Piezoelectric Element Impedance Question - Page 1
Forum section: electronics / projects

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 23, 2018, 07:55:49 pm
content: Does a piezoelectric element's impedance change with applied amplitude?I'm wondering if, all other factors being equal, a larger or smaller applied signal amplitude will cause resonant/anti-resonant frequencies to shift around, or the absolute impedance value to shift up or down.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: January 23, 2018, 07:58:33 pm
content: Yes, the materials are usually far from ideal. The harder materials (type 4 and 8 ) will do better. Also consider the fact that a pure reactance can't dissipate any power. If the PZT is going to do any work, there will have to be a loss component. That loss component will change with the mechanical loading. IOW, the dissipation factor will rise.

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 23, 2018, 08:06:03 pm
content: So is it any use using a standard bench RLC/impedance meter to get impedance curve with a 0.5Vrms-2Vrms range of signal, when the final application will have orders of magnitude higher excitation voltage?Or do I need to do a more manual impedance measurement using scope probe and current probe at the excitation signal level I need?

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: January 24, 2018, 04:37:02 am
content: It depends on how good an answer you need and the type of material. If you're going to be running under power at resonance, you'll need to measure at the actual signal levels. If you're just roughly characterizing the material, a low voltage measurement might be good enough. Need more info.

author: ejeffrey
author_rank: Super Contributor
timestamp: January 24, 2018, 05:44:16 am
content: Capacitance usually goes down at high voltage much like a high-k ceramic capacitor. They are frequently made of the same materials. A low voltage measurement with an LCR meter will give you a worst case load if you want to know how beefy of a driver you need. But it depends on how accurate of an answer you need.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 24, 2018, 07:48:31 am
content: You can change the natural resonance of piezoelectric devises by added or taking away mass. A quarts crystal at 1 MHz can be moved up to 1.1 MHz by filing bits off the edge of it. I use to do this for non critical pinballs for TV IF alignment. Those crystals frequencies were had to find so just file down a crystal that is close. You want to make real sure you do not file too much off as it is hard to go back , ha. I have a piezoelectric devise on my bench the resonates at 60 Hz. You hear me right 60 Hz! It was done by adding a sheet of fiberglass increasing its mass. The reason was to have a really quiet fan that is very reliable. I have a picture of it. It is hard to tell from the picture but if you look close you can make out the two fiber sheets waving back and forth at 60 Hz. It runs off 120 volts AC house wall plug with a 10 K resistor is series.

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 24, 2018, 12:46:39 pm
content: I'd be running high power, but not necessarily at the resonance point. There are LC components upstream of the piezo (think class-D amplifier filter), and I need to be able to simulate all the complex impedance interactions accurately. I need to know the impedance that the amplifier sees to ensure I don't present too low (in ohms) of an impedance that will cause over-currenting, and too high of an impedance that won't allow me to get the desired power output.

author: dmills
author_rank: Super Contributor
timestamp: January 24, 2018, 02:40:37 pm
content: I used to work in sonar, where the standard approach even for materials run at kV levels was just to use an HP 4192A to plot a G/B curve via GPIB, this was usually close enough.Temperature and voltage matter to a degree, but voltage tends to make the things non linear on a cycle by cycle basis, so you get intermod and harmonics, generally not a problem (Most of the time!).What we did find is that making sure the transducer face has fully wetted out is important before making these measurements, a handful of bubbles on the front face can cause chaos when trying to measure the admittance.In terms of matching, a tuned transformer can be your friend (Make the secondary inductance resonate the fixed capacitance of the transducer), and drive the primary (wound to have a sane voltage and magnetising inductance) with a RLC matching network (The R extends the bandwidth and mostly dissipates power near the band edges). Because of the resonant secondary, this is more of a coupled choke sort of situation so you often find you need a gap, and the driver has to work into a very low impedance at the start of a pulse until the resonant current builds in the secondary circuit.For sonar, a simple bridge drive with active clamp gets it done often as not.Regards, Dan.

