avalanche-breakdowndiodesled
I wonder this by pure curiousity. If we polarize a LED in avalanche mode by applying a very high reverse voltage (but keep the current low so that the component doesn't fry), is it possible that it also emits light when used this way ?
(The reason I don't "just try it and see" is because of the high voltages involved).
Re: Comments - the project book is titled the Arduino Project book. It comes with the Arduino Uno starter kit.
Re: Schematic - yea, I'm not quite at making those yet. Next time :(
Re: The answer -
I just figured it out, soooo...silly me. Optocouplers don't supply their own power. They're just a switch. So in the example video above there is nothing powering the 2nd LED.
The working model with both LEDs lit is here: http://www.youtube.com/watch?v=_CF1xPcXGNY&feature=youtube_gdata_player&desktop_uri=%2Fwatch%3Fv%3D_CF1xPcXGNY%26feature%3Dyoutube_gdata_player&nomobile=1
This link was super helpful in case anyone else stumbles on it: http://www.technogumbo.com/projects/Learning-to-Use-an-Optocoupler-or-OptoIsolator/
Ok, let's take this step-by-step up Wittgenstein's ladder.
Step 1:
Given the circuit:
simulate this circuit – Schematic created using CircuitLab
For the current to get from point A to point B it's going to go straight down the simple wire rather than take the more difficult route through the LED. So no current flows through the LED, it just goes straight past.
Step 2:
At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node
Ok, but on the circuit above point A and point B are connected directly together, so they are in effect the same point. The circuit is basically the same as:
(imagine the little link bit isn't there - the editor won't let you do diagonal lines).
The current I that flows in must equal the current I that flows out of point A. So if all that's going in is going straight out, there's none left to go up to the LED.
Step 3:
No wire has absolute zero resistance. The same with breadboards. So the actual circuit is more like this:
Ok, so we have an fixed voltage \$V_{CC}\$. Let's say this is, for simplicity. 5V. The LED has a fixed forward voltage drop. Let's assume for the sake of argument that it's 2V.
Ok. Let's take the LED out of the circuit initially and just work out the voltages dropped across the resistors R1, R4 and R5.
The total resistance for that section will be 100.002Ω (simply add them together). So the current through them would be \$\frac{5}{100.002} = 49.999mA\$.
Therefore the voltage dropped across R4 would be \$0.049999 \times 0.001 = 49.999{\mu}V\$.
Now if you attach the LED across that resistor it's only going to be getting 49.999µV, which is considerably less than the required forward voltage needed to turn it on. So it won't be conducting as it's not on, so the current through the resistors R2 and R3 will be zero.
Now there are more potential steps in Wittgenstein's ladder, but from here we're getting into the realms of subatomic physics, and even quantum theory, so we'll leave it there for now.
Best Answer
I'll go with a: In general, no, that's not the case.
Light emission in LED-type devices typically happens when electrons and holes recombine, and the energy that gets freed in that process gets converted to a photon with the resulting wavelength. That happens in the transition zone of a doted semiconductor junction, where there's gradient in the band structure.
Let's imagine a diode in reverse bias: In aforementioned transition zone, there's practically no free charge carriers (no holes and electrons), so the device would be a perfect isolator – I say "would be" if not spontaneous creation of such carrier pairs could happen due to thermal effects (and also, things like photon absorption).
Now, under avalanche breakdown conditions, the electric field across that isolating zone is so high that the charge carriers get accelerated very fast – and might "knock" out other charges from the non-conducting bands (to make this feel a bit more scientific: the electric field gives spontaneously created charges an impulse that is enough to transition further charges in k-space to the conduction band).
Now, these charges will just travel to the contacting areas and recombine there - usually nowhere where there's a) a well-defined bandgap to make emission of visible photons likely and b) no optical structures to couple out that light. You just heat up the substrate.
That is not to say there won't be light emissions in all this: purely from a stochastic point of view, some recombination with visible emissions might happen, and also, nothing says that over the temporal process of that avalanche breakdown, there won't be some times where the whole field configuration wouldn't lead to interesting band diagrams where recombination within the optically relevant parts of the LED might take place, at totally different photon energies than the LED was designed for.