ATP: Circuit Design with LEDs

We tend to over-simplify the act of dropping an LED into a circuit. But do you know what all of those specs in the data sheet mean?

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Welcome to July and thanks for stopping by! Today on ATP we’re going to look a bit more in-depth at LEDs. How complicated could that be? Turns out to be a bit more than you might think. What’s with the different voltages for different colors? What’s a candela or a lumen? Is this thing going to burn? While you can generally get away with using rules of thumb alone, you just might want to brush up on some of the finer points of LED selection and operation. I’ll pick out a few of the choice specs and give you the low-down.

Further reading? Oh, you know it. Check these out:

Band Gap

Candela

[Lumens](https://en.wikipedia.org/wiki/Lumen_(unit)

More lumens and candela

Interested in learning more about LEDs?

See our LED page for everything you need to know to start using these components in your project.

Take me there!


Comments 12 comments

  • NathanY / about 6 years ago / 1

    Hey Pete!

    So as another EE who fell down the rabbit hole that is color science a few months back, I think I can help you out here.

    To understand candelas and lumens, you can think of candelas as like a pressure and lumens as like the force behind it. You're only applying this pressure in a set area (well, solid angle) so you only need to apply the force to that area, not everywhere! This means the same amount of lumens over a smaller solid angle results in a higher intensity (candelas), like how the same force over a smaller area results in higher pressure.

    Another way to look at it is candelas are analogous to Watts per steradian in the equations, so they are an intensity. Lumens are then analogous to Watts in the equations and represent total light output, like how Watts represent total power output. They aren't exactly the same or fully proportional due to the luminosity function, but they are similar enough.

    So back to that equation, you start with your magic number of 683.002(lm/W), which knowing that lumens and Watts are similar, you can see it now as a sort of unit converter. Then have your luminosity function which weights the light based on how well people perceive it's frequency (the peak at 555nm means that eyes have the strongest response to green light). Then you've got this radiant intensity function, which is about how much actual light energy the source is putting out, per steradian.

    Now for a single wavelength LED, just find the value of the luminosity function at the wavelength, assume all power is at that exact wavelength, multiply by our magic number, and you've got your candelas. However, to use this equation on any sort of broader spectrum or white LED, you'll need some calculus and the spectral power density (which you can find among the pretty pictures on the datasheet).

    If you want to go further, take a look at the "radiation pattern characteristics" (angular distribution) of a wide angle LED to see why they tend to only have luminous flux data (lumens) rather than luminous intensity data (candelas). You can also find certain deep red LEDs give out a radiant flux/intensity (Watts) instead of a luminous one, because they fall off the high end of the luminosity function.

    Let me know if this helps any, or if you have further questions on this and/or anything else you couldn't quite wrap your head around!

  • FSJ Guy / about 6 years ago / 1

    I watched this last night (who knew you could fill 28 minutes with facts and information about LEDs??) and then coincidentally saw Pete walking out of SF when I went to pick up an order this afternoon!

  • Member #134773 / about 6 years ago * / 1

    Hi Pete!

    Another excellent ATP! You managed to teach me something for which I thank you, and that is that you explained the meaning of "steradian" in terms I can understand.

    Putting on my "physicist hat" for a moment, I have a couple of comments. First, there's a simple way to calculate the energy E of photons in electron-Volts (eV) based on their wavelength L (sorry I don't know how to tease a lambda out of this comment software) in nanometers (nm). It's

    E = 1239.8/L

    (I got this from a Wikipedia article which talks about it in um rather than nm, so I adjusted the constant.). Since what we usually want is the forward voltage of the LED, we have to toss in a "Finagle's constant" to account for such things as the internal resistance of the LED, etc. Note, too, that you can rearrange the formula to be

    L = 1239.8/E

    to get the (approximate) wavelength if you have the voltage drop.

    Another comment is that you didn't mention that there are also infrared LEDs (like in the TV remote) and ultraviolet LEDs that can be fun to play with.

    Yet another comment is that all junctions that are conducting (including both diodes and transistors) are LEDs. It's just that "common" junctions (say a normal rectifier diode or a transistor) with a drop of 0.7V gives off photons with around 1770 nm wavelength, solidly in the "thermal infrared" range. (I've had way more than my share of X-rays in my life -- an X-ray machine essentially uses a vacuum tube as an "LED" operating at several KV to generate the X-rays.)

    On a more practical note, I've used the trick of first using a potentiometer as a variable resistor (I generally try to. find a relatively low value linear one in the "junk box") to adjust the brightness on the LED, then being careful to not change the setting, power down and measure the resistance to use for the "permanent" fixed resistor. Yeah, it's a bit of a hassle, but quicker than itterating on fixed resistors.

    • Member #789164 / about 6 years ago / 1

      Yet another comment is that all junctions that are conducting (including both diodes and transistors) are LEDs. It’s just that “common” junctions (say a normal rectifier diode or a transistor) with a drop of 0.7V gives off photons with around 1770 nm wavelength, solidly in the “thermal infrared” range. (I’ve had way more than my share of X-rays in my life – an X-ray machine essentially uses a vacuum tube as an “LED” operating at several KV to generate the X-rays.)"

      That's not strictly speaking true, since the light from an LED is specifically the release of a photon via a quasi-electron's relaxation from the conduction band to the valence band in a direct-bandgap semiconductor. That fact leads to the semi-monochromatic light coming off an LED under normal operation (and also why you see a spectral shift when you run the LED at lower than its designed forward voltage). While it is true that all hot things radiate, and most room-temp things radiate most strongly in the far infrared, that's blackbody radiation. Blackbody is what e.g. incandescent light-bulbs radiate, and is not monochromatic light. If you built a diode out of an indirect-bandgap material like silicon or germanium, the relaxation happens non-radiatively. While that can be heat (which can in turn be radiated off as blackbody radiation), it doesn't have to be (nor does it have to be transported away as blackbody radiation). It could, for instance, lead to the creation of excitons, or any number of other interesting quasi-particles.

      You can look at e.g. Streetman & Banerjee's "Solid State Electronic Devices" for a breakdown of LEDs in particular.

    • Hey '773! Thanks for stopping by and for dropping some good knowledge on us! I taught you something? You honor me, sir.

      Yeah, iterating with fixed values blows. I've done it with a pot, too, But these days I'd much rather use PWM or an addressable. Sooo much easier.

      • Member #394180 / about 6 years ago / 1

        That's why I have a couple of these babies (and one for caps, too).

        • Member #134773 / about 6 years ago / 1

          That was part of the reason I got the SparkFun Decade Resistance Box about 3 years ago. Only problem is that they don't include a "round tuit" to help get it built, and there seems to be a severe shortage of "round tuits", especially in my house! ;-)

          • Member #394180 / about 6 years ago / 1

            Yeah, well, when I got my Heathkit boxes SparkFun wasn't around yet. In fact, not sure if Nate was even a gleam in his daddy's eyes yet. The next time I need one of these, I'll definitely get a SparkFun box, but Heathkit equipment doesn't seem to die (even though the company did).

  • Member #1274715 / about 6 years ago / 1

    Is there supposed to be a link to a video in this post like there is in others? I do not see where the article continues.

    • Chelsea the Destroyer / about 6 years ago / 2

      Boy you guys just want EVERYTHING handed to you, don't you? Jk it's fixed :)

    • Sembazuru / about 6 years ago / 1

      Until this blog is updated, here is the youtube link for the video: https://www.youtube.com/watch?v=ZaVQKxPAQfM

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