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Basic setup question, trying to use Or Gate

Discussion in 'Electronic Basics' started by WW, Oct 11, 2005.

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  1. WW

    WW Guest


    I have a bread board and a 4xAA battery pack, I'm just getting started,
    learning about electronics and wanted to try playing around with some
    simple gates and switches.

    I have connected the red lead to the top line of the breadboard, and
    the black lead to the bottom, testing with a multimeter i get about

    I then put the DM74LS08N chip on the board, it's a national
    semiconductor chip, so i went to their website and found the chip
    datasheet, it says pin 7 is ground and pin 14 is VCC so i assumed that
    pin 1 was the pin under the little circular marker and set power and
    ground accordingly.

    For fun, i thought i'd test setting the output from pin 11 to an Led
    but the Led lit up. I didn't expect this, i checked with the
    multimeter, at pin 14 the voltage is 5.03, at pins 13 and 12 (which
    aren't connected to anything), the voltage is 1.70v at pin 11 the
    voltage is 4.30 (this is one of the gate output pins), which is turning
    the Led on (I figure)

    If I connect pins 13 and 12 directly to ground the gate behaves as
    expected, but why is there voltage on these pins when the breadboard
    holes aren't connected to anything.

    Should the pins have voltage when not directly connected to anything?
    Am i missing something in my circuit setup?

    I tried hooking up a switch, it has 8 switches on it , i set the first
    two input to Vcc and their outputs to the input pins of the DM74LS08N
    but when the switch is outputing 0v the input pin to the chip still has
    1.7v (the output pin from the switch has 1.7v too, but if i disconnect
    it from the chip it has 0v.

    I hope my explaination makes sense,
    if there is a better forum/group to ask these questions in please let
    me know,

  2. Lord Garth

    Lord Garth Guest

    The notch in the IC package denotes the pin 1 end. The pins are numbered
    sequentially counter clockwise from the notch.

    There is no case in this logic family where an open circuit is defined as a
    logic zero. An open input is undefined and not allowed even on unconnected

    The allowable input voltage is 5 volts plus or minus 5% so 4.75V to 5.25
    is the correct range. Four 1.5V batteries is 6 volts.

    Finally, the necessary current to drive your LED should be compared to the
    current that is available from the gate. Typically, the transistor that
    current to ground, in the totem pole structure, is larger than the
    that pulls the output high. This means that a logic zero is often used to
    the LED. A driver is required when the LED requires more current than can
    be supplied directly.
  3. The inputs to the LS series of gates (or any other variant of TTL) is
    not an open circuit, but includes an internal pull up current that
    makes open circuit inputs act as if they were pulled up to logic
    one's. So you gate is acting normally. If you decide to experiment
    with 4000 or HC or HCT any of the CMOS kind of gates, their inputs do
    behave as floating lines when left unconnected, and tiny leakage
    currents or stored charge, so those inputs are indeterminate when left
    hanging. They also draw lots more supply current if the input
    voltages are not clean one's or zero's, so operating them this way is
    usually a bad idea. But your gate is perfectly happy with its inputs
    unconnected, as long as they are not subjected to strong noise that
    might bump them down to somewhere between a one and a zero.
    You would have to look at the internal circuit to see where that
    positive voltage is coming from. Take a look at page 2 of:

    The LS08 type has a 20k pull up resistor feeding a transistor base and
    there is a diode in series with the emitter, raising the base voltage
    above the normal 1 diode drop. This voltage gets out through the
    input pin through another diode.
    You activate the inputs of TTL chips by grounding the input, detouring
    the internal pull up current out through the input pin. A logic high
    in acts just like a disconnected pin, since both do nothing to the
    internal pull up current.
    Me too. :)
    This is the right one.
  4. Jasen Betts

    Jasen Betts Guest

    even though the pins are designated as input there are indirect connections
    to the VCC and ground pins inside the chip.
    the pins should always be connected to something, input pins left unconnected
    (or "floating") tend to pick up signals from other parts of a device and
    cause wierd effects.
    when it's turned off it's not connected to anything, if you want the chip to
    see 0V put a 10K (or less) resistor between the chip input pin and ground
    and then connect the switch..

    the resistor is called a pull-down resistor because when the switch is off
    it pulls the voltage on the pin down (close) to 0v,

    a different configuration would be to put the resistor to +5V
    and the switch between ground and the input pin. this time it would be a
    pull up resistor.
    I can't think of one.

    4xAA cells probably isn't the best way to power a 74LS series chip
    they're pretty fussy about getting the right voltage,

    generally this means using a higher input voltage (say 8-12V) and a
    LM7805 regulator and apropriate filter capacitors.

    The 74HCxx and 4xxx series chips are more forgiving about voltage,
    but even fussier about having all their inputs connected.

  5. Jasen Betts

    Jasen Betts Guest

    and four 1.25V rechargable cells is pretty close to 5V when they're
    reasonably fresh, but still probably not a good idea.

  6. ehsjr

    ehsjr Guest

    I'll mention two points. First, IC pin orientation. You correctly
    identified pin 1. It's always the same - heres a diagram: (Imagine
    the pins pointing away from you and into the monitor screen)

    ) Any DIP IC
    | | | |

    Next, your breadboard supply: If you want to power the breadboard
    from the AC supply to save batteries, get a 12 volt DC wall wart,
    and an LM7805 voltage regulator in a TO 220 package. Wire it up like

    +12 ---+---Vin| 7805 |Vout---+--- +5 volts
    | ------ |
    [C1] | [C2] C1 = .33 uF
    | | | C2 = .1 uF
    Gnd ---+---------+-----------+

    | o |
    | ||
    | 7805 ||
    | | |
    | | |
    Vin Gnd Vout

    Have fun!
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