[sdiy] paralleling piezos

[Ingo Debus]

Am 06.02.2011 um 22:47 schrieb Ian Fritz:

> If you have two devices in parallel and strike only one, you will  
> obtain half the expected voltage.  (The same charge but twice the  
> capacitance.)

Yes, but only if the two piezos are equal.
Today I went to the shop and bought another 20 piezos. I'll check  
their capacitance before using them.
Meanwhile I found a circuit called charge amplifier. It's basically  
just the piezo connected between ground and the inverting input of an  
opamp and a capacitor (with a bleed resistor parallel to it) from  
output to inverting input. Since the piezo sits at virtual ground,  
the additional capacitance of the other piezo won't matter. But this  
would require an extra opamp per key. I could just as well have  
separate buffer amps for the two piezos (which I plan to do anyway,  
to detect the position where the key has been hit).
Here's a nice writeup, I found this through Wikipedia:
<http://focus.ti.com/lit/an/sloa033a/sloa033a.pdf>

>   If you want to add the responses, I'm pretty sure you want to  
> connect them in series.

If they're viewed as voltage sources, yes. I thought they are more  
like current sources, given their very high DC impedance. But  
obviously that was over-simplifying.

I haven't tried to put them in series yet (to answer Harry's  
question). After replacing some piezos, 13 out of 16 keys are working  
ok-ish; so I don't want to make radical changes now.

> Careful, they can put out large voltages.  I have a couple of  
> experimental tap sensors where the tap bends the sensor. One of  
> them easily puts out 30-40 V

Yes, I have a 5V zener diode across each imput, just in case.
Hey, piezos are used in cigarette lighters to generate the spark that  
ignites the gas :-)


>   And there some cool folks online who have used these cheap  
> beepers to make sensitive seismometers!

Cool. I always wondered if they could be used as a heart beat monitor?


> Are you trying to detect the magnitude of the strike, or just its  
> occurrence?

The magnitude of course. And maybe roughly the position too (later).


As to Harry's suggestion:
> The piezos might load each other, the resistor would decouple them.  
> Maybe a resistor in series with
> each (a passive mixer) might work better.

Well, yes, with two unequal resistors, compensating for the unequal  
piezos...

Ingo