# Piezo transducer current consumption

is there any way how theoretically determine the current consumption of ultrasound piezo transducer?

I only know impedance for res freq = 50 ohms and excitation voltage around -170 V.

Does the Ohms law work in this case?

Thank you.

It depends on what kind of signal you are driving it with. Ultrasonic transducers are pulsed, how much you pulse it is up to the designer, which you are going to want to keep minimize, 1) Not to waste power and 2) Not to dissipate large amounts of power into the thing you are measuring. A piezoelectric transducer is capacative and cannot be driven with a DC waveform (and you wouldn't want to, because you wouldn't have any signal to observe from waves reflecting off materials)

$$P_{avg} = \frac{1}{T} \int_0^{t}p(t) \,dt$$

If you have strictly a sine wave then A would be the amplitude:

$$V(t) = A*sin(2\pi f t)$$

$$P_{avg} = \frac{1}{T} \int_0^{t}\frac{V(t)^2}{R} \,dt = \frac{1}{T} \int_0^{t}\frac{A*sin(2\pi f t)^2}{R} \,dt$$

or what ever function of voltage you are producing.

• I´m going to use pulse excitation with very short (dozens ns) unipolar (-170V) pulses. The consumption will not be continuous and I´m able to set few seconds between each pulse -> can work with long scanning cycle. I´m stuck, because I´m not able to figure out if my selected DC/DC (MAX1847) can handle this. – ToKra Mar 1 '16 at 22:32
• If your using a pulse, then you need to figure out what the current across the sensor is. Calculate the AC current on paper, use a spice package like LT spice. My point is, its your signal is time-varying. The DC/DC converter needs to source the average current, use caps to cover for the times when you need short term power. – Voltage Spike Mar 4 '16 at 19:52

If you're driving the transducer at resonance, the reactive terms will cancel and all you'll be left with is the resistance.

In this case, if the impedance looks like a pure resistance, then Ohm's law applies and the transducer will draw

$$I = \frac {E}{Z} = \frac{170V}{50\Omega} = 3.4 \text { amperes}$$

and dissipate

$$P= I^2 R = 11.56 \times 50\Omega = 578 \text { watts}$$