I've recently begun doing a school project. For one of the subsystems , I intend to mount a buzzer and vibration sensor on a walking stick and configure it such that the buzzer sounds when the walking stick falls and hits the ground.

I'm assuming the the impact of the walking stick hitting the ground will cause some deflection on the tip of the vibration sensor which will generate a voltage signal.

The problem is that I cannot seem to get the vibration sensor to work. What I have done was just to connect it to a 5V DC source and a 1K Ohm resistor and measure the voltage drop across it when I push the tip with my finger. However , the voltage drop across this vibration sensor does not change no matter how much I deflect the tip.

I'm using a MEAS MSP-1006 ND vibration sensor.

Thank you.

  • \$\begingroup\$ I link to the spec sheet would help. meas-spec.com/downloads/LDT_Series.pdf This I assume. It will output an AC voltage.. (not DC.) \$\endgroup\$ Oct 19 '15 at 16:02
  • \$\begingroup\$ I use this exact sensor in a product where I work. George is right in that it has an AC output (both pos and neg). For my purpose I rectified and boosted the output (it has a pretty small voltage response). If you are measuring with a multimeter you probably won't be able to see the voltage change. You'd need to use an oscilloscope. \$\endgroup\$
    – I. Wolfe
    Oct 19 '15 at 16:08
  • \$\begingroup\$ @GeorgeHerold Oh sorry, yes that is the spec sheet of the component. Uh , where does it say on the spec sheet that it would ouput an AC voltage? \$\endgroup\$
    – John
    Oct 19 '15 at 16:12
  • \$\begingroup\$ @I.Wolfe Oh I see. So if I connect the sensor to a resistance and a 5V dc source it should work right? Just that there will be an AC output at the 2 legs of the sensor. \$\endgroup\$
    – John
    Oct 19 '15 at 16:26
  • \$\begingroup\$ @John, I don't think you need (or want) a 5V source. I like to model (think of) a piezo as a voltage source with a capacitor in series, en.wikipedia.org/wiki/Piezoelectric_sensor. You need to think about the impedance of the capacitance(at the frequency of interest) and the resistor you use to monitor it. \$\endgroup\$ Oct 19 '15 at 16:36

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