Question on Tying the ununsed Reset Pin to ground

Discussion in 'Homework Help' started by wind_blast942, Oct 6, 2010.

  1. wind_blast942

    Thread Starter New Member

    Sep 7, 2010
    Hi all

    I heard that it is good practice to tie all the unused pins in an IC to GND, rather than to leave it hanging.

    However what if i have a reset pin on an IC (for example a shift register) that i am not using? Since if i connect it to ground, wouldn't the IC constantly reset itself?

  2. Georacer


    Nov 25, 2009
    If for the normal operation of your IC, Vcc is required, then connect it there.

    The meaning of what you heard is that if you leave a pin hanging you cannot be sure of its voltage. You may measure it in one instance and be happy with the result, but later, due to internal operations this voltage may change.

    So, in pins that affect the operation of your circuits, you should bind them in the voltage that guarantees no interference (Vcc or Ground).

    However, if a pin has no effect on your IC, you don't need to take such measures. For example, if you have deactivated the Load pin of a counter (by binding it on Vcc), you don't have to ground the Parallel Input Pins too.
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    It depends on the shift register. A 74LS94 needs a high in to reset, for instance, while a 74LS96 resets on a low.

    If the active level is low, then tie the pin high. For logic ICs, unused inputs must be tied high or low. The level you tie them to depends on the function.
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    All unused input pins should be connect to GND or to Vdd depending on the pin destination.
    For example if you input pin is "ACTIVE" at high state then you connect unuse pin to GND. But if input is active at low state then you connect them to VDD
  5. wind_blast942

    Thread Starter New Member

    Sep 7, 2010
    Thanks for the replies.

    On a slightly related question, in practice is it okay to assume that the outputs of ICs (for e.g i am using a 74194 shift register for a project) are initialised to fixed values (for e.g 0) on power up of the circuit, or do i have to manually clear the circuit everytime it is powered on?
  6. eblc1388

    Senior Member

    Nov 28, 2008
    Outputs are *NOT* initialized at power up. Some come up to certain state but there is no guaranty that it will do so every time.

    One must take active action to set/clear the output to a pre-defined state.

    Don't rely on luck to get a certain pattern on power up.
  7. Georacer


    Nov 25, 2009
    It is not guaranteed, but I guess you can suppose so. Usually your logic circuits do many times the job you want them to do everytime you switch them on. For example, if your register reads the amount of candy in a storage box, it will do it every second or less. Even if the first measurement is wrong, the others will be right.
    Experiment a little on your breadboard if you care to know what will happen.

    However, if precision in the first measurement is crucial, you can easilly clear all the IC's with a pushbutton before you hit the Clock ON.