I2C addressing (7-bit and 10-bit) in Embedded C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the difference between 7-bit and 10-bit I2C addressing?

7-bit addressing uses 7 bits to identify a device on the I2C bus, allowing up to 128 addresses. 10-bit addressing uses 10 bits, allowing more devices (up to 1024 addresses) on the bus.

beginner

How is a 7-bit I2C address transmitted on the bus?

The 7-bit address is sent in the first 7 bits of the address byte, followed by a single bit indicating read (1) or write (0).

intermediate

How does 10-bit addressing start on the I2C bus?

It starts with a special 7-bit code '11110XX' where XX are the two most significant bits of the 10-bit address, followed by a read/write bit, then the remaining 8 bits of the address.

intermediate

Why would you use 10-bit addressing instead of 7-bit?

Use 10-bit addressing when you need to connect more than 127 devices on the same I2C bus, as 7-bit addressing limits the number of unique addresses.

beginner

In embedded C, how do you typically specify a 7-bit I2C address?

You specify the 7-bit address shifted left by 1 bit, leaving the least significant bit for read/write control.

How many unique device addresses can 7-bit I2C addressing support?

A 64

B 128

C 1024

D 256

7-bit addressing supports 2^7 = 128 unique addresses.

What is the first part of the 10-bit I2C address frame?

A A special 7-bit code starting with 11110

B The full 10-bit address sent at once

C Only the lower 8 bits of the address

D A 7-bit address with read/write bit

10-bit addressing starts with a special 7-bit code '11110XX' to indicate extended addressing.

In 7-bit addressing, what does the least significant bit of the address byte represent?

A Read or write operation

B Device type

C Bus speed

D Error checking

The least significant bit indicates if the operation is a read (1) or write (0).

Why might you shift a 7-bit I2C address left by 1 in embedded C code?

A To match 10-bit addressing

B To encrypt the address

C To increase the address range

D To make room for the read/write bit

Shifting left by 1 leaves the least significant bit free for the read/write flag.

Which addressing mode allows more devices on the I2C bus?

A Both allow the same number

B 7-bit addressing

C 10-bit addressing

D Neither supports multiple devices

10-bit addressing supports up to 1024 devices, more than 7-bit's 128.

Explain how 7-bit and 10-bit I2C addressing differ in structure and usage.

Think about how many devices each can support and how the address is sent on the bus.

Describe how you would represent a 7-bit I2C address in embedded C code and why.

Remember the read/write bit is part of the address byte sent on the bus.

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