I'm working on an assignment where I need to draw a block diagram and the gate-level circuit of a 3-into-8 decoder with negative active inputs, a positive active
enable and positive active outputs.
I've drawn the block diagram, but before I draw the circuit, I wanted to do a truth table so that I made sure my logic was correct. And this is where I'm having some trouble. Normally, I think my table would look something like this. But with the inputs having negative active logic, I'm not sure how that will affect my inputs and change the overall truth table. I've always had a lot of trouble figuring out how N.A.L affects the overall circuit; could someone please help me out?
Best Answer
What I like to do for assignments is make a sanity check for at least 3 random cases and see if that checks out, do what I think is correct, then once I'm done, check again, with my first sanity check.
According to what you said, then these 3 expressions should be true:
\$\scriptsize EN = 0, A = B = C = 0 => D_7 = D_6 = D_5 = D_4 = D_3 = D_2 = D_1 = D_0 = 0\$ \$\scriptsize EN = 1, A = B = C = 0 => D_7=1,D_6 = D_5 = D_4 = D_3 = D_2 = D_1 = D_0 = 0\$ \$\scriptsize EN = 1, A = B = C = 1 => D_7 = D_6 = D_5 = D_4 = D_3 = D_2 = D_1 = 0, D_0 = 1\$
Let's continue with the rest.
$$\begin{array}{|c|c|} \hline E & C & B & A && D_7 & D_6 & D_5 & D_4 & D_3 & D_2 & D_1 & D_0 \\\hline \textbf{0} & \textbf{X} & \textbf{X} & \textbf{X} && \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} \\\hline \textbf{1} & \textbf{0} & \textbf{0} & \textbf{0} && \textbf{1} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} \\\hline 1 & 0 & 0 & 1 && 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\hline 1 & 0 & 1 & 0 && 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\hline 1 & 0 & 1 & 1 && 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\hline 1 & 1 & 0 & 0 && 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\hline 1 & 1 & 0 & 1 && 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\hline 1 & 1 & 1 & 0 && 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\hline \textbf{1} & \textbf{1} & \textbf{1} & \textbf{1} && \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{1} \\\hline \end{array}$$
I've bolded my sanity checks, which checks out. Inverting inputs is evil. And you got your diagonal wrong in your image.
Here's a sanity check for you that probably got you overthinking things. \$B = C = A = 0 => inverted => 111_2 = 7\$