Seven Segment Display Boolean Equations Example 2

This 7-segment display example shows how to derive the Boolean expressions to build a driver circuit. Suppose a counter provides a 3-bit output (XYZ) to count from 0 to 7, and your driver circuit has to display the numeric symbols from 1 to 5. You would have to make a truth table showing the segments that require lighting to display those numbers. The table then helps find the Boolean logic expressions required to light each segment. You will then need to find a way to simplify those expressions.

X	Y	Z	a	b	c	d	e	f	g	DISP
0	0	0	0	0	0	0	0	0	0	-
0	0	1	0	1	1	0	0	0	0	1
0	1	0	1	1	0	1	1	0	1	2
0	1	1	1	1	1	1	0	0	1	3
1	0	0	0	1	1	0	0	1	1	4
1	0	1	1	0	1	1	0	1	1	5
1	1	0	0	0	0	0	0	0	0	-
1	1	1	0	0	0	0	0	0	0	-

Here is the truth table showing the segments that require lighting to display the relevant numeric. If you get a question like this in GCSE then this is the easy part, which everyone can usually do. The next step is to determine the logic circuitry required for each segment. In this part, you have to think which logic circuit would provide the outputs given the inputs are XYZ. With two inputs, it would be simple to do, but when there are three inputs, it gets complicated and you need to work according to a methodology to solve it.

To solve this puzzle, you could use your skills in Karnaugh mapping; however, knowledge of Boolean algebra is even better, because there is a shortcut if you know how.

Segment "a" Boolean Expression

Here is the same table but this time we are looking at the outputs of segment "a" only. You have to derive a logic circuit, so that when it receives the inputs XYZ, the corresponding outputs will be those in column "a" to light segment a.

We first look for outputs in column "a" that have logic 1 state.
Then look at its corresponding inputs XYZ
For "X.Y.Z" to produce an output of 1, all must be 1’s, however some are 0’s, therefore you will have to invert the 0’s using the NOT function. This information helps to know which input variables will have a bar above it.
In the Boolean expression, place a bar above those variables that would require inverting.
You then simply make a sum of products as I have shown.

Once you have an equation for each segment, you use Boolean algebra to simplify it. Some may require simplification, whilst others will remain very long.