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stm32i2chalstm32cubeide
Mohsin Naik
Mohsin Naik

Reputation: 1

How to write the device address when using HAL_I2C_Mem_Write and HAL_I2C_Mem_Read functions?

I want to use the HAL_I2C_Mem_Read() function to read a control register of ZSC31050 and then configure the control register by using the HAL_I2C_Mem_Write() function.

The default device address of ZSC31050 is 78hex.

It is supposed to be shifted left by 1 bit.

So while using the write function is it supposed to be f0 and while using read function as f1?

Upvotes: 0

Views: 509

Answers (3)

code-lemur
code-lemur

Reputation: 7

You need to shift the address by 1 bit.

The HAL library then sets the read/write bit by itself.

Here a shot example how I used it on a STM32F0G71:

uint16_t i2CAddress{0x44 << 1};
HAL_StatusTypeDef result = HAL_I2C_Mem_Read(handle_, i2CAddress, cmd, sizeof(cmd), dataBuffer, sizeof(dataBuffer), 10);

And for only writing:

HAL_StatusTypeDef result = HAL_I2C_Master_Transmit(handle_, i2CAddress, txData, sizeof(txData), 10);

Upvotes: 0

B.Adlane
B.Adlane

Reputation: 41

You're right on your understanding of I2C addresses, but not right about STM32 I2C functions.

You do not need to shift the address manually. HAL_I2C_Mem_Write() and HAL_I2C_Mem_Read() already do that for you.

See I2C HAL module driver for STM32F4

Upvotes: 0

pmacfarlane
pmacfarlane

Reputation: 4317

Yes, that is basically correct. For I2C using 7-bit addresses, the first 7 bits of the first byte are the address, and the least-significant bit is the read/write bit.

So as you said, you need to shift the address left by one, and then set the least significant bit for a read, and leave it clear for a write.

See e.g. this good description of I2C from Texas Instruments. (Particularly section 3.)

Upvotes: 0

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