I created a circuit with some resistors in series with a led. I tried 1k, 2k, 10k, 20k, 32k but the voltage on the led always stays at 2.20V while of course, the voltages of the resistors change, since its a voltage divider.

Why is that?

 

What voltage is the power supply? It's best to post a schematic when discussing circuits so there's no ambiguity.

LED is an acronym and it's written in all caps.

 

Why is that?

Because the D in LED stands for diode.

One of the fundamental characteristics of a diode is that the forward voltage drop is relatively independent of the forward current (within limitations). Datasheets for signal diodes and many LEDs have a voltage-current plot that shows this.

ak

 

LEDs are NOT a resistive load. The Voltage drop across an LED is a constant. Therefore, a) you don’t have a voltage divider and regardless of any resistor, the LED is going to drop the same amount of voltage (it’s a parameter of an LED, and called forward voltage or Vf.) The resistor sets the LED current.

 

An LED has several PN (or NP) junctions in series. Think of a LED as several diodes in series. Since a diode's forward drop is, at least to a first order approximation, constant with respect to current, the same would apply to an LED.

Go look at a data sheet for an opto-coupler, and look at the forward voltage drop for the input diode (which is an LED). You'll see values somewhere between 1.5V and 2.0V.

 

An LED has several PN (or NP) junctions in series.

Where did you get that info?

Most LEDs have just one PN junction.

The difference in forward drop is due the different bandgap voltages of the junction material used, and that determines the LED light photon energy and thus the light color.

Gallium arsenide LEDs emit red and infrared light.
Gallium nitride LEDs emit bright blue light.
Yttrium aluminium garnet LEDs emit white light.
Gallium phosphide LEDs emit red, yellow and green light.
Aluminium gallium nitride LEDs emit ultraviolet light.
Aluminum gallium phosphide LEDs emit green light.

 

Here's an IV curve for Lx3341 LED's:

If you have the datasheet for the LED you're using, you can determine the typical forward voltage for a specific current. There will be some variance because the graph is for typical.

This is the parameter range for the LEDs mentioned above:

 

Hello,

What is the used supply voltage in your test?
At very low currents, the voltage will be rather low.

Different led colors will have different forward voltage as the others say.
Here is a page that shows some IV curves:
http://lednique.com/current-voltage-relationships/iv-curves/
Bertus

 

I seem to remember posting this for the TS...
http://www.gizmology.net/LEDs.htm

 

The supply is 5VDC. Ok with the graph that @dl324 posted it now makes sense. Since the resistors were actually changing the current slightly, the multimeter was always measuring about to 2.20V across the LED. And the graph shows that while the current increases the voltage changes slightly until you reach a very high current.

 

Hello,

Repeat your test with resistors of 150, 220, 270,330,390 and 470 Ohms.
Then the current will be higher and you will be higher in voltage accross the led.

Bertus

 

I know that I'm making stupid questions in this forum recently, but I'm quite new to this stuff. Even If at the university I completed 3 labs about analog and logic circuits, you really don't learn much from them. They just give the schematics without explaining what they do, and we just construct the circuit and do some measurements, that's it.

Btw I skipped class when the professor was talking about diodes

 

I know that I'm making stupid questions in this forum recently, but I'm quite new to this stuff.

They're not dumb questions unless you ask the same question more than once.

 

I created a circuit with some resistors in series with a led. I tried 1k, 2k, 10k, 20k, 32k but the voltage on the led always stays at 2.20V

Actually, if you measure the voltage with a digital multimeter, you should see a slight increase in voltage as the resistance goes down.

 

Where did you get that info?

Most LEDs have just one PN junction.

The difference in forward drop is due the different bandgap voltages of the junction material used, and that determines the LED light photon energy and thus the light color.

You're right, I stand corrected. I think my comment came from 1) the Vf of a LED relative to a standard Si diode, and 2) some of the graphics I've seen that showed more than 2 PN layers, something like this:

 

OK, with a regular multi meter you will not notice the difference in the forward voltage as the current is changed. Bout with a digital voltmetr having adequate resolution it will be obvious. Most of the diode voltage drop is due to the "band gap" voltage, which is rather constant with temperature, but there is also a resistive part and that does vary with current.