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I want to read this sensor(MPX4115AP) data accurately but I am not sure about using the given transfer function and the datasheet information good enough. The device is ratio-metric and temperature dependent.

Nominal transfer function is given as:

enter image description here

Above the sensor supply voltage Vcc is 5.1V but in my case it is less than 5V and it might vary so I also monitor it.

I have wired the the sensor as follows and reading the output via a micro-controller:

enter image description here

The micro-controller also reads the ambient temperature T and supply the voltage Vcc simultaneously.

So I have the ambient temperature data T, and the sensor's supply voltage data Vcc as well as the equation given in the datasheet. By using these, I need to write a single formula in C which can output the pressure data. I am afraid to make a mistake and measure the pressure wrong.

I am a bit lost in the datasheet to obtain a correct accurate equation. How can we obtain the transfer function considering the ambient temperature T and the sensor's supply voltage Vcc?

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  • \$\begingroup\$ You're not going to be far off with \$V_{out} = Vs(0.009\cdot P -0.095)\$. After all you don't know the error terms that apply - these need to be calibrated out. The data sheet contains maximum values for them. \$\endgroup\$
    – Andy aka
    Dec 3, 2020 at 16:08
  • \$\begingroup\$ Couple of things maybe I am repeating myself. In my case the Vs is not 5.1V. That formula should be corrected considering the device is radiometric. So if my Vs= 4.95V what will be the formula? And I cannot do anything about the temperature compensation? \$\endgroup\$
    – cm64
    Dec 3, 2020 at 16:17
  • \$\begingroup\$ You need to measure Vs. the result is ratiometric hence you need that value unless of course the ADC you use has Vs as its reference voltage. \$\endgroup\$
    – Andy aka
    Dec 3, 2020 at 16:22
  • \$\begingroup\$ In my case ADC reference voltage is independent of the sensor excitation voltage Vs, So if Vs is 4.95V, will the transfer function be Vout=4.95x(0.009⋅P−0.095) or Vout=(5.1/4.95)x(0.009⋅P−0.095)? \$\endgroup\$
    – cm64
    Dec 4, 2020 at 0:59

1 Answer 1

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The datasheet transfer function is valid for Vs=4.85V to 5.35V min ( ≈ ±5%). What is your expected variance over all environmental changes ?

If outside this, then the specs no longer are guaranteed, but it can be re-calibrated.

It is ratiometric with the 5V supply , i.e. proportional and linear but only within the +/-0.25V range of 5.1V

Vo = Vs * k where k= P*0.009 - 0.0095

If you have a higher voltage available then a 5.1V LDO could prevent changes to Vs for this 10mA max sensor and the verify environmental and component tolerances to your required and SPECIFIED design tolerances for accuracy which are TBD in your question. (hint) Since the output is not rail to rail, this gives a little more range near 5V if your ADC is 0 to 5V. That's my guess why they chose 5.1V using a BJT Op Amp inside.

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  • \$\begingroup\$ In my case ADC reference voltage is independent of the sensor excitation voltage Vs, So if Vs is 4.95V, will the transfer function be Vout=4.95x(0.009⋅P−0.095) or Vout=(5.1/4.95)x(0.009⋅P−0.095)? \$\endgroup\$
    – cm64
    Dec 4, 2020 at 0:59
  • \$\begingroup\$ It will be Vs=4.95 as in the formula \$\endgroup\$
    – Tony Stewart EE75
    Dec 4, 2020 at 16:21
  • \$\begingroup\$ Got it thanks a lot! \$\endgroup\$
    – cm64
    Dec 5, 2020 at 18:48

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