Pressure Sensor Signal Amplification

Discussion in 'The Projects Forum' started by Physnano, Aug 18, 2015.

  1. Physnano

    Thread Starter New Member

    Aug 18, 2015
    17
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    Hello,

    This is my first post. My electronics skills are rudimentary at best. I am trying to measure air pressure via MPS20N0040D-D DIP. (I actually think I have a chinese variant of this sensor as my pinout is slightly different) First I would like to prototype the setup using a NI-DAQ and LabView. I currently can measure a differential voltage in mV that verifies my sensor is working, roughly 30mV over 0 - 15 PSI. Now with thoughts towards miniaturization I would like to use an instrumentation amplifier to amplify this signal for reading with an Arduino (~0-5V). This is where I am running into issues. First I am trying to use the instrumentation amp reading the output using my DAQ, then I will move to the arduino. I am currently using a AD620 instrumentation amp with Rg = 500 ohm => G = 100.

    The sensor is energized using 5V as is the instrumentation amplifier. For the life of me I cannot get an amplified signal. I have three options for measurement in LabView: Differential, RSE, NRSE. I would like to measure via differential. I have attached a rudimentary circuit diagram. This is as far as I have gotten. Any suggestions on how to go about the measurement?


    upload_2015-8-18_15-16-53.png
     
  2. ericgibbs

    Senior Member

    Jan 29, 2010
    2,499
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    hi Phy,
    You are showing no connection to pin #5, Vref.??
    Look a this image for a single supply app with 5V supply.
    E
     
  3. Physnano

    Thread Starter New Member

    Aug 18, 2015
    17
    1
    So I am measuring differential between pin 6 and 5. Is this not correct? Is the ad705 opamp required? Could I measure the signal without it for the time being?
     
  4. ericgibbs

    Senior Member

    Jan 29, 2010
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    Hi P,
    The output is between pin# 6 and 0V.
    Pin #5 is Vref.
    E

    As a quick test, connect a 1K resistor from +5v to the anode of a 1N914 diode, connect the diode cathode to 0v.
    Connect the junction of the resistor and diode to pin#5, this will give approx 0.65V offset.

    EDIT:
    For the AD630, you will require two 1N914 diodes in series to give a +1.3v offset.
     
    Last edited: Aug 19, 2015
  5. adwina2

    New Member

    Jul 14, 2015
    18
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    For the AD620, the data sheet says you need -Vs + 1.9 for minimum input signal. Will this application need a 1.9V offset before you see any results since it is a single supply? Then the gain will be too high.

    Physnano, have you bench tested the AD620 using the mV inputs that a bridge will give you?
     
  6. ericgibbs

    Senior Member

    Jan 29, 2010
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    The AD630 on my bench works OK with two diodes, also other users are using the same method.
     
  7. Physnano

    Thread Starter New Member

    Aug 18, 2015
    17
    1
    OK so my new schematic is attached. I verified 1.3 - 1.4V between GND and junction of R2 with D2. On the DAQ I see a response but not much amplification. I inserted a Potentiometer as the Rg and I am only reading ~ 100 mV change for ~15 PSI difference (see attached). (When POT is optimized) Supposedly a 500 ohm Rg should give G = 100, but when I change the pressure I get no signal change. Any suggestions ?
     
  8. adwina2

    New Member

    Jul 14, 2015
    18
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    Are you sure on the pinout of your schematic? It looks to me like you are just reading the voltage across one of the bridge resistors, which would explain why the value does not change with pressure. All the schematics that I see for MPS20N0040D-D show the differential output to be between pin 3 and pin 1 or 6.
     
  9. ericgibbs

    Senior Member

    Jan 29, 2010
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    hi Phy,
    Without any pressure, IF the sensor bridge is balanced you should read only ~+1.3V on pin #6, Vout, not over 2Volts.?

    Try with a 250Rgain for ~ 200 times gain, also due to using only a 5V supply the Vout will limit at approx 3.8v

    E

    I agree with Adwina, the sensor wiring looks incorrect.
     
    Last edited: Aug 19, 2015
  10. Physnano

    Thread Starter New Member

    Aug 18, 2015
    17
    1
    Yes, but this is why I mentioned that I think I may have a variation of this pressure sensor. I bought it on amazon (it was advertised as the MPS20N0040D-D) and the package had Chinese writing all over it. According to the official data sheet, I should get a bridge resistance of ~4-6 kOhm. However when I measure resistance between pins 3 and 2 I get INF resistance. In fact, when I measure resistance with pin 3 and any other pin i get INF. This leads me to believe that pin 3 is NC. Additionally, I can measure a ~20 mV signal (that perfectly correlates with pressure) with the NI-DAQ (no instrumentation amplifier) which leads me to believe my sensor is wired at least somewhat correctly.
     
  11. adwina2

    New Member

    Jul 14, 2015
    18
    2
    Check out this data sheet. You may be on the right track, but backwards. This could possibly give you this result since you are working with a bridge. It is a MPS-2000-006GR. You can switch your inputs and outputs and see what happens.
     
