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Related

Problem with Honeywell ASDX Analog pressure sensor

Former Member
Former Member over 9 years ago

Dear friends

Few months ago I decide to built a multi channel pressure box, so I bought 16 Honeywell ASDX differential analog pressure sensor (Model: ASDX RRX005NDAA5). 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.

image

I design a PCB board to mount all 16 sensors on a single board. I use a 5Vdc power supply device for PCB board, now when I TURN ON the device and there is no applied pressure on sensors (just ambient pressure), the output voltages are almost 2.47 - 2.49 Volt (I checked Pin 1 voltage for each sensors and all of them is 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?  

2. when I applied an specific pressure (for example 300Pa) most sensors shows different voltage in their output and so it is impossible to calculate the Pressures. the datasheet said i can calculate pressure by below Eq.

 

image

 

3. by the changing of the pressure, change in output voltage of sensors are not equal. for example if I increase the Pressure 500Pa, the voltage change for one sensor is almost 1.5V but the voltage change for another sensor is 0.7V!!

I will be so appreciate if help me

Attachments:
imagehoneywell-sensing-asdx-series-analog-pressure-sensors-product sheet-008090-12-EN.pdf
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  • kulky64
    kulky64 over 9 years ago in reply to dougw +1
    I'm sure the correct interpretation is that the load needs to be between 5k and infinity. If the opposite were correct, then direct short from output to ground would be OK according to datasheet, which…
  • jw0752
    0 jw0752 over 9 years ago

    Hi Ashkan,

     

    I noticed in the data sheet that the sensor is fairly linear between 10 inches of H2O as a minimum up to 100 psi as a maximum. I have done the conversions and the minimum of 10 inches of H2O is actually 2,500 Pa while the 100 psi corresponds to 689,000 Pa. Your 500 Pa is below the minimum level of pressure for the sensor to be in its linear range. If I misunderstood your experiment let me know. Otherwise I would try putting 10 kPa on the sensors to see if they are more consistent in their outputs. It will be important to stay between Pmin and Pmax for accurate output.

     

    John

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  • Former Member
    0 Former Member over 9 years ago in reply to jw0752

    Dear John

    Thanks for your reply

    the sensor model that i use is -5\+5 inH2O

    And it is equal -1250/+1250 Pa

    i think the datasheet is for all ASDX series that contain ranges from 5inH2O to 100 PSI

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  • jw0752
    0 jw0752 over 9 years ago in reply to Former Member

    Hi Ashkan,

     

    Sorry, I saw the spec. on the Data sheet and thought it might be the problem. I will keep thinking but you might be able to contact Honeywell for some tech support if nothing else. I see that they have some phone numbers at the end of the data sheet. Did you remember the external bypass capacitor between pins one and three?

     

    John

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  • michaelkellett
    0 michaelkellett over 9 years ago in reply to Former Member

    A few other things I can think of:

    1) are you using the correct voltage for the sensor (they do 5V and 3.3V parts)

    2) is the load > 5k

    3) have some been damaged - the damage pressure for a 5"H2O device is only 15" which is  a very low pressure.

    4) that capacitor as John mentioned already

     

    If it's none of these try posting your circuit and even some pictures.

     

    MK

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  • dougw
    0 dougw over 9 years ago

    The pin numbering on this module does not follow standard IC numbering conventions. If you are seeing voltage variation with pressure you likely have it right but double check you are using the correct pins.

    One unusual spec is there is a minimum load needed of 5 kohms. Do you have at least a 5 K (or lower resistance) load on pin 2?

    The device can have a 2% of full scale error, so the output with zero differential pressure could be anywhere between 2.4 and 2.6 volts.

    If 2 identical devices measuring the same pressure give significantly different readings, there is probably something wrong with at least one of them. The other possibilities are the power supply - it could have noise that is not seen by your meter, or your meter could conceivably have something wrong with it.

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  • Former Member
    0 Former Member over 9 years ago

    Dear John, Michael and Douglas

    thanks for your quick reply

     

    the below picture shows my device, sensors and board

    image

    let me answer your question one by one:

    1) are you using the correct voltage for the sensor (they do 5V and 3.3V parts)

         I use adjustable -+5 Vdc and set this power supply on 4.99V

    2) is the load > 5k

        dear Michael and Douglas, I don't use any resistance on pin 2

    3) have some been damaged - the damage pressure for a 5"H2O device is only 15" which is  a very low pressure.

        I use this sensor in a low speed wind tunnel (the MAX speed is around 45 m/s and it is equal to  1113.7 Pa), so I don't think sensors damaged

    4) Capacitor

         I use one 100nF before any Vss pin of each sensor

     

    do you think the problem may be caused for the way i connect sensor to pneumatic connector? (see below picture)

    actually I am a mechanical engineering graduate and not very familiar with electronic details, can i use a 10 Kohm resistor? and do I use this resistor on output (Pin 2)

      imageimageimageimage

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  • jw0752
    0 jw0752 over 9 years ago in reply to Former Member

    Hi Ashkan,

    I believe that dougw and michaelkellett have solved your mystery. The minimum load of 5K would be necessary for proper operation. 10K would be outside of specifications and might not work properly just as no load does not seem to work properly. If you can't find 5K resistors to put between pin 2 and ground then perhaps the more common 4.7K will be available. Since this is a precision instrument is is probably a good idea to match the resistors or get 1% values. This may not make a big difference but it is good procedure. Keep in mind that whatever circuitry that you feed the outputs of the sensors into will also contribute to the load. Right now the only load is probably the 1 meg or so of load imposed by your meter.

    John

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  • michaelkellett
    0 michaelkellett over 9 years ago in reply to jw0752

    The way I read the data sheet is that the load resistance should not be less than 5k - so no load resistor should be OK.

     

    @ Ashkan - are your sensors the 5V type  (what's the part number written on them ).

     

    MK

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  • dougw
    0 dougw over 9 years ago in reply to Former Member

    That "Minimum load resistance" spec is definitely ambiguous.

    Normally an output would be specified to to drive a maximum load, where a smaller resistance is a bigger load, so you would want a larger resistance than what was specified.

    Here it seems they are saying the output needs a minimum load of 5K, which, if taken strictly, means the resistance from output to ground should be less than 5K. (because a smaller resistance is a bigger load)

    So if minimum refers to the output load it means one thing, if minimum refers to resistance it could mean the opposite.

    You may need to get clarification from Honeywell.

    I would try a 4.7K resistor to see if it helps. (you have already tried a much higher resistance)

    As to why the device would need a heavy load, it is not clear, maybe the output is unstable without a load, or maybe it gets quite non-linear without a heavy load.

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  • kulky64
    0 kulky64 over 9 years ago in reply to dougw

    I'm sure the correct interpretation is that the load needs to be between 5k and infinity. If the opposite were correct, then direct short from output to ground would be OK according to datasheet, which is nonsense. So the problem is in something else.

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