15k resistor to dim a LED??

Discussion in 'Electronics & Electrics' started by doubter, Feb 2, 2020.

  1. doubter

    doubter Member

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    So the LEDs on my keyboard (for caps, scroll, num lock) are super bright. I have measured that 5v is supplied for each LED.


    I bought some 3.2v 20ma LEDs (1000mcd brightness - lowest I could find in white) to swap in.
    Its stated to use a 91ohm resistor for 5v.. I bought a bunch of resistors to dim it, and even five 2k resistors (=10K) linked up, it was still quite bright. Everything I see online seems generally say 1k or less does the job though :/.

    I am thinking getting a 15k or 20k resistor.. thoughts?
     
  2. theSeekerr

    theSeekerr Member

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    OK, so 3.2V is the forward voltage of the LED - with less than 3.2V it won't light, and when lit it will drop around 3.2V

    20mA is the maximum rated current for the LED. With a 5V supply, and 3.2V dropped across the LED, the resistor will have 1.8V across it.

    Since V = I * R , R = V / I , where V is 1.8V and I is 20mA

    R = 1.8 / 20m = 1800 / 20 = 90 ohms, and the nearest E24 resistor is 91 ohms.

    So, the instructions are specifying a resistor for maximum rated brightness. Fair enough.

    Go buy a 50 kOhm potentiometer. It'll cost you about $3. That's a 3-terminal device. Hook it up with the middle terminal and either of the other two and you'll have a variable resistor. Adjust it until the brightness looks good, then turn off the power and measure the resistance across those terminals with a multimeter and buy resistors of that resistance.

    If you don't have a multimeter I guess just buy a bunch of resistors from 15k to 50k and keep trying until you find a level you like...
     
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  3. dakiller

    dakiller (Oscillating & Impeding)

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    Yea, if it is still too bright, keep adding resistors
     
  4. ArmoureD

    ArmoureD Member

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    Black permanent maker over the led?
     
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  5. merlin13

    merlin13 Member

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    What, you're measuring that directly across the actual LED itself? If so then that sounds a bit eyebrow raising/sus...

    Anyway, on the offchance you've got high efficiency LEDs and an actual 5 volt source try two in series for a quick laugh.

    Other trick if you're not playing with SMT LEDs, to drop the perceived light output use a bit of sandpaper or a fine file to frost or rough the top and even the sides of it - diffuses the emitted light pattern instead of all it squirting out in the normal 5/10/20 degree propagation pattern. Could tr tht on ST ones but that'd depend on how easy it is to get to the actual resin surface.
    Damn fine idea. Suggest grabbing a Linear one instead of a Log pot as well, a 10-turn if you really want to have a good play.

    Else post up your location and see if someone in here nearby can do the fiddlies for you.
     
  6. OP
    OP
    doubter

    doubter Member

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    Thanks for the replies.

    5v was measured with no LED installed.

    Some dumb questions.. does the 20ma LED mean it restricts the amount of amps to 20ma for everything else in that circuit?

    And is the the 90ohm resistor needed to use the remaining 1.8v (which is restricted to 20ma because of the LED?) to ensure the LED doesnt overload? Does that make the wattage of the resistor 20ma*1.8v = 0.036watts of heat??

    And say, instead of the 90ohm resistor, a 10k ohm resistor is used, will that make a lot more heat? essentially the circuit is 5v (the source) times 20ma (the amps of the led) so .1watt? that either has to be transformed into light (led) or heat (resistor)??
     
  7. theSeekerr

    theSeekerr Member

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    The answer to pretty much everything you just said is "no".

    The LED does not have a fixed current draw. It has a fixed voltage drop.

    The current draw is determined by the resistance in the circuit. Ohms law says that V = I * R , so I = V / R

    V, here, is (+5V +(-3.2V)) = 1.8V

    So with the original 91 ohm resistor, I = 1.8 / 91 = 20 mA

    If you put a much larger resistor in there, the current will drop significantly. Say with 10K : I = 1.8 / 10K = 0.18 mA

    Total power consumption of the circuit will be 5V * the loop current. So with the original 91R resistor: 5 * 20m = 100 milliwatts . Of this, the portion dissipated by the resistor is 1.8 * 20m = 36 mW.
    With less loop current, this drops proportionally, to essentially nothing: 5 * 0.18mA = 0.9 mW. Of this, the portion dissipated by the resistor is 1.8 * 0.18mA = 0.324 mW , or as near to nothing as makes no odds.

    EDIT: FIxed some maths errors caused by dropping a digit.
     
