making a piezo louder

They'll draw more current of course. The unit may not like that.

I'd simply go for the most effcient sounder you can get.

Graham

Hi, Mike. If you HAVE to get a bigger sound from that dinky piezo beeper, it's probably going to be a little more complicated than just buying another beeper. But the good news is you can probably cobble up what you need from the Radio Shack in the Pierre mall.

First, a little background. Bare piezo beepers are driven by AC -- the alternating voltage causes the piezo element to flex like a speaker. If you flex the piezo element at its mechanically resonant frequency, a lot of the electrical energy will be converted into sound.

Since 4KHz sine waves are kind of hard to generate, you can do something like this with two microcontroller output pins (view in fixerd font or M$ Notepad):

| | Piezo | Beeper | |.-.| | || || | |\\ || || /| | A o--| >O--|| ||--O< |--. | |/ || || \\| | | || || | | |'-'| | | | | A'o---------------------' (created by AACircuit v1.28.6 beta 04/19/05

If A' is the inverse of A, you will have one side at 5V (if that's the operating voltage of the uC -- could be 3V or 3.3V) and the other side of the piezo element at 0V. When A and A' switch, the second side has

5V and the 1st side has 0V. That's the same as applying a 10Vp.p. square wave from a function generator to the piezo element, and actually produces louder sound than a square wave, because the fast transitions cause the piezo element to move faster.

Now you could just replace the piezo element with a bigger one, but you're probably not going to get better results. That's because the piezo element is seen by the two digital outputs as a capacitive load (probably several thousand pF). The primary limitation will be how hard the uC outputs can drive that capacitive load. Not only that, but different piezo elements have different mechanical resonances. If it's the wrong frequency, you'll waste a lot of the energy from the two logic gates, which means it won't even be as loud as the original.

Now there is another way to do this. If you can pick off the logic signal, you can use it with an external power supply and a transistor to switch, to drive a separate self-driven piezo beeper. Here's the circuit: | | A o-------o---------o---o-----------o------------. | | |10K| | | | 1000pF | .-. | | | | --- | | | | | | --- | | | | | | | '-' |e .-----o-----. | | | ___ | |< | + | | | B o--|

How about a little transformer?

wow Chris .. thats a lot of info .. and really appreciated.

I really doubt that I will be able to find the logic pulse ... without a schematic .

I would think that I could use the 4khz signal as an input for a standard transistor amplifier ... pushing more power to another buzzer

another gentleman suggested a small transformer .. this actually sounds feasible ..

thanks mike

Beware of trying to drag too much power out of the existing connection. You could damage it.

Graham

Hi, Mike. In small piezo buzzers of this type, you're actually driving the piezo buzzer with two inverting logic signals, as shown above. If you get a 9V battery (floating), you can make one of the signals your common for the driver above, and use the other one to drive the transistor. Not too hard, especially if you get a beeper with a separate logic control signal. Although, you could probably turn on and off a lower current self-driving buzzer (10 to 30mA) with the transistor.

The primary limitation in your board (and the reason the beeper is probably so small and weak) is the limitations of the output pins driving capacitive loads. When the logic switches, the two outputs are basically discharging and charging in the reverse direction a capacitor. Even if the output pins of the uC are rated similarly to standard 74HC-type outputs, loading them down too heavily will stress them, and result in damaging the outputs. Don't want to go there. I've been there myself, using two 74HC14 inverter outputs to drive a piezo in an home-brew add-on annunciator for a several dozen pieces of factory test equipment. The HC outputs would work for a while (weeks to months), and then give up their precious smoke. Had to replace 'em all. Didn't look too good there. Durrr.

Also, having a transformer to pump the output up to a higher voltage doesn't change the point, as Graham said. There Ain't No Such Thing As A Free Linch (TANSTAAFL). A louder piezo beeper will require more power, no matter the voltage. Try drawing more power by using a transformer, and you just might lose the uC.

Using a separate battery, and treating the uC signal with kid gloves is usually a good idea, especially if repairs on your piece of equipment will be inconvenient or expensive.

Good luck Chris

snipped-for-privacy@yahoo.com (Chris) wrote in :

ok Chris .. you and Graham have given me lots of good advice .. let me roll up my sleves and see what happens ..

thanks again mike

1) Non-electronic in four delicious flavors:

a) Build a reinforcing acoustic cavity and put the piezo into it

b) Sandwich piezo between two of (a) and gather sound from _both_ sides of the piezo and pipe them to the same physical output/hole/whatever

c) A chamber with a resonant frequency the same as the output freq of the piezo

d) Combine (b) & (c): Pair of resonant chambers that feed the same output while _not_ canceling each other out. Hmmmmmm, the constraint means that one or both won't be able to be a perfect resonator ... might be able to use/hear the resulting beat frequency, tho.

2) Not-very-sexy, but electronic: Swap out the piezo with an old- style "high impedance dynamic earphone" and place said earphone into ear. I suspect the electronics that drive the piezo should be able to handle this kind of resistance just fine.

In case I'm wrong about any of these assertions, I'm sure some kind, gentle soul on the list will eventually call me a Nazi. :)

--jim

probably not, most probably not right away. but if it doesn't sound two times as loud you're overloading the chip that driving it so go back to using one.

yes, eg this: (use fixed font) +12V | +->|-+ | | | relay | coil | | +----+ | / .001 |/ A ---||-+-->|--+---[1K]-+--| 2n2222A | | | |\\| `-||-|-' | | | | | | B ---||-+-|

probably not, most probably not right away. but if it doesn't sound two times as loud you're overloading the chip that driving it so go back to using one.

yes, eg this: (use fixed font) +12V | +->|-+ | | | relay | coil | | +----+ | / .001 |/ A ---||-+-->|--+---[1K]-+--| 2n2222A | | | |\\| `-||-|-' | | | | | | B ---||-+-|