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Ok, so I know that the intensity of the light given by an LED is the minimum it can be at the turn-on voltage, however I don't know if the wavelength of the light emitted from the LED at the turn-on voltage will be below the peak wavelength (the wavelength at which the intensity of the light of the LED is at its maximum) or above it. My reasoning comes from the picture below:enter image description here

The picture shows that for a specific colour of LED, for any intensity instead of the peak intensity there can be two wavelengths associated with it. For instance, for the blue LED, there will two values for wavelength for when the intensity is 20%. So, my question is: Is the wavelength of the light emitted from an LED at its turn-on voltage greater than or less than the peak wavelength of the LED?

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I don't think you're reading this right. This graph shows the spread of the spectrum, meaning that the LED not only emits light at a single frequency, but in a range of frequencies centered around the middle, with a distribution as shown. It has nothing to do with the LED intensity. – pipe Mar 29 at 16:59
@pipe The values in the y-axis, I believe, represent the normalised intensity (%). I have changed the picture to a more suitable one. – user307397 Mar 29 at 17:02
But you still talk about "two values", there are no such values. What the graph shows (the new and the previous) is still the spread. The LED emits light at every wavelength shown under the curve. Simultaneously. It is not infinitely narrow-band. – pipe Mar 29 at 17:05
@pipe So, say for the intensity of the purple LED at 40%, will there be two values for the wavelength of the light? – user307397 Mar 29 at 17:07
@pipe From what I hear, do you mean that for any intensity of the light there will be a similar spectrum for the wavelenght of the colour of LED emitted? – user307397 Mar 29 at 17:10
up vote 13 down vote accepted

The wavelength distribution will look like the graph, pretty much regardless of current at a given temperature. LEDs are not purely monochromatic.

However, as the die heats (and it will tend to heat more at higher current) the center of the spectrum will shift toward the red for all LEDs (longer wavelength).

From this OSRAM document, you can see a typical change of about +70pm/K die temperature.

enter image description here

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So, are you saying that for any specific current going through any specific colour of LED, the distribution of the wavelenght of the colour of the LED will be the same? – user307397 Mar 29 at 17:08
At the same die temperature, yes. The dominant wavelength (peak of the spectrum) will remain constant. – Spehro Pefhany Mar 29 at 17:10
@SpehroPehfany Thanks a million! This makes my job sooooooooooooo much easier. – user307397 Mar 29 at 17:12
@SpenroPehfany Are you sure the dominant wavelenght is the peak of the spectrum? I thought they were different things. – user307397 Mar 29 at 17:24
Yes, you're right, that's not the technical definition when it refers to perception of color. It's technically "wavelength of the spectrum light that when combined in suitable proportions with the specified achromatic light yields a match with the light being considered" according to M-W. – Spehro Pefhany Mar 29 at 17:40

First off, as pipe said, you are misinterpreting the graph. The graph merely shows the composition of the the output spectra when operating in a normal regime, not as a function of relative output power.

However, to answer this:

Is the wavelength of the light emitted from an LED at its turn-on voltage greater than or less than the peak wavelength of the LED?

When an LED is "on" (operating in a normal regime, when the forward voltage is greater than the turn-on voltage and the forward current is less than the maximum), it emits a very narrow spectrum of light, centered around a wavelength that is determined by manufacture and temperature (and therefore, to a small extent, forward current).

By cooling an LED, the wavelength decreases. This link shows an LED going from orange to yellow when cooled in liquid nitrogen. Likewise, heating an LED increases the wavelength.

Consequently, when you apply a large current to an LED, it heats up and the wavelength increases.

When operating the LED at the turn-on voltage and a very small amount of current is passing through it, the LED is dissipating less energy than "normal" operating, so the temperature will be somewhat lower than usual. So it's possible that the wavelength is a teeny, tiny bit lower than typical, but realistically, it's exactly the same as when it's operating normally.

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LEDs are not monochromatic, meaning that their output emission is comprised of multiple different wavelengths of light. Your diagram is just showing the normalized amplitudes of the different wavelengths it produces.

As a comparison example, consider an incandescent lightbulb. White light as we see it is comprised of light of all visible wavelengths. A 3000K bulb looks orange because the most powerful emissions that we can see are those on the yellow-red side of the visible spectrum. Note how most power is converted into infrared energy, just radiating heat. Very large range of wavelengths.

enter image description here

On the other hand, a laser diode can be classified as monochromatic. Notice how narrow the range of emitted light is (0.5-1 nm) in comparison to the incandescent bulb. It's emission contains a much smaller range of wavelengths.

enter image description here


In the middle of the bulb and laser emission bandwidth are LEDs - a small enough wavelength range (about one hundred nm) for them to be considered a specific color but not small enough to be considered monochromatic. Your diagram appears to show the spectra of multiple different LED diodes, here is a diagram showing the spectra of three different LEDs - "common blue LED, a yellow-green LED and a high brightness red LED from the bottom of a Microsoft optical mouse" Image, Description Source and more info

enter image description here

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You don't actually include a graph for a LED, but it's closer to the laser one than the incandescent one :) – hobbs Mar 29 at 18:37
I didn't include for LEDs because the original question had a good one, but I'll find one specific to LEDs and edit it in when I get back to my computer. – MichaelK Mar 29 at 19:04
LED Graphs added – MichaelK Mar 29 at 19:17

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