I'm playing around with CMOS logic using CD#### dip packages. As far I understand input pins should never be left floating, always connect them to Vcc or ground. Unfortunately my digital source encodes "1" and "0" respectively by Vcc and floating pins. How can I properly connect such signal to CMOS logic?

Apologies for the incorrect terminology!


Use a pulldown resistor connected to ground that results in a current that is less than the drive from the previous stage but more than the input requirement for the next stage.

  • \$\begingroup\$ Thanks for the lightning fast and clear answer! I feel silly missing such obvious thing. Now going to figure out the right resistor value! \$\endgroup\$ – Anne Nov 11 '14 at 18:40
  • \$\begingroup\$ 10KΩ ....... :P \$\endgroup\$ – Majenko Nov 11 '14 at 19:03

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