  12. Physnano

    Thread Starter New Member

    Aug 18, 2015
    17
    1
    So the sensor works with a different configuration. However I am still unable to get any amplification with a 220 ohm resistor (or any other static resistor for that matter) I did wire up a Trim POT and could get 300 mV change from 0-15PSI which is the best I've seen yet. Any other suggestions on why my gain isn't higher? (with 220 ohm, measured voltage is ~4.3V) I have attached another pin diagram for what I think the sensor pin-out is. The sensor has this on the back: 040DRB1 201409
     
  13. adwina2

    New Member

    Jul 14, 2015
    18
    2
    I would start with measuring from -in to +out and then -in to -out. Both of these voltages should be close to 2.5V. The difference between these two is what is being amplified, so the next step would be to see if this difference is what you are seeing at the input of the AD620. According to the datasheet, 0 pressure, which I am thinking is atmospheric pressure, you should see around 20 mV difference here. But, if this difference is not offset higher than the -Vs + 1.9V it will not be amplified.

    If you directly measure the voltage from pin 5 to pin 2 of the sensor do you get a positive voltage change as the pressure increases?
     
  14. Physnano

    Thread Starter New Member

    Aug 18, 2015
    17
    1
    I do measure a positive voltage change (~20 mV) as the pressure increases between pin 5 and 2 directly from the sensor.
     
  15. adwina2

    New Member

    Jul 14, 2015
    18
    2
    How about from -IN to +OUT? ...which is the same as GND to pin 5 of the sensor
     
  16. Physnano

    Thread Starter New Member

    Aug 18, 2015
    17
    1
    Measuring between pin 1 (-IN) and pin 5 (-OUT) changes from +9.8 mV to +13.9 mV when I change pressure from 0 - 15 PSI
     
  17. adwina2

    New Member

    Jul 14, 2015
    18
    2
    I believe this should be closer to 2.5V, correct? It is just a bridge, and at a steady no pressure, the bridge should be close to balanced, and reading from ground to halfway through the divider, you should be reading half of your supply voltage. Is there any part of the circuit that ties your -IN and either +/- output together? If the voltage from ground to the input is not higher than the 1.9V threshold the instrument amp will not work.
     
  18. Physnano

    Thread Starter New Member

    Aug 18, 2015
    17
    1
    Sorry I made some sort of an error yesterday, I verified 2.54 V between pin 5 (+ out?) and pin 2 (-out?).

    Pin 1 is -IN and pin 4 is +IN
     
  19. adwina2

    New Member

    Jul 14, 2015
    18
    2
    Now I am more confused. You are reading ~2.5V between pin 5 and pin 2? Have you ever analyzed how a full bridge works? When you apply a voltage across across +IN and -IN, it makes the resistors into parallel voltage dividers. If all the resistors are the same value, it is in a balanced state. If you measure from -IN(GND) to +OUT you will be reading a voltage divider between your power supply and, with equal resistors, you will read half of the supply. Also, if you read from -IN(GND) to -OUT you will also read half of the supply for the same exact reason. So, if you read the difference between +OUT and -OUT, the voltage will read 0V. But, this zero volt difference is still offset by half of the supply. So, even though the difference is 0V between the points, they are not referenced to GND. This is why it will work on an OPA, which requires the minimum input to be higher than -Vs. Now, the way the sensor works, when you apply pressure to the IC at least one of the resistors changes value as pressure changes. Now the divider is no longer balanced. So, if you read voltage from -IN(GND) to -OUT, you may still read the same half Vs. But, if one of the resistors on the +OUT side of the bridge changed due to pressure, then you no longer have a perfect divider on that side. From -IN(GND) to +OUT might be .505 * Vs now with pressure. If you measure from +OUT to -OUT now you will get a small difference, usually mV. If supply is 5V, and the change in resistance from applied pressure gives you 2.525V drop from -IN(GND) to +OUT, the difference between +OUT and -OUT will now be 25mV. I hope this helps you understand how a full bridge works a little better. With this you may have better luck verifying the pinout of the sensor. You should not be reading ~2.5V between +OUT and -OUT. You should be reading that value between GND and +OUT and between GND and -OUT. You could also be reading ~2.5V between -OUT and +IN and between +OUT and +IN.
    There may still be a problem with the amplifier stage, but it would be nice to be convinced of the sensor pinout first.
     
    ebeowulf17 likes this.
  20. ericgibbs

    Senior Member

    Jan 29, 2010
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    hi Phy,
    If you needed to add an Offset trim pot, one end of the pot would got to pin 1 of the sensor and the other end of the pot to pin 6. the wiper of the pot would be one input to the OPA.
    For your initial test, the sensor Pins 1 and 6 could be linked together, as one input to the OPA, the other sensor output is pin 3.

    The 5V excitation for the sensor is across pins 2 and 5.

    When you apply pressure to the sensor, the output which goes slightly positive should be connected to OPA diff amp Non inverting input.

    Do you follow that OK.?

    E
     
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