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Simple TTL question

Discussion in 'Electronic Basics' started by Tim Klinger, Feb 5, 2004.

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  1. Tim Klinger

    Tim Klinger Guest

    I'm building something with a 7400N quad nand chip and don't
    understand the way it's behaving. If I just power the chip (pin 14 to
    +5, pin 7 to ground), connect pins 1 and 2 to ground, and check the
    voltage at pin 3, I see something around 3V. Shouldn't I be getting
    +5V?

    I'm using the 7400 to debounce a pushbutton switch and running the
    output to a JK flip flop to make a "stateful" switch that changes
    state when the switch is pushed and released and holds that state
    until it's pushed (and released) again. It's behaving unpredictably
    and I can't figure out why. I'm obviously new to this so I'd
    appreciate any tips or advice.

    Thanks very much,
    Tim Klinger
     
  2. John Fields

    John Fields Guest

     
  3. Tim Klinger

    Tim Klinger Guest

    OK, that's good. Will 3V or so be enough to count as "high" when
    connected to the clock line of a JK flip flop?

    It's a SPDT momentary switch. I'm connecting the common pole to
    ground and each of the others to a nand gate and also to one end of a
    2.2k resistor (the other end of which is connected to +5). I'm
    connecting the output (pin 3) to the JK flip flop.

    Thanks,
    Tim
     
  4. JeffM

    JeffM Guest

    7400N quad nand chip
    Even the unused gates can affect the results.
    Start by tying all unused inputs to a definate logic state
    (connect them to the supply voltage or to ground).
     

  5. If it is a TTL or LSTTL flip-flop.

    Bipolar TTL logic parts will consider anything above 2.4 volts as a
    high, and anything below 0.8 volts as a low.
     
  6. John Fields

    John Fields Guest

     
  7. w_tom

    w_tom Guest

    Voltages for logic levels is how logic families differ.
    Conventional TTL would output a logic 1 above 2.4 volts and a
    logic 0 below 0.7. Thresholds for TTL inputs defined logic
    one at something less than 2.4 and a logic zero at something
    above 0.7 volts. Sufficient margin for error.

    But if that 2.4 volt logic one from a TTL drove a CMOS
    (74C00) logic chip, then the CMOS could self destruct. CMOS
    required a logic one to be higher - something more like 3.8
    volts. A voltage sitting between 1 and 3 volts on a CMOS
    input could cause that IC to destroy itself.

    IOW verify logic one and logic zero voltage levels - both
    for inputs and outputs. These are a first difference between
    the many logic families.

    There are logic ICs designed just for debouncing switches.
    Even the famous NE555 does this function better; and using one
    chip instead of two. Just some better ideas. Finish your
    current design to better appreciate the suggestions.
     
  8. CFoley1064

    CFoley1064 Guest

    I'm building something with a 7400N quad nand chip and don't
    You've gotten quite a few answers on logic levels, but I'm wondering about your
    using the 7400 to debounce a pushbutton switch. Unless you do it right, your
    switch bounce will just pass right thru to the J-K FF, causing problems.
    Here's an alternative (credit due to Mr. Don Lancaster's classic TTL Cookbook
    -- view in fixed font or M$ Notepad):



    VCC
    | 6.8K
    .-. ___
    1K | | .----------|___|----------.
    | | |VCC VCC |
    '-' || __ | __ |
    T 10 | 470 |'---| | '---| | |
    --- ___ | ___ | |& |o--. |& |o----o-------o OUT
    .---o o--|___|----o---|___|-o----|__| '----|__|
    |
    |
    GND


    This will not invert, but it will work. If you need to invert, just add
    another 7400 gate. The two 7400 gates with the feedback resistor form a
    non-inverting schmitt trigger, which has hysteresis. That, along with the
    rapidly falling and slowly rising input, will get you a clean one-transition
    signal.

    Of course, you could just use 1/6 of a 7414 Hex Schmitt Trigger Inverter, and
    be done with it.

    Don Lancaster's Guru's Lair has the classic TTL Cookbook (example above from p.
    166, except he used a 7404), just about everything he's ever written, and a lot
    of great advice for all -- it's worth the time.

    http://www.tinaja.com/

    Good luck.
    Chris
     
  9. CFoley1064

    CFoley1064 Guest

    From: (Tim Klinger)
    You've gotten quite a few answers on logic levels, but I'm wondering about your
    using the 7400 to debounce a pushbutton switch. Unless you do it right, your
    switch bounce will just pass right thru to the J-K FF, causing problems.
    Here's an alternative (credit due to Mr. Don Lancaster's classic TTL Cookbook
    -- view in fixed font or M$ Notepad):




    VCC
    | 6.8K
    .-. ___
    1K | | .----------|___|----------.
    | | |VCC VCC |
    '-' || __ | __ |
    T 10 | 470 |'---| | '---| | |
    --- ___ | ___ | |& |o--. |& |o----o-------o OUT
    .---o o--|___|----o---|___|-o----|__| '----|__|
    |
    | | +
    GND ### 100uF
    ---
    |
    |
    ===
    GND



    This will not invert, but it will work. If you need to invert, just add
    another 7400 gate. The two 7400 gates with the feedback resistor form a
    non-inverting schmitt trigger, which has hysteresis. That, along with the
    rapidly falling and slowly rising input, will get you a clean one-transition
    signal.

    Of course, you could just use 1/6 of a 7414 Hex Schmitt Trigger Inverter, and
    be done with it.

    Don Lancaster's Guru's Lair has the classic TTL Cookbook (example above from p.
    166, except he used a 7404), just about everything he's ever written, and a lot
    of great advice for all -- it's worth the time.

    http://www.tinaja.com/

    Good luck.
    Chris
     
  10. JeffM

    JeffM Guest

    Posting the same question independently to multiple groups
    is a bad idea.
    How is someone answering it in 1 group
    supposed to know it has already been adequately answered in another
    (or to make corrections to incorrect responses)?
     
  11. Tim Klinger

    Tim Klinger Guest

    It was a mistake. I didn't see my response posted so I reposted.

    Thanks to all of you who responded. I understand better what's going
    on. I will try the "cookbook" circuit posted and see if it improves
    things...

    Best,
    Tim
     
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