bcd decimal

Discussion in 'General Electronics Chat' started by duxbuz, Apr 7, 2014.

  1. duxbuz

    Thread Starter Member

    Feb 23, 2014

    Just a question on setting up a logic circuit, to see if I am doing it correctly.

    I have a 74HC42N BCD to decimal decoder


    I am basically after probing its logic just to watch it working :)

    I dont really know how to figure these circuits out so I am trying to use the same resistor pull up values that I used for my other 74HC logic ICs.

    I add it to my breadboard with 5v regulator, I then add 4 buttons for inputs and on the buttons I use 10k resistors as pull ups.

    I do not know how this value is determined exactly, I was following a schematic when I made the last circuit I did, but since that I read this article:

    I am not rightly sure how to determine the input impedance then go a 1/10th less. But it does state 10k as a loose rule.

    I then decide to leave the outputs floating like I did with the 74HC00.

    With the 74HC00 I only had to pull down the inputs, so I am hoping this is normal practice?

    I am expecting the inputs to dictate each output pins state, so are the outputs even floating?

    Then I was hoping to be able to power on and test the state of the pins using logic probe and pressing the switches.

    Does this seem reasonable?

  2. ericgibbs

    Senior Member

    Jan 29, 2010
    hi dux,
    Its never a good idea let input pins float, it will make your circuits more susceptible to interference.

    I would use 10K pull up resistors and switch down to 0V.

    This pdf will help, its a little on the big size.! but very comprehensive.
  3. MrChips


    Oct 2, 2009
    Datasheets for digital logic will usually not show the input impedance.
    Look up the input current in the datasheet. You may have to go to a different manufacturer's datasheet such as the one from Texas Instruments.

    For 74HC00 series, input currents are typically around ±1μA max.

    Instead of using input impedance, you can do a circuit analysis as follows:

    Suppose you used a 1MΩ pullup resistor and the input pin draws 1μA.
    The voltage drop across the resistor is 1V. Hence the logic-high input voltage will be Vcc - 1V. For Vcc = 5V, the input voltage is 4V which is acceptable.

    Because the input current is very small the pull-up or pull-down resistor can be a value from a very wide range, 100Ω to 1MΩ.

    10kΩ pull-up or pull-down works fine.

    Totem-pole outputs do not float, i.e. you can leave output pins not connected.
  4. duxbuz

    Thread Starter Member

    Feb 23, 2014
    Great thanks.

    I will try to get my head around this stuff when I have a minute
  5. crutschow


    Mar 14, 2008
    For Active High inputs it may be easier to use a ≈10kΩ resistor pull-down resistor to ground on each input and connect the switches to +5V. That way the switches that are on will correspond to the BCD input value.