1
\$\begingroup\$

I have made this circuit :

enter image description here

Once I apply a shock to the sensor, I get this result on the scope :

  • CH1 (yellow) : piezoshock output
  • CH2 (blue) : LTC6240 output

enter image description here

Here is the spec of the piezo sensor : https://www.murata.com/~/media/webrenewal/products/sensor/shock/healthcare/pkgs_spce_00gxp1_en.ashx?la=en-us

What we know :

  1. The piezo has a 390pF capacitance (+/-) 30%.
  2. The piezo gives 0.35 pC/G (0 to 50G)
  3. The LTC6240 has a 3.3V+ to the Power+ and 0V to Power-

Now I am trying to understand how to calculate the gain of the integrator from the circuit above and figure out what are the G's from the mV at the output.

\$\endgroup\$
14
  • 1
    \$\begingroup\$ How did you get Trace 1? Was the sensor connected to this circuit at the time? \$\endgroup\$
    – Chris Knudsen
    Commented Dec 17, 2019 at 21:42
  • 1
    \$\begingroup\$ Does this answer your question? Basic of a charged piezo : pC/G, mV/G, Signal analysis \$\endgroup\$
    – Elliot Alderson
    Commented Dec 17, 2019 at 21:43
  • \$\begingroup\$ @ElliotAlderson it's already my question :p But now I made the circuit and the new question is more about the calculation of the integrator \$\endgroup\$
    – Christophe Gudlake
    Commented Dec 17, 2019 at 21:46
  • 1
    \$\begingroup\$ Did you duplicate that schematic in your circuit, including the circuit values? You want the piezo to be biased to 3.3V / 2, and you'd really like its voltage to be close to the voltage of the op-amp's + input -- the + input is grounded then Rf is too big or you've got leakage between the pins of your amplifier. \$\endgroup\$
    – TimWescott
    Commented Dec 17, 2019 at 22:53
  • 1
    \$\begingroup\$ You're probably right -- but I never accept "must be" until I know all the requirements. Sorry. I've seen knee-jerk "must be's" increase system cost unnecessarily, or even degrade the performance that actually matters, just because one design's "must be" is another one's "don't care" or "must not". You can't judge today's designs by yesterday's requirements. \$\endgroup\$
    – TimWescott
    Commented Dec 18, 2019 at 19:46

1 Answer 1

Reset to default
3
\$\begingroup\$

Here's what's going on, and what you can (maybe) do about it:

The circuit shown assumes positive and negative supply voltages on the op-amp, with the output having positive and negative excursions. If there were no bias, you'd get exactly half the signal.

The reason you have a signal out at all is because there's about 1V of bias, and it just happens to be in the "right" direction. You can't count on this.

The reason you have bias is because Rf is too big for the setup, or because you have leakage. One giga-ohm is BIG. If you had a perfectly insulating glass cylinder with wires sticking out, and you rubbed it all over with your bare fingers, you'd have a resistor with less than one giga-ohm resistance. I suspect there's a leakage path to something, but I can't figure out what (so comments from the peanut gallery are welcome). Looking at LTC6240 packages, and the fact that you're using a ground-referenced system suggests that, if anything, leakage paths would make the signal more negative, not more positive.

The reason you see activity on the bottom trace at all is because those peaks happen when the op-amp has saturated -- in normal operation the op-amp should keep it's inverting input equal to the non-inverting input.

You have not said what you can accept for a low-frequency cutoff. You can sacrifice low-frequency performance by using a lower value of Rf (for that value of Cf, 10M-ohm would put the low-frequency cutoff at around 7Hz). You can also sacrifice overall gain by using a larger Cf -- but you'll still have the offset that you'll need to calibrate out somehow.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.