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simplified
(source: electronics-tutorials.ws)

open version
(source: electronics-tutorials.ws)

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

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  • \$\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

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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

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  • 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

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