delay() function behavior in Arduino - Mini Project: Build & Apply

Understanding the delay() Function Behavior in Arduino

📖 Scenario: You are creating a simple Arduino program to blink an LED on and off. You want to understand how the delay() function works to pause the program for a certain time.

🎯 Goal: Build a program that turns an LED on, waits for 1 second using delay(), then turns the LED off, and waits for 1 second again. This cycle repeats forever.

📋 What You'll Learn

Use the delay() function to pause the program

Control the LED connected to pin 13

Turn the LED on and off with 1 second pauses

💡 Why This Matters

🌍 Real World

Blinking an LED is a basic way to test if your Arduino board and code are working correctly. It helps you understand timing and control in electronics projects.

💼 Career

Understanding how to control timing with functions like <code>delay()</code> is important for embedded systems programming, hardware testing, and creating interactive devices.

1

Set up the LED pin

Create a variable called ledPin and set it to 13. Then, write the setup() function to set ledPin as an output using pinMode().

Arduino

int ledPin = 13;

void setup() {
  // Set ledPin as output
  // Your code here
}

void loop() {
  // Your code here
}

Hint

Use pinMode(ledPin, OUTPUT); inside setup() to prepare the LED pin.

2

Turn the LED on and add delay

Inside the loop() function, turn the LED on using digitalWrite(ledPin, HIGH);. Then, add a delay() of 1000 milliseconds (1 second) to pause the program.

Arduino

int ledPin = 13;

void setup() {
  pinMode(ledPin, OUTPUT);
}

void loop() {
  // Turn LED on
  // Add 1 second delay
  // Your code here
}

Hint

Use digitalWrite(ledPin, HIGH); to turn the LED on and delay(1000); to wait 1 second.

3

Turn the LED off and add delay

Still inside the loop() function, after the first delay, turn the LED off using digitalWrite(ledPin, LOW);. Then, add another delay() of 1000 milliseconds to pause again.

Arduino

int ledPin = 13;

void setup() {
  pinMode(ledPin, OUTPUT);
}

void loop() {
  digitalWrite(ledPin, HIGH);
  delay(1000);
  // Turn LED off
  // Add 1 second delay
  // Your code here
}

Hint

Use digitalWrite(ledPin, LOW); to turn the LED off and delay(1000); to wait 1 second again.

4

Observe the blinking LED

Upload the program to your Arduino board and observe the LED connected to pin 13 blinking on and off every second. Use Serial.println() to print "LED ON" when the LED turns on and "LED OFF" when it turns off.

Arduino

int ledPin = 13;

void setup() {
  pinMode(ledPin, OUTPUT);
  Serial.begin(9600);
}

void loop() {
  digitalWrite(ledPin, HIGH);
  Serial.println("LED ON");
  delay(1000);
  digitalWrite(ledPin, LOW);
  Serial.println("LED OFF");
  delay(1000);
  // Your code here
}

Hint

Use Serial.begin(9600); in setup() and Serial.println() in loop() to print messages.

Practice

1. What does the delay(1000); function do in an Arduino program?

easy

A. Pauses the program for 1000 milliseconds (1 second)

B. Stops the program permanently

C. Speeds up the program by 1000 times

D. Restarts the Arduino board

Solution

1. Step 1: Understand the delay() parameter

   The number inside delay() is the time in milliseconds to pause the program.

2. Step 2: Interpret delay(1000)

   1000 milliseconds equals 1 second, so the program pauses for 1 second.

3. Final Answer:

   Pauses the program for 1000 milliseconds (1 second) -> Option A

4. Quick Check:

   delay(1000) = 1 second pause [OK]

Hint: delay(x) pauses for x milliseconds, 1000ms = 1 second [OK]

Common Mistakes:

• Thinking delay stops the program forever
• Confusing milliseconds with seconds
• Assuming delay speeds up the program

2. Which of the following is the correct syntax to pause an Arduino program for half a second?

easy

A. delay(0.5);

B. delay = 500;

C. delay(500s);

D. delay(500);

Solution

1. Step 1: Check correct function usage

   delay() takes an integer number of milliseconds inside parentheses.

