Reset signal for CD4017

Discussion in 'The Projects Forum' started by andreapg, Aug 5, 2010.

  1. andreapg

    Thread Starter Active Member

    Feb 2, 2010
    41
    0
    Hi,
    That circuit should drive the reset signal of a 4017 (reverse condition), but I'm not sure that it will work.
    When the output (pin 3) takes the high level Q1 switches on, and pin 15 is connected to the ground.
    When the output takes the low level Q2 switches on, and pin 15 is connected to +9V.
    What do you think about?
    Thanks

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  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    If you need to invert the signal from the 555, one transistor will do it. Use a 20K resistor on the collector to Vcc and ground the emitter. Attach the 4017's pin 15 to the collector of the transistor. Drive the transistor with a smaller resistor to insure good switching - about 1 - 2k ohms.
     
  3. andreapg

    Thread Starter Active Member

    Feb 2, 2010
    41
    0
    Thank you for reply, do you mean a circuit like that?
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  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    Yes, that should work fine.
     
  5. andreapg

    Thread Starter Active Member

    Feb 2, 2010
    41
    0
    What do you think about the first circuit?
     
  6. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    684
    92
    Then PNP and NPN collectors to Pin 15 should be drawn as a bridge for clarity and intermediate states related to slew rates from the 555 could be a problem. You will have bigger problems with the 555 output not getting close enough to rail to turn off the PNP.

    I doubt that you need that much drive so if you want more drive you could try a smaller pull up resistor and take the signal from a single NPN as others have already suggested.

    In theory your H-bridge will have lower waste power since for collector loads, bridging to a turned off transistor will give you extremely high resistance equivalent and a few microamps of waste. But that PNP might not turn off with the 555 peak volts out. If necessary put a diode between the Emitter and the 9V. A small Vf Schottkty should be enough to make a difference but a silicon diode .7 Volts will let you turn that PNP off. Even better is to use a Darlington PNP instead.

    For slew rate problems you might want a timing network to make certain that their complement transistor turns off before either transistor turns on.

    I still don't think you need that much drive power but that is the start at least of a decent high drive inverter from discrete components.
     
  7. andreapg

    Thread Starter Active Member

    Feb 2, 2010
    41
    0
    Thank you for suggestions and info. I asked that question because of curiosity. However I think that the best solution is taking the signal from a singol transistor, as beenthere suggested. Using a simple circuit I will have low chance of problems and failure.
     
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