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Original question:

LMP91200: ORP SENSING USING LMP91200 IC

LMP91200: maximum input voltage

Krzysztof Szmurlo
Prodigy 80 points
Part Number: LMP91200

Hi

I see this post:
https://e2e.ti.com/support/amplifiers/f/amplifiers-forum/657381/lmp91200-orp-sensing-using-lmp91200-ic

Which kind of suggests the input voltage is ±500mV around VCM. But this is for device switched off only isn't it?

So ±1.2V in reference to VCM should be ok?

Vs=3.3V VCM = 1/2*Vs

I'm trying to have bias current <1pA

Regards

Krzysztof

  • 0
    TI__Expert 8840 points

    Krzysztof,

      The first thing to remember is that this is a transimpedance amplifier. That means that the current developed in the probe is converted in the LMP91200 to a voltage.

    The part has a nominal VDD of  1.8V to 5.5V. You can apply the VDD directly to the VREF pin. If operating at 5.0Vdc, the VCM will be half of VDD or 2.5Vdc. This means that the IMP pin will be operating at 2.5V center +/- the voltage delta from the probe current. The –500 mV < (INP – VCM) < 500 mV, 85°C, VS = 0 V. –6.5 to 6.5 pA, means that as current flows in the probe, the IMP to VCM voltage delta will change as the current changes. The maximum current that the probe can vary will push the voltage +/- 500mV from the IMP voltage. 

      The actual current will depend on the probe current, and the rest of the system setting you choose. (VDD, VREF, Gain, ADC loading). 

      The Vs=3.3V VCM = 1/2*Vs is just your zero (7ph). (+500mV = 14ph and -500mV = 0ph. Theoretically speaking) It depends on the probe and gain and other factors. the +/-500mv is just the maximum current it can convert. adjust the gain to match your probe to get the dynamic range you need for your measurement.

  • 0 in reply to Gordon Varney
    Prodigy 80 points

    Thanks for reply Gordon. I still need some clarification, please:

    1. Does IMP means INP input pin 5?

    2. "will push the voltage +/- 500mV from the IMP voltage" - did you mean "from the VCM voltage" ?

    3. If you could confirm:
    As voltage will be pushed ±500mV and device will be off (Vs=0) this will not affect measurement accuracy cause it's just simply off so we are not measuring. So, when VDD=3.3V GND=0V Vs=VDD-GND=3.3V Vref=3.3V VCM=1.65V I should be fine measuring ±1.2V referenced to VCM with input bias current less than 100fA @25°C ?

    ps. I'm designing 4-20mA loop powered, insulated, pH meter. 

  • 0 in reply to Krzysztof Szmurlo
    TI__Expert 8840 points

    My apologies, Yes the INP pin 5. The VCM supplies the probe and the INP is the input to the PH Buffer. The delta represent the current in the probe. VOUT is the voltage that represents the current in the probe. VOUT is measured with your ADC.

    Yes to (3). You will be good to go.