# How to use a Piezo ShockSensor and an op-amp practical integrator?

I have made this circuit :

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

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

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.

• How did you get Trace 1? Was the sensor connected to this circuit at the time? Dec 17, 2019 at 21:42
• Does this answer your question? Basic of a charged piezo : pC/G, mV/G, Signal analysis Dec 17, 2019 at 21:43
• @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 Dec 17, 2019 at 21:46
• 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. Dec 17, 2019 at 22:53
• 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. Dec 18, 2019 at 19:46