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DAC7822: voltage at Iout1 and Iout2

Part Number: DAC7822

Hi,

Is it mandatory to set Iout1 and Iout2 to 0V from GND or it is possible to set them to different voltage (let it be V_virtual)?

With V_virtual not required to be 0, I understand it is possible to obtain: Vout - V_virtual = -(Vref - V_virtual) * CODE / 4095.

Thanks,

Gabriel

  • Howdy Gabriel,

    In typical applications IOUT2 is connected to analog ground and the positive terminal of the transimpedance stage, as displayed in Figure 38 of the datasheet.  The IOUT1 connection is held at virtual GND potential by the negative terminal of the transimpedance stage.

    Usually the maximum rating of pins is shown in the Absolute Maximum Ratings table of the datasheet, but unfortunately this information is not shown for IOUT1 and IOUT2.  I do believe they can go beyond '0V' as page 5 states that "IOUT2A should be tied to analog ground of the system, but can be biased to achieve single-supply operation".  Therefore, it should be possible to apply a non-zero value to this terminal, but will have to verify with design for the absolute maximum ratings of the pin.  I will update this post once I receive this information.

    Best Regards,

    Matt

  • Howdy Gabriel,

    The absolute maximum voltage of the IOUT1/2 pins is VDD+0.3V. However, this does not imply the part would function properly at this maximum level, only that the pin voltages could go up that high without incurring damage to the device. For reasonable single supply operation, and considering that the device is only 12-bit, design suggests that these pins (IOUT1/2) should be kept at a voltage with the following relation : (VDD - V_IOUT) >= 2V, where V_IOUT is the voltage applied to IOUT2 for single supply operation.

    Additionally, the transfer function you've include above has a small typo, which is corrected below.
    Vout - V_virtual = - (Vref - V_virtual) * CODE / 4096

    Best Regards,
    Matt
  • Thanks Matt.

    I'm assuming from your comments that V_IOUT cannot be negative for normal operation. Is this correct?

    Regards,
    Gabriel