delay() function behavior in Arduino - Step-by-Step Execution

Concept Flow - delay() function behavior

Start Program

↓

Call delay(ms)

↓

Pause execution for ms milliseconds

↓

Resume next instruction

↓

End or loop

The delay() function pauses the Arduino program for a set number of milliseconds, stopping all code execution during that time.

Execution Sample

Arduino

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

void loop() {
  digitalWrite(LED_BUILTIN, HIGH);
  delay(1000);
  digitalWrite(LED_BUILTIN, LOW);
  delay(1000);
}

This code turns the built-in LED on and off, pausing 1 second between changes using delay().

Execution Table

Step	Action	Time Passed (ms)	LED State	Notes
1	Start loop	0	OFF	Loop begins
2	digitalWrite HIGH	0	ON	LED turned on
3	delay(1000)	0 to 1000	ON	Program pauses 1000 ms
4	digitalWrite LOW	1000	OFF	LED turned off
5	delay(1000)	1000 to 2000	OFF	Program pauses 1000 ms
6	Loop repeats	2000	OFF	Back to step 1

💡 Loop repeats indefinitely; delay pauses execution for specified milliseconds each time.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6
LED State	OFF	ON	ON	OFF	OFF	OFF

Key Moments - 3 Insights

Why does the LED stay on during delay(1000)?

Because delay() pauses the program but does not change outputs; the LED stays ON until the next instruction changes it (see step 3 in execution_table).

Does delay() allow other code to run during the pause?

No, delay() stops all code execution for the given time, so nothing else runs during the delay period (see time passed column in execution_table).

What happens after delay finishes?

After delay ends, the program continues with the next instruction immediately (see step 4 in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table, what is the LED state during step 3?

A OFF

B Blinking

C ON

D Undefined

💡 Hint

Check the LED State column for step 3 in the execution_table.

At which step does the program pause for the first time?

A Step 3

B Step 2

C Step 4

D Step 5

💡 Hint

Look at the Action and Time Passed columns in execution_table to find when delay starts.

If delay(1000) was changed to delay(500), how would the Time Passed column change at step 3?

A 500 to 1000

B 0 to 500

C 0 to 1000

D No change

💡 Hint

Delay time controls the pause duration shown in Time Passed column in execution_table.

Concept Snapshot

delay(ms) pauses Arduino code for ms milliseconds.
During delay, no other code runs.
Outputs stay as last set until changed.
Use delay for simple timing but beware it blocks code.
For multitasking, use millis() instead.

Full Transcript

The delay() function in Arduino stops the program from running for a set number of milliseconds. In the example, the LED turns on, then delay(1000) pauses the program for 1 second while the LED stays on. After the delay, the LED turns off, then delay(1000) pauses again for 1 second. This cycle repeats forever. During delay, no other code runs, so the program is paused completely. After delay finishes, the program continues with the next instruction immediately. This behavior is shown step-by-step in the execution table and variable tracker.

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 <