- #1

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## Main Question or Discussion Point

Hello, I'm new to logic circuits and I was making a 4 bit binary adder and I thought I could maybe simplify it. However, I can't find anything simpler than what I already got. What I wanted to know is:

The circuit works as follows:

We have the 1st 4 bit number:

We have the 2nd 4 bit number:

The total sum with one bit at a time would be: (D + H) and (C + G) and (B + F) and (A + E). In other words:

ABCD

EFGH

---------

XXXX

where, (D + H) is the only one that doesn't receive any carry because it's the initial point.

So I need 4 outputs with their respective carries as illustrated:

As you can see the pattern of circuits start at the second sum. Meaning, (C + G), (B + F), and (A + E) are the same circuits. I tried to simplify 1 of them so I could simplify the rest of the adder. However, I can't find any way to simplify it any further. When I look at the Karnaugh Map I made out of the Truth Table of the circuit that repeats I don't know what to do from there because I don't have any mapping groups there.

Truth Table: [tex]((C \oplus G) \oplus CARRY) \wedge CARRY[/tex]

\begin{array}{|c|c|c|c|}

\hline

C & G & CARRY & F1 \\ \hline

0 & 0 & 0 & 0 \\ \hline

0 & 0 & 1 & 1 \\ \hline

0 & 1 & 0 & 0 \\ \hline

0 & 1 & 1 & 0 \\ \hline

1 & 0 & 0 & 0 \\ \hline

1 & 0 & 1 & 0 \\ \hline

1 & 1 & 0 & 0 \\ \hline

1 & 1 & 1 & 1 \\

\hline

\end{array}

Karnaugh Map: [tex]((C \oplus G) \oplus CARRY) \wedge CARRY[/tex]

\begin{array}{|c|c|c|c|c|c|}

\hline

& G \wedge CARRY & 00 & 01 & 11 & 10 \\ \hline

C & & & & & \\ \hline

0 & & & 1 & & \\ \hline

1 & & & & 1 & \\

\hline

\end{array}

Does all this mean that I cannot simplify my adder any further?

**is my circuit already in its most simple form or am I making something wrong in my attempts to simplificate it (meaning it can still be simplificated)?**The circuit works as follows:

We have the 1st 4 bit number:

**ABCD**We have the 2nd 4 bit number:

**EFGH**The total sum with one bit at a time would be: (D + H) and (C + G) and (B + F) and (A + E). In other words:

ABCD

EFGH

---------

XXXX

where, (D + H) is the only one that doesn't receive any carry because it's the initial point.

So I need 4 outputs with their respective carries as illustrated:

As you can see the pattern of circuits start at the second sum. Meaning, (C + G), (B + F), and (A + E) are the same circuits. I tried to simplify 1 of them so I could simplify the rest of the adder. However, I can't find any way to simplify it any further. When I look at the Karnaugh Map I made out of the Truth Table of the circuit that repeats I don't know what to do from there because I don't have any mapping groups there.

Truth Table: [tex]((C \oplus G) \oplus CARRY) \wedge CARRY[/tex]

\begin{array}{|c|c|c|c|}

\hline

C & G & CARRY & F1 \\ \hline

0 & 0 & 0 & 0 \\ \hline

0 & 0 & 1 & 1 \\ \hline

0 & 1 & 0 & 0 \\ \hline

0 & 1 & 1 & 0 \\ \hline

1 & 0 & 0 & 0 \\ \hline

1 & 0 & 1 & 0 \\ \hline

1 & 1 & 0 & 0 \\ \hline

1 & 1 & 1 & 1 \\

\hline

\end{array}

Karnaugh Map: [tex]((C \oplus G) \oplus CARRY) \wedge CARRY[/tex]

\begin{array}{|c|c|c|c|c|c|}

\hline

& G \wedge CARRY & 00 & 01 & 11 & 10 \\ \hline

C & & & & & \\ \hline

0 & & & 1 & & \\ \hline

1 & & & & 1 & \\

\hline

\end{array}

Does all this mean that I cannot simplify my adder any further?