    Last edited: Feb 3, 2020
  8. SLATYE

    SLATYE SLATYE, not SLAYTE

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    No. The LED doesn't restrict current at all - it'll use all you can give it (until it dies, which will occur at slightly more than 20mA).

    The point of the resistor is to limit the LED to only 20mA (or less, depending on the resistor). This will restrict everything else in a series circuit to the same current.

    Sort of; you're just thinking about it in an odd way.

    The LED will use 3.2V, more-or-less constant regardless of how much current goes through it. This leaves 1.8V available, and a resistor provides a convenient way to turn that 1.8V into a current limit. The current flowing will be I = 1.8/R.



    Yes. Which is a very small amount.

    No! This is a very common misconception.

    First you calculate the current:

    I = 1.8 / R

    For a 10K resistor, I = 0.00018A

    Now the power is 1.8 * 0.00018 = 0.000324W, which is essentially nothing.

    You can simplify the maths to be simply P = V²/R, where V = 1.8. The bigger R gets, the smaller the power gets. Obviously in the limiting case (R = infinity, as if the resistor is not connected at one end) P = 0, because no current can flow.
     
  9. ArmoureD

    ArmoureD Member

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    So would a simple black permanent marker on each LED not suffice in this scenario?
     
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  10. mtma

    mtma Member

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    A reminder that the parts need to be connected in series, such that each of their terminals connects to the next device and so forth without any kind of 'star' point occurring in the circuit.

    example

    Perhaps it's already obvious but 10k with 5V and a regular 20mA white led should be quite dull and it has probably colour shifted to blue as well.
     
  11. merlin13

    merlin13 Member

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    THAT explains a lot. As the Old Joke goes, unfortunately You're Doing It Wrong...

    Depends on the density of the ink and how evenly you can apply it to the tops of multiple LEDs ('coz ifyou can't/don't then you'll get uneven/mismatched lighting...).

    But if the OP is at the stage of fiddling with the hardware at Hot Stick levels then why not do it properly and adjust/add/change the series resistance.

    At least it's a learning exercise.

    And mucking about with the LEDs 'n resisters means the OP will get even lighting levels at what they want to see, rather than trial 'n error dabbing away.

    Should be, but it's really going to depend on the efficiency of the LEDs the OP has got. I've seen high quality/high output LEDs run dim at over 5 mA, I've seen cheap Yum Cha ones still put decent light out at 2 mA...
     
  12. OP
    OP
    doubter

    doubter Member

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    Thanks for all the knowledge guys. There is a lot of statements on the net saying that resistors convert energy into heat, so I think that influenced my poor thinking.

    So I bought a bunch of resistors and ended up settling with with a 42k per LED, and also sanded them to diffuse the light. The keyboard came with blue LEDs and now they are white. I didn't want to just texter them black, as I wanted to change colours and do it a bit more proper (plus it was a minor project).

    Some further questions if you are willing to answer..
    So a resistor reduces current, given the supply is a fixed 5v... but it doesn't do this by producing heat? its just the material properties of the resistor that does this?

    SLATYE - when you talk about power, do you just mean the amount of power that the resistor allows the circuit to have? not the amount of power it converts to heat (because this concept is irrelevant)?

    Also I know the LED is 3.2v, meaning 1.8v left over - but why dont we do the maths for the resistor using 5v (say the resistor is connected to the + side of the LED, wouldnt it be 'fed' 5v)? (maybe i have incorrect assumptions about electricity, electrons, voltage and if electrons are ''converted')
     
  13. theSeekerr

    theSeekerr Member

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    Oh, it produces heat - wherever I talk about a resistor dissipating power in my post above, that power is dissipated as heat.

    Where you've gone wrong is in assuming that a 5V source somehow equates to a fixed amount of power to be dissipated. It doesn't. If you run 5v through a large resistance, very little current flows and very little power is dissipated. If you run 5V through a small resistance, a lot of current flows and a lot of power is dissipated (i.e. it gets hot).

    Kirchhoff's Voltage Law tells us that the sum of voltages in a loop of series elements is 0. We know that the power supply element goes up from 0V to 5V, and we know that the LED subtracts 3.2V. The only other element in the series is the resistor, so we solve for the unknown and get 1.8V subtracted across the resistor.

    If you put the resistor on the high side of the LED, it "sees" 5V on one side and 3.2V on the other, for a difference of 1.8V
    If you put the resistor on the low side of the LED, it "sees" 1.8V on one side and 0V on the other, for a difference of 1.8V
     
    Last edited: Feb 4, 2020

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