Mcknight
Graham Armitage
Grant
GreySoul
Henry H
IamSmooth
In memory of Leo Powning
Jacob Cash
James Howells
James Pawson
Jeff Greenfield
Jeff Thomas
Jesse Frost
Jim Mitchell
jlr134
Joe Mastroianni
John Forcina
John Oberg
John Willcutt
Jon Newcomb
klugesmith
Leslie Wright
Lutz Hoffman
Mads Barnkob
Martin King
Mats Karlsson
Matt Gibson
Matthew Guidry
mbd
Michael D'Angelo
Mikkel
mileswaldron
mister_rf
Neil Foster
Nick de Smith
Nick Soroka
nicklenorp
Nik
Norman Stanley
Patrick Coleman
Paul Brodie
Paul Jordan
Paul Montgomery
Ped
Peter Krogen
Peter Terren
PhilGood
Richard Feldman
Robert Bush
Royce Bailey
Scott Fusare
Scott Newman
smiffy
Stella
Steven Busic
Steve Conner
Steve Jones
Steve Ward
Sulaiman
Thomas Coyle
Thomas A. Wallace
Thomas W
Timo
Torch
Ulf Jonsson
vasil
Vaxian
vladi mazzilli
wastehl
Weston
William Kim
William N.
William Stehl
Wesley Venis
The aforementioned have contributed financially to the continuing triumph of 4hv.org. They are deserving of my most heartfelt thanks.
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
Uspring wrote ...
BC: We're looking at the same diagram. The lower black data points seem to indicate a levelling off. From the data alone it's not completely convincing. There is a theoretical reason for levelling off, though. At lower currents, i.e. lower voltages, the fraction of electrons with enough energy for a light emitting transition is smaller. Correspondingly more electrons will lose their energy by non optical decay. At some voltage the ratio won't change much more. I'd expect levelling off at voltages (currents correspond to these) scaling with k*T.
It's a pity they didn't get more data for the 25C measurements at low currents.
I think there's probably a lot of common ground between our viewpoints. I think that an electron has to be excited enough by a combination of thermal and electrostatic energy to get over the barrier, "fall" down and emit light.
The data show that, at low voltages , not many electrons have enough energy to get over the barrier so there's not much current and not much light. Raising the voltage or the temperature increases then number that can make the jump.
All fascinating stuff, but there's still no clear threshold and if there is one, it's well below the voltage quoted for LEDs' "thresholds". As I said earlier (and produced the data to show it) the current vs voltage curve is smooth over that range, and follows the mathematical model for the process.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Bored Chemist wrote ...
I think that an electron has to be excited enough by a combination of thermal and electrostatic energy to get over the barrier, "fall" down and emit light.
I respectfully disagree. In my opinion, two electrons need to move before a photon can be emitted. first, a hole needs to be injected into the lattice. this requires the removal of an electron. Any electron with enough energy to get over the (possibly reduced?) barrier then has a hole to fall into. (I've put this as simply as possible)
Uspring wrote ...
Correspondingly more electrons will lose their energy by non optical decay. At some voltage the ratio won't change much more.
I think what you are suggesting is exactly what is observed in an electro-polishing tank. (I mentioned earlier I was working on a theory, as none of the textbooks I read could explain it). Above the plateau one process occurs, below the plateau another process occurs. I know the machanisms for each process must be different. I have a theory, but I'm trying to learn what I can from this thread before I try to put it into words.
EDIT: In the EP tank graph, it's current plotted against voltage, though. The current plateau at the transition.
All fascinating stuff, but there's still no clear threshold and if there is one, it's well below the voltage quoted for LEDs' "thresholds".
In a strict sense you are right. In this sense LEDs are really never off solely also due to their thermal radiation. The papers main point is about the region, where quantum efficiency is bigger than 1. There light output is 69pW and thermal output in the same wavelength range 40nW, i.e. about 3 orders of magnitude more. So a careful background subtraction had to be made. By any practical means, the LED is off there. Users of blue LEDs will generally agree, that their LEDs are "off" at 2 V, even though, in the above strict sense, they are wrong. Just a matter of definition.
Ash Small wrote:
I respectfully disagree. In my opinion, two electrons need to move before a photon can be emitted. first, a hole needs to be injected into the lattice.
