Weird THF effect on LED

357

02:47:17

Okay so here I'll lay a few mathematical examples of what's going on and you can tell me what you think about it:

- On the LEFT we've got a normal DC LED operation at 2V/20mA which makes a total power consumption of 40mW (P=U.I) So far so good.
- Then on the right we have an AC LED operation at F=10Ghz. At such a speed the LED would remain still as if it were powered by DC. This is where it gets interesting. The same LED as the green one is magically lit into full power with just 10mV of AC. While the total power consumption according to the Power formula P=Urms^2/Z is around 100uW. Yes uW! That's like 400 times less power requirement than with DC. 

Now let's get to the why - Both LED's have a 32pF junction capacitance. When it comes to DC that's not a problem. But with AC, especially at such high frequencies things work differently. The capacitance of the junction has a reactive resistance of around 0.5ohm at 10Ghz. This means that the LED is now acting more like a capacitor than a diode. Usually the diode's crystal wouldn't get energised at 10mV but since the capacitance is dominant now, rather than the PN junction, that makes the diode's resistance very low at such frequencies even to 10mV and the current through it actually becomes enough (20mA) to energise the LED and make it lit even with 400 times less power requirement. From 40mW to 100uW. 


Reference points to reasearch: 
-  Semiconductor Junction capacitance
- Reactance of Capacitance on AC

published 10 years ago

flowDAQ

10 years ago

Ok I think that is a good explanation, now the question is would that current (the capacitive current) light the LED or is that a EC bug.

thebugger

10 years ago

In practice it should, since the passage of a given amount of current is what energises the junction. The 2V is only the junction voltage drop that needs to be met with DC. With AC things run differently due to the capacitance effect on AC.

rich11292000

10 years ago

Its no bug, the extreme reaction is from the high F reacting on the capacitance. If you put a switch on it you will get the opportunity to see the first cycle, 70ma to get this kick started. To look at this situation, i need to see reality as the bugged program, as there is a lot of losses to overcome.

rich11292000

10 years ago

Also we need to consider the reaction of different wave types in this situation, as a sine wave might not be easily reproduced at 10G with out noise and distortion. Could a triangle wave surpass the sine wave in the prototype stage, or easier/cheaper to build. Who/when will know.

223kaua

10 years ago

Look at the arrow.

flowDAQ

10 years ago

Repost: since I see spelling errors and.. well stuff. From what I recall when a carrier crosses the band gap (from the conduction band to the valence band) it has to give up energy in some manner (light or heat) so light can only be emitted when current flows through the junction. https://en.wikipedia.org/wiki/Band_gap

flowDAQ

10 years ago

But the funny thing about capacitors, zero current flows in the gap (dielectric) between the plates, it flows onto the plates or off them but not in the dielectric material itself. The LED junction driven with 10mV at 10GHz is not seeing electron flow in the junction only the space charge region is changing (which houses the E-field and causes capacitance) so the electrons are not actually crossing the bandgap and releasing photons or heat.

flowDAQ

10 years ago

Another hint is that it is a capacitor and they don't dissipate energy (so where is the light from). https://en.wikipedia.org/wiki/P%E2%80%93n_junction#/media/File:Pn-junction-equilibrium.png

jpoulin0901

10 years ago

I agree with daq, the flow of current, although apparently significant would effectively be shunted past the junction without doing any work.

Addramyrz

9 years ago

Tonight at 9!

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