Programmable Control

The exam's microcontroller

• 8 pins: each one is an input or an output.
• The I/O table in the question says which.
• In our room: inputs on pins 2 and 3, outputs on pins 4 to 7.

Input (a switch or a sensor) → process (the microcontroller and its program) → output (a lamp, buzzer or motor).

Hard-wired or programmable?

Wiring inputs and outputs

• Input: a switch from 5 V to the pin. Pressed, the pin gets 5 V and the program reads it as on.
• Output: from the pin to 0 V. Switched on, the pin gives 5 V, current flows and the buzzer sounds.
• A motor needs a transistor. A pin gives a small current, about 20 mA. A motor needs much more.

The transistor driver: the base lead goes pin → resistor → base. The collector has the motor, with a diode across it. The emitter goes to 0 V.

Watch out: "driver" on its own scores 0. Say it is a transistor.

Lines and arrows

• Down or right: an arrow is not needed.
• Up or left: an arrow is needed.
• Every loop back: draw its arrow. The marking instructions mark it.

What goes in the box

The booklet writes the same boxes as Lamp (output, pin 13) ON, Wait 1 s (1000 ms) and count < 5?. Both score. Either way, every output has its pin and every delay has its unit.

How the marks are given

• Each output: on and off, in the right place (1)
• Each input decision: Y, N, a loop and an arrow (1)
• Each delay: right place, with a unit (1)
• x3?: Y, N, and an arrow back before the repeat (1)
• A loop to the start with an arrow, or an end (1)
• All the symbols right (1)

The method

1. Read the I/O table: which pins are in, which are out.
2. Draw one box per bullet, in order, top to bottom.
3. Turn each input into a decision, pin 1 on?, with Y and N.
4. Turn each repeat into x3?, with an arrow back.
5. Finish: loop to the start, or end. Check pins, units and arrows.

Build it in a simulator

NoSyntaxSim draws the same symbols. Each input or output box picks a pin from the hardware panel. Run Programme lights each box as it runs. It writes a delay as Wait 3000ms; on paper we still write pause 3 s.

Yenka builds the same flowchart with its own symbols. In class you use both, so you know both.

Describe the function of line 5 in the barrier program. 1 mark

Scores 0: "It goes to main." Say what that does.

Marking instructions, 2022 Q9(b).

Explain why lines 2 to 7 do not run 10 times. 2 marks

The word so joins the cause to its effect.

Marking instructions, 2022 Q9(c).

Your turn: program to flowchart

Draw the flowchart for this program on paper. Check every pin, every unit and every arrow. Then open the answer.

1. main:if input3 is off then main
2. switch on 5
3. switch on 4
4. pause 2000
5. switch off 5
6. pause 1000
7. switch off 4
8. goto main

Worked example 1: the flashing lamp

A warning lamp flashes on and off the whole time the machine is on.

void setup() {
  pinMode(4, OUTPUT);
}
void loop() {
  digitalWrite(4, HIGH);
  delay(1000);
  digitalWrite(4, LOW);
  delay(1000);
}

setup() runs once: pin 4 is an output. Everything in loop() repeats for ever: that is the continuous loop. delay(1000) holds the lamp for 1000 ms, 1 s.

Worked example 2: traffic lights

The UK sequence: red; red and amber; green; amber; back to red. Red on 4, amber on 5, green on 6.

void setup() {
  pinMode(4, OUTPUT);   // red
  pinMode(5, OUTPUT);   // amber
  pinMode(6, OUTPUT);   // green
}
void loop() {
  digitalWrite(4, HIGH);
  delay(3000);
  digitalWrite(5, HIGH);
  delay(1000);
  digitalWrite(4, LOW);
  digitalWrite(5, LOW);
  digitalWrite(6, HIGH);
  delay(3000);
  digitalWrite(6, LOW);
  digitalWrite(5, HIGH);
  delay(1000);
  digitalWrite(5, LOW);
}

Three outputs, so three pinMode lines, and each LED has its own 220 Ω. Two outputs HIGH at once give the red and amber stage.

Worked example 3: the darkness alarm

When it gets dark, the LED flashes 5 times, then the system keeps watching.

void setup() {
  pinMode(4, OUTPUT);
  // A0 is an analogue input: no pinMode
}
void loop() {
  int light = analogRead(A0);
  if (light < 300) {
    for (int i = 0; i < 5; i++) {
      digitalWrite(4, HIGH);
      delay(500);
      digitalWrite(4, LOW);
      delay(500);
    }
  }
}

analogRead(A0) gives a number from 0 to 1023. In the dark the LDR's resistance goes up, so A0 goes down. if (light < 300) is the branch. for (i = 0; i < 5; i++) is the fixed loop: i counts 0, 1, 2, 3, 4, which is five flashes.

Find and fix

Finding faults is called debugging. Change one thing at a time, then test again: change two things and you will not know which one fixed it.

1. What should it do?
2. What does it do?
3. Which line?
4. Change one thing.
5. Test it again.

Debug build 1 should flash pin 4, 1 s on and 1 s off. It has 2 faults.

void setup() {
}
void loop() {
  digitalWrite(4, HIGH);
  delay(1);
  digitalWrite(4, LOW);
  delay(1000);
}

Test, amend, re-test

A pedestrian crossing: red until the button is pressed; then green and the beeper for 5 s; then back to red. One test for each job:

Say what you saw, by part. "It didn't work" names no part and no result. "The beeper stayed on after 5 s" does.

Final version. After the re-test, save it under a new name. The assignment asks for the final version.

Model answer

Suggest one improvement to the crossing. Justify it. 2 marks

"Make it better" scores 0: it names no improvement and gives no reason.