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How Can I Lower Voltage More Efficiently?

  1. May 14, 2014 #1
    Hi all, I'm working on an automotive LED lighting project in which the supply voltage is from the battery/alternator and so varies 12 to 14V. I'm trying to find a way to drop the LED circuit supply voltage that would be better than using resistors. Here's the current design:

    circuit1.png

    R1 = 100Ω
    R2 = 3.9kΩ
    Vsupply = 14V

    LEDs: 20mA @ 2V


    The problem I see with the above circuit is R1 must dissipate

    P = I*I*R ≈ 0.120A * 0.120A * 12V ≈ 1.5 W

    which is rather large considering I would prefer to use 1/4 or 1/8 W rated resistors.

    So that got me thinking in a more general sense about the efficiency of voltage dividing circuits that use resistors. There must be some other circuit that I can use that would be more efficient and not that much more complex. Something like a transformer but for DC. I read a bit about linear regulators and SMPS but I'm quite sure how to use those. Any tips or ideas?

    I'm a newb with electrons so Thanks in advance for any help!
     
    Last edited: May 14, 2014
  2. jcsd
  3. May 14, 2014 #2

    berkeman

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    Yeah, you don't use resistors for LED lighting projects. You use constant-current buck regulators, with low-side current sensing. Maybe try doing some searching with those terms, and have a look at the typical circuits used. Let us know if you have questions about those circuits. :smile:
     
  4. May 14, 2014 #3
    couldn't you just put the six LEDs in series?
     
  5. May 14, 2014 #4

    berkeman

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    If they are well-matched, you can put LEDs in series and get uniform brightness. That is used in LED lighting applications, but you still would use a current-sensing buck regulator usually to maximize the efficiency. When you buy a bag of LEDs, they are generally not well-matched, though. If you buy them on a reel or Cut Tape, they should be matched fairly well.
     
  6. May 14, 2014 #5
    Thanks for the ideas, I'm reading up on these buck babies now.

    I actually was originally designing for series wiring, hence the six LEDs (12V / 6 gives 2V over each, in theory), but an EE friend advised against it for the reasons mentioned above. Just curious though, how does one "match" them? What characteristics exactly are matched?

    And in all honestly the higher priority for me is just getting the project done. I don't need 99% efficiency, I'm just looking for something better than wasting 1.5W to heat. Although that would be acceptable especially if it is much simpler and easier for a newb to do. That being said, what do you think about max power dissipation ratings for resistors? What goes into that? I mean what does that assume in terms of convection, radiation, conduction, etc.?

    Thanks for the ideas and thought-provoking discussion; that's why I love these forums!
     
  7. May 14, 2014 #6

    berkeman

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    I think that if you use the "Cut Tape" option when you buy the LEDs from Digikey, you will get LEDs that are from the same batch, so they will be matched fairly well. There is another level of matching that I think LED Lighting high-power LED manufacturers use (like tighter tolerances on the efficiencies), but if you are doing a fast one-off project, that may not matter.

    And a good compromise on complexity versus efficiency is to buck the 12V down to around 5V, and use a single resistor for each 2xLED string. Place as many 2xLED strings in parallel as needed to get you to your desired lighting level.
     
  8. May 14, 2014 #7

    berkeman

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    Or adjust the number of LEDs and the buck output voltage as needed -- you need to look at the tolerance on Vforward at your target current to decide what combination of Vout and Rseries you will use.
     
  9. May 15, 2014 #8

    meBigGuy

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    Basically, if you have 3V across the LED's (depends on the LED's spec, based on color, etc) you have 14-3 = 11V across 100 ohms which is 110ma. If the LED's are matched that's 110/6 = 18ma each.
    The 3.9K does nothing except draw about 1.3ma.

    I don't know whether matching is harder for parallel or series connections, or if there is any difference. Intuitively I think parallel matching may be more of an issue since 1 LED can hog a lot of current.

    Regarding matching when LED's are in series: all you are trying to match is the brightness at a given current.

