Writing data to I2C device in Embedded C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start an I2C transmission to device address 0x50.

Embedded C

I2C_Start();
I2C_Write([1]);

Drag options to blanks, or click blank then click option'

A0x50 << 1

B0x50

C0xA0

D0x28

Attempts:

3 left

💡 Hint

Common Mistakes

Using the device address without shifting left.

Using the read address instead of write address.

The device address must be shifted left by 1 bit to make room for the read/write bit in I2C communication.

2fill in blank

medium

Complete the code to write the data byte 0x3F to the I2C device.

Embedded C

I2C_Write([1]);

Drag options to blanks, or click blank then click option'

A0x3F

B0xA1

C0x50

D0x00

Attempts:

3 left

💡 Hint

Common Mistakes

Sending the device address instead of data.

Sending a wrong data byte.

0x3F is the data byte we want to send to the device.

3fill in blank

hard

Fix the error in the code to properly stop the I2C transmission.

Embedded C

I2C_Stop[1];

Drag options to blanks, or click blank then click option'

A{}

B;

C[]

D()

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting parentheses when calling functions.

Using incorrect punctuation after function names.

I2C_Stop is a function and must be called with parentheses.

4fill in blank

hard

Fill all three blanks to write a byte 0x7A to register 0x10 of the I2C device at address 0x60.

Embedded C

I2C_Start();
I2C_Write([1]);
I2C_Write([2]);
I2C_Write([3]);

Drag options to blanks, or click blank then click option'

A0x60 << 1

B0x7A

C0x10

D0x50

Attempts:

3 left

💡 Hint

Common Mistakes

Sending data byte before register address.

Not shifting the device address left.

First, send the device address shifted left for writing, then send the register address, then the data byte 0x7A.

5fill in blank

hard

Fill all three blanks to complete the function that writes a data byte to a register of an I2C device.

Embedded C

void writeRegister(uint8_t deviceAddr, uint8_t regAddr, uint8_t data) {
    I2C_Start();
    I2C_Write([1]);
    I2C_Write([2]);
    I2C_Write([3]);
    I2C_Stop();
}

Drag options to blanks, or click blank then click option'

AdeviceAddr << 1

BregAddr

Cdata

DdeviceAddr

Attempts:

3 left

💡 Hint

Common Mistakes

Not shifting the device address.

Swapping register and data bytes.

The function sends the device address shifted left, then the register address, then the data byte.

Practice

1. What is the first step when writing data to an I2C device in embedded C?

easy

A. Read the device address

B. Start the I2C communication

C. Stop the I2C communication

D. Send the data byte

Solution

1. Step 1: Begin communication with the device

   Writing data to an I2C device always starts with initiating the communication by sending a start condition. After starting, the device address and data can be sent, but the first step must be the start condition.

2. Final Answer:

   Start the I2C communication -> Option B

3. Quick Check:

   Start communication = first step [OK]

Hint: Always start communication before sending data [OK]

Common Mistakes:

• Sending data before starting communication
• Stopping communication too early
• Skipping the start condition

2. Which of the following is the correct way to specify the 7-bit I2C device address for writing in embedded C?

easy

A. device_address << 1 | 0

B. device_address >> 1

C. device_address << 1 | 1

D. device_address << 1 & ~1

Solution

1. Step 1: Understand I2C address format

   The 7-bit device address is shifted left by one bit to make room for the read/write bit as the least significant bit. For writing, the least significant bit must be 0, so OR with 0 is correct.

2. Final Answer:

   device_address << 1 | 0 -> Option A

3. Quick Check:

   Write bit = 0, address shifted left = correct [OK]

Hint: Shift address left and set LSB to 0 for write [OK]

Common Mistakes:

• Setting LSB to 1 for write operation
• Shifting address right instead of left
• Not shifting address at all

3. Given the following code snippet, what will be the result of the I2C write operation?

uint8_t device_addr = 0x50;
uint8_t reg_addr = 0x10;
uint8_t data = 0xAB;

I2C_Start();
I2C_Write(device_addr << 1 | 0);
I2C_Write(reg_addr);
I2C_Write(data);
I2C_Stop();

medium

A. Only register address is sent, data is missing

B. Data 0x10 is written to register 0xAB of device 0x50

C. Write operation fails due to wrong device address

D. Data 0xAB is written to register 0x10 of device 0x50

Solution

1. Step 1: Analyze the address and data sent

   The device address is correctly shifted and write bit set to 0. Then register address 0x10 and data 0xAB are sent in order. The sequence Start -> Address+Write -> Register -> Data -> Stop is correct for writing data to a register.

2. Final Answer:

   Data 0xAB is written to register 0x10 of device 0x50 -> Option D

3. Quick Check:

   Correct sequence = data written properly [OK]

Hint: Check address shift and write bit, then data order [OK]

Common Mistakes:

• Mixing register and data order
• Not shifting device address
• Forgetting to send stop condition

4. Identify the error in the following I2C write code snippet:

uint8_t device_addr = 0x68;
uint8_t reg_addr = 0x20;
uint8_t data = 0x55;

I2C_Start();
I2C_Write(device_addr >> 1 | 0);
I2C_Write(reg_addr);
I2C_Write(data);
I2C_Stop();

medium

A. Device address is shifted right instead of left

B. Data byte is sent before register address

C. Stop condition is missing

D. Write bit is set to 1 instead of 0

Solution

1. Step 1: Check device address shifting

   The device address is incorrectly shifted right instead of left, which corrupts the address sent on the bus. Register and data are sent in correct order, and stop condition is present, so no other errors.

2. Final Answer:

   Device address is shifted right instead of left -> Option A

3. Quick Check:

   Address must be shifted left for I2C write [OK]

Hint: Always shift device address left by 1 bit [OK]

Common Mistakes:

• Shifting address right instead of left
• Sending data before register address
• Omitting stop condition

5. You need to write multiple bytes to consecutive registers starting at 0x30 on an I2C device with address 0x60. Which code snippet correctly performs this operation in embedded C?

uint8_t device_addr = 0x60;
uint8_t start_reg = 0x30;
uint8_t data[] = {0x01, 0x02, 0x03};

// Which snippet is correct?

hard

A. I2C_Start();
I2C_Write(device_addr << 1 | 0);
for (int i = 0; i < 3; i++) {
I2C_Write(start_reg + i);
I2C_Write(data[i]);
}
I2C_Stop();

B. I2C_Start();
I2C_Write(device_addr << 1 | 0);
for (int i = 0; i < 3; i++) {
I2C_Write(data[i]);
}
I2C_Stop();

C. I2C_Start();
I2C_Write(device_addr << 1 | 0);
I2C_Write(start_reg);
for (int i = 0; i < 3; i++) {
I2C_Write(data[i]);
}
I2C_Stop();

D. I2C_Start();
I2C_Write(device_addr << 1 | 1);
I2C_Write(start_reg);
for (int i = 0; i < 3; i++) {
I2C_Write(data[i]);
}
I2C_Stop();

Solution

1. Step 1: Start communication and send device address with write bit

   The device address is shifted left and write bit set to 0, then start register address is sent once. After sending the start register, consecutive data bytes are sent to consecutive registers automatically by the device.

2. Final Answer:

   Start register sent once, then data bytes sent in loop -> Option C

3. Quick Check:

   Send start reg once, then data bytes [OK]

Hint: Send start register once, then all data bytes [OK]

Common Mistakes:

• Sending register address before each data byte
• Using read bit instead of write bit
• Not sending start register at all