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I am using a piezo transducer to launch an ultrasonic pulse in air, at a frequency of 125kHz. A picture of the received waveform is attached. My question concerns how I can (electronically) limit the long resonance tail in the emitter pulse which I assume is because of the relatively high Q of the device.

enter image description here

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    \$\begingroup\$ You are showing the received waveform, but asking about the transmitter? \$\endgroup\$ – Eugene Sh. Feb 19 at 15:57
  • \$\begingroup\$ Usually mechanically. Resonance is mechanical. Damper attached to back side. \$\endgroup\$ – user263983 Feb 19 at 16:01
  • \$\begingroup\$ It's mostly the construction of the transmitter (and receiver, if separate). Mechanical 'Q'. We found transducers designed for Doppler use were basically worthless for pulsed applications no matter the circuitry. \$\endgroup\$ – Spehro Pefhany Feb 19 at 16:04
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    \$\begingroup\$ How did you made sure that the ringing isn't happening in the receiver? In the transmitter you can try to connect a loading resistor in parallel with the transducer to eat the ringing energy. \$\endgroup\$ – user287001 Feb 19 at 16:17
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    \$\begingroup\$ Not sure of your circuit, but could you put a FET across the transducer so that when you are "done" transmitting, you activate the FET and place a short across the transducer? Or combo it with @user287001 idea and use the FET to insert a resistor. \$\endgroup\$ – Aaron Feb 19 at 17:06
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Yes, these transducers tend to have high Q, which makes them "ring" like a bell.

In the example below, a 40kHz piezo was measured and fitted to an electronic model that substitutes L-C-R components for the mechanically-resonant components. For this transducer, the electrical equivalents were:

  • Lm = 92.1 millihenries
  • Cm = 175.384 picofarads
  • Rm = 634 ohms

The value of parallel capacitance is real, measured with a capacitance meter: 1.3 nanofarads.

piezo driving schematic

The current through the series arm comprising Lm, Cm, Rm is proportional to physical motion of the transducer. When driven directly from a voltage source, current takes a long time to rise to full amplitude (purple trace, I(R3)).


An external inductor (L2=12.42 mH) along with a damping resistor (R1=5400 ohms) gives a significantly faster, critically damped rise time (green trace). The penalty paid is far less current - making the transducer less sensitive.
piezo transient waveforms

The simple approach of damping with a resistor (R1=5400 ohms) is wasteful of available transducer signal. An impedance-matching transformer would be a better, more costly alternative.

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  • \$\begingroup\$ What does that graph look like when you open V1 after the 1ms? \$\endgroup\$ – Aaron Feb 19 at 17:48
  • \$\begingroup\$ @Aaron Opening V1 would disable 5400 ohm damping - not something you'd want to do if damping the resonant "tail" is important. If you simply drive V1 to 0V after 1ms, then damping is similar to the rising waveform shown in the time graph above (green, I(R2)). \$\endgroup\$ – glen_geek Feb 19 at 18:11
  • \$\begingroup\$ "...drive V1 to 0V after 1ms..." that's what I was after, thanks! \$\endgroup\$ – Aaron Feb 19 at 18:34
  • \$\begingroup\$ What is the resonant frequency of the L in combination with the transducer C? \$\endgroup\$ – Dirk Bruere Feb 19 at 20:33
  • \$\begingroup\$ Dirk: The external L (12.42mH) resonates with the 1.3 nf capacitance inside the transducer at 39.6 kHz. \$\endgroup\$ – glen_geek Feb 20 at 2:11

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