I'm looking for a logic gate IC that implements a set of AND gates with one leg that is common to all AND gates and one that isn't. To create a "group enable" function where if the common input is low, all outputs are 0 and if it is high inputs on the non-common input are passed through to the output.

Anyone know if such a thing exists (before I go an implement it with AND gates)?

Thanks in advanceenter image description here

  • 1
    \$\begingroup\$ What's wrong with using a standard AND gate IC and connecting the relevant inputs together externally? \$\endgroup\$
    – LMS
    Apr 23, 2018 at 17:59
  • \$\begingroup\$ This would be an N-by-1 multiplier. I guess a PAL22v10 would let you do 10 bits in one chip without having to buffer the common line. \$\endgroup\$
    – stark
    Apr 23, 2018 at 18:18

1 Answer 1


It sounds like you're looking for a basic line driver with enable pins.

Something like CD74HC241 or MC74HC244A should behave almost as you described (except with active-low enables).

If you really need the outputs to be low when not enabled, you could use pull-down resistors.

logic diagram

  • \$\begingroup\$ @someguythatcodes Though, you're really not saving yourself any work if you need to pull the outputs down. I only proposed this route in case Z states will suffice, in which case this would be the simplest. \$\endgroup\$ Apr 23, 2018 at 19:17
  • \$\begingroup\$ Yeah that looks like the right sort of thing! I'll have to use pull-down resistors on the outputs as this application is safety critical. Thanks! \$\endgroup\$ Apr 23, 2018 at 19:17
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    \$\begingroup\$ Sorry accidentally deleted my first comment and re-added above. I'll cost up whether just using AND gates or octal line buffers with pull downs would make the most financial sense. My guess is that the line buffers would still be cheaper as I've got 32 outputs to cover \$\endgroup\$ Apr 23, 2018 at 19:21
  • \$\begingroup\$ @someguythatcodes Should simplify your signal routing too. \$\endgroup\$ Apr 23, 2018 at 19:32

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