LED Magic Trick

442

05:23:37

Look!  The LED is powering itself!  

What in the world is going on here?

The idea of this circuit is not original to me.  Its concept has been done at least once before here:  https://everycircuit.com/circuit/5276667463860224

However, I changed some of the values and components to emphasize the effect.  Flip the switch up first to connect the LED to the voltage source, then flip it back down.  Current will suddenly start flowing through the LED and resistor by itself.  It looks like the LED turns itself ON!

What's actually happening is that the capacitance of the PN junction is discharging.  The 30V source charges up this hidden capacitor.  Then, flipping the switch will cause the capacitor to discharge through the resistor and back to the other side of the LED.  EveryCircuit will treat this current as powering the LED.  (NOTE, this may not be a totally acceptable explanation.  See comments below.  This is to the best of my knowledge...)

This does not work in real life.  But it's a swell trick in EC, and it may wow your friends.

Check out this circuit where I do nearly the same thing, but with a 7-segment display.  https://everycircuit.com/circuit/5861787383365632

My next post uses this technique for an oscillator flashing circuit.  See here for details:  https://everycircuit.com/circuit/4699313716068352

published 2 years ago

jason9

2 years ago

Actually, the capacitance is not powering the PN junction. The LED is modeled as an ideal diode in parallel with a capacitor. However, because the capacitor is built into the model and not a separate component, EC erroneously considers current passing through it and not the PN junction to still generate light. If EC was a bit smarter about this it would only show light from the current that actually passes through the PN junction part of the model, not the capacitor.

592azy2circuitdude

2 years ago

I didn't realize that. You're probably correct. Let me go try and research this a bit...

592azy2circuitdude

2 years ago

Hmm, this topic is somewhat beyond my understanding. So are you saying the current is flowing through that capacitor part and not the PN part? But isn't it correct that the depletion capacitance caused by a reverse biased PN junction is part of that PN junction? So, wouldn't charge flowing through that be flowing through the PN junction and emit light? (Theoretically)

jason9

2 years ago

Yes, the capacitance is a byproduct of the PN junction, and current can be said to be flowing through the junction, but there is no carrier recombination and consequently no light. When reverse biased, the junction is depleted of charge carriers and becomes insulating. This insulating region creates a capacitor in the same way that the insulating film between the two electrode of a real capacitor create a capacitor. And in the same way, current can be said to move through that film (through the capacitor), even though no charge carriers are actually moving in the film since it is not a conductor. As such, there is no movement of charges in the junction itself, even if current is flowing because of the capacitance. The proper term for this current is displacement current, and is a product of the changing electric field.

592azy2circuitdude

2 years ago

Oh, that's right. I should have remembered that. Current doesn't pass through the capacitor. It's excess charge moving off the plates. It makes sense that this is the case in a PN junction, too.

592azy2circuitdude

2 years ago

But now I'm wondering if any current would actually flow at all in real life. Do you think the charge from the depleted region would actually flow around to the other side (through the resistor), or would it just slosh back internally when the external reverse bias voltage is removed?

jason9

2 years ago

It will hold a charge like any capacitor. This charge will slowly deplete due to thermal production of charge carriers (i.e. reverse leakage current) but it will nonetheless hold charge for a short while at least. In order to slosh back internally, it needs an external source of charge carriers to fill the depletion region, and that comes from current through the resistor.

592azy2circuitdude

2 years ago

I think that helps. Considering I don't know much (or anything) about semiconductor physics, I think your explanations are very helpful. I tried to modify the description to be a little more accurate. What do you think?

jason9

2 years ago

Looks good.

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