Hi,
I think what you are measurinhg  are DC characteristics.
What you have is a transducer with complex impedance.  The peizo should
show up as a resistance in series with capacitance.  You probably have the
phase of the current different to the voltage that you are measuring. The
 current leads the  voltage, phase wise for a capacitive transducer.  Try
looking at the current and voltage with an oscilloscope. Use a small pure
resistance (metal film not wire wound) resistor with a small value (0.1
ohm) in series with the peizo to measure the current to the transducer.
measure the voltage coming straight from your signal generator.  You should
be able to see the V-I phase relation ship.

BTW dont use a wire wound shunt resistor (the 0.1 ohm) because that will
introduce inductance and that will introduce errors in your measurements.

Use complex (x + jy)  for V & I to calculate the actual power.

Cheers ,
 george

------------------------------------------
George Raicevich    (PhD in prog.)   M. Eng  (elec),    B.E. (elec)
Research Scientist
National Acoustic Laboratories
126 Greville St.
Chatswood   NSW  2067
Australia
+612 9412 6838  w
+612 9411 8273  fax
email:  [email protected]




> Hi, everybody:
> Does anyone knows or has experience to measure or calculate the power
> consumption of ZnO piezoelectric transducers?
>
> Now, I'm trying to measure it and compare with commercial
> electromagnetic dynamic acoustic transducers. But,
> strangely, piezo acoustic transducers give more power consumption as
follows.
> But normally, considering the advantage of piezoelectric acoustic
> transducers, shouldn't the power consumption of piezoelectric transducer
> bellow than the one that of dynamic transducers?
>
> Is there any misunderstanding in my way to measuring the power
> consumption ?
>
> 
> Dc resistance = infinity
> AC current = 0.36 mA
> AC voltage = 7.1Vrms
> Therefore, R= 7.1/0.00036 = 1.97E4 ohm
> Therefore, P= square(7.1V)/(1.97E4 ohm) = 2.5 mWatt
>
> 
> Dc resistance = 28 ohm
> AC current = 4.4 mA
> AC voltage = 0.035Vrms
> Therefore, R= 8.03 ohm
> Therefore, P = square(0.035V)/(8.03 ohm) = 0.15 mWatt
>
>
>
>