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confused about hookup of a gate.

Discussion in 'General Electronics Discussion' started by measuretwice, Nov 23, 2013.

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  1. measuretwice


    Nov 23, 2013
    new poster solidly in the beginner quadrant, I was trying to hook up an xor gate (7486) on a breadboard. This should be simplicity itself, right?

    supply = 5 volts. I place a led and 330 ohm resistor in series from pin 3 to ground, and a 10k pull down resistor from each of 1 and 2 to ground. What I think is supposed to happen is i bring either 1 or 2 high (connect to 5 volt rail) and the led lights up.

    It doesn't

    I measure resistance accros each resistor and get 15M ohms. I don't understand that. it out the circuit it is clearly a 10k resistor. I measure voltage at each pin and its 2.6 volts. If I short one pin and power the other, the xor works as expected....but why the huge resistance and why are these pins at 2.6 volts?

    this seems a very simple circuit but i'm stumped and need a push.

    Last edited: Nov 23, 2013
  2. Six_Shooter


    Nov 16, 2012
    What do you mean by "Short one pin"? You can't short a single pin.
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Firstly, is it really TTL?

    Is it a 7486, or 74LS86, or 74HC86, or something else.

    Secondly, Have you connected up the power supply pins?

    Thirdly, TTL cannot source much current. If it is really TTL, you shouldn't be trying to drive a LED like that. Can you read the output voltage using your voltmeter?

    I can't understand how you can measure 15M across 10k resistors.

    oh, and TTL inputs will float up to around 2.6V and can actually source current!
  4. BobK


    Jan 5, 2010
    If you are trying to measure resistance in circuit with power on, don't do that.

    You mention connections to Vcc, both inputs and the output, but no connection to ground. Is it connected?

  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    If you're using a genuine 7486 (TTL) or 74LS86 (LS-TTL) device, you would be better to use active low inputs and outputs. Connect your LED between VCC and the output, and put the input resistors from the inputs to VCC, not GND. Ground an input to test the circuit.

    The reason for this is the asymmetrical characteristics of TTL and LS devices. Outputs pull low strongly, but pull high only weakly. Connecting a 10k resistor from an input to 0V will not necessarily pull the input low because TTL/LS inputs source current; this could be why you measure 15 megohms if you measure the input resistors with power applied.

    These problems can all be avoided by using a 74HC86 instead of a 7486/74LS86.
  6. measuretwice


    Nov 23, 2013
    thanks for the responses.

    what I meant was connect one of input pins to ground

    its a 74LS86

    yes, as well as ground

    thanks using the output to drive a tansistor that turns on power to the led would be a better approach?


    yes ground is connected.

    I think I understand this...inputs then have a pull up resistor instead of pull down resistor. both inputs are high until a signal puts one to ground. will do

    I really appreciate all the help. My Q's probably seem ridiculously simple but its easy to get confused as a beginner
  7. measuretwice


    Nov 23, 2013
    I tried using the 74ls86 with pull up resistors, worked perfectly....also had a 74HC86 and tried it with pull down resistors. as you say, it worked. I didn't realize this difference in the chips. thanks!
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