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INA118: Input pull down resistors

GREG SHIPLEY
Intellectual 290 points

Part Number: INA118

I am using an INA118 in a simple transducer application.  The transducer is a Wheatstone bridge type sensor running at 10V.  The INA118 is running single supply mode at 12V and the gain is set to 50 with a 1.02K resistor.

I have installed 50K 1% resistors from each differential input of the INA118 to ground to keep the inputs from floating when the transducer is not connected.  When I connect the transducer, the inputs both go to 4.46V.  Am I saturating the inputs?  If I remove the 50K resistors, the problem goes away.

Is there a better way to pull the inputs weak to ground?  Better to put a large resistor between the inputs and not to ground?

  • 0
    TI__Guru 73891 points

    Greg,

    Connecting 50k resistors to ground on 12V single supply violates the linear input common-mode voltage range (V-)+1.1V < Vcm < (V+)-1V  - see below.  A better solution would be to set the Vcm at mid-supply by connecting two 50k resistors from each differential input to each supply (ground and 12V so the Vcm=6V).

    Using Vref=0 allows only positive input differential voltage from 4mV to 192mV to be measured - see below. 

    Thus, if you need to be able to measure both polarity of Vin_diff, you must use Vref = 6V - see below.

  • 0
    Intellectual 290 points

    I understand the violation now.  By doing the 50K to ground, I was trying to pull the signals near each other when nothing was connected to keep them from floating, Can this be accomplished by putting a resistor between the two inputs say a 100K?

  • +1 in reply to GREG SHIPLEY
    TI__Guru 73891 points

    Adding 100k resistor between the inputs does NOT prevent them both from floating to one of the rails - assuming the bridge resistance driving the inputs is low (~1kohm) you must add the voltage dividers using 100k resistors as shown below so the inputs Vcm ~5V when the Wheatstone bridge type transducer is not connected.