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Wattage rating on a resistor

  1. Dec 17, 2014 #1
    Hello, I am new to any type of engineering and just have been working with Arduinos and other microcontroller lately for fun.

    I am creating a prototype for a project that requires me to use a 9V power supply with 1Amp to power the Arduino(note, the arduino only needs around 300mA to run). I want to use some LED's(probably around 30) to be controlled by the arduino, BUT, I want the 9V, 1A power supply to power it. I have done some research and found that for controlling lets say, 30 LEDS, I would need the positive power to run through a 33Ohm resistor per 4 LEDs.

    LED uses a max of 2.2V and a current draw of 20mA.

    Okay. Easy, I just need a 33Ohm resistor now, but what wattage rating? I was told wattage can be calculated with:
    P= I * R
    or
    Wattage = 9 * 1

    Now doesn't this mean that the output of the power supply is going to be 9 Watts? The calculator I used said a 1/8th Watt resistor would work, but if the power supply is really supplying 9 Watts, wouldn't it burn up?
    Thanks,
    Bryan(HeyAwesomePeople)
     
  2. jcsd
  3. Dec 17, 2014 #2

    Averagesupernova

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    The wattage rating of the resistor needs to be at least as large as the number of watts it will dissipate. Rule of thumb is approximately double. Your formula is wrong. P = I^2 * R. So, your resistor will dissipate about .132 watts. A quarter watt resistor is fine.
     
  4. Dec 17, 2014 #3
    Isn't I the voltage and R the amps? Wouldnt that turn out to be 81 watts? I know that is wrong, but how exactly did you calculate the dissipation?
     
  5. Dec 17, 2014 #4

    NascentOxygen

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    Hi HAP. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    What do you mean by "run through a 33Ohm resistor per 4 LEDs."? How did you calculate that you need 330Ω?
     
    Last edited by a moderator: May 7, 2017
  6. Dec 17, 2014 #5

    Averagesupernova

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    Yes good question. I had missed that.
     
    Last edited by a moderator: May 7, 2017
  7. Dec 17, 2014 #6
    33 Ohms, not 330.

    I am trying to find a solution to connect 30 LEDS with a 9v 1.5Amp supply and an arduino. If I use one resistor to power all LEDs, it consumes like 600mA. The arduino has a recommended amp input of 300mA+. That's pushing it.

    If I connect both the LEDs and the arduino to a 9v 1amp power supply, would the total consumed amps be the LED draw + arduino, or do they both get 1amp?
     
    Last edited by a moderator: May 7, 2017
  8. Dec 17, 2014 #7

    NascentOxygen

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    that.

    The 1A capability of the supply must not be exceeded.
     
  9. Dec 17, 2014 #8

    NascentOxygen

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    Oops.

    So my question becomes how did you calculate 33 ohms?
     
  10. Dec 17, 2014 #9
    Used a calculator. http://ledcalc.com/

    First Option: To make it a lot easier to work with, I could wire it sorta like this, but it uses a lot of mA. http://prntscr.com/5i4249
    This option would be easier to work with and less messy, but requires more mAmps to run. Does this affect how many mA will go to the Arduino?

    Second Option: For less of a mA usage, I could do it like this: http://prntscr.com/5i42je
    This would be more of a mess on the board, but only consumes 157 mA.

    So here are my questions:

    1. What is used to calculate the dissipation by resistor?
    and
    2. Say I split a 9volt 1amp power supply into two lines. One for an Arduino, which requires at least 7 volts and 250mA+, and one for the LEDs, let's say the first option. If the arduino used up say 400mA at one time, would that only leave 600mA for the LEDs? And vice versa?

    Thanks

    I love your immediate abbreviation of my name. I've only seen a few people automatically do that, most I must tell. But that is what I go by. :)
     
    Last edited by a moderator: May 7, 2017
  11. Dec 17, 2014 #10

    NascentOxygen

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    I still don't see the value 33 Ω appearing anywhere. :olduhh:

    When operating LEDs in series, you need to allow plenty of volts across their current-limiting resistor. 3 is the max you can use here.

    When operating LEDs in parallel, you can't operate all 20 off one resistor. Individual differences mean that some will hog the current, some will get little. 3 or 4 in parallel should be okay if you don't run them too hard, but if any individual LED seems noticeably dull then swap it out.

