I2C addressing (7-bit and 10-bit) in Embedded C

Introduction

I2C addressing lets devices on the same wire know who is talking. It uses 7-bit or 10-bit addresses to identify each device uniquely.

When connecting multiple sensors to a microcontroller using I2C.

When you need to communicate with a device that uses a 10-bit address.

When writing firmware to read or write data over I2C bus.

When debugging I2C communication to check device addresses.

Syntax

Embedded C

/* 7-bit address example */
#define DEVICE_ADDR_7BIT 0x3C

/* 10-bit address example */
#define DEVICE_ADDR_10BIT 0x1F3

/* Sending address on I2C bus */
// For 7-bit: address shifted left by 1, last bit is R/W
uint8_t address_byte = (DEVICE_ADDR_7BIT << 1) | read_write_bit;

// For 10-bit: send special header and two address bytes

7-bit addresses are most common and fit in one byte with a read/write bit.

10-bit addresses require sending extra bits and are less common.

Examples

This shows how to prepare the 7-bit address for sending with read or write bit.

Embedded C

/* 7-bit address example */
#define SENSOR_ADDR 0x50

// To send address with write bit (0)
uint8_t addr_write = (SENSOR_ADDR << 1) | 0;

// To send address with read bit (1)
uint8_t addr_read = (SENSOR_ADDR << 1) | 1;

Shows how to split a 10-bit address into two bytes for I2C communication.

Embedded C

/* 10-bit address example */
#define DEVICE_10BIT_ADDR 0x305

// First byte: 11110 + 2 MSBs of address + R/W bit
uint8_t first_byte = 0xF0 | ((DEVICE_10BIT_ADDR >> 7) & 0x06) | (read_write_bit & 0x01);

// Second byte: lower 8 bits of address
uint8_t second_byte = DEVICE_10BIT_ADDR & 0xFF;

Sample Program

This program shows how to prepare and print the I2C address bytes for both 7-bit and 10-bit addresses with read and write bits.

Embedded C

#include <stdio.h>
#include <stdint.h>

#define SENSOR_ADDR_7BIT 0x3C
#define DEVICE_ADDR_10BIT 0x1F3

void print_7bit_address(uint8_t addr, uint8_t rw) {
    uint8_t address_byte = (addr << 1) | (rw & 0x01);
    printf("7-bit address byte (with R/W=%d): 0x%02X\n", rw, address_byte);
}

void print_10bit_address(uint16_t addr, uint8_t rw) {
    uint8_t first_byte = 0xF0 | ((addr >> 7) & 0x06) | (rw & 0x01);
    uint8_t second_byte = addr & 0xFF;
    printf("10-bit address first byte (with R/W=%d): 0x%02X\n", rw, first_byte);
    printf("10-bit address second byte: 0x%02X\n", second_byte);
}

int main() {
    print_7bit_address(SENSOR_ADDR_7BIT, 0); // Write
    print_7bit_address(SENSOR_ADDR_7BIT, 1); // Read

    print_10bit_address(DEVICE_ADDR_10BIT, 0); // Write
    print_10bit_address(DEVICE_ADDR_10BIT, 1); // Read

    return 0;
}

Important Notes

Remember that the read/write bit is always the least significant bit in the address byte.

7-bit addresses are shifted left by 1 to make room for the read/write bit.

10-bit addressing is more complex and less common, so check your device datasheet carefully.

Summary

I2C uses 7-bit or 10-bit addresses to identify devices on the bus.

7-bit addresses are shifted left by 1 bit and combined with a read/write bit.

10-bit addresses require sending two bytes: a special header and the address split into two parts.

Practice

1.

What is the size of the standard I2C address used to identify devices?

easy

A. 7 bits

B. 8 bits

C. 10 bits

D. 16 bits

Solution

1. Step 1: Understand I2C addressing basics

   I2C commonly uses 7-bit addresses to identify devices on the bus.

2. Step 2: Recall standard address size

   The standard I2C address size is 7 bits, while 10-bit is an extended mode.

3. Final Answer:

   7 bits -> Option A

4. Quick Check:

   Standard I2C address = 7 bits [OK]

Hint: Remember standard I2C uses 7-bit addresses [OK]

Common Mistakes:

• Confusing 7-bit with 8-bit addresses
• Thinking 10-bit is the default
• Assuming 16-bit addresses exist in I2C

2.

