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PCM4222: FSD Voltage for PCM4222 Clarification

shadders
Expert 1065 points
Part Number: PCM4222
Other Parts Discussed in Thread: OPA1632

Hi,
I am implementing the PCM4222 in a design and i have a question in respect to the maximum analogue input voltage on pins VinL+/VinL- and VinR+/VinR- for FSD.

The datasheet states on page 5 that the full scale voltage is 5.6v pk-pk differential input. This is repeated on page 17 too.

With VCOML/VCOMR at 1.95volts, and the OPA1632 used as the differential output opamp for the input circuit as per page 31, then the stated full scale input is stated to be 2.8v pk-pk as per page 5 and page 17, described as per pin.

Figure 37 on page 17 shows that the input voltage range is 2.8v pk-pk centred on a mean DC of 1.95volts. This translates to a minimum input voltage 0.55volts to a maximum input voltage 3.35volts to achieve 2.8volts pk-pk.

Therefore, how is it that a 5.6volts pk-pk is possible as a differential input voltage, if the Vcc analogue supply is +4volts ?.

This would mean that the analogue input voltage is 0.75volts greater than the Vcc supply, or 0.85volts lower than analogue ground.

As such, what is the correct differential voltage across the pins VinL+/VinL- and VinR+/VinR-, to achieve FSD ?

Thanks and regards,
Shadders.

  • 0
    TI__Guru 62475 points

    Hi Shadders,

    This is a common question and has to do with the fundamentals of differential versus single-ended signals.  With two 2.8Vpp signals that are 180 degrees out of phase and applied differentially when you subtract the two voltages at the peak where VIN+ is higher there is a differential voltage of +2.8V and on the opposite peak there will be -2.8V.  So, with the positive full-scale signal at 2.8V and the negative full-scale signal at -2.8V:  2.8Vpk - (-2.8Vpk) = 5.6Vpp.

  • 0 in reply to Collin Wells
    Expert 1065 points

    Hi Colin,

    Thanks for the reply. What is nagging me is that a +/-2.8volts pk cannot be applied to any of the input pins. A voltage of +/-1.4volts centred on 1.95volts can be.

    I simulated the circuit on page 31, Figure 50, and used an LTC6403 (as i did not have the OPA1632 as standard) and the difference voltages that would be applied is still 2.8volt pk-pk (centred on 1.95volts). To achieve a 5.6volts differential, then you would be driving the Vin pins higher than supply, and lower than ground.

    I suppose we all have different ways of viewing things, so if you can confirm that a +3.35volts on VinL+ and a 0.55volts on VinL- is the max differential which generates a positive FSD, and that a +3.35volts on VinL- and a 0.55volts on VinL+ is the max differential which generates a negative FSD, then it will answer my question.

    Thanks and regards,

    Shadders.

  • 0 in reply to shadders
    TI__Guru 62475 points

    Hello,

    I put together a quick simulation that matches with your expectations.  With a 1.95V DC offset, and a 1.4Vpk signal, each input signal swings from 0.55 to 3.35V, or 2.8Vpp.  Since both inputs are 180 degrees out-of-phase and swing 2.8Vpp,the total differential signal is double that at 5.6Vpp.

    /cfs-file/__key/communityserver-discussions-components-files/6/PCM4222_5F00_Inputs.TSC

  • 0 in reply to Collin Wells
    Expert 1065 points

    Hi Chris,

    Thanks for confirming and much appreciated. I now see the logic behind the differential statement and specification.

    Regards,

    Shadders.