SBOS925A December 2020 – January 2021 OPA391

PRODUCTION DATA

- 1 Features
- 2 Applications
- 3 Description
- 4 Revision History
- 5 Pin Configuration and Functions
- 6 Specifications
- 7 Detailed Description
- 8 Application and Implementation
- 9 Power Supply Recommendations
- 10Layout
- 11Device and Documentation Support
- 12Mechanical, Packaging, and Orderable Information

- DCK|5

- DCK|5

First, determine the V_{REF} voltage. This
voltage is a compromise between maximum headroom and resolution, as well as
allowance for the minimum swing on the CE terminal because the CE terminal generally
goes negative in relation to the RE potential as the concentration (sensor current)
increases. Bench measurements found the difference between CE and RE to be 180 mV at
300 ppm for this particular sensor. To allow for negative CE swing, footroom, and
voltage drop across the 10-kΩ resistor, 300 mV is chosen for V_{REF}.

Equation 2. V_{ZERO} = V_{REF} = 300
mV

where

- V
_{REF}is the reference voltage (300 mV). - V
_{ZERO}is the concentration voltage (300 mV).

Next, calculate the maximum sensor current at highest expected concentration:

Equation 3. I_{SENSMAX} = I_{PERPPM}
* ppmMAX = 69 nA * 300 ppm = 20.7 µA

where

- I
_{SENSMAX}is the maximum expected sensor current. - I
_{PERPPM}is the manufacturer specified sensor current in amperes per ppm. - ppmMAX is the maximum required ppm reading.

Then, find the available output swing range greater than the reference voltage available for the measurement:

Equation 4. V_{SWING} = V_{OUTMAX} –
V_{ZERO} = 2.5 V – 0.3 V = 2.2 V

where

- V
_{SWING}is the expected change in output voltage. - V
_{OUTMAX}is the maximum amplifier output swing.

Finally, calculate the transimpedance
resistor (R_{F}) value using the maximum swing and the maximum sensor
current:

Equation 5. R_{F} = V_{SWING} /
I_{SENSMAX} = 2.2 V / 20.7 µA = 106.28 kΩ (use 110 kΩ for a common
value)