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I have a circuit that has a 24V power rail and I put an LED and series resistor on it for power indication. Even at 10mA forward current, over 200mW of power is dissipated, over 90% of it in the series resistor. This is more than many (but not all) SMD resistors can even provide. This seems like a lot of power for a simple LED.

Neon indicators might work at the higher voltage end, but they aren't commonly available in SMD packages and they're bulky, expensive and can have limited life.

I also considered an astable multivibrator to do a low-duty cycle flash to keep average power down, but that needs 4 resistors, 2 capacitors and 2 transistors, or an IC (which would be unlikely to work at 24V) and some passives.

Is there a very simple (low component count, cheap, low-power and low-board-area) way to show power indication in circuits with a rail voltage more than about 10 times the LED \$V_f\$ (say around 15V up to 48V).

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  • \$\begingroup\$ which is more than many SMD resistors can even provide There are SMD resistors with a higher maximum power dissipation. They are larger (in size) though. \$\endgroup\$ – Bimpelrekkie Jun 25 '18 at 13:27
  • \$\begingroup\$ @Bimpelrekkie sure, even in the same package sometimes, especially if you go from thin to thick film. But then you have to start specifying different power ratings from your normal stock. \$\endgroup\$ – Inductiveload Jun 25 '18 at 13:54
  • \$\begingroup\$ Befiore you accept an answer, can you please specify your acceptance criteria for power dissipation. 100mW? 10mw? or ? and your budget 0.xx$ \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 25 '18 at 14:26
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    \$\begingroup\$ What about using two LEDs in series and lower current? Say, 5mA through two LEDs instead of 10mA through one LED? There are SMD packages with two LEDs that can be connected in series ("bi-color LEDs"). \$\endgroup\$ – Jukka Suomela Jun 25 '18 at 17:57
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    \$\begingroup\$ Be careful with your terms: "medium voltage" usually refers to ranges from 600V-1000V and up. With 48V you're still in the "extra-low" range. \$\endgroup\$ – Agent_L Jun 26 '18 at 11:00
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Use a different LED. High brightness LEDs should still be plenty bright enough for an indicator at 1-5 mA.

The problem you have is whatever component(s) you use to do this linearly will be dissipating the excess energy as heat.

The only way to perform this more efficiently if you wish to run from a high voltage, and put 20 mA thru the LED is to use some kind of switching device but that goes against one or more of the simple, cheap and low component count requirements.

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  • \$\begingroup\$ +1 For what it's worth there are 240VAC indicator LEDs that use relatively small size series dropping resistors. Modern LEDs are that good. \$\endgroup\$ – Spehro Pefhany Jun 26 '18 at 14:10
  • \$\begingroup\$ A solution would be a very low current LED chip with several diodes connected in series. The light beam should look like that of a single LED. I got no information if such a LED chip is manufactured. It will be more expensive than the same number of individual LED chips. \$\endgroup\$ – Uwe Jun 26 '18 at 14:11
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Use multiple resistors in series!

At the cost of a little board space, you can stick with standard-sized parts. If you use the same value resistors the power dissipation will be (approximately) equal across them...

So for a \$3V\$ LED at \$10mA\$, you need to drop \$21V\$ so you need \$2.1k\Omega\$; if you use 2x \$1.05k\Omega\$ (\$\le\$2%) resistors that's \$2.1k\Omega\$ which allows the requisite \$10mA\$ across the LED, and the resistors will dissipate \$(0.01A \times 21V) \div 2 = 105mW\$ apiece. If you're using 125mW+ resistors you should be fine.

If you're using 5% resistors, just use 2x \$1.1k\Omega\$ resistors, which will give \$2.2k\Omega\$ combined resistance for about \$9.5mA\$ instead, and that will be \$(0.0095A \times 21V) \div 2 = 99.8mW\$ apiece.

Since they're all in series (both resistors and the LED) you can get creative with placement. Put both resistors before the LED, or one before and one after, or both after!

