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Vocm appears at input of OPA1632

Niek ten Brinke
Prodigy 170 points
Other Parts Discussed in Thread: PCM4220, OPA1632, NE5534, OPA227, PCM4222

I have designed an ADC Circuit with the PCM4220 buffered by the OPA1632 but strangely enough, when I supply a Vcom at the OPA1632 this voltage also appears at the + and - input on the XLR connector causing plops when connecting a device to the input.

I followed the example designs in datasheet of the PCM4220 and the OPA1632 closely, the only differences are the decoupling that I used and that the Vcom is buffered by a NE5534 instead of an OPA227.

Can someone help me in figuring out how to solve this problem?

Regards,

Niek ten Brinke

  • 0
    Prodigy 170 points
    Of course this probably has to do with the feedback loop around it but I am suprised that this is by design and that no solution/workaround was suggested in the datasheets.

    I can solve it by puting a DC-blocking capacitor between the xlr-connector and the OPA1632 but this is a less than ideal solution in a high-end audio system.
  • 0 in reply to Niek ten Brinke
    TI__Genius 12631 points

    Hello,

     I assume you are using the circuit shown in the PCM4222 EVM user guide (Figure 4) for this?

    If the signal driving the XLR cables is DC coupled then that driven signal will have its own common-mode voltage which will not be dictated by the voltage on the  Vocm pin. Can you please describe the signal that is being driven at the input XLR cables (AC and DC).

    The way a fully differential amplifier (FDA) like the OPA1632 works is that the voltage at the Vocm pin forces the common-mode output of the amplifier at pins 4 and 5 to be equal to Vocm. Lets assume that is 1V.

    Now assume that the common-mode voltage on the XLR cables (dc coupled scenario) is say 2V. In that case the input common mode voltage (Vicm) at pins 1 and 8 of the OPA1632 can be determined by a simple voltage divider through the 560ohm and 270ohm input and feedback resistors. The calculation will look like:

    (2V - Vicm)/560 ohm = (Vicm - 1V)/270

    solving the above equation gives Vicm as 1.326V.This is how an FDA is supposed to function.

    Now if the signal being driven by the XLR cable is an AC coupled signal then there will be no dc voltage drop across the feedback network so you will see the Vocm voltage (1V) at pins 1 and 8 and also at the XLR cable outputs.

    Hope this makes sense.

    -Samir

  • 0 in reply to Samir Cherian
    Prodigy 170 points

    Yes, that's indeed the schematic I use.

    The signal that comes in from the XLR-connectors comes from a pre-amp where the DC is servo-controlled (not sure if that's the correct english word) and kept at zero volts. In other words, there are no DC-blocking capacitors involved in the signal path.

    It might be good to mention that I used an single ended connector like shown here: (since the pre-amp involved has an RCA output)

     

    So following your explanation it is apparently correct that I get the Vocm voltage at the OPA's input pins. Is there a way to work around this (without having to introduce a DC-blocking cap)?

  • 0 in reply to Niek ten Brinke
    TI__Genius 12631 points
    If the signal coming in is at 0V, and the voltage at Vocm = x volts then the voltage at the OPA1632 input pins (pins 1 and 8) will be x*560/(270+560) = 0.67x and not the same as Vocm.
  • 0 in reply to Samir Cherian
    Prodigy 170 points

    Excuse me, it was the same when the XLR cable was disconnected from the pre-amp. And I measured from the XLR connector, (so before the 560 ohm resistor) not directly at the IC pins.
    The pre-amp wasn't designed to be able to compensate such a high DC offset so the input signal at the XLR is not exactly at 0V anymore after connecting the XLR cable. ((iirc a few hunderd millivolts)

  • 0 in reply to Niek ten Brinke
    TI__Genius 12631 points
    Something else is going on here.... If the previous stage is driving a common mode of 0V then you should see exactly 0v on the left hand side of the 560 ohm resistor. Is it possible that the previous stage is unable to drive the 560 ohm load?
  • 0 in reply to Samir Cherian
    Prodigy 170 points
    I expect it to be able to drive such load but to be absolutely sure I need to look that up with the engineer that designed the pre-amp (which unfortunately won't happen before tomorrow, european time)
  • 0 in reply to Samir Cherian
    Prodigy 170 points

    I talked about de problem with the other engineer today and he didn't understand what the DC that's fed into the output of the pre-amp has to do with the ability to drive a 560 ohm load. Could you elaborate a bit more?

  • 0 in reply to Niek ten Brinke
    TI__Genius 12631 points
    Hello Niek,
    I will assume that the previous stage is also an amplifier stage. In order for an amplifier to hold its output voltage it should be capable of sourcing or sinking the necessary load current at the given frequency. If it cannot do so (source/sink the Iload) then its output voltage will change to point where it can deliver the necessary Iload.

    The reason it changes its output voltage is because it is losing its open loop gain (under heavy loads) and thus no longer follows the ideal amplifier equations. All amplifiers have such an open-loop gain vs load current relationship. Again I am not sure if this is the issue in your circuit, however it is something to be aware of.

    One thing you can do is to simulate just the OPA1632 circuit in TINA Spice. Both TINA and the OPA1632 macromodel are available for download from the TI portal.