This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

LM9044: Differential input and Common mode input range

Part Number: LM9044

Hi,

As our customer indicated that sometimes O2 sensor gives output less than 0 volts in case if gets contaminated.

And Vdiff is always mentioned as 0 to 1V in datasheet. if Vdiff is less than 0V then the output is always zero as LM9044 is not powered with dual power supply. Please confirm.

 .

The common mode range mentioned for LM9044 is -1V to 1V.

Is below interpretation of common mode voltage is correct?

Case1:

Vin(+) = 1V

Vin(-) = 0V

V.cm = [Vin(+) - Vin(-)]/2 = (1-0)/2 = 0.5

V.diff = Vin(+) - Vin(-) = 1-(0) = 1

Case2:

Vin(+) = 0V

Vin(-) = 1V

V.cm = [Vin(+) - Vin(-)]/2 = (0-1)/2 = -0.5

V.diff = Vin(+) - Vin(-) = 0-1 = -1

Regards,

Sunney

  • Hello Sunney,

    Regarding your LM9044 questions:

    And Vdiff is always mentioned as 0 to 1V in datasheet. if Vdiff is less than 0V then the output is always zero as LM9044 is not powered with dual power supply. Please confirm.

    • Yes, the Electrical Characteristics specified VDIFF is 0 ≤ VDIFF ≤ 1V

    Case1:

    Vin(+) = 1V

    Vin(-) = 0V

    V.cm = [Vin(+) - Vin(-)]/2 = (1-0)/2 = 0.5

    V.diff = Vin(+) - Vin(-) = 1-(0) = 1

    • Yes, that is correct

    Case2:

    Vin(+) = 0V

    Vin(-) = 1V

    V.cm = [Vin(+) - Vin(-)]/2 = (0-1)/2 = -0.5

    V.diff = Vin(+) - Vin(-) = 0-1 = -1

    • Yes, that is correct. Even though the LM9044 has a common-mode range of −3V ≤ VCM ≤ +1V, a +1 V voltage applied to the inverting input would try to drive the output to a negative voltage which can't be done with a single positive supply device. The 0 ≤ VDIFF ≤ 1V range assures that won't happen.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • Thanks Thomas for clarifications.