AD5687R reaback

Hi NKokubo131 ,

Command 1001 is reserved for the readback function. This command, in association with selecting one of address bits, selects the register to read. If more than one or no bits are selected, DAC Channel A is read back by default thus the 0x900000 command stated on the datasheet.

Best Regards,

Den

Hi Den 

Thank you for your reply.

In case of  DAC B read back, What code should be written to AD5687R?

0x910000?

Best regards

N.Kokubo

Hi NKokubo131 

As the per datasheet the readback command can be invoked in association with selecting one of the address bits, DAC B or DAC A (note that only one DAC register can be selected during readback.) and looking at the address bits for the DAC's channels provided on the datasheet's table 8, DAC B's address bits are 1000 (8)

Best Regards,

Den

Hi Den 

Thank you for your reply.

According to table8, in case of  DAC B read back, 0x980000 should be written to AD5687R.

According to table8, DAC A Address is 0001(1).

Despite table8, in case of  DAC B read back,  should 0x900000 be written to AD5687R?

Best regards

N.Kokubo

Hi NKokubo131 ,

The readback instructions provided under the "Readback Operation" where 0x900000 will be used is just an example where you will perform a readback on Channel A as default. For DAC channel B readback you are correct to use 0x980000 (Command bits + Address bits).

Best Regards,

Den

Hi Den 

Thank you for your reply.

I understand you.

In case of  DAC A read back, AD5687R should be written 0x900000.

In case of  DAC B read back, AD5687R should be written 0x980000.

I close this question.

Best regards

N.Kokubo

Hi NKokubo131 ,

You're welcome, just a clarification, for DAC Channel A you may also directly use the address bits indicated on table 8 which is 0001 (If more than one or no bits are selected, DAC Channel A is read back by default).

Best Regards,

Den