Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

LED Circuit

  1. Apr 17, 2012 #1
    Hi im making a project in my spare time and i am getting quite confused with the whole situation.

    I have 100 LED's that need to be wired up, my first question is what is best to wire them up in, series or parallel?

    My power source is a 9V alkaline battery and i need to know if a resistor/or several resistors are needed and what size is needed if neccesary?

    Ive read several sites across the web and still cant get my head around the situation.

    Can someone help me on this situation please.

    Tom
     
  2. jcsd
  3. Apr 17, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Hi SummersLCFC! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]
    Are the LEDs all of identical type? What is the forward voltage of each LED? What LED current do you need for the desired brightness? How many minutes would you like to run them from a single 9v alkaline? What is your purpose of using these LEDs?
     
    Last edited by a moderator: May 5, 2017
  4. Apr 17, 2012 #3
    Hi yes all the LED's are identical type

    The forward voltage of each is 3 - 3.4 V

    LED Current is 20 mA

    Im not to fussed about the time that they are running for and my purpose for them is to create a sign.

    My project is a LED sign that is connected up to a PIR sensor that when the sensor detects movement the LED's will light up and display my sign
     
  5. Apr 17, 2012 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    As a test, you could connect a pair of those in series with a 150Ω 1 watt resistor and confirm that you are happy with that current & brightness. But it is not the most efficient arrangement, and to make matters worse, the more LEDs you connect to that battery, the less voltage it will deliver. A LED driver module would be better, but you'll need too many if each drives only 2 LEDs. Could you go to using 5 or 6 of those 9V batteries?

    EDIT: fixed resistor value
     
    Last edited: Apr 17, 2012
  6. Apr 17, 2012 #5

    psparky

    User Avatar
    Gold Member

    You definietly want to wire them in parralel....series will not work!

    Since you have 9 volts available...and you only want 3 volts.......you need to add a resistor in series with each LED to drop 6 volts. Since each LED has a current of 20ma......and we know V=IR........

    6=.02*R

    Solve for R and you get 300 ohms. This will have 6 volts across the resitor then....

    Since your LED is 3 volts at .02 Amp....it obviously has a resistance of 150 ohms. Add this to your 300 ohm resistor.....

    9=450*I

    I equals 20 ma. Should work.

    Or you could put all LED's in parralel and figure their equivalent resistance. Then put one resistor in series with your voltage source and drop the voltage that way. This way should work and sounds simpler and cheaper.
     
  7. Apr 17, 2012 #6

    vk6kro

    User Avatar
    Science Advisor

    You could put two LEDs in series with a 120 ohm 0.5 watt resistor.

    If the LEDs are at full brightness at 3.4 volts and then draw 20 mA, then there will be 6.8 volts across the two LEDs and 2.2 volts across the resistor.

    2.2 volts / 120 ohms = 18.3 mA

    Try this with one resistor and two LEDs before you wire up any more.

    So, you would need 50 of these series arrangements, in parallel, and they would draw nearly 1 amp.

    This is too much for a 9 volt battery, but you could use a small power supply to supply 1 amp.
     
    Last edited: Apr 17, 2012
  8. Apr 17, 2012 #7

    psparky

    User Avatar
    Gold Member

    Which leads us to one obvious question. Why not just put 3 LED's in series.....all in paralell.

    3 volts drops across each one....have a nice day.
     
  9. Apr 17, 2012 #8

    NascentOxygen

    User Avatar

    Staff: Mentor

    120 ohms. Arggh! 47Bek.gif I'll go back and fix my mistake.
     
  10. Apr 17, 2012 #9

    vk6kro

    User Avatar
    Science Advisor

    You have to have some resistance to limit the current.

    So, having only 2 LEDs per series string allows some voltage across the resistor.

    You also can't put all the LEDs in parallel since this would result in some of them burning brightly but others not lighting at all. This is because all LEDs are not created equal and the voltages tend to vary a bit in a sample of LEDs.
     
  11. Apr 17, 2012 #10

    psparky

    User Avatar
    Gold Member

    OH....that's right...been a while.

    What is the actual resistance in a LED? There must be some~!
     
