Alternating flasher question...

Discussion in 'General Electronics Chat' started by critiera119, Feb 4, 2009.

  1. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    Hi, I want to use 9v instead of 12v for this circuit.

    1. Can the parts stay the same?
    2. Are the TIP122 necessary? If so, can anything else be used in it's place?
    [​IMG]
    Circuit is from: http://that.homepage.dk/EVS/
     
  2. Wendy

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    Mar 24, 2008
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  3. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    Ok, just read your write up. I could be wrong, but I gathered that only when going to 12v auto do you have to use something like TIP122 or LM317 regulator because of the radical variation in voltage. Since I am not using it in a car, but with only a 9v battery do I need the regulators?
     
    Last edited: Feb 4, 2009
  4. mik3

    Senior Member

    Feb 4, 2008
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    Regulator for what?

    How many leds you want to drive?
     
  5. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    16, 8 in each bank.
     
  6. Wendy

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    Mar 24, 2008
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    But did you understand Vf? The voltage across each diode adds, the sum dictate the resistor used for current limiting. If you have a current limiter it doesn't matter as much, but the sums of the LEDs Vf and whatever the current limiter takes still can't exceed the power supply voltage.
     
  7. Audioguru

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    Dec 20, 2007
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    The high power TIP122 was used because there were 25 strings of LEDs which is a current of 0.5A. The TIP122 is a darlington so its input current is low enogh to be driven from the Cmos logic IC.

    If you use 3.8V blue or white LEDs then two in series is 7.6V and a little 9V battery will quickly have less voltage so the LEDs won't light.

    Maybe with a single LED with its own current-limiting resistor as a string and have as many strings in parallel as your little battery can power.

    18 LEDs at 20mA each is a total current of 360mA. A little 9V alkaline battery will have its voltage drop to 7V in 10 minutes. But if the LEDs are flashing then the battery might last for 20 minutes.
     
  8. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    Yes.

    So if I have 8 blue LEDs in each bank, 9 volt will not work.

    9 - (3.8 + 3.8 + 3.8 + 3.8 + 3.8 + 3.8 + 3.8 + 3.8) = -21.4

    -21.4v / .02ma = -1070resistor = obviously won't work = exceeds the power supply voltage.

    Right? And that is why a regulator will come in play?

    So back to the original question...can I use K2098 in place of TIP122 in this circuit? I only ask because it is what I have on hand. I have all the parts except for TIP122s. Of course I can order or get to Radio Shack for the TIP122. Sorry for being such an amateur here guys.
     
    Last edited: Feb 5, 2009
  9. mik3

    Senior Member

    Feb 4, 2008
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    You can use two leds in each bank with a current limiting resistor.

    9-3.8-3.8=1.4V

    If the current is 20mA the resistor as to be 1.4/20mA=70 ohms 1/4W
     
  10. Wendy

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    Mar 24, 2008
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    Regulators do not increase voltage. You can not use 8 LEDs in a chain, but you can parallel them. I suggest you read the article, this is explained, regulators and all. Truth, you don't really need a regulator, just a resistor per chain. You did the math with the LED Vf, then reached a strange conclusion. Something is not connecting here.

    I don't have the data sheet on your transistor, but I suspect a 2N2222 would work. It is rated for a max of 0.6A, which means 0.3A is a good derated value. First diagram the LEDs, figure the current they will use (it isn't difficult), then see if the transistor will handle it.

    Repeating myself, the sum of the LEDs Vf can not exceed the power supply voltage.
     
  11. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    Bill, sorry to confuse. I was figuring out what resistor value would be for each of those 8 LEDs on each bank using the formula Bertus told me about at 7:30am in this post http://forum.allaboutcircuits.com/showthread.php?t=18091 and my result is strange, because it is saying a -1070ohm resistor would be called for (which of course doesn't exist) because the number of LEDs here exceeds power supply voltage. Am I connecting now?
     
    Last edited: Feb 5, 2009
  12. Audioguru

    New Member

    Dec 20, 2007
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    What are you talking about?
    You connect only two LEDs in series. A 3.8V white one and a 3.6V blue one. Then their total voltage is 7.4V and a 9V battery will need an 82 ohm current-limiting resistor for the LEDs.

    But the voltage of a 9V alkaline battery quickly drops below 7.4V so the LEDs will not light for long.

    If you re-design the circuit using darlington transistors and a bigger battery then the darlingtons can drive 4 strings of the LEDs when each string has its own current-limiting resistor.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    Your K2098's are N-channel power MOSFETS. They won't turn on and off properly with the current values of R4, R5. If you changed them to 22k it would likely work.
     
  14. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    Thank you SgtWookie.

    If I change R4 and R5 from 2k2 to 22k (to use the K2098 MOSFETs at Q1, Q2), what effect would this have on the circuit? Are R4 and R5 are only related to what is used at Q1 and Q2? I am thinking that is so. Just trying to make sure I understand it.

    Please forgive me with my amateurish questions.
     
  15. SgtWookie

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    R2 and R3 limit the gate charge current. The 4017 can only supply a few mA current without damage, so R2 and R3 safely limit the current.

    Since there are diodes between R2/R3 and the 4017's outputs, the 4017 can only source current to the gates of the MOSFETs; they cannot sink current.
    R4 and R5 serve as pull-down resistors to ensure that the gates discharge.

    The K2098 MOSFETs are maximum overkill for this project, as they are rated for 150V, 20A - but they should still work.
    Why use a 2x4 when a 6x12 will work just fine? :)
     
  16. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    Because the K2098 MOSFETs are what I have on hand. I do not need heat sink on these do I? I am using 9v. I am eager to put this together now. :) I will let YOU know how it comes out.
     
    Last edited: Feb 5, 2009
  17. Wendy

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    Mar 24, 2008
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    The funny thing is, I'm into the final stages of the article, where I discuss the emergancy flasher and cylon circuit.

    If all else fails you might think of a second 9v battery in series. It would allow more LEDs in series.
     
  18. critiera119

    Thread Starter Active Member

    Nov 21, 2008
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    *going back to read Bill's article now*
     
  19. SgtWookie

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    Careful about using a pair of 9v batteries in series; you may exceed the maximum voltage for the 555 timer and/or the 4017 Johnson counter.
     
  20. italo

    New Member

    Nov 20, 2005
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    LM555 MIN IS 4.5V DC TO MAX 18V Forget all about everything else with people came out with. add regulator it just to insure that the circuit will operate linearly until the battery gets to 4.5v +2v if timing is critical then yes add a regulator for just the lm555 the counter can care less of voltage dropping to 6-8v it will function as is.
     
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