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I can't transform this circuit in an exclusively NOR and another NAND circuit. Everytime I tryed to simulate both circuits, neither of them gave me the correct result.

enter image description here

F=(notB)+AC

enter image description here

English is not my native language and I couldn't put it in better words, sorry.

Thank you.

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closed as unclear what you're asking by Trevor_G, brhans, Voltage Spike, m.Alin, Bimpelrekkie Oct 31 '17 at 10:26

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ The bolean F equation does not depicts the actual logic diagtam above, it is just an internal signal. The whole equation is : > F= B'+ AC + (B+C)' \$\endgroup\$ – aluis.rcastro Oct 16 '17 at 19:10
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    \$\begingroup\$ That's the simplified equation, sorry I didn't specify. \$\endgroup\$ – André Oct 16 '17 at 20:36
  • \$\begingroup\$ This is not a good question. You're not stating what you have tried nor how it is failing. \$\endgroup\$ – akohlsmith Oct 29 '17 at 18:24
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You really should show some work in your question. You don't, except that you do know that you optimized the gate diagram's resulting boolean expression. I suppose that's good enough for now. In any case, others may benefit. So I'll use your question as a foil of sorts.

If you have NANDs, then you want equations of the form: \$\overline{A \cdot B}\$. If you have NORs, then you want equations of the form: \$\overline{A + B}\$. It's not complicated.

You've already worked out the simplified form: \$\overline{B}+A\: C\$. But you need a large bar over all that which covers everything. So to start out you just double-bar everything.

$$\begin{align*} F &= \overline{B}+A\: C\\\\ &=\overline{\overline{\overline{B}+A\: C}}\\\\ &=\overline{B\cdot \overline{A\: C}} \end{align*}$$

That is already in perfect form for NANDs, as you should be able to see:

schematic

simulate this circuit – Schematic created using CircuitLab

That's just how easy it is.

Now, can you do the same for NOR? You could follow the above recipe. Or you could just cheat and replace the NANDs with equivalent NORs:

schematic

simulate this circuit

So that you get:

schematic

simulate this circuit

That one can be optimized a little by replacing two consecutive inverters with a wire:

schematic

simulate this circuit

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  • \$\begingroup\$ Thank you for your help. Since yesterday I managed to deduce the expression from both the diagram and the truth table (which I wasn't able to before). Today I had the class and the teacher said it is correct with just the AND and OR gates. I didn't had my work because the camera on my phone can't focus on what's written in the paper in graphite (Huawei ascend g610), sorry if it showed that I put in no work as I did and actually tried to put it here but couldn't. Once again thank you. \$\endgroup\$ – André Oct 17 '17 at 13:26
  • \$\begingroup\$ @André Glad to hear it helped. If you are satisfied this is an answer, you can check it off as one. I think you get points for that, as well. If you still want other answers, then don't check it off. \$\endgroup\$ – jonk Oct 17 '17 at 19:35

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