Resistor & LED Power Dissipation Calculations in DEPTH

Discussion in 'General Electronics Chat' started by Guest3123, Jan 7, 2016.

  1. Guest3123

    Thread Starter Member

    Oct 28, 2014
    312
    18
    *** LED's IN SERIES ***
    So.. To Calculate Resistor Value for LED's in Series..
    LED Data Sheet : 2V Vf, 0.02A If, Two LED's of the same kind, with a supply voltage of 5vdc.

    With LED's in series, you add the forward voltage of all the LED's in series, and the LED's forward current stays the same.

    Step 1. Add LED's total Forward Voltage (VF). 2 x 2 = 4vdc.
    Step 2. Subtract LED's Forward Voltage from Supply voltage to find the amount of voltage to drop across the resistor.
    4Vdc - 5Vdc = 1Vdc.

    Step 3. To find the resistor value for the LED's in series, divide the Voltage to be dropped across the resistor by the forward current of the LED's.
    1Vdc / 0.02A = 50Ω (I will need a 50Ω resistor)


    *** LED's IN PARALLEL ***
    So.. To Calculate Resistor Value for LED's in Parallel..
    LED Data Sheet : 3.5Vdc Vf, 0.035A If, Twenty LED's of the same kind, with a supply voltage of 5Vdc.

    With LED's in Parallel, you add the forward current, and the LED's Forward Voltage Stays the same.

    Step 1. Add LED's Forward Current ( IF). 0.035A x 20 = 0.7A
    Step 2. Subtract LED's Forward Voltage from Supply Voltage, to find how much voltage is to be dropped across the resistor.
    3.5Vdc (VF) - 5Vdc = 1.5Vdc

    Step 3. To find the resistor value for the LED's in parallel, divide the voltage to be dropped across the resistor by the forward current of the LED's.
    1.5Vdc / 0.7A = 2.142Ω (I will need a 2.15Ω resistor)

    Doesn't exist..? Sure it does.. Mouser Electronics Part # 660-MF1/4DCT52R2R15F (2.15Ω, 250mW, 1% Tolerance, KOA Speer, Metal Film Resistor, $00.10 ea, & 1,754 Can Ship Immediately)


    If that's all correct.. then how do you calculate the Power dissipation of the LED's each, LED Total, and the Resistor Power Dissipation in Watts?
    Knowing that, you will be able to reinsure that 1. The Resistor you picked out 1/4 Watt, is good enough for your simple circuit, and 2. It also shows that you have a good understanding of what's going on with your LED and Resistor. Not know it, will cost you more money.

    1. You will not only be charged $00.10 per resistor, but also S&H charges again.
    2. You might also have to buy more LED's because you didn't know what you were doing in the first place.
    3. You will have to wait again.. For your electrical components to come in the mail.. So.. If you were working on this for someone, you've not only wasted your time and money, but you've wasted the clients time, and you spent more than you had to on the project.

    NOTE : Since we choose a higher resistor value, the current for the LED has went down.


    *** LED's IN SERIES, POWER DISSIPATION ***
    So.. Let's calculate LED and Resistor Power Dissipation, shall we. :)


    To Calculate Power Dissipation of each LED using the Data we already calculated in the above text.

    LED's In Series. SIngle LED Power Dissipation.
    Step 1.
    Using Ohm's Law, find the Power Dissipated in Watts. (LED Forward Voltage x LED Forward Current = Watts)
    2Vdc x 0.02A = 0.04 Watts, or.. Since there's 1000 Milliwatts in 1 Watt, it's a power dissipation of 40mW for a single LED.

    LED's in Series. Total LED Power Dissipation.
    Step 1.
    Using Ohm's Law, find the Power Dissipated in Watts. (LED Forward Voltage x Number of LED's in series) x LED Forward Current = Watts
    (2Vdc x 2ct.) = 4Vdc x 0.02A = 0.08 Watts. or.. Since there's 1000 Milliwatts in 1 Watt, it's a power dissipation of 80mW for all the LED's total.

    Resistor for LED's in Series Power Dissipation.
    Step 1.
    Using Ohm's Law, find the Power Dissipated in Watts for the Resistor. (LED Total Forward Current Squared divided by the Resistor Value)

    (0.02A x 0.02A) = 0.0004A * 50Ω = 0.02 Watts. or.. Since there's 1000 Milliwatts in 1 Watt, it's a power dissipation of 20mW for the Resistor.


    Can someone please verify that the above calculations are correct ?


    [​IMG]


    *** LED's IN PARALLEL, POWER DISSIPATION ***
    LED's in Parallel.. Single LED Power Dissipation.
    Step 1. To find the Power dissipated in watts for a single LED in Parallel, you multiple the LED Forward Voltage by the LED Forward Current.
    3.5vdc x 0.035A = 0.1225 Watts.

