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Interfacing with a piezo electric belt

Discussion in 'General Electronics Discussion' started by wjousts, May 24, 2011.

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  1. wjousts

    wjousts

    17
    0
    Apr 26, 2011
    I have a respiration belt that is basically a piezo electric sensor embedded into an elasticated belt that you put around your chest. As you breath, the belt stretches and this produces a small voltage from the piezo electric (of the order of a few tens of millivolts). I've been trying to put together a circuit that will amplify the signal to somewhere in the region of 0 - 5V that can be connected to a USB powered board that will let me read the signal programatically.
    I tried a circuit with a simple op-amp (MC3407) which worked, but was rather noisy and unstable. So I tried out a instrumental amplifier (INA126) in the hope that it would work better. My current circuit is attached. The 5K pot is used to adjust the gain, the 500K pot is used as a voltage divider to offset the signal. Ideally it would set the baseline to 2.5V. The 5V power comes from my USB board and the output goes back to the board. Some research on this suggested that I should pass the voltage divider through an op-amp setup as a buffer in order to have a high impedance reference for the instrument amp.
    I think this circuit is a little more stable than the op-amp circuit, but it still is more noisy and less stable than I'd like. I'm also finding it very difficult to get a good magnitude of signal centered around 2.5V since adjust either pot seems to have an effect on the other. I also see drift in the baseline. For example, a particularly deep breath might spike the signal and then normal breathing after that takes several seconds to settle back to where it was.
    Does anybody have any tips, comments or suggests on what I might do to make this work better?

    For reference: INA126, MC34071
     

    Attached Files:

  2. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    A piezoelectric sensor easily outputs several volts, you only need to provide impedance buffering; i.e. a high input impedance buffer (>>10M Ohms).
    Your circuit has a 47k ohm input impedance which basically short-circuits the signal from the sensor.
     
  3. wjousts

    wjousts

    17
    0
    Apr 26, 2011
    Thanks for the reply Resqueline.

    This belt puts out a few millivolts at best (usually 10-20 mV). Even if I tug it hard I can't get above about 200 mV from it. The 47K resistors was trying to follow the suggestion on the datasheet which has a circuit on page 8 (figure 3) which suggests that you need a path for "input bias current return" which I'll admit I don't quite grasp. I've tried with and without those two resistors and with seems to work.
     
  4. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Ok, but measured with what? A DMM usually has a 10M ohms input impedance.
    A piezo transducer is in principle a small capacitor and the lower the load impedance is the higher the cutoff frequency gets. Respiratory frequencies are low.
    Your amplifier is not so much measuring the transducer voltage, but more the transducer current.
    I still say you need to make a high impedance voltage follower for the transducer input. After that you can follow up with an ordinary amplifier stage if needed.

    Yes, different op-amps have different input bias (leakage) currents which puts different limits as to how high resistances you can connect to their inputs.
     
  5. poor mystic

    poor mystic

    1,067
    31
    Apr 8, 2011
  6. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Yes, that's the kind of circuit needed. Now, the INA126 is actually (almost) exactly that; an instrumentation amplifier. It has an input offset current of 10nA.
    That allows you to use resistors around 47M ohms (instead of 47k). 0.47V would be set up over that resistance, which is acceptably within the common mode range.
    Actually, it would be beneficial to exploit that current to set up 2.5V over the input resistors (= 250M ohms) to keep the inputs in the middle of the common range.
    You now have only 0.47mV at the inputs, which might explain why the amplifier easily saturates (spikes). A dual supply would have prevented that effect btw..
     
  7. wjousts

    wjousts

    17
    0
    Apr 26, 2011
    Measured with my multimeter. I also tried directly connecting to my board just on the off chance that it would magically work. The board has an ADC and gives a value between 0 - 1024. Directly connected I can get that value to change by 1 or 2 units max. Hence the need for amplification.
     
  8. wjousts

    wjousts

    17
    0
    Apr 26, 2011
    Ok, thanks. For dual supply, the board I'm trying to interface with supplies three leads, a 5V line, a ground line, and the signal line. The signal line is obviously connected to the output of my instrument amp and I would prefer to use the 5V supplied by the board to power it and eliminate the need for an external power supply. I'm trying to turn my 5V into +/- 2.5V by using the voltage divider connected to the op amp (the op amp is used as a buffer because the instrument amp has an internal 40K resistor connected to reference pin). Originally I had equal fixed resistors, but I decided that using a pot would give me the ability to shift the signal around at will. So ideally, the voltage at pin 7 of the instrument amp (V+) is 2.5V above the reference (pin 5) and the voltage at pin 4 (V-) is 2.5V below the reference.
    This setup does make things a little confusing though. There is what the instrument amp considers ground (2.5V) and what the interface board considers ground (0V) which makes it a little difficult to figure out where, for example, the resistors connected to the inputs of the op amp should be connected? To 2.5V or to 0V?
     
  9. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    The inputs of the op-amp only needs to stay somewhere between the supply rails.
    It's called the Common Mode Range, and it's doesn't matter exactly what voltage the inputs are at, but the input signal will be clipped if trying to go outside these limits.
    The inputs have a 10nA leakage current coming out of them, going toward ground. 250M ohm resistors to ground will then place the inputs at around 2.5V.
    If you connected the resistors to 2.5V the inputs would then end up at 5V which would introduce clipping right there.

    Yes, it's wise to buffer the divider with an op-amp like that. The ground reference confusion shows itself with the gain & offset pots seemingly interacting.

    With this suggested 500M ohm input impedance and a 1Hz cutoff frequency the transducer can have as little as 320pF capacitance btw..
     
  10. poor mystic

    poor mystic

    1,067
    31
    Apr 8, 2011
    Hi
    is the circuit attached of value here?
     

    Attached Files:

  11. wjousts

    wjousts

    17
    0
    Apr 26, 2011
    Thanks poor mystic. I think the circuit you've post is mostly what's contained within the instrument amplifier (at least the first two op amps - the voltage divider and final op amp are external). The resistor to set the gain is also external.
     
  12. wjousts

    wjousts

    17
    0
    Apr 26, 2011
    Thanks Resqueline. I think I'm understanding the system a little better now. The biggest resistors I have on hand are only 1M, but I did swap them for the 47K resistors and it seemed a little better. I have to look into getting hold of some bigger resistors but 250M seems like a monster!
     
  13. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Yes, they may be bigger/costlier than usual, and they're certainly not commonplace. If you can just get as much as 47M you could use a number of them in series.
    You can use your DMM to measure the exact input leakage current. Placing it in the Voltage range in series with the 1M resistors you'll get 100nA/V (so 0.1V = 10nA).
    Did you study figure 4 on page 9 in the datasheet btw.?
     
  14. poor mystic

    poor mystic

    1,067
    31
    Apr 8, 2011
    Yes wjousts
    It's most of an instrumentation amp.
    I've been advised in the past to run such circuits single-ended so as to reduce noise.
    I have attached a drawing of a guard ring, which might be worth building into the pcb design. It reduces unwanted leakage from the input across the pcb by removing the attractive emf. The guard ring surrounds the input terminal with a voltage equal to the signal voltage.
     
  15. poor mystic

    poor mystic

    1,067
    31
    Apr 8, 2011
    here's the omitted attachment
     

    Attached Files:

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