2. Step 2: Validate each option

   A uses delay(0.5); which is a float and incorrect. B assigns delay which is invalid. C uses delay(500); which is correct for 500 milliseconds. D uses '500s' which is invalid syntax.

3. Final Answer:

   delay(500); -> Option D

4. Quick Check:

   delay(500) is correct syntax [OK]

Hint: Use delay(milliseconds); with integer inside parentheses [OK]

Common Mistakes:

• Using decimal numbers instead of integers
• Assigning delay like a variable
• Adding units like 's' inside delay()

3. What will be the output on the Serial Monitor after running this Arduino code?

void setup() {
  Serial.begin(9600);
  Serial.println("Start");
  delay(2000);
  Serial.println("End");
}
void loop() {}

medium

A. Start and End printed immediately together

B. Only Start is printed, End never prints

C. Start immediately, then End after 2 seconds

D. No output because delay stops Serial

Solution

1. Step 1: Analyze Serial prints and delay

   "Start" prints immediately, then delay(2000) pauses 2 seconds before next line.

2. Step 2: Understand delay effect on output

   After 2 seconds pause, "End" prints. Both lines appear but with 2 seconds gap.

3. Final Answer:

   Start immediately, then End after 2 seconds -> Option C

4. Quick Check:

   delay pauses program, output delayed [OK]

Hint: delay pauses code, so output after delay appears later [OK]

Common Mistakes:

• Thinking delay stops Serial output completely
• Assuming both prints happen instantly
• Believing delay affects only loop(), not setup()

4. Identify the problem in this Arduino code snippet:

void setup() {
  pinMode(13, OUTPUT);
}
void loop() {
  digitalWrite(13, HIGH);
  delay(1000)
  digitalWrite(13, LOW);
  delay(1000);
}

medium

A. pinMode should be in loop()

B. Missing semicolon after delay(1000)

C. digitalWrite needs delay before it

D. delay() cannot be used in loop()

Solution

1. Step 1: Check syntax line by line

   delay(1000) is missing a semicolon at the end, causing a syntax error.

2. Step 2: Validate other statements

   pinMode is correctly in setup(), digitalWrite and delay usage is correct except missing semicolon.

3. Final Answer:

   Missing semicolon after delay(1000) -> Option B

4. Quick Check:

   Every statement must end with ; [OK]

Hint: Check for missing semicolons after delay() calls [OK]

Common Mistakes:

• Putting pinMode inside loop unnecessarily
• Thinking delay() can't be in loop
• Forgetting semicolons after statements

5. You want to blink an LED connected to pin 9 exactly 3 times with 0.5 second ON and 0.5 second OFF, then stop. Which code snippet correctly uses delay() to do this?

hard

A. for(int i=0; i<3; i++) { digitalWrite(9, HIGH); delay(500); digitalWrite(9, LOW); delay(500); }

B. while(true) { digitalWrite(9, HIGH); delay(500); digitalWrite(9, LOW); delay(500); }

C. digitalWrite(9, HIGH); delay(1500); digitalWrite(9, LOW); delay(1500);

D. for(int i=0; i<=3; i++) { digitalWrite(9, HIGH); delay(1000); digitalWrite(9, LOW); delay(1000); }

Solution

1. Step 1: Understand blinking requirements

   LED must turn ON and OFF 3 times, each ON and OFF lasting 0.5 seconds (500 ms).

2. Step 2: Analyze each option

   A uses a for loop 3 times with 500ms delay ON and OFF, matching requirements. B loops forever, no stop. C delays 1500ms which is too long and only blinks once. D loops 4 times (i<=3) with 1000ms delays, wrong timing and count.

3. Final Answer:

   for(int i=0; i<3; i++) { digitalWrite(9, HIGH); delay(500); digitalWrite(9, LOW); delay(500); } -> Option A

4. Quick Check:

   Loop 3 times with 500ms delay ON/OFF [OK]

Hint: Use for loop with delay(500) ON and OFF, repeat 3 times [OK]

Common Mistakes:

• Using infinite loops instead of fixed count
• Wrong delay times for ON/OFF
• Looping one extra time with <= instead of <