9 volt alternating flasher help....

Discussion in 'The Projects Forum' started by critiera119, Jan 15, 2009.

  1. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    [​IMG]
    Current Parts:
    Caps = 22uf
    Transistors = 2N3904
    Resistors = (2) 1k brown-black-red
    Resistors = (2) 15k brown-green-orange
    LEDs = (1st side) r, r, w, r, w, b
    LEDs = (2nd side) r, r, w, r, w, b
    Power = 9 volt batt.

    My problem is that as it stands, I can not get more than 2 of the LEDs to alternate as you see them connected in series. All of the LEDs I am using are from a Christmas light set (set of 60 for $5.00 was awesome from BigLots after Xmas clearance sale), none of which have resistors and that is where I am guessing I need the help or within the circuit possibly. I am also open to changing the power source - not to exceed 9 volts though. So if anyone can help me - I'd love it! Thanks. I am a newbie to this obviously.

    P.S. - I also have a couple 555 and 4017 IC's on hand if needed.
     
    Last edited: Jan 15, 2009
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Can you give the schematic ?
    A schematic tells more than 1000 words.

    Greetings,
    Bertus
     
  3. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    It is simliar to this circuit but with my values and needs:

    [​IMG]
     
  4. bertus

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  5. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    I do know that already - I was hoping for a little more detailed help. I think I need a resistor on each LED. It occurs to me that I may only be able to light 3 LEDs per side with 9volt no matter what resistors are used - I wonder if this is correct. Remember, I am a novice. But thank you anyway.
     
    Last edited: Jan 15, 2009
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You can calculate the resistor.
    Take the voltage from the power supply, subtract the voltage drop of the leds.
    Divide this difference by the desired led current.

    Example : powersupply = 9 Volts, led voltage = 2.2 for the red and 3.8 for the blue led in the string.
    So the voltage difference is 9 - (2.2 + 3.8) = 9 - 6 = 3 volts.
    You want 20 ma as led current. So the resistor is 3 / 0.02 = 150 Ohms.

    I do not have the voltage drop of the transistor (1 volt max) in the calculation, so the current will be a little lower.

    Greetings,
    Bertus
     
    Last edited: Jan 15, 2009
  7. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    Ok I gather what you are saying. I want to just get two LEDs lit for now. The blue and the white. So 3.8 + 3.6 = 7.4 ... so 9 - 7.4 = 1.6 ... and 1.6 / .02 = 80ohm. I am going to try this.....
     
  8. bertus

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    Apr 5, 2008
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  9. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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