Understanding Voltage Drop and LEDs

Other Hardware General Electronics
1

I am having a bit of a hard time wrapping my head around the concept of voltage drop. I see there are a lot of different explanations online, but the concept still doesn't seem to be sinking in.

If we take a situation where we are calculating the resistor value to use with an LED, the math usually looks a little something like this (assuming a 5 volt source and an led with a 3 volts forward voltage):

5 - 3 = 2

2 / 0.01 = 200 Ohms

What I don't understand is why the LED seems to be dictating how much voltage is left over for the resistor, and not the other way around. What is it about the LED that uses 3 volts?

2

What I don't understand is why the LED seems to be dictating how much voltage is left over for the resistor, and not the other way around. What is it about the LED that uses 3 volts?

Actually, you've got a pretty good grasp of what's going on...

What is it about the LED that uses 3 volts?

LEDs are non-linear (like all diodes). That means the resistance changes when the voltage changes. At low voltages the LED has very-high resistance and almost no current flows. At the "breakdown voltage" the resistance drops dramatically. That means it's "hard" to go above the LED's rated voltage. If you go above the breakdown voltage, the resistance heads toward zero and if there's nothing to limit the current you'll fry the LED.

Ohm's Law is ALWAYS TRUE and you can calculate the resistance of an LED at a particular voltage and current. But, Ohms Law is not really useful with LEDs (or any diode) because the resistance is not constant and you don't know what resistance to plug-into the formula.

3

What I don't understand is why the LED seems to be dictating how much voltage is left over for the resistor,

Because the LED is a non linear device. Look at this:-
http://www.thebox.myzen.co.uk/Tutorial/LEDs.html

4

Ohms law only applies to uniform conducting materials in the solid or liquid phase,
not to semiconductor devices, gasses, vacuum, insulators or super conductors. Ohms law
is definitely not always true, its an approximation, just happens to be a good approximation
in ordinary conditions.

The physics is that in such media electric currents consist of moving charge-carriers,
and these charge carriers interact thermally with the material, dissipating the kinetic
energy of their movement, thus converting electric energy into heat. This doesn't
happen in a gas (ionization), a superconductor (no interaction with the material),
a vacuum (no material) or an insulator (no free charge carriers).

In a semi-conductor device different parts of the material have different potentials
and different charge-carriers (which can combine for instance) - a lot more complex.

[ Oh yes, and in a battery there are several layers of different material, so again
more complex ]

5

One has to make the difference between Ohm's law as, what I would call, its litteral meaning ; and its usefulness as a model.
When you say "LED's resistance varies with voltage applies to it", we're not talking about a resistance as a physical property of an object. We're talking about Ohm's law's extension which is the voltage divided by the current mathematically speaking.

That's a mathematical model to explain things in a clear manner... And more simply, to model physics. Because physics is all about MODELING nature.

But people dealing with electrical engineering at a certain level have to make the difference between the most basic definition of a "resistance" and its extension which is the mathematical model R = U/I whatever the circumstance are.

PS: Yeah, I know, I like to argue on details xD

6

The energy of the light emitted comes from the electrons dropping into holes.

N type material means Negative, because it is doped to have "free" electrons. P type material means Positive, because it is doped to have "free" holes, which is the lack of electrons. Both materials have equal numbers of electrons and protons, but it is a valence band thing.

For a given material, it takes a certain amount of voltage to push an electron across the depletion layer. A photon is emitted when the electron drops into a hole on the other side.

The wavelength of this photon is dependent on the voltage gradient that the electron dropped across. At 1.77V, a red 700nm photon is released. At 3.1V, a violet 400nm photon would be released. An IR LED requires about 1.5V.

That is why if you take an IR, red, amber, yellow, and green LED, you'll find they drop increasingly higher voltages. Things are a bit up in the air for blue, ultraviolet, and white because other colors of LED may be used with a phosphorescent coating or frequency doubling method to generate the light. For white, a combination of red, green, and blue LEDs may be used in one package.

Some glass cased diodes are made to drop more voltage for some special electronic purpose, and as a side effect they glow red.

Since most digital cameras are IR sensitive, you can even point your camera at a glass diode like a 1N914 or 1N4148 with 50mA or so flowing, and see it glowing in the camera. It helps if you turn the lights off, because it is far IR (infrared) because it is only about 700mV dropped across the diode.

7

Some very informative answers! I won't say that I feel I completely understand still, but I have a much better idea. I will need to read through these materials and supplement the pieces I don't understand with further reading.

This is what I was looking for, though. I now know enough to start building towards a deeper understanding of the question at hand.

8

The quoted voltage across the LED is the what you'll find when the LED is being fed the correct current.

If you feed the LED the WRONG current then the voltage across the LED will NOT be the value quoted in it's datasheet. and since it's a non linear device, you will not be able to easily calculate what voltage will be across it at all.

So you have to work out how to give it the ideal current dictated by it's datasheet. Since you know what the voltage will be at this current you can deduct that before you work out what remaining voltage will be dropped across the resistor that you need to use.