Powering led's, 9v battery not enough.

Discussion in 'The Projects Forum' started by hashmaster, Oct 17, 2013.

  1. hashmaster

    Thread Starter Member

    May 30, 2012
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    Trying to add led lighting to my rc car. Ive averaged the pwm, got my comparator working, fed that into the base of a 2n3904, emitter to ground, collecter to the led cathode, anode to positive rail. Yay led lights. Added more led's in series, no more light. Well figured that was wrong, so I tied all the cathodes together and set all the anodes to their own positive voltage. Yay they all light, but for all of about 10 seconds or so. Im using a 9v battery for the source of the comparator, and the led's. With two bright white led's, two red led's, and eight green led's a new 9v battery lasted like I said about 10 seconds if that. I an using a fighre of about 20mA for each led, sincei dont have datasheets for them, I figured a 9v should last longer.
     
  2. #12

    Expert

    Nov 30, 2010
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    The word, "resistor" was not to be seen in that post. If you didn't use any resistors, you might have smoked something. A part! I meant you might have smoked a part!
     
    Last edited: Oct 17, 2013
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Try two 9 V batteries in parallel.
     
  4. #12

    Expert

    Nov 30, 2010
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    That would help reveal the presence of runaway current failure.:p
     
  5. hashmaster

    Thread Starter Member

    May 30, 2012
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    Hahaha yeah forgot resistors somehow in the first test, second test I used a few various resistors from 330ohms and 470 ohms with 4 less green leds, had just a little bit more light time but my white leds disnt light. I dont quite wanna start stacking 9v in parralel for more capacity because 9v batteries are pricey, and im cheap.
     
  6. #12

    Expert

    Nov 30, 2010
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    Use your 470 ohm resistor and a 9V battery and measure each LED for its personal forward voltage. Then report back here with how many and how much and we'll see if we can design something that works.

    ps, you can't run all of them in parallel because 12 x .02 amps is a quarter of an amp. Way out of the range of a 9V battery.
     
    Last edited: Oct 17, 2013
  7. hashmaster

    Thread Starter Member

    May 30, 2012
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    Alright 9v battery voltage is 7.74V. The green leds measure at 2.045V, 2.05v, 2.10v, 2.07v, 2.07v, 2.08v, 2.08v, 2.1v. The white leds are 2.71v, 2.75v. The red leds are 3.73v, 3.81v.
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Do you have them all in parallel with the battery?
     
  9. shteii01

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    Why I suggested a second battery.
     
  10. hashmaster

    Thread Starter Member

    May 30, 2012
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    They were all measured independently. 1 resistor 1 led.
     
  11. #12

    Expert

    Nov 30, 2010
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    When the failure time is 10 seconds, and everything is right about the circuit, a second battery will only change the failure time to 20 seconds. Not enough to satisfy the customer.

    Personally, I design for 7 volts = end of battery. Now I will go play with the calculator for a while and tell you what the results are.

    BRB.

    ps, Red LEDs are supposed to use less voltage than any other color. Could you double check with a fresh battery?
     
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  12. shteii01

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    Sorry. I did not mean your measuring technique. I mean the final circuit, all LED will be parallel to each other and parallel to battery, right?
     
  13. hashmaster

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    May 30, 2012
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    Kinda my orignal thoughts
     
  14. #12

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    I must get a second measurement on the reds. I doubt your answer and I can't even get 2 reds in series to work at the voltages you listed.
     
  15. #12

    Expert

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    This is the first shot. If your numbers were right, this will work. (Every LED that is not labeled is green.)
     
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  16. hashmaster

    Thread Starter Member

    May 30, 2012
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    Although the led light xant quite be seen it is lit.
     
  17. hashmaster

    Thread Starter Member

    May 30, 2012
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    Shall go give it a try. Thanks.
     
  18. #12

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    Nov 30, 2010
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    ps, 7.74V is pretty used up for a 9V battery. Watch for the battery voltage to fail as you get toward 100% "on" time.
     
  19. hashmaster

    Thread Starter Member

    May 30, 2012
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    Well had to go to radio shack since I only had one resistor fromthat schematic. Also using a 680 ohm resistor on the base, and got a fresh 9v battery. Its at least 6x more efficient then before. Left it running for about a minute before I pulled the power. Battery still metered 9.25v. Thanks #12. Could possibly give me a brief explanation on how you calculated that out too?
     
  20. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Resistor for LED==(Supply Voltage - LED Vforward)/20 mA

    Example:
    (9 V - 2.71 V - 3.73 V)/20 mA= 128 Ohm (he rounded down to 120 Ohm)
    Power for resistor == I*I*R == 20 mA*20 mA*120 Ohm=0.048 W (he round it up to 1/8 W which is 0.125 W for extra safety margin)
     
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