In my DIY project I need to decide do I want to use piezo buzzer (for sonar-like device) or not. To decide this I need to calculate current consumption of this device. Parameters I was able to find in datasheets for many different piezo buzzer discs include resonant resistance, resonant frequency and static capacitance, max voltage and SPL for some sample volt/freq (e.g. R=150 ohm, f=3.6 kHz, C=90 nF, SPL at 3.6 kHz = 91 dB). If I understand it right piezo buzzer should consume current same as resistor with resonant resistance if run at resonant frequency. I want to estimate what current it will consume on different frequencies and what SPL it will produce. Is there a way to estimate it theoretically using above information from datasheet? Will it be easier to just buy few discs and estimate it experimentally?

Update: I found that some piezo modules have SPL/freq diagram included in it's datasheet, so it is possible to estimate SPL in dB using this diagram.

  • \$\begingroup\$ First determine if you are working with an actual piezo element or a module. The basic element is just the capacitive piezo piece, and with a DC input it doesn't make a sound and draws no current, you drive it with an AC voltage. A piezo module will have an oscillator circuit inside to drive the element, and is operated with a DC input, the input current will be less related to the output frequency. \$\endgroup\$
    – Nedd
    Jan 15, 2015 at 13:49
  • \$\begingroup\$ I'm talking about piezo element, not module. And I'm planning to drive it by PWM. \$\endgroup\$ Jan 15, 2015 at 18:24

1 Answer 1


If you're running it at resonance the current won't have a lot to do with the Xc because the mechanical energy is being changed back into electrical energy as the ceramic and backing disc flexes, like a tank (LC) circuit where energy circulates back and forth between the L and the C.

It will have more to do with the non-useful losses (series equivalent resistance, mechanical losses at the flexure, mechanical losses in the materials, damping losses in the sides of the Helmholtz cavity etc.) and the intentional transfer of power to the air (or whatever medium it is).

Then there are losses in the driving circuit.

Measuring it under representative conditions may be your best bet to get an approximate answer quickly, though I'm sure it could be modeled in a multiphysics package like Comsol given the time and inclination.

  • \$\begingroup\$ If I understand it right, current consumption on resonant frequency is optimal and current will increase when I'm moving driving frequency away from resonance so I can estimate current consumption to be more than consumed by equivalent resonant resistance? \$\endgroup\$ Jan 15, 2015 at 18:28
  • 1
    \$\begingroup\$ No, it's not that simple. See this link on behavior of a piezo around resonance. \$\endgroup\$ Jan 15, 2015 at 18:38

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