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 24, 2018, 03:03:48 pm
content: Dan,That was a very helpful response. Here's a few more questions.Wetting during impedance measurement - does it matter if the transducer is in little bucket of water vs. "infinite" bucket? Effect of reflections on the impedance measurement?Matching - I don't necessarily want to operate at resonant point, so how would you suggest handling the matching in this case?Modeling - is the C in parallel with series RLC model of piezo that I see only applicable around the resonance point?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 24, 2018, 05:05:23 pm
content: Quote from: Evan.Cornell on January 24, 2018, 03:03:48 pmMatching. By their nature they are high voltage low current. I would run with an flyback transformer from an old TV and use the high voltage business end of it to drive the piezoelectric. With both the piezo and transformer off resonance you could drive it with a function generator to get a feel of what you need for a driver. With a little luck it could be as easy as a off the self audio amp, nice and cheap.

author: dmills
author_rank: Super Contributor
timestamp: January 24, 2018, 05:09:52 pm
content: Little buckets don't work at all well, you need to be at least out into the far field, which for a large transducer at high frequency can be a LONG way (Think 10m or so), for smaller things that are less directional you need less distance, for some measurements (particularly things like beam patterns) temporal gating on the analyser is your friend.Some companies offer test cells with absorptive foam lining that claim to emulate a large body of water, but I have no experience with these.To get the transducers to wet out, spray with a dilute solution of washing up liquid before dropping it into the water.RX off resonance is not a problem, just use a high impedance preamp close to the transducer, but I wish you luck getting much power into the thing more then about an octave off resonance, that is quite the trick (Note that many transducers have two usable resonances, sometimes with very different radiation patterns (Ring transducers with both hoop and thickness modes, looking at you). The acoustic matching between the ceramic and the water is often a 1/4 wave impedance match done with something like a carefully chosen thickness of FR4 in a simple device, or either a 1-3 piezo composite or something like a Tonpills element in low frequency devices.Matching is matching, the standard RF methods apply, and usually you are trying to get a flatish response between the two usable resonances and maybe just a bit outside them.There are two capacitors in the standard model of such things, the series one (Motional capacitance) and the parallel one at the input of the network (The 'Fixed' capacitance, which as its name suggests does not much change ,well apart from temperature and ageing).They are actually not all that low current, you can have 10A circulating at resonance with a kV across the thing, hence the gapped core in the shunt inductor.Regards, Dan.

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 24, 2018, 06:31:25 pm
content: If my amplifier can drive the element off-resonance (whatever voltage is required at that particular impedance level to achieve the desired power numbers), will the piezo actually vibrate? Or is there something about needing to have the piezo present as a resistive load rather than reactive?

author: dmills
author_rank: Super Contributor
timestamp: January 24, 2018, 06:42:20 pm
content: Have a look at the GB curve for your transducer, you will find that producing any real level in the water off resonance requires the sort of voltages (And thus currents in the fixed C) that make it ah, "difficult").Can it be done? Sure, if you only want tiny levels in the water, but you would find it difficult if you were after the sort of 200dB SPL (Ref 1uPa, standard for underwater work) that is considered normal for a sonar transmitter.Note that not only is the electrical match difficult off resonance, but there will be significant losses in any acoustic matching layer which will now be the wrong thickness to produce the desired transformation.Usually you can manage a sort of flatish octave of bandwidth on transmit without the efficiency going too horrible, more is a big ask.Regards, Dan.

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: January 24, 2018, 06:53:34 pm
content: Dan,Understood. Just to clarify, by GB curve, do you mean impedance vs. frequency, or perhaps you mean SPL(in the water) vs. frequency?

author: dmills
author_rank: Super Contributor
timestamp: January 24, 2018, 08:28:01 pm
content: GB is Conductance (G) and Susceptance (B) vs frequency, for various reasons this is generally used for sonar transducers instead of a complex impedance, (which is clearly just the reciprocal of G+jB.Regards, Dan.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 24, 2018, 11:46:32 pm
content: Quote from: Evan.Cornell on January 24, 2018, 06:31:25 pmA off the shelf amp and possibly a step up transformer to match the impedance.I was stewing on your problem at work today. You indicated an interest in Knowing L R C properties of the piezoelectric while it was running dynamically. This brings to mind a dirty trick HP used a long ago to quickly test components on a PC board. They would send out an sine wave impedance 0 then monitor current at - 90 , 0 , + 90 degrees chopped up clean with a square wave to FET switches. By measuring AC at -90 , 0 and + 90 degrees you have the C R L impedance in one shot without fancy calculations for a given frequency. As luck would have it all computers and lap tops have the required test equipment in them in the sound card standard. 2 function generators with FM AM and most importantly in this case phase change between function 1 and function generator 2. In short you will only have to sort out the software details as the hardware is already there in any computer for C R L details of the piezoelectric unit while it is running.

