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Minimizing power consumption while attenuating signal and driving an ADC

Karen
Intellectual 940 points
Other Parts Discussed in Thread: THS4131, ADS1271, THS4521

I am currently using the THS4131 as a driver for the ADS1271 much like figure 70 in the datasheet.  However I now want to reduce power consumption and was looking at the THS4521, but the THS4131operates from a ±15V supply, the input signal is ±10V while the THS4521 operates from a single +5V supply.  Essentially I need to attenuate my signal and drive the ADC while reducing power.   Is it feasible to use the THS4521 in place of the THS4131 with the ±10V signal?  If so, are there other component changes I need to make? 

  • 0
    TI__Expert 4115 points

    Karen,

    Yes, this is a feasible design.  The THS421 operated from a single 5V power supply and with < 1/10th the quiescent current can be used to convert the ±10V input to the appropriate voltage levels for the ADS1271.  The circuit shown in  figure 70 of the datasheet ADS1271 will work, however I would suggest increasing the resistor values so the feedback resistors (RF) are 1k ohm and the input resistors (RG) are 4.02k ohm. For best phase margin, place a 2.67k ohm resistor (RT) directly accross the input pins of the THS4521 (pins 1 & 8).  For further information on this circuit see TI Analog Applications Journal article Using fully differential op amps as attenuators, Part 1: Differential bipolar input signals .

    Another design consideration for this circuit is the physical size of the input resistor can impact performance.  I worked on a  circuit with similar attenuator configuration to take a 32Vpp differential input and reduce the signal level to 5Vpp differential for input to an ADC. Originally we used 0402 size resistors and saw bad THD and noise performance. I found that increasing the physical size of the input resistor to 1206 size fixed the problem (I used resistors made by Rohm , but I think other brands will also work). My theory is there is a non-linearity due to the voltage coefficient of the input resistor or there is a non-linearity due to heating of the input resistor. 

     

    Respectfully,

    Jim Karki