Problem with Vout of Honeywell ASDX Analog pressure sensor

Question

Few months ago I decide to bought an absolute pressure measure system using a Honeywell ASDX differential analog pressure sensor (Model: ASDX ACX 030PAAA5). The datasheet of this type of sensor attached.

Base on datasheet, this sensor has 3 active pin, pin No.1 is for voltage supply, Pin No. 2 is Analog output and Pin No. 3 is GND. The manufacturer suggest to use a load resistace > 5K ohms (I used a 4.7K RL) and an external bypass capacitor is required across the supply voltage (Pins 1 and 3). I used 0.1 uF and 0.001 uF capacitors between Vsupply, Vout and GND respectively.

I design a circuit on a breadboard to mount the sensor. I use a 5Vdc power supply device for the circuit, now when I TURN ON the device and there is no applied pressure on sensor (just ambient pressure), the output voltage is almost 2.47 - 2.59 Volt (I checked Pin 1 voltage for the sensor an itś same 5Vdc). the problems are here:

Problems: 1. I expect when there is no applied pressure, the output voltage be 2.5V but as i mentioned above it is a little less, what is the reason?

This is the schematic.

I will be so appreciate if help me.

Answer 1

Why do you expect the voltage to be exactly equal to 2.5V? From the datasheet:

http://www.farnell.com/datasheets/1850579.pdf

Given Pmin = 0 (since this is an absolute pressure sensor), Pmax = 30psi (since you are using a 030PAAA5 model), Vsupply = 0.5V and applied pressure = 14.7psi (760 mmhg approx.), we get Vout = 2.46 volts. For a slightly higher pressure of 770mmhg = 14.9psi you will get 2.47 volts out and for 811mmhg = 15.7psi approx. you will get 2.59 volts output.

If you want to measure not the absolute pressure, but the difference between the atmospheric pressure and some other input, either get a second sensor and subtract the output from both sensors from one another or get a gage sensor (XXXMGAA5 in your case).

By the way, you did not include the datasheet.