# Truth Table for 3-into-8 decoder with N.A. inputs, P.A outputs and enable

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?

• "I've drawn the block diagram", pics or it didn't happen. – Harry Svensson Oct 19 '17 at 1:15
• I've added the appropriate tag and attached my work to the original post. – gbm0102 Oct 19 '17 at 1:17
• I don't see your problem. I think you are overthinking this...If you can do the positive inputs... do so.. then invert all the zeros and 1s in the ABC column. The logic is the same, only the labelling changes on the inputs and outputs. – Trevor_G Oct 19 '17 at 1:30

3-into-8 decoder with negative active inputs, a positive active enable and positive active outputs.

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. $B = C = A = 0 => inverted => 111_2 = 7$