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I previously believed there was a gap between the plates inside an LED (between the post and the anvil).

However, I recently saw the following drawing on wikipedia (https://en.wikipedia.org/wiki/Light-emitting_diode#/media/File:LED,_5mm,green(en).svg ) : illustration of LED

I noticed that it shows the "Wire bond" which seems to indicate a wire from the post to the anvil.

Question 1

Is there actually a physical wire there?

I always thought the electrons jumped the gap and that there was no wire and that was part of why LEDs lasted so long.

Question 2 - for clarity

If there is a wire there, is it the part that lights up? Or, do the electrons still jump the gap? (I'm assuming they are conducted on the wire, but just attempting to clarify.)

Question(s) 3

If the wire is there, is this a (more) recent design change to LEDs? Did LEDs of the past have or not have this wire? Is this a special type of LED with the wire and there are some that don't have it?

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    \$\begingroup\$ Why would it show a wire if there wasn't one? \$\endgroup\$
    – Finbarr
    Nov 7, 2017 at 14:43
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    \$\begingroup\$ Nothing modern about this design. There has always been a gap between the two leads - they'd short together otherwise. Where did you think the actual semiconductor material was located? \$\endgroup\$
    – Finbarr
    Nov 7, 2017 at 14:56
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    \$\begingroup\$ This is why I constantly read and learn. Everyone has to learn things at some point. I'm not an electronics engineer -- I'm a software developer -- so I have never examined the up-close physics of how an LED works. And, realistically, probably 90% of people who use LEDs in their projects didn't know this. I think I made up the idea of jumping the gap, because I couldn't see the wire. Isn't that cool? :| Recently I wanted to explain the basics of how a LED works so I investigated before I explained. Something else a lot of people don't do. :) \$\endgroup\$
    – raddevus
    Nov 7, 2017 at 15:17
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    \$\begingroup\$ If 2V was enough to jump a gap that size we'd all be in big trouble. \$\endgroup\$
    – Finbarr
    Nov 7, 2017 at 15:54
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    \$\begingroup\$ The "gap" that the electrons jump is sub-atomic - between energy levels in the atoms themselves, It is not a real physical space between electrodes. \$\endgroup\$
    – Peter Bennett
    Nov 7, 2017 at 17:02

4 Answers 4

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Answer 1:

Yes, there is actually a physical wire there. However, the wire does not short out the anode and cathode, it connects from the anode onto part of the semiconductor die.

Answer 2:

The wire is just a wire, it does not contribute to the light output of the LED. (Other than providing a necessary electrical connection)

Answer 3:

This is not a new LED design.

Further comments:

All of the light emitting function happens on the small part labeled "Semiconductor Die" in the figure. In fact, that is the only part necessary. You could produce an LED with just the die, and it would work just fine. Although it would probably be quite difficult to use, and fragile.

LEDs are a special type of diode that emit a significant amount of light under forward bias. There is no "gap" that electrons jump across.1 The "Wire Bond" in the diagram connects the anode wire on the LED package to the anode connection on the semiconductor die. It is called a "wire bond" because "wire bonding" is the method used to connect these wires. The purpose of everything else in the diagram is mechanical stability, or light focusing/defocusing.

1 Electrons do transition across the energy band gap, which is what causes them to emit photons, but this is not a physical gap, it is all within a single block of semiconductor material. The band gap is a part of the energy band model of semiconductors.

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    \$\begingroup\$ Give the LED enough current and the wire might actually glow. :) (If the die fails short after an overload.) But it would also be a SED (smoke emitting diode), and probably a FED (flame emitting diode.) \$\endgroup\$
    – Oskar Skog
    Nov 7, 2017 at 18:26
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    \$\begingroup\$ @OskarSkog I once connected an SED directly to a 9VAC power supply (which, to make matters worse, was a linear transformer without the slightest sort of regulation). \$\endgroup\$
    – Doktor J
    Nov 7, 2017 at 18:29
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    \$\begingroup\$ But is that wire breaking a common failure mode? It seems very thin based on the picture. \$\endgroup\$
    – JPhi1618
    Nov 7, 2017 at 19:00
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    \$\begingroup\$ @JPhi1618 I dont know for sure but I doubt that wire bond wire breaking is a common failure if you stay within the specified operating conditions. Bond wires are from around 15 um to a few hundred um in diameter. But they are used in basically every integrated circuit on the market today. If bond wire failure was a common problem, all ICs would have the problem. \$\endgroup\$
    – Matt
    Nov 7, 2017 at 19:04
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    \$\begingroup\$ @DaveM I assume they have some kind of package or lens. The bare semiconductor is quite fragile. Especially the III-Vs used in LEDs \$\endgroup\$
    – Matt
    Nov 8, 2017 at 2:50
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To show a good image of what's going on, here is an LED in cross-section (taken from this nice article on LED failure analysis)

enter image description here

Here you can see the semiconducor device sitting in the cone well of the cathode. This well is there to work just like the reflector in a normal flashlight. The light emission from the device comes from the edge of the chip at the junction between the p-type and n-type semiconductors and the reflector deflects the light to exit the top of the LED device.

enter image description here

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  • \$\begingroup\$ Fantastic additional info. That view / diagram is really instructive. \$\endgroup\$
    – raddevus
    Nov 8, 2017 at 14:08
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Here is an example of a different style LED package.
The active part of the LED is the die which one side is the lead, and the bond wire attaches to the other connection.

LED in a surface mount package

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LEDs create light (photons) by electroluminescence. Light sources that create light from a heat glow wire are called incandescent.

LEDs are fabricated with a diode p-n junction where the p-type and n-type material is separated by a bandgap material. When a diode is forward biased, donor free electrons on the n-type side diffuse over to the p-type side where they encounter many p-type holes with which they recombine. This recombination can be either non-radiative (diodes) or radiative (LEDs).

enter image description here

When a diode is forward biased, donor free electrons on the n-type side diffuse over to the p-type side where they encounter many p-type holes with which they recombine. This recombination can be either heat generating non-radiative (diodes) or radiative, light generating (LEDs).

In a radiative recombination event, one photon with energy equal to the bandgap energy of the semiconductor is emitted.

LED layers

The bandgap is commonly called a multi-quantum well (MQW)



LED structure

enter image description here

Textured Surface



enter image description here

The light emitting surface is no longer flat.

enter image description here



In current LEDs the light emitting surfaces are structured in columns.

enter image description here

Columns can be horizontal or vertical.

enter image description here

enter image description here

enter image description here

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