I searched in the "List of 7400 series integrated circuits" on Wikipedia for a open-collector OR-gate, but couldn't find any.

The list looks complete, so I assume there is no such chip in the 7400 series.

Why is there no open-collector OR-gate in the 7400 series?

  • \$\begingroup\$ There was never sufficient commercial demand to create such a special-purpose part. \$\endgroup\$
    – Dave Tweed
    May 27, 2017 at 23:26
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    \$\begingroup\$ As @DaveTweed said there is very little demand. Open-collector parts are only used rarely - they are mainly for interfacing to non-logic gates. OR gates are also not used very much - in 40 years of designing digital circuitry I have never designed them in. The combination of the two is more rarely required. \$\endgroup\$ May 28, 2017 at 0:28
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    \$\begingroup\$ An application of open-collector gates was driving a bus structure, where an idle bus was pulled high by resistors. NANDs have an advantage over NORs, because one input is used as enable while the other input is used as data. Tri-state bus buffers were introduced later. \$\endgroup\$
    – glen_geek
    May 28, 2017 at 2:39
  • \$\begingroup\$ Only the people that created the 7400 series eons ago could answer this. Everyone else is just guessing at reasons they might have made the decisions they did. Closing this pointless opinion-based question. \$\endgroup\$ May 28, 2017 at 13:11

1 Answer 1


I don't genuinely know the answer to your question, but I can speculate that it's because there are (quad) open-collector NAND and NOR gates, which are universal gates, and you can relatively easily synthesize an open collector OR gate from these. For example:


simulate this circuit – Schematic created using CircuitLab

Because, of course, A NOR 0 = ~A and ~(A NOR B) = A OR B


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