1. Home
  2. Embedded Programming
  3. Piezoelectric beeper steering - newbie problem.

Piezoelectric beeper steering - newbie problem.

Dear all,

I have a very simple problem to solve. I have PIC12F675 powered from 3V lithum battery. I'd like it to play simple audible tones (just a few frequencies). When I tried to drive a standard buzzer (without built-in generator) directly from PIC output pin via serial resistor it sounded very, very silently. How can I make it louder? I have tried to drive it form generator with 12V output voltage and it sounded loud enough. Do I really need so high voltage? How should I implement audible sounds in portable device with 3V battery keepeing in mind that cost and current consumption are critical. How is it done in hand watches - they sound loud enough and are powered from small batteries? Many thanks for any advices and redirections to any usefull web sites with piezoelectric buzzer practical solutions.

Wlad

Reply to
Wlad
Loading thread data ...

Ensure that you are driving it differentially ie two outputs driven with opposite logic states. This will give you 6V p-p. If you already use this arrangement and it is not loud enough consider creating a charge pump using two other output pins. That would almost double the battery voltage and give you close to 12V p-p at the piezo.

Reply to
Joseph Casey

Differential driving as JC suggested is a good thing, not sure about that 6V though....

Perhaps an RS232 driver IC, the ones with the built in charge pumps, MAX232, MAX202, etc, etc.... That'll get you about +-9V. A real neg 9V WRTE making ~18Vp-p.

The piezo buzzers rely(sp?) heavily on resonance, being the element size, housing, and the hole in the front. Usually loudest at one frequency and not so good at others but if driven hard enough it'll play your tunes. If you avoid the resonant frequency in the tones you want to play then they should sound equal in volume. If that makes sense....

HTHs Greg the Grog

Reply to
Grog

I am open to correction on this but I believe there is no need for for a series resistor. The piezo device is equivalent to a small capacitance of < 1nF. In the absence of a series resistor there will be current peaks at the instant of switching but the MOSFETS in the PIC can abosorb this energy without complaining.

If you are using a resontant piezo (the type with one small hole in the top plate) the hole plus the cavity inside form a Helmholtz resonator tuned to one frequency. By cutting off the top plate you would decrease the resonance and give better venting at the other frequencies. There will still be some resonances in the piezo disk itself.

Regards, JC

Projects ruined at no charge -just ask.

Reply to
Joseph Casey

I am open to correction on this but I believe there is no need for for a series resistor. The piezo device is equivalent to a small capacitance of < 1nF. In the absence of a series resistor there will be current peaks at the instant of switching but the MOSFETS in the PIC can abosorb this energy without complaining.

If you are using a resontant piezo (the type with one small hole in the top plate) the hole plus the cavity inside form a Helmholtz resonator tuned to one frequency. By cutting off the top plate you would decrease the resonance and give better venting at the other frequencies. There will

still be some resonances in the piezo disk itself.

Regards, JC

Projects ruined at no charge -just ask.

Reply to
Joseph Casey

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.