    LED's in Parallel.. Total LED Power Dissipation.
    Step 1. To find the total power dissipated in watts for ALL THE LED's in Parallel, you multiple the Total Forward Current of all the LED's in parallel by the LED's Forward Voltage.
    (0.035A x 20ct.) x 3.5Vdc = 2.45 Watts

    LED's in Parallel.. Total Resistor Power Dissipation.
    Step 1. To find the total power dissipated in watts for the Resistor. It's LED's Total Forward Current Squared multiplied by the Resistor Value used.

    In this case, we went with a 1.5Ω Resistor.


    So it's LED's Total Forward Current Squared x Resistor Value used.
    (0.7A x 0.7A) x 2.15Ω = 1.0535 Watts

    So it looks to me that the part number that I picked out for my project just isn't going to cut it..

    That's ok.. I'll just use a 2.2Ω, 2 Watt Resistor from Mouser Electronics. Mouser Part # : 594-5083NW2R200JA100
    Or.. I could also use.. Mouser Part # : 594-5093NW2R200J , which is also a 2.2Ω Resistor, but with a power rating of 3 Watts.

    So.. the calculations for the Resistors Power dissipation would be something like this..

    (0.7A x 0.7A) x 2.20Ω = 1.078 Watts. PERFECT.. Right..?


    Or.. Using Ohm's Law yet again.. Instead of using the current.. I'll use the voltage to be dropped across the resistor and the resistors value, to find out exactly how much power is to be dissipated on the resistor.

    So.. according to Ohm's Law.. There's going to be 1.5Vdc dropped across the resistor. Using a 2.2Ω resistor the formula is as follows.

    Power = Voltage Squared divided by the resistance of the resistor.

    (1.5Vdc x 1.5Vdc) / 2.2Ω = 1.0227 Watts. PERFECT

    Using Ohms Law once again, using the same information.. 1.5Vdc & 2.2Ω, let's find out how much Current there will be.


    Ohm's law says that Current = Voltage divided by Resistance.

    1.5Vdc dropped across the resistor x 2.20Ω = 0.68 Amps.

    THAT'S AWESOME !!! Our LED will be at..

    0.68A / 0.7A = 0.97402 x 100 = 97.402% Brightness !!!

    Awesome right..?
     
    Last edited: Jan 7, 2016
    shiva007nand likes this.
  2. Veracohr

    Well-Known Member

    Jan 3, 2011
    552
    76
    You should use a resistor on each LED in the parallel scenario, you can't be sure the current will be shared equally.
     
    Guest3123 likes this.
  3. Guest3123

    Thread Starter Member

    Oct 28, 2014
    312
    18

    Np, I'll look into it. Would you like to explain why I can't be sure the current will be shared equally? So it's available for people that look at this post..?
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    Because if you grab a bag of 100 LEDs from the same maker, and carefully measure their Vf and plot the results, the Vfs will distribute on a bell (probability) curve. If you wire them in parallel with a single current-limiting resistor, the ones with the lowest Vf hog all the current (are overloaded), and the ones with the highest Vf get almost no current. As the ones with the lowest Vf start to burn out (and they will), the current shifts to the ones with progressively higher Vfs...

    The Chinese build crap products with parallel LEDs, but it is just that: Crap!
     
    anhnha and Guest3123 like this.
  5. GopherT

    AAC Fanatic!

    Nov 23, 2012
    6,061
    3,824
    It's not just the Chinese.
     
    Guest3123 likes this.
  6. Guest3123

    Thread Starter Member

    Oct 28, 2014
    312
    18
    Ok, and if I buy my LEDs from Mouser..? and they're all the same Vf..?
     
  7. dl324

    Distinguished Member

    Mar 30, 2015
    3,249
    626
    Doesn't matter where you buy the LEDs from, forward voltage will vary even from the same manufacturing lot.

    Here's the spec for an OSRAM white LED (LW3333):
    upload_2016-1-7_20-57-36.png

    and a Vishay blue LED (TLHB5800): upload_2016-1-7_21-0-14.png
     
    Guest3123 likes this.
  8. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,542
    1,251
    Something about Pennsylvania you want to share...?
     
    Guest3123 likes this.
  9. GopherT

    AAC Fanatic!

    Nov 23, 2012
    6,061
    3,824
    Nothing about PA. I haven't been into electronics very long. When I did start, I remember hearing the warning early on about LEDs in parallel so I watched out for things like this as I reverse-engineered and salvaged small motors and other bits from appliances, printers and the like. I've seen things from major manufacturers (German and Japanese) that violate the parallel LEd rule. It is a bad design but, is anything going to burn out if you have two or three In parallel with a 1k resistor on a 5V circuit? They would have to have really different forward voltages before problems start.
     
    Guest3123 likes this.
  10. Guest3123

    Thread Starter Member

    Oct 28, 2014
    312
    18
    Ok.. Do I calculate the resistor value for each LED, even though I buy 20 of the same LED's from Mouser or Digi-Key..? Thanks for the reply.. I'm from Pennsylvania as well. Just moved here actually about 3 years ago.. Much better than New York State.. I guess. things are a little different, but overall, pretty good people so far.
     
