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Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Dr. Slack wrote ...
Ash Small wrote ...
I think it's also safe to assume that, if the LED is conducting, it must be emitting photons, just not in sufficient quantities to be detected by the human eye.
No, I think that's an unsafe assumption. I have no difficulty in believing, expecting even, that a current can flow across a junction without doing the 'thing' you hoped it would, expected it to do at a higher voltage drop. I'd called that a leakage current, a lot of other people would as well.
You say so yourself a few lines later
Ash Small wrote ...
I'm currently of the opinion that not all the electrons which pass through the LED emit photons.
come on, get a grip, argue consistently.
Maybe I didn't pick the best example, but you were arguing that no current will flow before the threshold voltage is supplied. We've shown that some current can flow with less than this voltage applied.
Regarding the point I made that I don't think all the electrons which pass through the LED emit photons, I did clarify that by saying that I think it's only the electrons that form what would be the base current in a BJT.
Those electrons that flow entirely within the conduction band, I think, don't emit light. These can't flow unless 'base current' electrons are also playing a part (according to conventional semiconductor theory), which led me to the conclusion that an efficient LED must have very low gain, so that few electrons flow entirely within the conduction band.
Obviously, in a diode, the base and collector are combined into a single terminal.
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
Dr. Slack wrote ...
Hi BC.
Thanks for quoting me correctly.
It's an interesting link. I'd imagine the physics is much the same as for evaporative cooling, so in a population with a thermal spread of energy, the most energetic ones can surmount a barrier which is above the average energy, so resulting in a net loss of energy to the population as the energetic ones escape.
I wonder if you can help me with a problem I'm having. I am building a bench power supply, with a blue LED on the front panel as an indicator light. I have 2v and 5v supplies available. I've connected the LED to the 2v supply with a 100ohm resistor in series, but there doesn't seem to be any light coming out. Can you suggest any way I can make it visible please?
Sure, switch off the lights and wait for your eyes to get accustomed to the dark. Then wait. Eventually you will get enough photons to see. (this may take some time but it will be quicker if you warm the LED up a bit.).
I must admit I ran out of patience, but I could still see the glow from a blue LED ( A rather old RS 577-617 if you are interested) when it was driven at 2.6 volts where it drew about 700 nA. On a slightly more practical note, it lit up quite nicely at 3 volts: drawing about 20mA. So (making the rather dubious assumption that yours is that same - you should probably test it first) you might be able to light the LED nice and brightly by connecting it between the 2v and 5V lines and save the cost of the resistor (don't spend all the money at once). Of course the people who believe in thresholds won't expect it to light- but, as you point out, LEDs don't read, so they don't know they have a threshold. It seems nobody told them they need 3.4 volts so they run OK at less than that- even down to just 2.6.
Registered Member #230
Joined: Tue Feb 21 2006, 08:01PM
Location: Gracefield lower Hutt
Posts: 284
OK The .4 volts I was referring to is the operating point at which a certain luminousity is expected. I have used a setup involving a PMT (hamamatsu blue sensitive) a SensL solid state PMT and a blue LED to characterise the the solid state PMT. This setup sensitive down to single blue photons. I do not remember exactly where I could detect single photons but it was above 2 volts. Remember the diode part of the LED is still working and current is flowing but not enough eV to excite the quantum well into producing light photons
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
johnf wrote ...
OK The .4 volts I was referring to is the operating point at which a certain luminousity is expected. I have used a setup involving a PMT (hamamatsu blue sensitive) a SensL solid state PMT and a blue LED to characterise the the solid state PMT. This setup sensitive down to single blue photons. I do not remember exactly where I could detect single photons but it was above 2 volts. Remember the diode part of the LED is still working and current is flowing but not enough eV to excite the quantum well into producing light photons
two points. I saw the photons- so there's no way you can sensibly say "Remember the diode part of the LED is still working and current is flowing but not enough eV to excite the quantum well into producing light photons". Also, did you try repeating the experiment with the PMT keeping the LED at different temperatures?
Registered Member #230
Joined: Tue Feb 21 2006, 08:01PM
Location: Gracefield lower Hutt
Posts: 284
BC no thereare two differing things happening in a LED 1 it is a diode with its characteristic CV plot 2 light emitance curve due to the quantum well
no I have not done this particular experiment at differing temperaters
I have excited red LED's at 70 K with extreme current pulses and have got them to lase at a peach colour @ 100, 000 times their normal light output --and no CV curve at this point-but 40 amps forward conduction albeit in short pulses gives continuous light output- go figure!!!!!
