How to power pressure sensor MPX2050 correctly with input voltage?

I want to use the pressure sensor MPX2050 (data sheet: https://docs.rs-online.com/65ac/0900766b814a74eb.pdf)

These are some values from the data sheet:
Supply Voltage: typ 10 V, max 16 V
Supply Current = 6 mA
Input Impedance: min 1000 Ω , max 2500 Ω
Output Impedance: min 1400 Ω , max 3000 Ω

I have a few questions about how to feed this sensor correctly with input voltage:

  1. Can I directly connect a battery (9V) (or a battery pack 12V) to the pressure sensor MPX2050 (Pin 1 GND, Pin 3: Vs)?

  2. Does "supply current = 6 mA" mean that it is kept constant by the pressure sensor circuit, or is it necessary to limit the correct by an external resistor? (e.g. 2k Ω if Vs = 12 V) 2b) Can I damage the battery if I do not limit the current with an external (of the pressure sensor) resistor?

  3. As the level of the output voltage depends on the applied pressure and the input voltage, should I then keep the input voltage constant? I.e. should I use a voltage regulator? (e.g. LM217L or LM317L)


1 Answer 1

  1. Yes you can directly connect a battery to the pressures sensor. The voltage of the battery must be less than the max input voltage of the pressure sensor. The max voltage of the pressure sensor is 16V so a 9V or 12V battery will work.

  2. The supply current of 6mA is the typical current consumption of the pressure sensor when operated with a input voltage of 10V. A higher input voltage will draw more current and a lower input voltage will draw less current. You do not need to limit the current with an external resistor for the pressure sensor if you provide the correct input voltage.

  3. Regulating the supply voltage is a good idea. The output voltage is ratiometric with the supply voltage, meaning the output voltage span will change depending on the input voltage. For example, 10V supply voltage will have a 40mV output at 50kPa and a 5V supply voltage will have a 20mV output at 50kPa. As the battery discharges, the supply voltage will change which will cause the output voltage to change and can interfere with your measurements.


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