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I've moved this to another thread, since I'm getting too OT in the other thread. Ash Small wrote:
I agree that thermal effects themselves won't excite many electrons sufficiently for them to jump, but any potential difference accross the junction is going to raise the average energy levels of the electrons waiting to 'jump'.
As the voltage approaches the 'threshold voltage', a greater percentage of the 'jiggling electrons' will attain sufficient energy.
Looks like a sensible idea to me. Moreover, the Shockley equation, which describes the current of a diode wrt to the voltage, has exactly the same exponential form as the energy statistics of particles in a warm environment., i.e. the number of particles at or above some energy. So it looks plausible, that it your idea is underlying Shockleys equation. Also there is a negative temperature coefficient to the voltage drop along a diode, i.e. you need less voltage for a given current at a higher temperature. A larger temperature will have more electrons have a high energy, which would cause just this effect.
But it seems, that in this way one could utilise thermal energy to some extent to power light emission in a LED. For me I see a contradiction to the second law of thermodynamics, which says, that thermal energy cannot be used in this way. I haven't been able to reconcile this.
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
One can And it rather destroys the idea that there's a threshold for LEDemission- unless you are working at 0K
Does anyone remeber what Dr Slack said about this idea? Oh yes, I remember now "I hope The Chemist is working on a paper entitled 'anomolous sub-threshold photon emission in LEDs, and their application to dark-looking halloween costumes'. I think the paper should include arguments that you can use to persuade your fellow trick-or-treaters that those dark plastic things are in fact emitting light because thresholds are mythical."
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
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?
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
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Are you sure? According to the expert in LEDs on this forum, there's no such thing as a threshold voltage.
on 23rd September, he posted in two different posts in the orginal 14vf LED thread
The "threshold" voltage is a myth
and
So, why is this bizarre myth that LEDs have a "threshold"?
and then at the start of this thread
Bored Chemist wrote ...
And it rather destroys the idea that there's a threshold for LEDemission- unless you are working at 0K
so I must conclude that LEDs can run on any voltage. My red LED will run on 2v through a 100 ohm resistor. Why won't the blue one? Just to confirm, I am working at a standard lab ambient of 298K.
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Joined: Sun Nov 14 2010, 05:05PM
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Neil, Wikipedia describes one electronvolt as 'one electron volt is equal to 1.6021766208(98)×10−19 J', or approximately 160 zeptojoules.
It's true that each electron has to achieve ~3.4 times this energy before it can jump (I'l get onto exactly which electrons jump first later). That's ~544 zeptojoules.
Maybe using these units helps to understand that it's the thermal energy of the electrons that initiates the jump?
(I apologise for not answering your point directly, I'm still working on it )
Udo, thanks for taking the time to check the maths. I appreciate it
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
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Ash Small wrote ...
Neil, Wikipedia describes one electronvolt as 'one electron volt is equal to 1.6021766208(98)×10−19 J', or approximately 160 zeptojoules.
It's true that each electron has to achieve ~3.4 times this energy before it can jump (I'l get onto exactly which electrons jump first later). That's ~544 zeptojoules.
Maybe using these units helps to understand that it's the thermal energy of the electrons that initiates the jump?
(I apologise for not answering your point directly, I'm still working on it )
Udo, thanks for taking the time to check the maths. I appreciate it
Do you mean if I teach the blue LED to read, and show it the wikipedia article, it will say 'Ah!, now I can emit at 2v'. Or should I just give it more than 2v? And if so, how much? Bear in mind, I'm just trying to get a LED to light, I'm not a materials physicist trying to design a better LED material, or a photoemissive cooler. If you recall, in the earlier thread, a self-identified noob was trying to get his Halloween costume to light.
Do you think johnf was on to something?
3.4 volts or there about is needed to excite the quantum well in blue Leds sorun it from the 5 volt rail
I did spot this link which might have some bearing on the problem, but the authors seem to be of the deranged opinion that LEDs have threshold voltages, so they obviously know nothing about them.
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Joined: Sun Nov 14 2010, 05:05PM
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With all due respect, Neil, I was responding to the point made here:
Dr. Slack wrote ...
With a 9v battery, and a 14v LED, no current will flow.
We've established this is not the case, and that thermal energy plays a part here.
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.
