I found two ways of working with piezo like a buzzer.

In the first case, we use only a resistor. When we turn ON the circuit, I have my transistor off, With Vc=VDC. When the piezo sends a feedback voltage, we have saturation in the transistor; my Vc = 0.2V ( approximated), and the piezo doesn't stretch and feedback voltage = 0V.

But, how does it work using an inductor? When we saturate the transistor, the coil tries to maintain the current throughout the circuit, and a -EMF voltage appears and polarizes my base in reverse and cuts off the transistor. Does this cause an oscillation in my piezo?

Is there any difference in efficiency of each circuit?

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2 Answers 2


The inductor stores energy while the transistor is on, releases it when the transistor is off, giving something like twice the supply voltage (peak-peak) across the piezo.

The resistive version gives a peak-peak voltage that's only equal to the supply.

So the inductive one should be 6dB louder.

  • \$\begingroup\$ You Can explain why the inductor will give a twice supply voltage ? \$\endgroup\$ Oct 12, 2016 at 22:02
  • \$\begingroup\$ I know that the peak voltage will be higher. But how do you know it will be twice the input voltage? \$\endgroup\$ Apr 19, 2017 at 4:41
  • 1
    \$\begingroup\$ @Chupacabras, Coil+Piezo form an LC-tank, which AC oscillation voltage have peaks of equal amplitude at both maximum and minimum around power supply voltage. Minimum voltage may reach 0 V, so maximum may reach 2*supply. \$\endgroup\$
    – Eugene K
    Dec 4, 2017 at 22:28

The left side diagram was posted by me in one of my buzzer related articles, and here I will try to present perhaps a better explanation regarding how the system might be working.

Please note that the piezo element will not create a significant amount of sound if the frequency and the voltage are low. The coil consists of around 500 to 1000 turns giving rise to perhaps many 100s of volts across the piezo. I have not practically measured it but because I could sometimes see severe arcing happening within the piezo element tearing apart the piezo central white portion, allowed me to conclude that the voltage from the coil was significantly high depending on its number of turns.

Now how does the circuit create a high frequency?

It probably does in the following manner:

When power is first switched ON, the transistor conducts and shorts the coil to ground, allowing the coil to store an amount of voltage equivalent to its number of turns. This also shorts the piezo connections triggering a reverse voltage pulse to the base of the transistor switching it OFF, which prompts the inductor to kick back the stored energy on the piezo. However as soon as the transistor switches OFF, the negative feedback from the piezo disappears, forcing the transistor to conduct back, and this process creates rapid oscillations across the coil, which in turn causes the high frequency high, high voltage pulses on the piezo element, inducing the required sharp beeping sound through it.


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