I have to create a two level circuit NAND only gates for the Fibonacci from 1-8. After creating a truth table and K-Maps, I got the Function F=A'B + B'C. Then I drew the AND-OR circuit and tried to convert it NAND only circuit. After implementing it on our bread board, I did not get the correct output, so I am guessing something is wrong with my circuit diagram. I am not sure. For the truth table that we were given, the directions given were:

Each row in the truth table is identified by a decimal number. For each row that corresponds to a Fibonacci number, specify the output to be 1. Otherwise, specify the output to be 0.

Truth table


Circuit Diagram

  • \$\begingroup\$ Do you understand what the Fibonacci sequence is..... \$\endgroup\$ – Trevor_G Mar 7 '18 at 20:44
  • \$\begingroup\$ What are 'A' 'B' and 'C' in this context? How do you expect the same input to give different answers without feedback to create latches? \$\endgroup\$ – Warren Hill Mar 7 '18 at 20:48
  • \$\begingroup\$ In States 4 to 7, A should be 1. Which makes your truth table so wrong. But you have a correct term \$\bar B C\$, even though K-map is wrong. \$\endgroup\$ – StainlessSteelRat Mar 7 '18 at 21:34
  • \$\begingroup\$ The truth table is the one we received, we had to fill in the Fib.Num column. So when the minterm number is within the Fibonacci sequence, then we put a 1 in that column, else put a 0. The original truth table was from 0-15 so the last 8 rows were all 1s, but for this one, it has to be from 0-7. We just cut the table in half \$\endgroup\$ – Nasim Ahmed Mar 7 '18 at 21:48
  • \$\begingroup\$ A,B,C are the outputs of the binary bit counter IC we have which goes into the input of our NAND and NOR gates \$\endgroup\$ – Nasim Ahmed Mar 7 '18 at 21:51

You can use the De Morgan theorems to convert any OR logic to AND and vice-versa:

!(A * B) = !A + !B


!(A + B) = !A * !B


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