Which of the following is the correct way to prepare a 7-bit I2C address for a write operation in embedded C?

uint8_t address = 0x3C; // 7-bit address
uint8_t write_address = ?;

easy

A. write_address = address << 2;

B. write_address = (address >> 1) | 1;

C. write_address = address | 0x80;

D. write_address = (address << 1) | 0;

Solution

1. Step 1: Shift 7-bit address left by 1

   The 7-bit address must be shifted left by 1 to make room for the read/write bit.

2. Step 2: Set read/write bit for write operation

   For write, the least significant bit is 0, so OR with 0.

3. Final Answer:

   write_address = (address << 1) | 0; -> Option D

4. Quick Check:

   Shift left 1 + write bit 0 = correct address [OK]

Hint: Shift left by 1, OR with 0 for write [OK]

Common Mistakes:

• Shifting right instead of left
• Setting read/write bit incorrectly
• Using wrong bitwise OR values

3.

Given the following code snippet, what is the value of header for a 10-bit I2C address 0x305?

uint16_t addr = 0x305;
uint8_t header = 0xF0 | ((addr >> 7) & 0x06) | 0x00; // write operation

medium

A. 0xF6

B. 0xF0

C. 0xF2

D. 0xF4

Solution

1. Step 1: Analyze addr = 0x305

   Binary 0b1100000101 (bits 9-0: bit9=1, bit8=1, bit7=0). Shift right 7: 0b110 (6).

2. Step 2: Mask and compute header

   (6 & 0x06) = 0x06. Then 0xF0 | 0x06 | 0x00 = 0xF6.

3. Final Answer:

   0xF6 -> Option A

4. Quick Check:

   0xF0 | ((0x305 >> 7) & 0x06) = 0xF6 [OK]

Hint: Shift address right 7, mask with 0x06, OR with 0xF0 [OK]

Common Mistakes:

• Not shifting address before masking
• Incorrect mask value
• Confusing read/write bit in header

4.

Identify the error in this code snippet for sending a 10-bit I2C address:

uint16_t addr = 0x3FF;
uint8_t header = 0xF0 | ((addr >> 7) & 0x06) | 0x01; // read operation
uint8_t low_addr = addr & 0xFF;
// send header and low_addr over I2C

medium

A. The read/write bit in header should be 0 for read

B. The low_addr should be shifted left by 1 before sending

C. The header calculation is correct; no error

D. The mask 0x06 should be 0x07 to include all bits

Solution

1. Step 1: Check read/write bit in header

   For read operation, the last bit is set to 1, which matches the code.

2. Step 2: Verify mask and low address

   Mask 0x06 correctly extracts bits 7 and 8; low_addr is the lower 8 bits, no shift needed.

3. Final Answer:

   The header calculation is correct; no error -> Option C

4. Quick Check:

   Header and low_addr correct for 10-bit read [OK]

Hint: Check read/write bit and mask carefully [OK]

Common Mistakes:

• Confusing read/write bit value
• Incorrect mask for bits 7 and 8
• Shifting low address unnecessarily

5.

You need to write a function in embedded C that takes a 10-bit I2C address and returns the two bytes to send: the header byte and the low address byte for a write operation. Which of the following implementations is correct?

void get_i2c_10bit_bytes(uint16_t addr, uint8_t *header, uint8_t *low_addr) {
    // ?
}

hard

A. header = 0xF0 | ((addr >> 7) & 0x06) | 0x00; low_addr = addr & 0xFF;

B. *header = 0xF0 | ((addr >> 7) & 0x06) | 0x00; *low_addr = addr & 0xFF;

C. *header = 0xF0 | ((addr << 7) & 0x06) | 0x00; *low_addr = addr >> 8;

D. header = 0xF0 | ((addr >> 7) & 0x07) | 0x01; low_addr = addr & 0xFF;

Solution

1. Step 1: Use pointers to assign output values

   The function parameters header and low_addr are pointers, so use * to assign values.

2. Step 2: Calculate header and low_addr correctly

   Header is 0xF0 ORed with bits 7 and 8 of addr shifted right 7, and OR 0 for write; low_addr is lower 8 bits.

3. Final Answer:

   *header = 0xF0 | ((addr >> 7) & 0x06) | 0x00; *low_addr = addr & 0xFF; -> Option B

4. Quick Check:

   Pointer dereference + correct bit ops = correct function [OK]

Hint: Use * to assign pointer outputs, shift right 7 bits [OK]

Common Mistakes:

• Not using * to assign pointer values
• Shifting bits incorrectly
• Setting read/write bit wrong