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INA333 Instrumentation Amplifier Current on Reference Pin

Andrew Gibson
Intellectual 640 points
Other Parts Discussed in Thread: INA333

Hi - I understand that the source impedance driving the reference input on the INA333 must be less than 15 Ohms. 

How much current is that pin likely to sink or source?  From the equivalent circuit in the data sheet it seems as if it should be less than 5 V/300 KΩ i.e. 17 uA.  Is this the right order of magnitude or are there some surprises?


Thank you in advance for your thoughts.

  • 0
    TI__Guru 68541 points

    Andrew,

    Your calculation of maximum ~17uA current on REF pin is correct for G=1 but this is not the main reason for keeping the source impedance driving the REF pin below 15ohm.  Since any added source resistance at REF pin will cause a mismatch with other three 150kohm resistors around A3 (see below), resulting in Gain and CMRR errors, it is critical to keep this source impedance below 15ohm in order to limit the maximum resistor mismatch to 0.01% (15ohm/150k*100%) and thus assure max Gain error of 0.1% and minimum CMRR of 80dB as specified in the PDS.

     

     

  • 0 in reply to Marek Lis
    Intellectual 640 points

    Hi Marek,

    Thank you.  I am looking at circuit options other than an op amp to drive the REF pin.  That is why I would like to know how much current my circuit must source or sink while providing less than 15 Ohms to the pin.  Please verify my reasoning:

      - For a single ended 5.5 Volt supply circuit using the INA333 the maximum current out of the REF pin will be 5.5 V/150 kΩ = 37 uA.  At 0 Volts output...0 uA.

      - For a bipolar supply at +/- 2.75 Volts the maximum current into or out of the REF pin should be 2.75 V/150 KΩ = 18 uA.

    If not correct, please let me know the actual numbers.  Thank you.

  • 0 in reply to Andrew Gibson
    Intellectual 640 points

    (Please also confirm my reasoning if it does seem accurate - I don't want to assume that silence means agreement and later find out that silence meant vacation)

  • 0 in reply to Andrew Gibson
    TI__Guru 68541 points

    Andrew,

    The maximum current on REF pin does NOT depend on single vs dual supply application but rather on where do you connect the reference pin.  If you connect the REF pin to a negative rail (GND in single supply, or (V-) in dual supply), the maximum sink current will occur when the output of op amp A2 (see below) is at the positive rail (+ 5.5V in a single supply, or +2.75V in a dual supply configuration); thus the maximum voltage across the two 150k internal resistors will be 5.5V resulting in max sink current of 5.5V/300k=18.3uA (REFpin current will always flow thru both resistors).  On the other hand, if you connect the REF pin to mid-supply (GND in dual supply or +2.75V in single supply, with Vs=+5.5V), the maximum current on REF pin will be cut in half - with the output of op amp A2 at  (V+), the REF pin current of 2.75V/300k=~9uA must be sunk by REF pin while in the case of A2 output being at (V-) around 9uA current need to be sourced by REF pin.  All in all, the best solution to minimize the error related to REF pin series resistance is to connect it directly to either GND (in dual supply) or a negative supply.