App note: Green LED replaces LDO regulator

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App note(PDF) from MAXIM using a green LED as LDO for low power application

Insystems in which a microcontroller must communicate with peripheral devices that operate at different supply voltages, a level translator may be necessary. This design idea explains how for lowpower loads, a single green LED can replace a 1.8V LDO regulator in a level-translator circuit.

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9 Comments

  1. That is a horrendously dangerous suggestion – far more worthy of an “Instructables” page than a major manufacturer. Sure, an extremely competent designer with properly controlled conditions and component supply may be able to make it work well enough, but the people who are most liable to see it and use it are those who do not know enough to design properly. The potential (pun noticed) for sorrow or disaster is significant.
    They are suggesting that the Vf of a Green LED is close enough to 1.7V under conditions which are liable to apply here, to keep the 1.8V rail within specification. They do note that ” … the slave devices should operate on low power and draw reasonably constant supply currents.” No data sheets from reputable manufacturers, that I have ever seen, major in terms like “reasonably constant” and maximum current draw of “low power”.
    The app note should be retired asap. Or just burnt.
    Russell McMahon

  2. Oh, this is just one of those trick circuits that EEs send to EE Times, y’know, like Reader’s Digest jokes. Cute, but no need to take it seriously. It’s contrived, 3.5V on the left, really? But then writer assumes a 1.9V green LED in the doc so the right side will actually be 1.6V, hilarious… Also, remember that a Maxim engineer managed to get some free advertising for the MAX1840 part. A party trick, not much more.

    In real life, the main MCU will very likely be on the lower voltage side and as such any production or proper board will have a power supply for the MCU that will meet tolerances and specs. A small MCU running a 1.8V IC process will have its own LDO regulator (which one may be able to steal some power from the regulator capacitor pin if available), while larger MCUs running 1.8V will use much more power and are more valuable, thus it is highly unlikely that one will ever need to apply this trick anywhere.

  3. This is really from Maxim? Holy crap, this is a terrible idea. For one, Vf of an LED varies all over the place. Sure, there’s a typical value, but even at exactly the same forward current, the typical LED datasheet lists a large range of allowable forward voltages. And, of course, as the 1840’s current draw changes — on clock edges, etc. — the LED V will change even more.

    Maxim should pull that app note; it’s horribe advice. If you need 1.8V and you’ve got 3.5V (3.5?), use a volage regulator. For low-current applications, a 20 cent linear regulator in a tiny 5-pin package will work great, cost next to nothing, take up little board space, and will actually work, unlike the LED approach.

    1. Plenty of this kind of oddities at EE Times, send it in and get it published. The quality, er, varies, er, wildly. Often it looks more like giving some ICs a little free adversiting. But unlike Maxim, I wouldn’t have put this as a company app note, heh heh.

    2. this particular application might be far fetched, but using an LED to set a voltage isn’t always a terrible idea, sometimes the voltage isn’t that critical and LEDs doesn’t vary that much.

      and why use a 20cent regulator if a 2 cent LED will do the job

      1. Yep. I’ve used LEDs before as a voltage reference. I didn’t need high accuracy and it did double duty as the power LED.

  4. I wish to tell you that I only looked at this because it was in my email as “popular” in my twitter network and it seemed to me, and is, utterly preposterous. I deal with enough amateurs doing stupid things, we don’t need to give them ideas.

  5. I do remember using a similar trick back in the day – inserting several diodes into the common anode / cathode circuit of LED displays to drop the voltage / reduce the current instead of resistors on each segment. Well, it worked just fine every time, but I never claimed it was a proper solution… :)

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