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Output/input of inverter

  1. Sep 4, 2005 #1
    This is a prelab i am working on:
    we have a 74HC04 inverter, whose pin 1 is connected to the DIP switch (switch connected to 5V source and ground), pin 2 is connected to
    1) a load (R + LED) that connects back to the source
    2) a load (R + LED) that is properly connected to the ground
    pin 7 is connected to the ground.
    And the LEDs light up as the prelab describes, so i assume i built the circuit correctly.
    So, it asks to measure Voh, Vol...
    but i don't know where the output is! I drew the circuit. If pin 7 is output, my multimeter does not detect a thing! what is the output in this case and how do i measure it?
    So, generally if you deal with LCs, are pin1 and 2 usually considered inputs or is there no convention?

    Thanks sooo much again and again!
     
  2. jcsd
  3. Sep 4, 2005 #2

    berkeman

    User Avatar

    Staff: Mentor

    Check out the datasheet:

    http://focus.ti.com/docs/prod/folders/print/sn74hc04.html

    The 74HC04 is a hex inverter. Pin 14 is power, pin 7 is GND, and the inverters are like this:

    IN --> Out
    1 --> 2
    3 --> 4
    5 --> 6
    13 -->12
    11 -->10
    9 --> 8

    (Assuming that I remembered them correctly...I didn't look on the datasheet, but you'll end up memorizing simple logic parts like that if you use them a lot.) If you drive pin 1 high, you should definitely see pin 2 go low. If you drive pin 1 low, you should definitely see pin 2 go high. Since pin 1 is the input for your switch & LED circuit, pin 2 is the output. Measure the Voh and Vol there as you flip your switch. And measure the voltage across the resistor to find the current in both cases (there will be no current when the LED is off...).

    BTW, when working with CMOS logic, you should never leave a CMOS input pin floating (unconnected). You should tie it either high or low -- you can either tie with a resistor, or hard tie the input to either 5V or GND. Do leave unused *outputs* open. CMOS inputs are high impedance and not self-biasing, so if you leave a CMOS input floating, it can float near mid-rail, which turns on both the pullup and pulldown FETs inside, which can increase the Idd for the chip a lot. It can also create buzzing noise in your circuit, if the floating input picks up noise and floats up and down near mid-rail. For the circuit as you've described it, just ground the unused inputs on pins 3, 5, 13, 11, and 9.
     
  4. Sep 5, 2005 #3
    thanks for all your help and the datasheet, now i can check myself.
    sorry if I am too hard-headed (this is my first real lab in electronics) for measuring Voh and Vol, across what do i need to measure it, since pin2 is connected to two different R+LED combinations (one is grounded and the other is hooked up to the source)? edit: across the grounded one?

    yeah, there is no current across the resistor when the LED is off, but ... there is current across the LED!!!! i am pretty sure my circuit is fine, the lab has a photograph attached showing what it should look like.
    with this circuit i don't get the same values as the sheet, although our Vcc is +5V
    Attached is the circuit.
    green LED is connected to the ground and yellow to the + (source?) and in the picture the input is HIGH (switch is on).

    ok: edit#3: here's what i think after looking at it for a while:
    when the switch is in Low i have a sourcing current, so I should be measuring V and I across the grounded R+LED (green) and when the switch is High i have a sinking current, so i should be measuring V and I across the R+LED that is hooked up to the source, right? and i should measure it across R+LED, right?
     

    Attached Files:

    Last edited: Sep 5, 2005
  5. Sep 5, 2005 #4

    dlgoff

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    Science Advisor
    Gold Member

    " for measuring Voh and Vol, across what do i need to measure it..."
    Measure voltage at pin 2 (where both resistors come together) relative to the ground pin 7 for the high (Voh) then the low (Vol) state.
     
  6. Sep 5, 2005 #5
    OK, thanks, that sorta makes sense.
    For input H, i get Vol in vicinity of what the datasheet says, but for input L Voh is 5V which is same as the source and supposed to be around 3.84V, since the load is not a CMOS device. I measured voltage between pin 7 and the wire that connects to both resistors.
    So ... where am I measuring it wrong?
     
  7. Sep 6, 2005 #6

    berkeman

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    Staff: Mentor

    I think your green LED is backwards. That would make it look like an open circuit when the output of the inverter is high, so that would let the HC04 output go to the 5V rail. Flip the green LED so that its cathode is to ground, and then you should get a Voh that is a little lower.
     
  8. Sep 6, 2005 #7
    @berkeman: i think it's correctly plugged in, that is: exactly as you say.
    What happens i think is that i need a digital multimeter, i can't read analog very precisely :yuck: it is a tiny bit less than 5V.
    Thanks all for help!
     
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