Writing HIGH and LOW to output pins in Embedded C - Step-by-Step Execution

Concept Flow - Writing HIGH and LOW to output pins

Setup pin as OUTPUT

↓

Write HIGH to pin

→Pin voltage set to HIGH

↓

Write LOW to pin

→Pin voltage set to LOW

↓

End

First, configure the pin as an output. Then write HIGH or LOW to set the pin voltage accordingly.

Execution Sample

Embedded C

pinMode(13, OUTPUT);
digitalWrite(13, HIGH);
delay(1000);
digitalWrite(13, LOW);

Set pin 13 as output, turn it ON (HIGH), wait 1 second, then turn it OFF (LOW).

Execution Table

Step	Action	Pin Mode	Pin State	Output Voltage
1	pinMode(13, OUTPUT)	OUTPUT	LOW (default)	0V (LOW)
2	digitalWrite(13, HIGH)	OUTPUT	HIGH	5V (HIGH)
3	delay(1000)	OUTPUT	HIGH	5V (HIGH)
4	digitalWrite(13, LOW)	OUTPUT	LOW	0V (LOW)
5	End of code	OUTPUT	LOW	0V (LOW)

💡 Code ends after setting pin 13 LOW.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
Pin 13 Mode	INPUT (default)	OUTPUT	OUTPUT	OUTPUT	OUTPUT	OUTPUT
Pin 13 State	LOW (default)	LOW	HIGH	HIGH	LOW	LOW
Pin 13 Voltage	0V (LOW)	0V (LOW)	5V (HIGH)	5V (HIGH)	0V (LOW)	0V (LOW)

Key Moments - 3 Insights

Why do we need to set the pin mode to OUTPUT before writing HIGH or LOW?

Because the pin must be configured as OUTPUT to control voltage levels; otherwise, writing HIGH or LOW won't change the pin voltage (see Step 1 in execution_table).

Does writing HIGH immediately change the pin voltage?

Yes, writing HIGH sets the pin voltage to 5V right away (Step 2 shows pin state changes to HIGH and voltage to 5V).

What happens if we write LOW after writing HIGH?

The pin voltage changes back to 0V (LOW), turning the output off (Step 4 shows this change).

Visual Quiz - 3 Questions

Look at the execution_table, what is the pin state after Step 2?

A LOW

B HIGH

C INPUT

D UNKNOWN

💡 Hint

Check the 'Pin State' column at Step 2 in the execution_table.

At which step does the pin voltage become 0V after being HIGH?

A Step 1

B Step 2

C Step 4

D Step 3

💡 Hint

Look at the 'Pin 13 Voltage' in variable_tracker after Step 4.

If we skip setting pinMode to OUTPUT, what will happen when writing HIGH?

A Pin voltage stays LOW

B Pin voltage changes to HIGH

C Pin mode automatically changes to OUTPUT

D Program crashes

💡 Hint

Refer to key_moments about why pinMode must be set before writing.

Concept Snapshot

pinMode(pin, OUTPUT) sets pin as output
Use digitalWrite(pin, HIGH) to set voltage HIGH (5V)
Use digitalWrite(pin, LOW) to set voltage LOW (0V)
Must set pinMode before writing
Writing changes pin voltage immediately
Delay can keep state before changing

Full Transcript

This example shows how to control an output pin in embedded C. First, the pin is set as OUTPUT using pinMode. Then digitalWrite sets the pin voltage HIGH or LOW. The execution table tracks each step: setting mode, writing HIGH, waiting, then writing LOW. Variables track pin mode, state, and voltage changes. Key moments clarify why setting pin mode is necessary and how writing changes voltage immediately. The visual quiz tests understanding of pin state and voltage changes at each step. The snapshot summarizes the commands and rules for writing HIGH and LOW to output pins.

Practice

1. Which function is used to set a pin as an output before writing HIGH or LOW to it?

easy

A. setPinMode(pin, HIGH)

B. digitalWrite(pin, OUTPUT)

C. pinMode(pin, OUTPUT)

D. writePin(pin, OUTPUT)

Solution

1. Step 1: Understand pin configuration

   Before writing HIGH or LOW, the pin must be set as an output to control voltage.

2. Step 2: Identify the correct function

   The function to set pin mode is pinMode(pin, OUTPUT), which configures the pin for output.