    Yes.
    Good luck! http://thumbnails111.imagebam.com/37333/a62855373324849.jpg [Broken]
     
    Last edited by a moderator: May 7, 2017
  12. Dec 17, 2014 #11
    Okay well let me tell you what I am planning to do. It is actually quite a simple idea that I came up with. My family just moved into a new house and my mom still gets out of the car before turning it off to check if she is far enough inside the garage so the door will not hit the car. I said I could create a stoplight type system to help her park. Using an ultrasonic distance sensor I figured it would be a fairly easy project.
    Red = Pull up
    Yellow = Go slowly
    Green = Inside garage fully

    Seems pretty simple. But I want to run this all off of a 9v power supply. I said 30 LEDs because I want 10 per red/yellow/green. So what would you recommend I do?
    Use a 120Ohm resistor for every 3 LEDs?

    Thanks
     
    Last edited by a moderator: May 7, 2017
  13. Dec 17, 2014 #12

    NascentOxygen

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    Grouping series LEDs in threes would work. How much current will 120 Ω give?
     
  14. Dec 18, 2014 #13
    Used the calculator. http://prntscr.com/5i4b3g
    How would I calculate the current with the Voltage and Resistance?
    V / R = I
     
  15. Dec 18, 2014 #14

    NascentOxygen

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    for 3 LEDs, current = ( 9 - 6.6 ) / 120

    ⅛ watt resistors should be okay
     
  16. Dec 18, 2014 #15
    Now lets say I used a 12Volt power supply, with 1A. How would I calculate the resistor ohms I would need between the power and leds?

    R = V / I

    Would "I" be the current recommend for an LED and "V" be input voltage? This would mean that to power from a 12 Volt supply, I could have 5 LEDs(11Volts) with a 60 Ohm resistor(or 68 Ohm because that is the closest rounding up)?

    EDIT: Wait, wouldn't I only be able to have a max of 4 LEDs, because after the resistor of 68Ω I would have only roughly 10.2 Volts to work with? Is this math correct?
    Final Voltage = ( 12 / .2 )*( 12 / 68) = 10.2 Volts

    I am using
    Final Voltage = ( supplyvoltage / loadamperage )*( supplyvoltage / resistorvalue)
    to find the voltage after a resistor.
    Supply Voltage = 12v
    Resistor Value = 68Ω
    Load Amperage = Total current of combined LED's? (In this case, for 4, 0.08A?)
     
    Last edited: Dec 18, 2014
  17. Dec 18, 2014 #16

    NascentOxygen

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    5 LEDs would use about 11V, leaving roughly 1V across R and that is not quite enough for comfort, unless you make an effort to match the LED strings for their voltage drops.

    Looking at 4 LEDs, using 8.8V leaves 3.2V across R.
    For 20mA, choose R = 3.2/0.02
    = 160 ohms
    Then choose the nearest preferred value, 150 ohms
     
    Last edited: Dec 18, 2014
  18. Dec 18, 2014 #17
    So to find the resistance to use, I would use:
    Resistance = (supplyvoltage / totalcurrent)

    So if I got this right, then to power 3 LEDS on a 9 volt supply, I would use a resistor of 150 Ohms(or the closest one)?
     
    Last edited: Dec 18, 2014
  19. Dec 18, 2014 #18

    NascentOxygen

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    EDIT No, not now that you edited your post and made it wrong. Yes

    Different colour LEDs have different voltage drops. You'll probably find that 4 LEDs of each colour will be ample. Poking out of a scrap of black plastic there will be such good contrast that they can't be missed.
     
    Last edited: Dec 19, 2014
  20. Dec 18, 2014 #19
    Great! Thanks!

    One last question:
    I know that splitting the supply(1 amp) into 2 parts, and using up 700mA leaves 300mA for the other split, but does that apply for volts as well? Say for example, the Arduino requires at least 7.5 volts, and no more than 12V. If I use a 12Volt power brick, would 12V be dedicated to just the arduino, or does a split keep voltage same across both splits?
     
  21. Dec 19, 2014 #20
    Also, I seem to calculate that powering 5 LEDs that require 25mA and 2 volts to run with a Resistance of 82Ω has a dissipation wattage of ~1.2W. This would mean I need a resistor rated for more than 1.2 watts, correct? Like a 5 Watt resistor would work, right?
     
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