I found a piezo buzzer in an old circuit board. It's (literally) a black box, size of a big coin, maybe 0.8 cm thick, 3 cm across, and a red and a black wire coming out. When I short the red and black wire, the piezo makes a clicking sound. The longer I wait between shorts the louder the click.

It has been disconnected from any power supply for a long time. Does anyone know what's going on? I wouldn't think it has a battery inside. How does it generate the energy for the clicks?


1 Answer 1


A piezo transducer is a capacitor, and capacitors exhibit an effect called dielectric absorption, in which a capacitor which has been held charged for a long time, then discharged, may then recover some voltage across its terminals.

Dielectric absorption could account for your observations. I don't know whether it would be of sufficient magnitude to be audible.

If you're wondering why the circuit would be holding the piezo element at a nonzero voltage: for a given supply voltage, the maximum volume is obtained by not just driving one side high and low, but by driving both sides with opposite polarity (much like a motor reversing circuit), thus achieving a peak-to-peak voltage of twice the supply voltage. If the circuit driving the piezo did not have a separate "off" state (which there would be no reason to need) then it would have been held charged as long as the device was powered.

By the way, a point of terminology: a “piezo buzzer” is specifically a device which contains an oscillator circuit, so it makes a (constant) tone if you apply DC. A piezo transducer (piezo disc, piezo element) without this circuit requires an actual audio waveform — if you apply DC then you'll only get a click on applying power (like the ones you heard).

The dielectric absorption theory is more likely if there is no oscillator circuit and you have a bare transducer (since the oscillator would likely drain the tiny amount of stored energy or otherwise cause what you're seeing not to happen).

  • \$\begingroup\$ The piezo element is sensitive enough to detect the electrical gradient in free air (journals.ametsoc.org/doi/pdf/10.1175/…) or charges on your body compared to the circuit board. \$\endgroup\$ Aug 4, 2019 at 22:54
  • \$\begingroup\$ I was also thinking dielectric absorption (I used to know the effect as dielectric memory) to be the most likely cause. Good to hear it from somebody else. Regarding the terminology: As it was packaged and quite thick, I thought it would be a buzzer, not just the plain piezo-crystal, especially as the wires are red and black, which hints at the polarity mattering. Unfortunately I don't have the means to test it here. I was baffled since the circuit was not in use for months or maybe even years, and I would have thought any charge would have long disappeared. \$\endgroup\$
    – Georg E.
    Aug 4, 2019 at 22:57
  • \$\begingroup\$ @DrMoishePippik I had a look at the paper, and didn't see where they made use of a piezo transducer. As far as I know valve electrometers are not based on piezoelectricity. Can you say how it is related to the (inverse) piezoelectric effect? \$\endgroup\$
    – Georg E.
    Aug 4, 2019 at 23:05
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    \$\begingroup\$ @GeorgE. The plastic package at a minimum allows convenient mounting without any chance of preventing the disk from vibrating, and might be acoustically beneficial otherwise. \$\endgroup\$
    – Kevin Reid
    Aug 4, 2019 at 23:19
  • \$\begingroup\$ @GeorgE., The reference was only to explain the atmospheric electrical gradient, not widely known. To demonstrate it yourself, put an insulated wire on a pole a few meters tall, with the ends stripped. Put the piezo element between the bottom of the wire and a ground, and you should hear a click. Now remove the the piezo device, wait a few seconds for the wire to acquire a charge, reverse the piezo sounder and expect a second click. \$\endgroup\$ Aug 5, 2019 at 16:29

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