(source: electronics-tutorials.ws)

open version
(source: electronics-tutorials.ws)

I want make this 3 input device 4 input legged. Any suggestions?

  • \$\begingroup\$ What is the boolean expression for what you want? \$\endgroup\$
    – Andy aka
    Mar 15, 2015 at 11:09
  • \$\begingroup\$ You can make it using an inverting 4-to-16 decoder (one where the outputs are active LOW) and a 4-input AND gate... \$\endgroup\$
    – Majenko
    Mar 15, 2015 at 11:48
  • \$\begingroup\$ @Majenko: If one and only one input high makes a high output, then a 4-to-16 decoder with active low outputs would have to be followed up with a four input NAND, yes? \$\endgroup\$
    – EM Fields
    Mar 15, 2015 at 17:45
  • \$\begingroup\$ No, because it's XNOR so you want a low output not high. \$\endgroup\$
    – Majenko
    Mar 15, 2015 at 17:49

1 Answer 1


If all of the inputs are the same (all high or all low) the output will go high, otherwise the output will g0 low.

enter image description here

  • 1
    \$\begingroup\$ That's correct for the "other" type of XNOR gate, but the one pictured has "=1" in it, so it's a "one and only one input HIGH" style XNOR gate. \$\endgroup\$
    – Majenko
    Mar 15, 2015 at 11:46
  • \$\begingroup\$ @Majenko: AARGHHHH!!! \$\endgroup\$
    – EM Fields
    Mar 15, 2015 at 11:57
  • \$\begingroup\$ That's the problem with XNOR / XOR with more than 2 inputs - what does it really mean? \$\endgroup\$
    – Majenko
    Mar 15, 2015 at 12:06
  • \$\begingroup\$ I was confused, so I went looking for clarification and found this. They use =1 also, but their truth table shows that when A is low, B,C, and Y make an XOR, but when A is high, B,C, and Y make an XNOR, so that: "one and only one input High" rule doesn't seem to be hard and fast. :) \$\endgroup\$
    – EM Fields
    Mar 15, 2015 at 12:39
  • \$\begingroup\$ We had a big long discussion about it all here not so long back: electronics.stackexchange.com/questions/93713/… \$\endgroup\$
    – Majenko
    Mar 15, 2015 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.