There is really no need to transform the function. If you check the data sheet on a 74138 decoder/demultiplexer, you'll see that the outputs are active LOW.Transform your function to a complemented product. The terms of the product will be the maxterms that you need to extract from your DEMUX and drive to the NAND gate. I don't see the need for more than one NAND gate.
Transform your function to a complemented product. The terms of the product will be the maxterms that you need to extract from your DEMUX and drive to the NAND gate. I don't see the need for more than one NAND gate.
No, pretty sure about the expression.We have:
f=abc'+ab'+a'b'c
=( (abc')' (ab')' (a'b'c)' )'
=( (a'+b'+c) (a'+b) (a+b+c') )'
We see now that the desired function is built from a NAND gate that combines the terms (a'+b'+c), (a'+b) and (a+b+c').
Thus, f=1 when one of these terms is 0. This happens when abc=110 or 10X or 001. You can take these signals from a 3-to-8 DEMUX. With inputs abc, the outputs 6,4,5 and 1 will be set to LOW for each of the above cases and can be used as inputs to the NAND gate.
However, this implementation requires one OR gate, which isn't available and can't be constructed from the two remaining NAND gates. Are you sure the initial expression isn't abc'+ab'c+a'b'c?
I don't see any question mark here.No, pretty sure about the expression.
BTW here we have a 74138 IC, and you are using it as a decoder.
Thanks for help.
Better look again!Your circuit won't work. Here is the analysis of the output function:
F=(Y3Y5Y7)'
=( (A2'A1A0)' (A2A1'A0)' (A2A1A0)' )'
=A2'A1A0+A2A1'A0+A2A1A0
=c'a(bc)'+ca'(bc)'+ca(bc)'
=c'a(b'+c')+ca'(b'+c')+ca(b'+c')
=ac'+ab'c'+a'b'c+ab'c
=ac'+b'c
which is different from the required boolean function.
Your circuit works correctly too, except for the fact that you used the OR gate (as you already pointed out). Also, you left G2B (pin 5) open; it should be grounded.
by Jake Hertz
by Duane Benson