teaching machines

CS 352 Lecture 8 – 7-Segment Logic

September 22, 2016 by Chris Johnson. Filed under cs352, fall 2016, lectures.

Dear students,

Today we’re round out our understanding of Karnaugh maps in the ultimate test. We’re going to design the logic to drive a 7-segment display! What is that, you ask? A 7-segment display is one of these:

It’s called “7-segment” because it’s decomposed into seven independent LEDs, each identified by a letter, A through G:

The big question that the circuit must answer is, “Do I light segment X?” In pseudocode, we might craft the logic like so:

if segment A should be on
  power segment A

if segment B should be on
  power segment B

if segment C should be on
  power segment C

...

What are the conditions for those if statements? We could write them sloppily:

if n == 2 || n == 3 || n == 5 || n == 6 || n == 7 || n == 8 || n == 9
  power segment A

...

But the logic would not be minimal. Instead, we can look at the underlying bit patterns for numbers 0 through 9. We need four bits to represent numbers in this range. We’ll call the most significant bit D3 and the least D0. We can then sketch out a truth table, where the inputs are D3, D2, D1, D0, and the outputs are whether or not segment X is illuminated for the given bits:

Let us now begin our challenge:

1. Form a team of three.
2. Await your segment assignment: a letter in [A, G].
3. Complete your column of the truth table.
4. Create a Karnaugh map with D3/D2 on one axis and D1/D0 on the other.
5. Share the minimized expression with your instructor.

If the entire class gets the logic working by the end of our 50 minutes together, your instructor will award a participation point to all present! We will test on a real live Arduino board hooked up to a real live 7-segment display.

See you next class!

Sincerely,

P.S. It’s Haiku Friday!

My first watch display
1×2 + 3×7
Was good at the time

P.P.S. Here’s the code we wrote together in class…

segment.cpp

#include "Arduino.h"

void setup() {
  pinMode(0, OUTPUT);
  pinMode(1, OUTPUT);
  pinMode(2, OUTPUT);
  pinMode(3, OUTPUT);
  pinMode(4, OUTPUT);
  pinMode(5, OUTPUT);
  pinMode(6, OUTPUT);
}

int i = 0;

void loop() {
  int d0 = (i >> 0) & 1;
  int d1 = (i >> 1) & 1;
  int d2 = (i >> 2) & 1;
  int d3 = (i >> 3) & 1;
  
  // A
  if (!d3 && d1 ||
      !d1 && !d0 && !d2 ||
      !d1 && !d2 && d3 ||
      !d3 && d2 && d0) {
    digitalWrite(0, HIGH);
  }

  // B
  if (!d3 && !d1 && !d0 ||
      !d2 && !d1 ||
      !d3 && d1 && d0 ||
      !d3 && !d2) {
    digitalWrite(1, HIGH);
  }

  // C
  if (!d3 && d2 ||
      !d3 && !d2 && !d1 ||
      d3 && !d2 && !d1 ||
      d0 && d1 && !d3) {
    digitalWrite(2, HIGH);
  }

  // D
  if (d1 && !d2 && !d3 ||
      !d0 && !d1 && !d2 ||
      !d0 && d1 && !d3 ||
      d0 && !d1 && d2 && !d3) {
    digitalWrite(3, HIGH);
  }

  // E
  if (d1 && !d0 && !d3 ||
      !d1 && !d0 && !d2) {
    digitalWrite(4, HIGH);
  }
 
  // F
  if (d3 && !d2 && !d1 ||
      !d3 && !d1 && !d0 ||
      !d3 && d2 && !d1 ||
      !d3 && d2 && !d0) {
    digitalWrite(5, HIGH);
  }
  
  // G
  if (d3 && !d2 && !d1 ||
      !d3 && d2 && !d1 ||
      !d3 && !d2 && d1 ||
      !d3 && d1 && !d0) {
    digitalWrite(6, HIGH);
  }

  delay(1000);

  digitalWrite(0, LOW);
  digitalWrite(1, LOW);
  digitalWrite(2, LOW);
  digitalWrite(3, LOW);
  digitalWrite(4, LOW);
  digitalWrite(5, LOW);
  digitalWrite(6, LOW);

  i = (i + 1) % 10;
}