I can go into the mechanism that I think is responsible (I've touched on this before, in a thread from a couple of years ago, linked to in the other thread).
My ideas still aren't complete, though, for example, I'm currently of the opinion that not all the electrons which pass through the LED emit photons. I'm currently of the opinion that only those electrons that re-combine in the depleted region emit photons. The ones that pass through in the conduction layer , in my opinion, don't (my current thinking).
To put it another way, if you compare a diode to an NPN BJT, the electrons flowing through it are composed of 'base current' and 'collector current'. My current thinking is that it's only those electrons that comprise the base current that emit photons. I'm guessing that an efficient LED has a very low gain (comparing it to a BJT).
Again, these are just my ideas. I don't have any references I can link to.
I've looked over the article quoted by Bored Chemist in a very skeptical mood, but I've convinced myself, that it is indeed possible to tap thermal energy for the production of light. Think e.g. of a filament light bulb. 100% of its electrical input is converted into heat e.g. at 3000K and radiated away as light and at lower infrared wavelength.
Now instead of using electrical current, one could imagine using a heat pump feeding heat from room temperature (300K) to 3000K. In principle an ideal Carnot engine would then provide 110% of thermal energy for 100% power input. Technically perhaps possible but in practice not worth the 10% gain in efficiency.
One could run the light bulb at 600K. That would allow a much higher heat pump efficiency. In this case power output would be doubled wrt to power input. The drawback is, that a 600K light bulb would be extremely dim, radiating most of its power in the infrared. In the article, the high efficiency was reached at sub nW light output and I don't think, that this can be scaled up to higher levels. Basically I think, that when you lower the voltage on a LED, that it will never actually turn off, but that there is a smooth transition from useful light output down to a level, where it coincides with that of a black thermal radiator of room temperature. At this level not even electrical input is necessary. There is still some emission of light, although at a completely useless strength.
Dr. Slack wrote:
I must conclude that LEDs can run on any voltage. My red LED will run on 2v through a 100 ohm resistor. Why won't the blue one? Just to confirm, I am working at a standard lab ambient of 298K.
You're not looking hard enough. You should be using a photomultiplier detector capable of observing single blue photons at hourly intervals.
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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.
Dr. Slack wrote ...
With a 9v battery, and a 14v LED, no current will flow.
In the context, and I emphasise context, of the OP of the 14vf LED thread, what does 'no' mean? It clearly means an amount of current that will not result in enough light to impress his fellow trick-or-treaters. It doesn't mean a leakage current 8 orders of magnitude (or whatever) below the rated current.
What will light the blue light on the front panel of my power supply? Supplying more than the threshold voltage required to get the LED conducting of course.
I have already apologised for my mis-assumption in picking an inappropriate level of abstraction for the OP, by being lazy and not looking up the data sheet before posting a reply. I assumed that the 14v spec came from the series connection of several LEDs (like the efficient '12v' LEDs in lighting fixtures), rather than the series connection of a LED and resistor. I therefore yacked on initially about 'supply enough voltage', rather than 'connect 9v and see if it lights'.
Here are some levels of abstraction that will be useful to different people, from costume illuminators to theoretical physicists. I am not proposing these as 'the' levels, you may think more or fewer are appropriate. The point is that there is a hierarchy of increasing detail as you try to explain the operation of the level above. And you will notice, as I've listed it, it is 'turtles all the way down' (look it up).
It is an error to work at the wrong level of abstraction. It is even more of an error to insist that the jargon of your level is the right level. It is an even more egregious (look it up) error that having been pointed out that the context, the abstraction level, is not shared, to continue arguing without accepting the need to sort out the context, and to get it agreed.
a) a commercial lighting device - which has a spec of 12v or 14v or whatever b) ... is constructed from? several LEDs in series, or a series LED + resistor c) ... and a LED is a PN junction d) ... made from suitably doped materials e) ... that have bandgaps/energy levels and a spread of thermal energy for the charge carriers f) ... understood by quantum mechanics of solids g) ... built from particles understood by QCD h) ... but we're not really sure about the standard model yet i) ... or what governs the laws of the universe j) ... or what the universe is or why it bothers to exist k) ... ??
So I will stand by every post I have made in these few threads that, in the context of the OP, LEDs have a threshold voltage below which they don't light.
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.
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Was 14V LED question
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