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ADS1015: Will this circuit work to drive the ADS1015 without issue?

Darren (FAE)
Expert 5483 points
Part Number: ADS1015
Other Parts Discussed in Thread: OPA314

I have a question on the stability of a design using the OPA314 and ADS1015.
There will be a differential signal to the OPA314 inputs, and the output will go to ADS1015.

Schematic

Do you see any concerns with this circuit?

  • 0
    TI__Mastermind 31655 points

     Hi Darren,

    I am more familiar with the ADS1015 in particular. The input impedance of the ADS1015 will vary based on the FSR that you select, but it could be between 3 and 100 Mohm for a single-ended measurement. The stability of the circuit will fall more on the capacitive load to the opamp. There is quite a large capacitive load in this circuit, so I am not entirely sure how the opamp will behave. I recommend taking this question to the opamp team for a better response regarding the stability of the amplifier circuit.


    Regards,

    Scott

  • 0 in reply to Scott Cummins
    TI__Expert 5483 points

    Hi Scott, 

    I asked the question about OPA314 here. 

    Could you help me with the following then?
    The concern I have is with kickback, and how the 1kΩ/1uF will work as a charge-bucket filter.

    1) What is the Csh value for the sampling caps in the ADS1015? How many pF?

    2) How long is the acquisition phase? (How long is the switch on the input connected to AINx and sampling cap for?)
    This will determine both how much charge is pulled from 1uF, and what the equilibrium voltage is, and then given the time required for OPA314 to charge the 1uF back up via the 1kΩ resistor, if this meets the 1/2LSB requirement for the acquisition time period.

    I know the conversion period depends on the DR, but what about the acquisition phase?

    3) Would you say that with the 1kΩ and 1uF, the kickback from ADC sampling is negligible? (For one, I understand if 1uF is substantially larger than the sampling capacitor, only a tiny voltage drop occurs, as the 1uF needs to transfer a lot more charge for the voltage to drop considerably...)

    Darren