The doping on the p side of the semiconductor creates vacant states into which the electrons can fall into. The electrons there eventually do have to be carried out to the diodes connectors. Otherwise a charge would build up there.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Uspring wrote ...
The doping on the p side of the semiconductor creates vacant states into which the electrons can fall into. The electrons there eventually do have to be carried out to the diodes connectors. Otherwise a charge would build up there.
Exactly. These sites have to be vacant before another electron can 'fall in'. In some ways this process can by compared to the base-emitter channel of a BJT, which, it can be demonstrated, takes time to 'switch off'. In a BJT, collector current continues to flow , presumably until all these 'sites' are full?
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
Ash Small wrote ...
Bored Chemist wrote ...
I think that an electron has to be excited enough by a combination of thermal and electrostatic energy to get over the barrier, "fall" down and emit light.
I respectfully disagree. In my opinion, two electrons need to move before a photon can be emitted. first, a hole needs to be injected into the lattice. this requires the removal of an electron. Any electron with enough energy to get over the (possibly reduced?) barrier then has a hole to fall into. (I've put this as simply as possible)
Uspring wrote ...
Correspondingly more electrons will lose their energy by non optical decay. At some voltage the ratio won't change much more.
I think what you are suggesting is exactly what is observed in an electro-polishing tank. (I mentioned earlier I was working on a theory, as none of the textbooks I read could explain it). Above the plateau one process occurs, below the plateau another process occurs. I know the machanisms for each process must be different. I have a theory, but I'm trying to learn what I can from this thread before I try to put it into words.
EDIT: In the EP tank graph, it's current plotted against voltage, though. The current plateau at the transition.
Re."In my opinion, two electrons need to move before a photon can be emitted. first, a hole needs to be injected into the lattice. this requires the removal of an electron" Actually, an awful lot of electrons need to move- one leaves the negative terminal of the battery and it takes the place of one that has moved along a bit, which takes the place of one that has moved along a bit... and so on for a zillion electrons, all round the circuit. However, the "two" electrons you are talking about don't flow past the same place at the same time, so they are part of just 1 electron's worth of current.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Bored Chemist wrote ...
However, the "two" electrons you are talking about don't flow past the same place at the same time, so they are part of just 1 electron's worth of current.
Yes, but one has to move first. The hole can't be injected until the current starts to flow. The site must be vacant before the second electron can be attracted to it, surely?
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
It's being "attracted" to the positive pole of the battery. The point is that it's still like having a string of beads. If you move the string so that 1 bead passes a point each second then it's 1 bead per second, no matter how long the string is.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Bored Chemist wrote ...
It's being "attracted" to the positive pole of the battery. The point is that it's still like having a string of beads. If you move the string so that 1 bead passes a point each second then it's 1 bead per second, no matter how long the string is.
I get that. I also know that most textbooks are pretty vague and simplistic, to the point that they use words like 'threshold'
I'm just trying to understand exactly where the various forces involved are actually applied. for example, any electrons with sufficient thermal energy will be able to jump backwards and forwards over the junction, if there is no attractive force from a hole in the valence band. The energy released upon re-combination is dependent on the forward voltage of the LED.....I think I'm getting a better picture now.
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
There is an old joke about a man running from the police who jumped over a low wall in his attempt to escape; and found that, on the other side it was the rather high wall of the local prison. The police arrested him in the prison hospital. Only one side of the pn junction is at ground potential.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Bored Chemist wrote ...
There is an old joke about a man running from the police who jumped over a low wall in his attempt to escape; and found that, on the other side it was the rather high wall of the local prison. The police arrested him in the prison hospital. Only one side of the pn junction is at ground potential.
Is there no conduction band on the other side, like in a BJT? Don't they have to cross the junction THEN fall from the conduction band to the valence band?
This site is powered by e107, which is released under the GNU GPL License. All work on this site, except where otherwise noted, is licensed under a Creative Commons Attribution-ShareAlike 2.5 License. By submitting any information to this site, you agree that anything submitted will be so licensed. Please read our Disclaimer and Policies page for information on your rights and responsibilities regarding this site.
Was 14V LED question