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  • \$\begingroup\$ You can do it in parallel as well - use \$n\$ resistors of \$R \times n\$ and divide the power between them that way. \$\endgroup\$ – Inductiveload Jun 25 '18 at 16:51
  • \$\begingroup\$ Another question is, how high the 24V line can get? If it maxes at 24V then powering LED via divider is enough, but if it's 24V-or-more like up to 100V/etc then simple divider powered from 1 source will just blow up the LED. \$\endgroup\$ – quetzalcoatl Jun 26 '18 at 11:39
  • \$\begingroup\$ @Inductiveload the problem with paralleling LEDs is that (unless you use an individual resistor on each LED, which is more components) if the LEDs are not perfectly identical, one may conduct more current than the others, and will burn out quicker -- then resulting in an overload on the remaining LEDs that share a common resistor (and their imminent burnout). \$\endgroup\$ – Doktor J Jul 23 '19 at 14:52
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One option you may have is to take a look at your circuit and where the supply current is used. It may be possible to simply place the LED in series with some other circuit that:

  1. Does not mind having a voltage drop equivalent to the forward voltage drop of the LED.
  2. Limits the current in the series circuit to a value that is compatible with the current ratings of the LED.

If you can find this in your circuit you would basically be moving that wasted 200mW of power from the resistor to another part of the circuit where it may be put to better use.

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If you have any lower voltage rails available, you can use a single transistor and a voltage divider.

The voltage divider drives the 24V down into the gate/base of a transistor, then your LED draws power from the lower voltage rail. Something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ The downside of this is that it effectively creates an "and" gate, the LED will only light if both the 24V and the 5V rails are up. \$\endgroup\$ – Peter Green Jun 25 '18 at 17:54
  • \$\begingroup\$ That's an excellent point. I hadn't considered that. I've used this topology mostly when I have a low voltage microcontroller toggling high voltage loads, and want to indicate when the high voltage stuff is active \$\endgroup\$ – Chris Fernandez Jun 25 '18 at 19:59
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    \$\begingroup\$ Is that actually a downside? \$\endgroup\$ – Russell Borogove Jun 26 '18 at 2:32
  • \$\begingroup\$ @RussellBorogove Of course. Someone sees 24V is off, dives in with a screwdriver, shorts something and it all blows up. How's that not a downside? \$\endgroup\$ – Dmitry Grigoryev Jun 26 '18 at 10:50
  • \$\begingroup\$ Well, nothing prevents one from increasing Rlimit and connecting to 24V instead.. but then you could drop the transistor and just use a voltage divider tailored for the LED itself. The real question is, how high the 24V line can get? If it maxes at 24V then powering LED via divider is enough, but if it's 24V-or-more like up to 100V/etc then simple divider powered from 1 source will just blow up the LED. \$\endgroup\$ – quetzalcoatl Jun 26 '18 at 11:39
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I looked for a high efficiency, high voltage step-down DC/DC switching converter and found the LTC3632, but you need the chip, two capacitors, two resistors and an inductor. May be used for input voltages up to 50 V.

Efficiency better than 70 % for a current of 0.1 mA. Two resistors to set output voltage, one condensator at input and output and one inductor. Of course one more resistor to set LED current.

Less components would be difficult. May be the condensator at input is not necessary when placed closed to the condensator of the power supply.

Did not look for the price of the chip. There may be other DC/DC switching converters from other manufacturers.

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Drop the voltage with silicon.

Your problem is the voltage across the resistor. If you can reduce that, you'll cut your power dissipation.

The obvious solution for dropping voltage would be a Zener diode in series with the LED and current-set resistor. With an 18V Zener and maybe 2V across the LED (to use round numbers), you've got 4V across the resistor. A 400R resistor will give you 10mA, and you've only got 40mW power dissipation on the resistor.

Zeners aren't perfectly stable voltage-wise, of course, but for just lighting a LED they're plenty good enough.

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  • \$\begingroup\$ Instead of a zener, multiple LEDs in series would also work, especially blue ones. If the are very close together you could reduce the current even further. \$\endgroup\$ – Pelle Jun 26 '18 at 7:58
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    \$\begingroup\$ You'll cut your power dissipation on the resistor, but you're still dissipating the same power overall, it's just spread over more components. In your example, the zener is dissipating 0.18 W. \$\endgroup\$ – Colin Jun 26 '18 at 7:58
  • \$\begingroup\$ What happens to the electrical power defined by Zener diode voltage drop and current? What is the difference to power dissipated within a resistor? \$\endgroup\$ – Uwe Jun 26 '18 at 14:04

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