  12. Apr 17, 2012 #11
    LEDs are non ohmic devices. Parasitic resistance of ohmic contacts should be negligible.
     
  13. Apr 17, 2012 #12

    psparky

    User Avatar
    Gold Member

    Oh ya....that would explain why they don't use much power and they last so long.

    Yet....we need to use a power burning resistor in series to "limit" the current.

    Strange...indeed.
     
  14. Apr 17, 2012 #13
    No, they use power. The voltage drop times the current is the power they use. Its just efficiently converted to light rather than heat.

    Edit: they last so long because thermal damage is the main fatigue or wear that semiconductors experience, aside from transient voltage damage that LEDs probably don't see much of, and LEDs don't generally experience temperature shock or large amounts of heat dissipation.
     
    Last edited: Apr 17, 2012
  15. Apr 17, 2012 #14

    NascentOxygen

    User Avatar

    Staff: Mentor

    It might be feasible to arrange 2 LEDs in parallel, esp when OP is prepared to pick and choose to get matched pairs to share the current equally.
     
  16. Apr 17, 2012 #15

    vk6kro

    User Avatar
    Science Advisor

    The problem with LEDs, and most diodes, is that their resistance drops rapidly with increasing forward voltage.

    So, a LED with 3.5 volts across it and drawing 20 mA has a resistance of 175 ohms. (3.5 volts / 0.02 amps)

    But give it 4 volts and it could draw 1 amp and then its resistance would be 4 ohms and it would be dissipating 4 watts. Enough of this 4 watts will be heat to melt the LED and it will go dark and stay dark forever.

    If you could be sure the voltage never went above 3.5 volts, you wouldn't need a resistor.

    However, if the voltage then went below 3.2 volts, the LED would turn off. Not much of a range.

    With a resistor, the voltage could go up a bit, or down a lot, (as it does over a battery's life) and the LED would still be alight, although the brightness would vary.
    So, a resistor is good insurance.
     
  17. Apr 17, 2012 #16

    vk6kro

    User Avatar
    Science Advisor

    Yes, you can do this.

    Assume each LED needs 3.5 volts at 20 mA and the supply is 9 volts.

    If the LEDs are in parallel, the resistor will have 9 volts - 3.5 volts or 5.5 volts across it.
    So, it will be 5.5 volts / 0.04 amps or 137.5 ohms
    and dissipating 5.5 volts times 0.04 amps or 220 mW.

    Now, if you put the LEDs in series, the resistor would have 9 volts - 7 volts or 2 volts across it.
    It will be 2 volts / 0.02 amps or 100 ohms
    and it will be dissipating 2 volts times 0.02 amps or 40 mW.

    So, it is 5.5 times more efficient to put them is series. (220 mW / 40mW = 5.5)
     
  18. Apr 18, 2012 #17

    NascentOxygen

    User Avatar

    Staff: Mentor

    I meant 4 diodes sharing the one driver/resistor. Two parallel strings, each of two LEDs.

    But a resistor won't be practicable as these 9v batteries have poor regulation. An active driver can compensate for this and maintain a fixed current. By taking the time to match each 4-diode combo he can halve the number of drivers needed.
     
  19. Apr 18, 2012 #18

    vk6kro

    User Avatar
    Science Advisor

    This circuit isn't going to work with a 9 volt battery.

    Even if the drivers were 100 % efficient, the power used would be (3.4 volts times 0.02 amps times 100) or 6.8 watts.
    This is 755 mA from a 9 volt battery. It may work for 10 minutes or so....Maybe if the battery was made up of D cells?

    The added complexity of using switch mode drivers isn't going to make a battery feasible, so it is better to use a little more current from a mains power source and keep it simple. It is already 75% efficient.
    (6.8 volts* 0.02A / 9 volts*0.02A) * 100
     
  20. Apr 18, 2012 #19

    NascentOxygen

    User Avatar

    Staff: Mentor

    You missed my suggestion about using 5 such batteries?
    I suspect he wants it relocatable or portable, i.e., no power connection.
     
  21. Apr 18, 2012 #20

    vk6kro

    User Avatar
    Science Advisor

    I see that Radio Shack (US) have 9V alkalines for 4 for $9.89. That seems like a pretty expensive way to do it.

    You may be able to get plastic holders that will hold 6 "D" cells, or it may be better to get a small 12 V battery that can be recharged.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: LED Circuit
Loading...