LED Resistors

Discussion in 'General Electronics Chat' started by englewood, Jul 24, 2015.

  1. englewood

    Thread Starter New Member

    Jul 24, 2015
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    Oaky I have 11 LEDS.
    I have 9V supply
    I have 1V Drop Across a Transistor

    I have 11 LEDS
    All LEDS have a fV of 3.2V
    I want all LEDS run around 20mA
    To make a letter

    I want one current resistor for the 10 LEDS
    I want one current resistor for the 1 LED

    I have 5 STRINGS OF 2 LEDS in para to make the 10.

    8-3.2/0.02/240/10 = 24 OHMS (27)

    Will one 27OHM resistor be enough for 10 LEDS?

    8-3.2/0.02 = 240 OHMS SINGLE LED

    SEE ATTCHED IMAGE
     
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  2. englewood

    Thread Starter New Member

    Jul 24, 2015
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    27 OHM
     
  3. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    Nope... LEDs in parallel without their own current limiting is a NO-NO as they do not share current nicely and will eventually die..
    Since your LEDs have a Vf of 3.2 you can do 5 strings of 2 on each string with a resistor for each string.
    Then the single LED with its own resistor too.

    150 ohm 1/4W resistor for each series string.
    330 ohm 1/4W for the single LED
     
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  4. englewood

    Thread Starter New Member

    Jul 24, 2015
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    So then

    8v - 3.2 / 0.02 = 240 / 2 = 120 (Closet highest resistor = 150) One for each string of 2 LEDS.

    Why 330 ohm for the single? :-S

    8v - 3.2 / 0.02 = 240 (Closet Highest Res = 270)
     
  5. KJ6EAD

    Senior Member

    Apr 30, 2011
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    For the calculation of the limiting resistor, you keep showing 8V - 3.2V but series strings of 2 would be 8V - ( 2 * 3.2V ) = 8V - 6.4V = 1.6V.

    1.6V / 0.02A = 80Ω. The preferred value would be 82Ω.

    For the single LED, ( 8V - 3.2V ) / 0.02A = 240Ω.

    Your total current is going to be 120mA which is not going to be sustainable for long with a standard 9V battery. The brightness of the LEDs will drop continually over a period of approximately 4 hours.
     
    Last edited: Jul 24, 2015
  6. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    Oh sorry.. I used 9V.. forgot about the 1v drop part..
     
  7. englewood

    Thread Starter New Member

    Jul 24, 2015
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    So whos right ha. Im confused now.

    That does make sense:

    8 - 6.4 (2x3.2 VD) / 0.02 = 82

    8 - 3.2 / 0.02 = 240

    Its running from a 9V supply :).
     
  8. KJ6EAD

    Senior Member

    Apr 30, 2011
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    I'm right, of course. ;)

    The BU806 data sheets I've seen don't show Vce for Ic less than 500mA so your 8V figure may be a little bit off, but if you supply the circuit with a single battery, the voltage drop at the source will be the most noticeable factor. If your supply is sustainable without drop (not a battery), no problem.
     
  9. englewood

    Thread Starter New Member

    Jul 24, 2015
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    Haha, I think you are just done the maths myself.

    Oh sorry, ive changed that its a BC517 Darlington
     
  10. englewood

    Thread Starter New Member

    Jul 24, 2015
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    How can I make a 4017, skip Q2 -Q8?
     
  11. KJ6EAD

    Senior Member

    Apr 30, 2011
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    So your transistor voltage drop will be 0.8V now. It probably won't make much difference but you should run the numbers to be sure.
     
  12. KJ6EAD

    Senior Member

    Apr 30, 2011
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    Now you're hijacking your own thread with this off-topic post. o_O You can't "skip" any of the states on the counter but you can decode the output or shorten the sequence by resetting before it completes all ten counts.
     
  13. englewood

    Thread Starter New Member

    Jul 24, 2015
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    Hold on,

    Why does the calc say this for 5 x 2 LEDS?

    Oh sorry, I want to do this should I make a new thread?

    8 - 6.4 / 0.02 = 80 / 2 cause there 2 LEDS = 40 OHMS Makes Sense.
     
  14. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    Where do you get the 6.4 voltage drop? Parallel circuits have the same voltage across each leg. If the LED Vf is 3.2 V, the voltage drop across the two legs is also 3.2V.
     
  15. englewood

    Thread Starter New Member

    Jul 24, 2015
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    Each LED has a 3.2 VD, 2 x 3.2 = 6.4.

    8-6.4VD / 0.04mA = 40 ( Next res 47 OHM)

    2 LEDS WITH 3.2 VD
    2 LEDS WITH 0.02mA

    This is want the end result is.

    14.63mA through 270 OHM Res
    14.55mA through 47 OHM Res cant get any closer than the :).



    The calculations are correct, I have checked the current flow in Proteus through each resistor and its almost the same :)

    Try this

    http://www.quickar.com/bestledcalc.php?session=
     
    Last edited: Jul 24, 2015
  16. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    You add the voltage drops in a SERIES circuit. Your LEDs are in a PARALLEL circuit where the voltage is the same in each leg. In your case 3.2V.

    The calculations are correct; you are using the wrong equation.

    To add to the confusion, LEDs in a parallel circuit need a resistor in each leg...
     
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  17. englewood

    Thread Starter New Member

    Jul 24, 2015
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    Oh idiot,

    They are in Series in 2 LEDS in my Schematic. Doh!!!!!!

    But why does the software show the correct mA flowing through each resistor?
     
    Last edited: Jul 24, 2015
  18. englewood

    Thread Starter New Member

    Jul 24, 2015
    11
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    I think ive messed up on the first Letter DOH
     
  19. englewood

    Thread Starter New Member

    Jul 24, 2015
    11
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    All sorted :), I think.
     
    Last edited: Jul 25, 2015
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