Would the sum of products expression look like so: \$\overline{A}\cdot B+A\cdot\overline{B}=Z\$ ? And the corresponding circuit should look like so? enter image description here

Sorry for the bad drawing. I'm just wondering if this is how this concept works, but surprisingly this is not in any obvious google results.

Thanks for any help!

  • \$\begingroup\$ Does the circuit produce the truth table it should? You can check that yourself just be seeing what would happen as you put in each pair of inputs. \$\endgroup\$ – jramsay42 Feb 9 '18 at 3:08
  • \$\begingroup\$ Yes, it produces the truth table that it should. If my circuit is correct and I drew that from the SOP expression then that also means my SOP expression is correct. Thank you. \$\endgroup\$ – JustHeavy Feb 9 '18 at 3:10
  • \$\begingroup\$ AKA Exclusive OR \$\endgroup\$ – AlmostDone Feb 9 '18 at 3:25
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    \$\begingroup\$ Just let you know, you can notate negation like so: \$\overline{A}\$ using this syntax \overline{A} and for the multiplication dot, you can do \cdot. \$\endgroup\$ – KingDuken Feb 9 '18 at 4:57

Your circuit diagram is correct.

Side note however, if you're designing logical transistors (i.e. CMOS logic), then you are able to reduce the amount of transistors it will take to produce an XOR gate. The amount you have above would take 16 NMOS and PMOS transistors. However, I contend that you can reduce this down to 10.

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