Calculate what resistance needed for an LED

Discussion in 'General Electronics Chat' started by Tonyr1084, Aug 27, 2016.

  1. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    How would I calculate the appropriate resistance needed for a specific voltage and a specific LED?

    OK, you're going to ask what voltage and what LED. My question is hypothetical. However, in the future if I take a known LED, how do I calculate the appropriate resistor to prevent it from burning up? Sure, there are known resistors for known voltages; and give or take a little resistance for adjusting desired brightness. I don't want a "Pat" answer, I'd like to learn how to figure this out for myself.
     
  2. dl324

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    Mar 30, 2015
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     \small R = \frac{V}{I} = \frac{(V_{sup}-V_{LED})}{I_{LED}}

    Datasheet will give typical voltage/current relationship.
     
  3. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    I wasn't sure, but I suspected you subtracted the diode voltage from the supply voltage before doing the calc's.

    Theoretically speaking: I have an LED with 2 volts forward and a current of 30 mA. Powering from a 12 volt supply, I'd calculate 10/0.03 and get 333.3Ω I'd use a 330Ω resistor in series.

    Wrong ? ? ?
     
  4. wayneh

    Expert

    Sep 9, 2010
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    That's correct. However in practice I usually aim for, say, 5mA and only crank it up when more light is absolutely required. I never run an LED at the specified maximum current. They age.
     
  5. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    Good point. Thanks all.
     
  6. dl324

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    Mar 30, 2015
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    Your calculation is correct.

    One thing to be aware of is that luminous intensity is a function of current. It isn't a straight line, and isn't the same for all LEDs. Here's some info from some Kingbright LEDs from the L934 family (high efficiency red, green, and yellow, with max DC currents (in mA) of 30, 25, and 30, respectively):
    upload_2016-8-27_9-45-11.png upload_2016-8-27_9-44-45.png upload_2016-8-27_9-46-29.png
    From the graphs, it's apparent that there's no point in driving the high efficiency red at a current above about 15-20mA. The other two show increased intensity with increasing current, up to the maximum DC current.

    The peak currents for these LEDs are 140-160mA. In multiplexed applications, you want to increase the average power dissipation of the LEDs for increased perceived brightness. The datasheet for these LEDs states the conditions for applying peak current are 0.1 duty cycle, 0.1mS pulse width.
    EDIT: Previous paragraph was poorly worded. To maintain the same brightness as an LED driven continuously, the average power dissipation of a multiplexed LED should be about the same as one that is driven DC.
     
    Last edited: Aug 27, 2016
  7. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    Of course, for each LED I have I need to consult the data sheet to get the specific numbers. I have a bag full of odds and ends LED's and can only assume very low currents for them until I have a chance to get an idea of their voltages and currents. But thanks for the pointers that the RED LED does not handle high currents very well. I sort of knew that. Your charts proved that out.

    LED's are things I like to mess with from time to time. I've built a cart for carrying picnic stuff from the truck to the camp sight. I've installed some LED's to act as headlights and tail lights. Work lights would be nice too. Of those already installed, they're battery powered and market bought. Anything additional I'll need to know how to power them without burning them out.

    Parallel LED's with individual resistors, one for each LED.

    Thanks all again.
     
  8. ian field

    Distinguished Member

    Oct 27, 2012
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    To add to what has already been said - the bigger resistance you end up with; the more tolerant it is of supply voltage variations - but at the expense of wasting more power.

    If your supply voltage comes from a 3 terminal regulator - you can cut it fine.

    If your supply is a battery - the terminal voltage may start off a little higher than the stated nominal value and gradually sink to as low as you'll put up with. You have to take this higher voltage into account and calculate the resistor to prevent excessive current under these conditions. The LED will get dimmer as the battery runs down. This is less noticeable if the voltage across the resistor is large relative to Vf - but as I said; it wastes more power.
     
  9. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    Doesn't a higher resistance mean a lower amperage? Thus, a lower wattage of wasted power?
     
  10. dl324

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    Mar 30, 2015
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    If you don't know part numbers or can't find datasheets, 10-20mA is a fairly safe current. You can tell if you're pushing them if the color changes due to the junction getting hot. A few decades ago, 30mA would have been high for DC, but there are some high power LEDs now.

    Unfortunately it isn't that simple. Here's data for OSRAM 3341 LEDs; all show very similar intensity characteristics:
    upload_2016-8-27_16-16-23.png
     
  11. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    Is there a reliable way to determine the forward voltage of an LED?
     
  12. tindel

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    Sep 16, 2012
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    Yes google search "load line led"
     
  13. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    Gee thanks Tindel. :(
     
  14. tindel

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    Sep 16, 2012
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    You don't know how to google search?
     
  15. dl324

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    Measuring is the only reliable method because there will be variation within die from the same wafer. Manufacturers will often sort LEDs in a lot according to brightness, but not by using Vf. Any graphs the manufacturer provides are for typical parts.
     
  16. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    Tindel: I once searched for the sex offender website to see if there were any predators living near my grand daughter. I got TONS of porn websites. Yeah, google is good for all kinds of stuff. And even after searching based on YOUR recommendation I still had to review several sites until I found one that answered the question. I guess it's easier for you to look smart and say "Google it" than to give an actual answer. Maybe it's because you didn't know the answer or weren't sure if you knew the correct answer. At least I had the good sense to ask.

    Do me a favor - if ever I ask a question that makes you indignant toward me - just don't answer it. If I'm that ignorant then why would you even want to answer one of my questions?
     
  17. ian field

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    Oct 27, 2012
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    You have missed the point entirely.

    A higher voltage was implied by a higher resistance to set the specified current.
     
  18. tindel

    Active Member

    Sep 16, 2012
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    Well I wasn't indignant toward you. Now i think you're just a complany pants because i didn't spoon feed you. I have a bsee and have been doing this for nearly 10 years. I knew the answer, gave you the correct search terms, and the first result from the search gives a link to a wikipedia site that explains the concept well. What more do you want from me? I give up.
     
  19. Tonyr1084

    Thread Starter Active Member

    Sep 24, 2015
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    Complanypants? BSEE? OK. I suppose this has to end somewhere. I already got my answer long time ago. I wish I could close this thread and avoid any more answers who claim to spoon feed me with an empty spoon.

    Peace y'all.
     
  20. tindel

    Active Member

    Sep 16, 2012
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    Adding one to the ignore list... Doubling the total there... Oh brother. I was sincerely trying to help.
     
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