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I see that there are a number of questions in a similar vein but I specifically want the simplest method to implement an LED to indicate whether power is present when a device is turned on using 0603 resistors. I have a number of different voltage rails but the highest is 24V. I want to use a basic red 0603 SMD. A typical example of the LED that I intend to use is here: http://www.farnell.com/datasheets/2046023.pdf

Now, with a voltage drop of approximately 2V, and if I go half brightness with 15mA then I need a resistance of 1466Ω. Taking the approximate power dissipation of 0.33W, I think that typical 0603 SMD parts have a maximum heat dissipation of 0.1W? If this is the case then I need at least 4 resistors. The solution I have gone for is here:

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I'm using standard value resistors and this reduces the current to 12mA. I also stuck to using 4 resistors to reduce the power dissipation across each resistor further. It's ugly but I think it should work fine? Any comments as to why this may not work or is a bad idea would be appreciated. Thanks.

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  • \$\begingroup\$ How bright do you want the led? My gut feeling is that 5mA should be sufficient. Using a number of resistors is a common method to decrease the voltage dropped across each resistor for spreading the load or higher voltage applications. \$\endgroup\$
    – Kartman
    Jun 1, 2021 at 12:07
  • \$\begingroup\$ I want the LED to be fairly bright, it may need to be viewed outside in the sun. I don't think that it will need to be the full 30mA though, I assume that will be blindingly bright. \$\endgroup\$
    – ChrisD91
    Jun 1, 2021 at 12:12

3 Answers 3

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For 24V and a single LED, assuming say 2mA LED current, I'd use a single resistor and a zener. The LED will drop about 2.5V (depending on what colour you want, but I'm not trying to be very precise here), so you have 21.5V. I'd choose a 20V zener, so you want a resistor for 1.5V at 2mA, or 750R. Here's your circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Note the zener runs at 40mW, because of the current. You could get away with 1mA, depends how bright you want the LED. The LED will go out as you get down to about 23V or so - you can make it a bit more tolerant of voltage variation by using, say, an 18V zener and choosing another resistor.

NOTE : The datasheet talks about 20mA forward current, but you probably don't need anything like this amount of current to get a decent indication. For this kind of thing I find it's best to make it and do a bit of trial and error. But with this circuit you can achieve a good range of "off" voltages and brightness.

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  • \$\begingroup\$ Thanks for the answer. What is the advantage of your suggested circuit against mine other than it will only light the LED when 24V is present? Just trying to understand the numbers for my case. Say I want 12mA as the LED needs to be sunlight readable, and with a drop of 2V. Sticking with your 20V zener, I'd need 2V/12mA = 167R. But then I'd need a much higher wattage Zener? \$\endgroup\$
    – ChrisD91
    Jun 1, 2021 at 13:08
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    \$\begingroup\$ You already pointed it out. You don't want the LED on if you have, say, 18V due to some malfunction. So a zener is the way to do this. Have you actually determined the amount of current needed to make the LED visible in the conditions you anticipate? I very much doubt you need 15mA. The most I ever designed to was 10mA, and they were visible in bright lighting. You also need to try a selection of colors, as others have pointed out below. Once you figure out the current and actual LED, you can adapt the design suggested here. If you do need high current you may indeed need two zeners in series. \$\endgroup\$
    – danmcb
    Jun 1, 2021 at 13:15
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    \$\begingroup\$ "But then I'd need a much higher wattage Zener?" Yes. No matter what the components are, you are creating a linear (as opposed to switching) current regulator. For the same Vcc, Vf, and If, the power dissipation will be exactly the same for any series combination of components. You can move the heat here and there, but at 22 V and 12 mA, 0.264 W is gonna go somewhere. \$\endgroup\$
    – AnalogKid
    Jun 1, 2021 at 13:20
  • \$\begingroup\$ No, I haven't determined anything as of yet. This is all preliminary R&D. I'll take your suggestion and add the additional zeners and resistors to the schematic anyway in case higher current is needed later. \$\endgroup\$
    – ChrisD91
    Jun 1, 2021 at 13:22
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    \$\begingroup\$ I'd buy the actual LED you think of using, and any old through hole resistor and a variable supply just to get an idea of current. That's the thing you need to determine it seems. \$\endgroup\$
    – danmcb
    Jun 1, 2021 at 13:27
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You say you need good indication even in sunlight, with low current the simplest is a LED with:

  • A bright visible color like orange or green instead of red

  • Higher millicandela rating so you can run it at low current

  • No diffuser.

If you can use thru hole parts, a 5mm high millicandela orange LED will be bright even in sunlight with 1 mA if you look directly into it, so you can avoid wasting power and use only one resistor. In fact these LEDs are annoyingly bright when used indoors.

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  • \$\begingroup\$ Red LED is for aesthetic/branding purposes although I would generally agree with going for green. Curious as to why you suggest orange though? \$\endgroup\$
    – ChrisD91
    Jun 1, 2021 at 13:15
  • \$\begingroup\$ In case you don't like green for aesthetic reasons ;) The eye is more sensitive to orange that red so you still get more visibility per watt... But, if the LED is behind a diffuser with a brand logo, maybe you can put several LEDs in series? \$\endgroup\$
    – bobflux
    Jun 1, 2021 at 15:01
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It will work, but it will start indicating dimly at a voltage way below 24 V. Putting a zener diode in the string will give a more useful indication.

About that power dissipation... At over 80% of its rated power dissipation, the resistors might unsolder themselves.

The maximum rated power listed on a datasheet is not a good design point for any electronic component. If you are going to literally push a part to its limits, resistors can survive this better than anything else. Still, a much better, and more common, design point for long-term reliability is 50%. Do not operate a resistor age greater than 50% of its power rating, not a transistor at more than 50% of its voltage and current ratings, 50% for capacitors (especially electrolytics), etc.

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