  11. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,810
    834
    No, you just calculate the resistor for the nominal value of Vf. Just like the special case of series LEDs, where there is only one LED.

    Nitpicking, your formula in LEDs in Series, Step 2 is wrong. We know what you mean, just that is not what you say. You state subtract the total Vf (4v), from the supply voltage (5v), but you write it backward. As written, the Vr should be -1v.
    4V - 5V = -1v
    I think you meant
    5V - 4V = 1V
     
    Guest3123 likes this.
  12. Guest3123

    Thread Starter Member

    Oct 28, 2014
    312
    18
    Hunted000647.jpg
    Like that..?

    Each LED is the same. Red LED's, 2Vdc VF, 0.02A IF, 10ct.
    Each Resistor is calculated based on 1 LED.

    So each resistor is 150Ω, with the ability to handle 1/4 Watt . Even though 60mW is being dissipated for each resistor.
    Each LED is dissipating 40mW, 400mW total.

    Did I mention I made an Ohm's law calculator for Windows in VB.NET..? lol, not to mention an LED Calculator. So the calculations are pretty easy, except when I actually write them out, like you pointed out with my negative number.. I have a habit of doing that.
     
    Last edited: Jan 8, 2016
  13. GopherT

    AAC Fanatic!

    Nov 23, 2012
    6,061
    3,824
    Carbon county? Eat a nice dinner at The Power House in White Haven. Get there early if you go tonight, Fridays are busy. Quite a following for a restraunt in the middle of nowhere.
     
  14. Guest3123

    Thread Starter Member

    Oct 28, 2014
    312
    18
    Thanks.. about 45 minute drive from where I am.. Powerhouse Eatery, 60 Powerhouse Road, White Haven, PA 18661

    Don't go out much.. the only places I've been a few times is Olive Garden, Red Lobster, All you can eat chinese, Red Robbins, & Shady Maple Smorgasbord located 129 Toddy Drive, East Earl, PA 17519, which is an 1h 29m drive.. :) I'll keep the powerhouse eatery in mind.. thanks Chief RB.
     
  15. dannyf

    Well-Known Member

    Sep 13, 2015
    1,825
    364
    Way too complicated for something this simple.
     
  16. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,542
    1,251
    Agree. And as more and more consumer products are super-stripped for low cost and volume production, what some call bad design actually is a conscious, cost/benefit business decision, one of the dozens or thousands made during any product design.

    ak
     
    Guest3123 likes this.
  17. hp1729

    Well-Known Member

    Nov 23, 2015
    1,957
    219
    Why do you use that 20 mA rated value? If you need maximum brightness I guess that makes sense but I often run then at 2 mA to 5 mA, not needing full brightness. If I am just monitoring a point in logic 1 mA may work. I use the current limited high side driver as the ballast resistor.
     
    Guest3123 likes this.
  18. Guest3123

    Thread Starter Member

    Oct 28, 2014
    312
    18
    Ok yeah cool. I actually heard about people doing that, and tried to remember what the value was under the rated maximum. So your saying 2mA to 5mA under it's maximum current is acceptable? That sounds reasonable. I'd probably go with 2mA under it's max, which also, obviously, I'd have to calculate the resistor value for 2mA under it's rated Forward Current (IF).
     
  19. hp1729

    Well-Known Member

    Nov 23, 2015
    1,957
    219
    Not 2 mA under its rated value. 2 mA total. I can drive an LED from a CMOS gate as long as I don't try to drive other logic gates with it, too. That rated output voltage and current is stated "to stay within stated levels". We can use it outside those rated values to some degree. We can drive a 1.6 Volt LED. We just can't drive to a good logic high to drive other logic inputs, too. Try it sometime.
     
  20. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Generally you want to work with percentages (though hp1729 is choosing a value based on drive capability).

    If you have an LED rated for a maximum current of, say, 20 mA, then it is common practice not to exceed about 85% of this in operation. The LED will then last virtually forever. A lot of people will run it at 50%. A change in current by a factor of two in an LED is often hard to notice unless they are right next to each other.

    Also, one thing I didn't see mentioned is that the thing that makes running LEDs directly in parallel a bad thing is a phenomenon known as thermal runaway. Yes, when you hook them all up there will be some differences in the forward current at the applied forward voltage, even if all of the LEDs came from the same wafer. So one LED will be carrying a bit more current than any of the others. As the LED junction warms up, it's forward voltage drops at the same current or, saying it in a way that is more applicable to this situation, the current will increase at the same forward voltage drop. Since the LED with the most current is dissipating the most power, it will also be the one that heats up the most quickly and so its current will increase the quickest. But the total current into all of them is being held pretty must constant (due to the single resistor) so the current in the others will decrease. This crowding out effect results in one LED hogging most of the current. If you only have two or three LEDs in parallel, it is quite possible that having one hog all the current will not lead to its rapid failure. But if you have ten or twenty in series, then it will likely be so overloaded that it burns out very quickly. That starts a chain reaction in which another LED will become the weak link and burnout even quicker. And so on and so on.
     
Loading...