impedance calc

I'm trying to find out the impedance of a piezo, it isn't logged in the datasheet and I'd like to know.

The circuit i'm using is an amp feeding a series resistor and inductor that go into the piezo, see pic.

From the amp I input a 12Vpp square wave, across the 1 ohm resistor is 4.5V, so it appears I'm inputting 12Vpp and 4.5A.

The inductor and piezo capacitance are matched to work at resonance, it is a dynamic piezo and rings like it should.

Across the piezo I have a 40Vpp sine wave.

My questions...I think there is enough info here to calculate a) the current draw of the piezo and b) the impedance of the piezo.

I will do any other tests if more information is needed.

Cheers.

To quote Sherlock Holmes, "When you have eliminated the impossible, whatever remains, however improbable, must be the truth." In this case you are trying to find the impedance of the capacitor, so if you know the frequency, of the signal, and the capacitance then you should be able to find it with the equation 1/2πfC. That will give you the capacitive reactance in ohms.

So with frequency = 18KHz and c=680nF

Reactance = 13 ohms

I was thinking because it is a resonant LC circuit the reactance of the L and C cancel so I would be working with only the resistive element. I need to think a bit more to understand. Thanks.

Any time you apply an AC signal across a cappacitve or inductive load, it creates a reactance regardless of whether they are in series or not. You should be able to test your calculations by measuring the voltage between the inductor and capacitor using ohms law to find the voltage drops. Just remember that the measured voltage will be in RMS not peak-to-peak voltage.

I think I understand a bit better now. I will try some calcs.

4.5A square wave ~ 6.75Arms
40Vpp sine ~ 15Vrms

Capacitor reactance of 13 ohms more or less cancels inductor reactance of 11 ohms (will ignore difference for simplicity)

R=V/I = 15/6.75 = 2.2 ohms impedance of piezo at resonance.

It feels wrong but the calcs look right! It is an interesting exercise.

Another question: Does it make any difference if I put the inductor in parallel with the piezo as opposed to series? I've done this but couldn't see any difference but it is easy to miss something.

Yes, if you put the inductor and the peizo in parallel, then you have basically made a tank circuit. The set of calculations changes completely, the new set of calculations is Ω(total)=LC/L+C. The resonant frequency of the circuit is given by 1/(2π√LC).

I also got it a little wrong earlier (very tired, a nap helped). when the 2 are in series, their impedance is added, not canceled. As the frequency increases, the reactance of the inductor increases, but the reactance of the capacitor decreases so that is something that you need to consider. But the voltage drop across the 2 will remain proportional because of the voltage divider. You should also consider the resistor that you have placed in the circuit.

Hi,

Its funny how I've had in mind all day (till I could come back to this) that I had the tank circuit with the L and C in series. And I was wrong all along! Just because I couldn't see any difference between the L in series and parallel didnt mean the difference wasn't there, it is easy to err and better to know the theory you've mentioned.

I put the resistor there as a means of measuring current, I noticed they crackled a bit so must have been pretty close to their power rating, now I know it is better to use the voltage drops across the L and C themselves I have removed it.

I've done some more testing with the LCR circuit. I was inputting 11Vrms from the amp and getting these voltage drops L=17Vrms C=23Vrms R=2.5Vrms

It looks like I need to be able to draw 2.5Arms from the amp.

The impedance of the circuit is 11+13 reactances + 1 resistor + 5 piezo (that is a guess) = 30 ohms

The current being dissipated is I=V/R =11Vrms/30 ohms =0.3A.

I also tested the circuit with the same inductor in parallel. I had to input 15Vrms, less and it wouldn't ring. I had these volts across the components L=14.7Vrms C = 14.5Vrms R= 1.2Vrms

The current draw looks to be 1.2A.

Impedance of that circuit is (LC/L+C) + R + Piezo = 12 ohms

The circuit dissipates I=V/R = 15/12 = 1.2Arms.

I think the numbers got muddled somewhere....

Generally it looks like the inductor in series uses much more power, it requires lower volts to start the piezo ringing, but it needs lots of current. The inductor in parallel uses less power but needs more volts to start and less current.

If your input voltage is 12Vpp, then the RMS is 8.484V RMS. If your input is a square wave, and you want it to become a sine wave, then you need to have the inductor and piezo in series. Depending on the amplifier that you are using, you may want to use a bigger resistor because the inductor stores electrical energy in a magnetic field and when that field collapses it can deliver a voltage many times larger than the original voltage. This may be why your calculations don't match your measurements. It's almost impossible to determine what the fly-back voltage will be... without more calculations. Since I'm short on time, I'm going to give you a link to a site that should help: https://www.electro-tech-online.com/custompdfs/2010/12/slup127.pdf

If you have anymore questions, I will certainly try to help.

Hi,

The circuit works best with the inductor in series, the series LCR circuit seems to be a resonator and the parallel circuit an anti-resonator. As the piezo is dynamic/moving it is better work with series although I could use both. The calcs look to be:

Z (impedance) = -13 +11 + 1 + 5 (piezo) = 4 ohms

I=V/Z= 11/4 = 2.75Arms

This more or less tallies with the measured 2.5Arms and I'm still guessing at the resonant impedance of the piezo.

It is interesting that you pointed me to some power inductor/transformer theory I have made up some transformers to test with a similar but not identical piezo and they didn't work! I will try them again and post back with what happens, it would be useful to have some feedback. Thanks.