(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 '15 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 '15 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 '15 at 17:45
  • \$\begingroup\$ No, because it's XNOR so you want a low output not high. \$\endgroup\$ – Majenko Mar 15 '15 at 17:49

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

|improve this answer|||||
  • 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 '15 at 11:46
  • \$\begingroup\$ @Majenko: AARGHHHH!!! \$\endgroup\$ – EM Fields Mar 15 '15 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 '15 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 '15 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 '15 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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