Hello to every one, I hope you doing well.
I need some help about AD1938 TDM input.At page 15 of AD1938 we have TDM signal diagram. According to datasheet we have to set DBCLK to 256xfs that if we want to have 48Khz audio we have to set DBCLK to 12.28Mhz. Now I have a data set for a audio signal and break them to every 4mS. We have 8 DAC and each of them is 24 bit and sampling frequency is 48Khz so for each 1mS we have to send 8*24*48=9216 bit data or 1152 byte data. So for 4mS we have 1152 Byte x 4 = 4608 byte data. Now according to DBCLK frequency calculation that mentioned in datasheet, we have send this data with 12.28Mhz frequency that its so faster than 4mS.Now my question is that what will happen for DAC output at part of TDM signal that DSDATA is zero. What will be the output value of the analog signal when TDM signal is zero? As shown in the picture, what will be the output of the analog signal during period of circle A?As you see yellow circle is near to 36000 bit of data that send out with frequency of 12.28Mhz.The total duration of the two circles is approximately 4 milliseconds.Now we have a zero state of DSDATA with duration of circle A. Is it correct way for sending data of each 4mS of a signal with TDM?Is there any buffer for DACs?If yes how can we know its size?Should we set a delay other than one bit offset between sending each TDM frame?How is this difference between data transmission (1 millisecond) and the audio volume that should be played (4 milliseconds) justified? Only in one case is it possible to have a buffer inside the converter and receive a certain amount of data through TDM and then convert it with a frequency of 48 kHz.The reason of 4mS is that I receive audio signal from another protocol and that send me data of signal each 4ms. the picture is my TDM output.
thanks for your helps.
I added some questions
[edited by: ELEC23 at 3:56 PM (GMT -4) on 20 May 2024]