    Try it and see if they are uniform enough, and swap out the ones that don't match until you get acceptable results. Nothing to lose.
     
    Last edited by a moderator: May 15, 2014
  10. May 15, 2014 #9

    berkeman

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    It's pretty much impossible to match them in parallel as shown in the OP. Series matching is a much better bet, and is done routinely in LED lighting applications.
     
  11. May 15, 2014 #10
    OK so an alternative circuit (assuming 12V supply) would be to just wire the 6 LEDs in series, and swap outliers? Would that distribute 2V across each LED evenly? or close enough to an even distribution?
     
  12. May 15, 2014 #11

    berkeman

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    You still need a current-defining resistor. Drop about 2V across the resistor, and put 5 LEDs in series. Use the value of the resistor to set the series current.

    What current are your LEDs rated for?
     
  13. May 15, 2014 #12
    They're 20mA at 2V and the max anticipated supply voltage is 14V. So I can keep the 6 LEDs, drop in a current limiting resistor of R = 2V / 0.120A = 17 Ohms and I should be all good, no?

    I've read its not a good idea to run LEDs at their max rated continuous current, but I figure I'll be fine as this circuit is on a momentary switch and will be relatively low duty.

    Thanks for all your help berkeman!
     
  14. May 15, 2014 #13

    berkeman

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    What is the minimum anticipated supply voltage...?
     
  15. May 15, 2014 #14
    It's for a motorcycle brake light, so battery voltage which ranges ~12V to 14V
     
  16. May 15, 2014 #15

    berkeman

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    So you're back to 5 LEDs, correct? :smile:
     
  17. May 15, 2014 #16
    6... why?
     
  18. May 15, 2014 #17

    berkeman

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    Because if the LEDs are 2V apiece and the supply is 12V, there is no voltage drop across the resistor and the LED current is undefined. You always want to have a voltage across the current-setting resistor.

    And you can see the tradeoff/disadvantage of using just a resistor to set the current when you power supply voltage can vary. To get a fairly well-defined current, you need to drop more of the PS voltage across the resistor (like 6V across the resistor and the rest across 3 LEDs, so you don't get a 2:1 variation in LED current over the PS range of 12-14V). But doing that wastes a lot more power in the resistor. When the power supply can vary, going with the DC-DC buck with current sensing makes more sense...
     
  19. May 15, 2014 #18
    Ok so my question of the voltage distribution arises again... intuitively I imagine the battery/supply voltage being divided evenly (or close to that) amongst each of the elements in the circuit, which must be wrong if what you're saying is true.

    I suppose what I am assuming about the LED forward voltage must be incorrect. I am thinking that the forward voltage is the "ideal" supply voltage, rather than a fixed value.

    So then to reconcile my understanding with what you are saying, the LEDs exhibit a voltage drop equal to the forward voltage no matter what? if that is the case, then you are right and I must go back to 5 LEDs and use a higher resistance...
     
  20. May 15, 2014 #19

    berkeman

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    No, please have a look at the datasheet. It will give the V-I characteristic (typical and tolerances). The V-I characteristic is non-linear (not linear like for a resistor). You need some mechanism to set the LED current, so that small variations in the Vf of the LED do not cause large changes in the LED current.
     
  21. May 15, 2014 #20

    meBigGuy

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    LED current increase logarithmically with voltage across the diode. The standard solution is to use a constant current LED driver. A resistor will work also, but it's hard to calculate the correct value for a given string. Also, the voltage varying from 12 to 14V would significantly increase the current if there was 2V across the resistor..

    An lm317 voltage regulator makes a very nice current source with 1 sense resistor. You can google the circuits. Nice thing about it is that the current won't vary with voltage, which would be a significant variation with a resistor. The to-220 package is cheap ($0.72)and has good thermal dissipation. Read the data sheet.

    There might be other regulators that need less headroom and have a lower reference voltage. There are other LM317 packages also.
     
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