Displaying sensor data on screen in Arduino - Mini Project: Build & Apply

Displaying sensor data on screen

📖 Scenario: You have a temperature sensor connected to your Arduino. You want to read the temperature and show it on a small screen so anyone can see the current temperature.

🎯 Goal: Build a simple Arduino program that reads temperature data from a sensor and displays it on the screen.

📋 What You'll Learn

Create a variable to store the sensor pin number

Create a variable to store the temperature value

Read the sensor value from the correct pin

Convert the sensor reading to a temperature

Display the temperature on the screen using Serial.println()

💡 Why This Matters

🌍 Real World

Displaying sensor data on a screen is common in weather stations, home automation, and health monitoring devices.

💼 Career

Understanding how to read sensors and show data is a key skill for embedded systems and IoT developers.

1

Setup sensor pin variable

Create an int variable called sensorPin and set it to A0 to represent the analog pin where the temperature sensor is connected.

Arduino

int sensorPin = A0;  // Your code here

Hint

The temperature sensor is connected to analog pin A0. Use int sensorPin = A0; to store this.

2

Setup temperature variable

Create an int variable called temperature and set it to 0 to hold the temperature value.

Arduino

int sensorPin = A0;
int temperature = 0;  // Your code here

Hint

We need a variable to store the temperature reading. Use int temperature = 0;.

3

Read sensor and convert to temperature

In the loop() function, read the analog value from sensorPin using analogRead(sensorPin) and store it in an int variable called sensorValue. Then convert sensorValue to temperature in Celsius by using the formula temperature = (sensorValue * 5.0 / 1023) * 100.

Arduino

int sensorPin = A0;
int temperature = 0;

void setup() {
  Serial.begin(9600);
}

void loop() {
  int sensorValue = analogRead(sensorPin);  // Your code here
  temperature = (sensorValue * 5.0 / 1023) * 100;  // Your code here
}

Hint

Use analogRead(sensorPin) to get the sensor value. Then convert it to Celsius with the formula given.

4

Display temperature on screen

Use Serial.println() to print the text "Temperature: " followed by the temperature variable and the degree symbol "\xB0C" inside the loop() function. This will show the temperature on the serial monitor.

Arduino

int sensorPin = A0;
int temperature = 0;

void setup() {
  Serial.begin(9600);
}

void loop() {
  int sensorValue = analogRead(sensorPin);
  temperature = (sensorValue * 5.0 / 1023) * 100;
  Serial.println("Temperature: " + String(temperature) + "\xB0C");  // Your code here
}

Hint

Use Serial.println("Temperature: " + String(temperature) + "\xB0C") to show the temperature with the degree symbol.

Practice

1. What is the main purpose of displaying sensor data on a screen in an Arduino project?

easy

A. To store data permanently

B. To see real-time measurements easily

C. To power the sensor

D. To connect to the internet

Solution

1. Step 1: Understand sensor data usage

   Sensor data is collected to monitor or measure something in real time.

2. Step 2: Purpose of displaying data

   Showing data on a screen helps users see these measurements immediately.

3. Final Answer:

   To see real-time measurements easily -> Option B

4. Quick Check:

   Displaying sensor data = real-time viewing [OK]

Hint: Displaying sensor data means showing live values on screen [OK]

Common Mistakes:

• Confusing display with data storage
• Thinking display powers the sensor
• Assuming display connects to internet

2. Which Arduino command correctly clears the screen before showing new sensor data?

easy

A. lcd.reset();

B. lcd.clean();

C. lcd.erase();

D. lcd.clear();

Solution

1. Step 1: Identify the correct LCD clear command

   The standard command to clear an LCD screen in Arduino is lcd.clear();.

2. Step 2: Check other options

   Commands like reset(), erase(), or clean() are not valid for LCD clearing.

3. Final Answer:

   lcd.clear(); -> Option D

4. Quick Check:

   Clear screen command = lcd.clear(); [OK]

Hint: Use lcd.clear() to refresh display before new data [OK]

Common Mistakes:

• Using non-existent lcd.reset() or lcd.erase()
• Forgetting parentheses in lcd.clear()
• Confusing clear with reset functions

3. What will be displayed on the LCD if the following code runs and the sensor value is 512?

int sensorValue = analogRead(A0);
lcd.clear();
lcd.setCursor(0, 0);
lcd.print("Value: ");
lcd.print(sensorValue);

medium

A. Value: 512

B. sensorValue

C. Value: A0

D. 512

Solution

1. Step 1: Understand variable assignment

   The variable sensorValue stores the analog reading from pin A0, which is 512.

2. Step 2: Analyze LCD print commands

   The code clears the screen, sets cursor to start, prints "Value: " text, then prints the number 512.

3. Final Answer:

   Value: 512 -> Option A

4. Quick Check:

   lcd.print("Value: ") + lcd.print(sensorValue) = Value: 512 [OK]

Hint: lcd.print(variable) shows its value, not the name [OK]

Common Mistakes:

• Printing variable name instead of value
• Confusing pin name with value
• Missing lcd.clear() causing old data

4. Identify the error in this code snippet that tries to display sensor data:

int sensorValue = analogRead(A0);
lcd.setCursor(0, 0);
lcd.print("Temp: ");
lcd.print(sensorValue);
lcd.clear();

medium

A. lcd.clear() should be before lcd.print()

B. analogRead() needs a semicolon

C. lcd.setCursor() is missing parameters

D. sensorValue must be declared as float

Solution

1. Step 1: Check lcd.clear() placement

   lcd.clear() clears the screen, so it must be called before printing new data to avoid erasing it immediately.

2. Step 2: Verify other code parts

   analogRead() has semicolon, setCursor() has correct parameters, and sensorValue as int is fine for analog values.

3. Final Answer:

   lcd.clear() should be before lcd.print() -> Option A

4. Quick Check:

   Clear screen before printing = correct order [OK]

Hint: Always clear screen before printing new data [OK]

Common Mistakes:

• Clearing screen after printing erases output
• Misplacing lcd.setCursor() parameters
• Wrong variable type assumptions

5. You want to display temperature in Celsius on an LCD using a sensor that outputs voltage. The sensor voltage is read by analogRead(A1) and converted by tempC = (sensorValue * 5.0 / 1023) * 100;. Which code snippet correctly displays the temperature with one decimal place?

hard

A. lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(int(tempC)); lcd.print(" C");

B. lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC); lcd.print(" Celsius");

C. lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC, 1); lcd.print(" C");

D. lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC + " C");

Solution

1. Step 1: Understand temperature conversion

   tempC is a float value representing temperature in Celsius with decimals.

2. Step 2: Display with one decimal place

   lcd.print(value, 1) prints float with 1 decimal. Adding " C" after shows units clearly.

3. Step 3: Check other options

   lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC); lcd.print(" Celsius"); prints full float but no decimal control; lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(int(tempC)); lcd.print(" C"); truncates decimals; lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC + " C"); tries to add string to float incorrectly.

4. Final Answer:

   lcd.clear(); lcd.setCursor(0,0); lcd.print("Temp: "); lcd.print(tempC, 1); lcd.print(" C"); -> Option C

5. Quick Check:

   lcd.print(float, 1) = one decimal place [OK]

Hint: Use lcd.print(value, 1) to show one decimal place [OK]

Common Mistakes:

• Not controlling decimal places
• Trying to add string directly to float
• Using int() truncates decimals