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PCM1798 output DC offset

Brian Oppegaard
Prodigy 30 points
Other Parts Discussed in Thread: PCM1730, PCM1798

I am replacing a PCM1730 with PCM1798 as recommended by TI. The PCM1798 seems to have a DC offset current that the PCM1730 does not. Look at the PCM1798 data sheet reference circuit Fig 24. My pcb has only the I/V converters, the differential filter is on another pcb which I do not make. With the 1730 I could change R1 and R2 (the feedback resistors of the I/V converter) to any value I like and the output stays centered with regard to DC. However, with the values shown in Fig 24 of PCM1798 sheet I get about -1.0 VDC at the output of U1 and U2 that increases with larger R1 R2 values until the signal is slammed against the negative rail.   That is the problem. I have to able to adjust the gain of the I/V converters without compromising output swing.

  • 0
    TI__Mastermind 23685 points

    Brian,

    I took a look at the schematic you sent me off line. Your I/V stage is little different than what is recommended in the data sheet but should act similarly. I will have to confirm that.

    But I would expect the voltage out of U1 and U2 to increase with the value of the feedback resistors (R1 and R2). On page 18, the data sheet says that the voltage output level of the I/V convert will equal (4mA * Rf ).

  • 0 in reply to Michael Gambrill
    Prodigy 30 points

    Yes I am familiar with calculating the gain of an inverting amplifier. The problem is the DC Offset at the output of the PCM1798 it is amplifying.

    PCM1730 does not have DC offset

    PCM1798 has DC offset. Why and what to do about it.

  • 0 in reply to Brian Oppegaard
    Prodigy 30 points

    OK how about this. I can connect a 4K resistor from the PCM1798 output to -12V and offset the offset. Is there any down side to this? Since I don't know what is going on inside your chip, I sure would like a design engineer to chime in on this.

  • 0 in reply to Brian Oppegaard
    TI__Mastermind 23685 points

    I have asked two of my colleagues to look at your schematic and both have recommended following the measurement circuit in the data sheet more closely. This is the same circuit used on several of our EVMs. Could you try the I/V conversion section and see what your results look like?

    The op-amp you selected looks to be OK, but try using the same circuit as in figure 24.