author: james_s
author_rank: Super Contributor
timestamp: January 25, 2018, 12:13:44 am
content: I remember seeing ads for those piezoelectric fans, thought they seemed cool but I never had one. I've never seen one in any kind of equipment either, I wonder if they ever achieved any degree of commercial success?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 25, 2018, 06:19:52 am
content: Quote from: james_s on January 25, 2018, 12:13:44 amI think I found it in a junk electronic shop. Being unusual I had to have it. I tested it's natural resonance by driving it directly with a function generator. It turned out to be surprisingly high Q only operating from 58 to 61 Hz. Off frequency is only 1/3 the movement. The nice thing about this fan is there are no bearings to dry out. To this day I do not know how the fan was used nor have I seen it in any electronic equipment. It's a mystery.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: January 25, 2018, 03:19:30 pm
content: Running off-resonance is no problem, it can just take a lot of voltage to accomplish anything. Look at the D31 and D33 constants for the material to get an idea of the sensitivity. We used to tune optics with stacks of disks, usually about 0.05" thick, running at 1 kV. Motion was a few to a few dozen micrometers at low frequencies. Arcing was sometimes an issue.

author: dmills
author_rank: Super Contributor
timestamp: January 25, 2018, 04:32:58 pm
content: Yup, sounds about right, you quickly learn a lot about the virtues of Kapton tape and vacuum degassing of potting compounds when playing with that stuff.DC for micropositioners is a rather different problem to driving power into an acoustic load of course.Regards, Dan.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: January 25, 2018, 09:19:52 pm
content: True. Of course off-resonance not only do you not get much motion output, but you still have to drive the bulk capacitance at possibly a high frequency and high voltage. You can resonate it with a low loss inductor, but the circulating currents can still be surprising. Inductor losses become an interesting topic, depending what frequency is chosen. I once worked on a system that had to run at 1 kHz and a very high power level (100W+). The customer just didn't believe the acoustic output would be a problem. Even with hearing protection it was hard to be in the same room with it!

author: dmills
author_rank: Super Contributor
timestamp: January 25, 2018, 09:50:44 pm
content: I did a kW over ~2-4kHz for a marine mammal deterrent system, that thing was pigging loud even above the water, I would have hated to be a diver anywhere in the area. Incidentally, you know you have too much Q when the free flooded ring about a foot across keeps failing due to internal arcing.The fixed cap does tend to make for large currents in that loop, the saving grace for sonar is that you usually run **WAY** less then 1% duty cycle.Regards, Dan.

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: February 01, 2018, 08:57:37 pm
content: So I'm running into an issue where I can easily get a couple hundred volts peak-to-peak and an amp or two peak-to-peak into the piezo element, but it's below resonance, and current is close to 90deg (maybe 77-85deg) out of phase with voltage. Obviously, voltage * current = power, and so even my peak power is rather lower than I'd like. Does this mean I'm not actually doing work (i.e. converting signal to sound) in the piezo element?Do I need to have matching network to make the impedance mostly real in the frequency range of interest, in order to actually get mechanical output for all my voltage?Thanks!

author: John Heath
author_rank: Frequent Contributor
timestamp: February 01, 2018, 10:53:01 pm
content: Quote from: Evan.Cornell on February 01, 2018, 08:57:37 pmQuoteDo I need to have matching network to make the impedance mostly real in the frequency range of interest, in order to actually get mechanical output for all my voltage?End quoteWow that was a lot of information in a succinct intuitive way that does not need a calculator to understand and act on . Nicely done. Have not heard it put that way but I know exactly what you mean. Matched impedance is the max energy transfer point with 50 percent efficiency and 100 percent real power.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 02, 2018, 12:13:53 am
content: You don't need a matching network, if your amplifier can supply all the reactive power that must come with the desired output power.Acoustic as well as electrical matching would seem to be desirable here!Tim

author: Evan.Cornell
author_rank: Regular Contributor
timestamp: February 02, 2018, 12:47:25 am
content: John, thanks very much for the compliment! I've been doing quite a lot of thinking on this topic lately, glad to know it's sinking in somewhat!Tim, understood that the amplifier requires to supply both reactive alongside real power. Does that imply that only the real power is doing work in the transducer?Thanks,Evan

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 02, 2018, 02:09:09 am
content: Quote from: Evan.Cornell on February 02, 2018, 12:47:25 amYes, more or less by definition.(Also, whether that's work "in" the transducer, or transmitted as useful acoustic work, is another matter, but hopefully it's reasonable efficiency.)Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: February 03, 2018, 01:31:45 am
content: Yes the days when a speaker impedance should match the old tube type amplifiers is long over. A solid state amplifier with negative feedback will electrically simulate an impedance of Z = 0 ohms so the only relevance of speaker impedance is how low before the amplifier gets into trouble.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 03, 2018, 01:35:11 am
content: The small-signal output impedance of the amplifier doesn't matter very much. The impedance matching theorem strictly applies to linear Thevenin/Norton equivalent circuits. Amplifiers are not linear (at some point or another).Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: February 07, 2018, 02:14:26 pm
content: Quote from: T3sl4co1l on February 03, 2018, 01:35:11 amThere is an interesting way of thinking about a condenser coupled amplifier. With a 10 volt power supply the output goes goes from ground to 10 volts or somewhere in the middle. When the output goes to 10 volts the electrons are going from ground through the speaker through the condenser then discharge the power supply. However when the output goes from 10 volts to ground it is not power supply electrons that are being used to drive the speaker. .It is the electrons that are stored in the output coupling condenser that are used in this case. The 10 volt power supply is not being discharge. It is the output coupling condenser that is discharging not the power supply. You have the chance to use the same electron current twice . On the way up the power supply pays the price for energy. But on the way down it is free energy as the power supply is not being discharge only the output coupling condenser. It is not free energy as that one is a no no but it is interesting to think of it as using the same current twice.