If anybody wants to delve into the mechanism by which light is emitted, the proportion of carriers that exceed some energy threshold, whether LEDs behave like filament lamps when you reduce the current (really Uspring, really? at least you say 'think', I'd say that's the extraordinary claim that requires the extraordinary evidence, not the photon energy threshold model) then that is just fine and dandy, but it means diddly squat to people who buy LEDs and wire them up to make light.
Emissions of light from a filament and from a LED are not fundamentally different. Both rely on the decay of excited atomic states. In a filament, there is a continuuum of states so there is no lower transition energy at where it begins. In a LED, there is a band gap, above which electrons have to be pushed in order to fall down across it.
Wr to thresholds: The Shockley equation shows a strong temperature dependence. At absolute zero temperature current would show a sharp rise at e.g. 2V. Probably even Bored Chemist would agree, that there'd be a threshold. At higher temperature, this rise is smeared out due to thermal energy. At room temperature, this smearing is still small, since average thermal energy is then only about 25mV, much less, than the energy required for a transition in the visible region. It's much up to the eyes of the beholder if a smeared transition and at what amount of smearing still qualifies as a threshold.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Uspring wrote ...
Emissions of light from a filament and from a LED are not fundamentally different. Both rely on the decay of excited atomic states. In a filament, there is a continuuum of states so there is no lower transition energy at where it begins. In a LED, there is a band gap, above which electrons have to be pushed in order to fall down across it.
There's only one type of person in the world - those that see a benefit in drawing distinctions between different classes of behaviour, and those that see everything as a continuum.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
@JohnF, Thanks for mentioning 'quantum wells'. The Wikipedia areticle on quantum wells isn't particularly informative. Can you suggest any other suitable links? I'm more interested in the mechanism itself, rather than mathematical models which oversimplify the process, but make the maths easier.
I had assumed that all the photons emitted from an LED were due to electrons from the conduction band re-combining with 'holes' left in the valence band by 'base current' electrons.
@Uspring, I was going to use the zero kelvin example, but then it occured to me that maybe other mechanisms come into play at these low temperatures, so it might not be such a good example. Also, once you apply a potential difference you start to excite electrons, I think, thus introducing thermal energy, so maybe it's not quite that simple. I think you've effectively made the point, though.
EDIT: This Stanford paper on quantum wells, for example, only really introduces mathematical models, and doesn't (from what I've read) really go into the mechanism.
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
johnf wrote ...
BC no thereare two differing things happening in a LED 1 it is a diode with its characteristic CV plot 2 light emitance curve due to the quantum well
The ratio of the light emission to the current (measured in photons per second and electrons per second) is the quantum yield. It's in fig 3a of the paper I cited and it flattens out below about a milliamp. So the number of photons per electron does not suddenly drop to zero at some small current.
Below about a milliamp, the light emission vs voltage curve looks the same as the current vs voltage curve- that's what the data in that paper tells you.
So, there may be two ways of looking at it, but there's only one "thing happening in an LED".
The efficiency does, however, depend on the temperature in the way one would expect if it's relying on thermally excited electrons having the enrgy to get over the barrier, fall down, and emit light.
It's not that the light gets faint because the electrons do a worse job of producing light, it's just that there are fewer of them
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Bored Chemist wrote ...
The efficiency does, however, depend on the temperature in the way one would expect if it's relying on thermally excited electrons having the enrgy to get over the barrier, fall down, and emit light.
This is what the conventional, simplified, mathematical models tell us happens, BC, however I look at it slightly differently.
I argue that the 'threshold energy' is the energy required to 'pull' an electron from the already depleted region on the collector side of the junction (the 'base region'), this electron then becomes 'base current'. The presence of of the 'hole' left behind 'neutralises' the potential gradient at the junction, allowing electrons to cross the junction. This flow continues, in the conduction band, until an electron 'falls' into the hole left in the valence band, releasing a photon, and stopping the flow across the junction, thus re-introducing the 'potential gradient'.
I base this reasoning on the fact that no conduction band electrons flow until the base current starts flowing (it possibly becomes easier to visualise as ' charge flow' (coulombs) at this point)
I accept that my reasoning doesn't include 'quantum wells' at this point, so I'm expecting to be proved wrong (how else do you learn anything?)
EDIT: just to clarify, once an electron grosses the junction into the conduction band in the 'base region', this probably also re-established the potential gradient across the junction, until it either leave the base region in the conduction layer and flows to the collector, when, presumably, another electron can cross the junction. This stops when an electron falls from the conduction band into the 'hole' left in the valence band. The more 'holes' in the base region, the greater the current.
Again, just my ideas, but this reduces the process to effectively the same mechanism as a thermionic valve.
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