Digital Design Fundamentals·Combinational Circuits

Sum of Products

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The core problem

You can describe any combinational circuit with a truth table: list every input combination, write the output you want. The question this lesson answers is the reverse. Given the table, what gates produce it?

There is a mechanical recipe that always works. It is called sum of products, and it reads a circuit straight off the rows where the output is 1.

A	B	Y
0	0	0
0	1	1
1	0	0
1	1	1

Sum of products, step by step:

Find every row where $Y = 1$ . In the table above, that is $AB = 01$ and $AB = 11$ .
For each of those rows, write one product (an AND) that equals 1 only for that row.
Row $01$ gives $\overline{A} \cdot B$ , and row $11$ gives $A \cdot B$ .
OR all the products together: $Y = \overline{A} \cdot B + A \cdot B$ .

Every 1 in the table becomes one AND gate. One OR gate collects them. That is the whole method.

AND gates into an OR

This circuit is the sum of products $Y = (A \cdot B) + (\overline{A} \cdot C)$ : two product (AND) terms feeding one sum (OR).

Try it:

Set a = 1, b = 1. The first product turns on and y goes high.
Set a = 0, c = 1. Now the second product carries the output instead.
Notice the shape: AND gates on the left, one OR gate collecting them on the right. Every SOP circuit looks like this.

A table that shows simplification

Figure: A 2 by 2 Karnaugh map for a two-input function

A Karnaugh map (K-map) is a truth table folded into a grid, arranged so neighbouring cells differ by one input bit. That layout makes simplification visual: adjacent 1s that form a rectangle of size 1, 2, 4 or 8 collapse into a single smaller product.

The map above holds the function from step 3. Its two 1s in one column group together, and the answer simplifies from $\overline{A}\,B + A\,B$ down to just $B$ .

Cancelling a variable

Group two adjacent 1s and one input changes across the pair while the output stays 1. That input cannot matter, so it drops out.

\overline{A}\,B + A\,B = (\overline{A} + A)\cdot B = 1 \cdot B = B

$A$ appears once true and once complemented, so it cancels. Bigger groups cancel more variables: a group of 4 removes two inputs, a group of 8 removes three. Larger groups mean fewer, smaller gates.

Fewer gates, faster chip

Simplification is not tidiness for its own sake. Every product term you remove is an AND gate you do not build, and every literal you cancel is a wire and a transistor saved.

Fewer gates in series also means a shorter critical path, so a simplified circuit is often faster as well as smaller. K-maps are how designers found these savings by hand before software did it automatically, and reading one still tells you at a glance whether a function is as small as it can be.

Sum of products builds a circuit from the rows where the output is 1. The next lesson does the mirror image, building from the 0-rows instead: product of sums.

The core problem

The core problem

Two words, two gates

One product per 1-row

AND gates into an OR

A table that shows simplification

Cancelling a variable

Fewer gates, faster chip