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I have a small circuit working on 2xAA batteries (3V) with an MCU and a buzzer for alarm sound.

I've already successfully used an active buzzer (with build-in driver oscillator) drived by an NPN transistor in on-off, and it worked fine.

Now I've changed the board and its case, and I need to change the buzzer to a piezo one of this kind:

enter image description here

enter image description here

I know that I need to drive it with a waveform because it doesn't have the driver circuit and all the stuff, so I started using the circuit on the right, which worked, but was not very loud, so I started document how to make it louder.

I'm already feeding the BJT with a PWM with duty cycle 50% around the resonant frequency of the buzzer, so I can't do better about the frequency. Since the maximum peak to peak voltage in the datasheet is 30V I tried to use a 25V source (only for the BJT circuit, not the digital part), and in fact it becomes significatly louder, so basically the goal is to step-up the voltage around the buzzer.

Since the piezo is basically like a capacitor, putting an inductor instead of R2 makes a resonator and makes it louder, but since my power supply is very low compared with the 30V maximum of the buzzer, that's still too little.. I want more :)

One option is to use an H bridge for doubling up the peak to peak voltage, but since the starting voltage is low, the buzzer will be louder but not as much as I wish. Another option is to insert in the board a circuit to step-up the supply of that part. I suppose it would work, but finding something simpler would be better.

While I was searching, I took a look at a small device I had at home which makes a loud noise. I opened it and I realized that was using a buzzer very similar (bigger in radius which makes it work at a different frequency, but that's not the point) and a circuit very similar to the one posted above, but instead of R2 there's an inductor-like thing. I started googling and I'm pretty sure the part is this, connected as in this figure:

component circ1
Where two pins are connected as R2, and the third (don't know which of the three) is connected to Vcc.
edit: I've inserted the circuit because the previous sentence was not correct/clear. Also I would say that in this circuit I actually don't know what Vin is (square wave or something else).
I found the component on alibaba/aliexpress from China, and there it's always called something like "alarm boost three pin inductor" (if you try to search on alibaba those exact words you'll find a lot of them).

These are my questions:

  1. What is that component that seems (based on the name) specifically made for this purpose?
  2. Why does it make the buzzer make so much more noise with it (also the circuit that I saw the component on works at 3V, like mine, so no difference there)?
  3. I managed find that component only on alibaba/aliexpress, possibly I can't find it on RS-Components/Farnell/Mouser/etc. because I don't know what to search for and, eventually, how to dimension it.
  4. Other suggestions about how to make the buzzer louder?

The goal is to understand what's going on there, eventually understand how to dimension it, and find a couple of those for try this way.

edit: After the @marko answer, I understood that the mysterious component is an autotransformer, is it correct to say that it's like a transformer with two pins connected? If it is correct, the circuit I saw should be one of the three cases forward, possible? Which one could it be?

enter image description here

Say that my Vcc is 2.5V and I want todrive the buzzer with 25V (so x10). My Vin is a square wave at whatever frequency and duty cicle are needed (the resonant frequency of the buzzer is 6.5kHz so I guess that's the best choice), how can I broadly calculate the values of the autotransformer? Could the stepped-up voltage given by the inductor damage the source power? Would be better to put a diode before the Vcc?

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  • \$\begingroup\$ If you remove the piezo element from an enclosed buzzer (naked, like your piezo photo) it will sound feeble - the stiff enclosure into which it is mounted boosts amplitude significantly. The enclosure likely also affects the frequency of maximum audio output. \$\endgroup\$
    – glen_geek
    Dec 29 '20 at 22:42
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    \$\begingroup\$ Could be an autotransformer. \$\endgroup\$ Dec 29 '20 at 22:43
  • \$\begingroup\$ This makes the disk resonate at it's resonance frequency i think. That could be a reason of being much louder. \$\endgroup\$
    – fifi_22
    Dec 29 '20 at 22:54
  • \$\begingroup\$ @glen_geek that's right, i didn't mention this in the answare but i was aware of it, but was not the reason of the weaker sound i expected \$\endgroup\$
    – EttoreP
    Dec 30 '20 at 8:02
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    \$\begingroup\$ It is case 3, which is the autotransformer, which MarkoBuršič explained both above and in his answer. Use your multimeter to connect the autotransformer correctly -- the lowest resistance measured gets connected between battery-positive and the NPN-collector. The piezo transducer gets connected across the highest resistance. The center-tap gets connected to battery-positive as shown in case 3. \$\endgroup\$ Mar 31 at 22:23
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It works like an autotransformer. It's a transformer without galvanic isolation, primary winding is also a part of secondary. The purpose of using a transformer is to step up the voltage, thus increasing power.

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

$$V_{piezo}= V_{signal}\frac{L_1+L_2}{L2}$$

Beware that there is a maximum on pulse duration, before the inductor saturates. Beyond this time the inductor will loose the inductance and it will become like a short circuit. Therefore you would have to find out which is the minimal frequency (in case you use a square pulse of 50% DT) that you can use this circuit.

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  • \$\begingroup\$ thanks for the useful hints, if you can elaborate a little bit your answer after my edit would be perfect. Also, maybe the schema you put in your answer is a little bit misleading since it's not clear from there that the two inductors ar coupled? \$\endgroup\$
    – EttoreP
    Dec 30 '20 at 20:03
  • \$\begingroup\$ @MarkoBuršič -- Other than your CircuitLab schematics, your answer is great. Please consider using a transformer instead of two uncoupled inductors. Thanks. \$\endgroup\$ Mar 31 at 22:38
  • \$\begingroup\$ in "How to make a piezoelectric transducer buzz louder?" an answer contains a similar idea "I have seen piezo transducers in fire alarms which are driven by 1:10 transformers, and those are indeed pretty loud." said by Dmitry Grigoryev. \$\endgroup\$ Apr 1 at 4:55
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This is often achieved with a simple two-lead inductor. When the transistor closes, the inductor will create a reverse spike that is used to drive the piezo element. There are several considerations to make in order to implement this successfully:

  • The pulses must be narrow enough to not oversaturate the core leading to the loss of permeability.
  • The pulses need to terminate sharply so that the energy isn't dissipated in the resistance of a half-open transistor.
  • The coil needs to have a high Q, so the reverse spike still contains most of the input energy. In practice, you must pick enclosed ferrite coils as opposed to the open bobbin design.
  • The transistor needs to withstand the higher voltage.

I've seen this work right off of a single 1.5V button cell.

Of course, the disadvantage to this design is that it introduces a lot of harmonics. But this might be fine if you want a piercing sound.

If you haven't done it already, make sure you have a Helmholtz chamber in front of your piezo element. They can never be loud without it.

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  • \$\begingroup\$ Lots of good info in here, You explained how to do the same thing with a simple inductor, which I appreciate, but you did not really answer the question, which is necessary to do well here. Welcome to Electrical Engineering Stack Exchange! \$\endgroup\$ Mar 31 at 22:11
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    \$\begingroup\$ @MicroservicesOnDDD Sorry, you got me, I guess, but how much does an answer "it's an autotransformer" add to the topic? Perhaps I should mention that it helps match the impedance of the output to the piezo element... I have to say: Life has taught me to infer the real answers, not to follow the questions to the letter. \$\endgroup\$
    – Zdenek
    Apr 2 at 9:28

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