3. Final Answer:

   pinMode(pin, OUTPUT) -> Option C

4. Quick Check:

   Set pin as output = pinMode(pin, OUTPUT) [OK]

Hint: Set pin mode with pinMode before writing HIGH or LOW [OK]

Common Mistakes:

• Using digitalWrite to set pin mode
• Confusing HIGH/LOW with OUTPUT/INPUT
• Forgetting to set pin mode before writing

2. Which of the following is the correct syntax to turn pin 5 ON (HIGH) in embedded C?

easy

A. digitalWrite(5, 1);

B. digitalWrite(5, ON);

C. digitalWrite(5, OUTPUT);

D. digitalWrite(5, HIGH);

Solution

1. Step 1: Identify the function to write pin state

   The function digitalWrite(pin, value) sets the pin voltage HIGH or LOW.

2. Step 2: Use correct constants

   Use HIGH to turn the pin ON. The correct syntax is digitalWrite(5, HIGH);.

3. Final Answer:

   digitalWrite(5, HIGH); -> Option D

4. Quick Check:

   Turn pin ON = digitalWrite(pin, HIGH) [OK]

Hint: Use HIGH constant, not ON or numbers, for clarity [OK]

Common Mistakes:

• Using ON instead of HIGH
• Using OUTPUT instead of HIGH
• Using numeric 1 instead of HIGH constant

3. What will be the output state of pin 3 after running this code?

pinMode(3, OUTPUT);
digitalWrite(3, LOW);
digitalWrite(3, HIGH);

medium

A. Pin 3 will be LOW

B. Pin 3 will be HIGH

C. Pin 3 will be in input mode

D. Pin 3 will be undefined

Solution

1. Step 1: Set pin 3 as output

   The code sets pin 3 as output with pinMode(3, OUTPUT);.

2. Step 2: Write LOW then HIGH to pin 3

   First, pin 3 is set LOW, then immediately set HIGH, so final state is HIGH.

3. Final Answer:

   Pin 3 will be HIGH -> Option B

4. Quick Check:

   Last digitalWrite sets pin state = HIGH [OK]

Hint: Last digitalWrite sets the pin state [OK]

Common Mistakes:

• Assuming pin stays LOW after writing HIGH
• Forgetting pinMode sets output mode
• Thinking pin state is undefined without delay

4. Identify the error in this code snippet:

pinMode(7, OUTPUT);
digitalWrite(7, HIGH)
digitalWrite(7, LOW);

medium

A. Missing semicolon after digitalWrite(7, HIGH)

B. pinMode should be called after digitalWrite

C. HIGH and LOW are not valid constants

D. Pin number 7 cannot be used

Solution

1. Step 1: Check syntax line by line

   The second line digitalWrite(7, HIGH) is missing a semicolon at the end.

2. Step 2: Confirm other lines are correct

   pinMode(7, OUTPUT); and digitalWrite(7, LOW); are correct syntax.

3. Final Answer:

   Missing semicolon after digitalWrite(7, HIGH) -> Option A

4. Quick Check:

   Every statement must end with semicolon [OK]

Hint: Check each line ends with a semicolon [OK]

Common Mistakes:

• Forgetting semicolon after function calls
• Calling pinMode after digitalWrite
• Misusing HIGH/LOW constants

5. You want to toggle pin 10 ON and OFF every second in a loop. Which code snippet correctly sets pin 10 HIGH then LOW with a 1-second delay between states?

hard

A. pinMode(10, OUTPUT);
digitalWrite(10, HIGH);
delay(1000);
digitalWrite(10, LOW);
delay(1000);

B. digitalWrite(10, HIGH);
delay(1000);
digitalWrite(10, LOW);
delay(1000);

C. pinMode(10, INPUT);
digitalWrite(10, HIGH);
delay(1000);

D. pinMode(10, OUTPUT);
digitalWrite(10, LOW);
digitalWrite(10, HIGH);

Solution

1. Step 1: Set pin 10 as output

   Use pinMode(10, OUTPUT); to configure pin 10 for output.

2. Step 2: Write HIGH, wait, then LOW, wait

   Use digitalWrite(10, HIGH); then delay(1000); to wait 1 second, then digitalWrite(10, LOW); and delay again.

3. Final Answer:

   pinMode(10, OUTPUT); digitalWrite(10, HIGH); delay(1000); digitalWrite(10, LOW); delay(1000); -> Option A

4. Quick Check:

   Set output, write states with delay between [OK]

Hint: Always set pinMode before toggling with delays [OK]

Common Mistakes:

• Forgetting to set pinMode as OUTPUT
• Missing delay between HIGH and LOW
• Setting pinMode to INPUT instead of OUTPUT