author: John Heath
author_rank: Frequent Contributor
timestamp: February 07, 2018, 02:47:55 pm
content: Quote from: Evan.Cornell on February 02, 2018, 12:47:25 amThere is a way to measure real vs imaginary energy. Monitor both voltage and current going into the piezo on a scope. With the two wave forms of current and voltage on your scope match up their amplitudes so that they are the same and overlapping . What is in phase for voltage and current is real power. What is out of phase is imaginary power. Am example is a piezo on resonance no load , high Q. In this case the voltage and current will be mostly out of phase , imaginary power. It could break the piezo the same way a wine glass in resonance will shatter into pieces.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 07, 2018, 06:22:41 pm
content: Quote from: John Heath on February 07, 2018, 02:14:26 pmBack in the day, you wouldn't even pay for electrons, the power company would just sell you the same old batch, brightened up a little each time.Since the introduction of AC power, they don't even do that, they just make 'em wiggle!Tim

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: February 07, 2018, 06:59:45 pm
content: Mogami has some fun electronic stuff here-http://www.mogami.com/e/puzzle/I think this one is the best-http://www.mogami.com/e/puzzle/pzl-03.html

author: John Heath
author_rank: Frequent Contributor
timestamp: February 07, 2018, 10:44:46 pm
content: Quote from: T3sl4co1l on February 07, 2018, 06:22:41 pmWell described. With a electron drift velocity of only a few m meters per second the electrons are barely moving. Just vibrating a bit. Now that I think of it I should get a discount from hydro as they are not giving me new electron rather vibrating the old ones. Either that or run some DC for a while so I at least have some new electrons for the money I am paying them.

author: John Heath
author_rank: Frequent Contributor
timestamp: February 07, 2018, 11:17:03 pm
content: Quote from: Conrad Hoffman on February 07, 2018, 06:59:45 pmNice link. Not the first one as I followed your recommendation to check out the second link. Let me repeat it so it is not lost.http://www.mogami.com/e/puzzle/pzl-03.htmlYes indeedee we can not be violating energy conservation laws so how do we dig ourselves out of this hole , Hmmm?There has to be a way.I have some thoughts on this but not with confidence . I am going to weeny out and ask for your take on this.

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 08, 2018, 08:49:35 am
content: Quote from: John Heath on February 07, 2018, 11:17:03 pmCharge is conserved, but the loss of energy is not a mystery. Some of it is radiated away, some is converted into heat and perhaps sound and light.See:http://www.hep.princeton.edu/~mcdonald/examples/twocaps.pdfand:https://www.physicsforums.com/threads/capacitors-paradox.687927/

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 08, 2018, 12:13:19 pm
content: There are no EM paradoxes, only incomplete questions.One cannot ask about E dynamics without necessarily including M.That is, a problem with C's changing must necessarily account for L as well.R is not necessary, but helpful.An ideal (no R) solution to the problem is simply the energy difference remains in the oscillating field. Peak input current is I = Vin / Zo, and peak voltage at the far end of the line is 0 and 2Vin.So you see, the problem is posed specifically to confound beginners to the subject. Another old question: "A ship has 26 goats and 10 sheep onboard. How old is the captain?" The correct answer is "insufficient information", but people are not taught to question premises, rather to forge ahead and guess an answer no matter how unreasonable it might be.Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: February 08, 2018, 02:47:36 pm
content: Quote from: T3sl4co1l on February 08, 2018, 12:13:19 pmI managed to get myself in trouble with this one and not sure how to navigate back to safe waters. I have two cables that have an impedance of 1 ohm. I will terminate these cables at the end with 1 ohm resistors. I will put a 1 second 1 volt pulse into both cables . My energy cost to do this is 1 watt for each cable. Later in time this 1 second pulse will terninate at the resistors leading to 1 watt of heat. So far we seem to be okay.I will reverse the polarity of one of the pulses and join the end of the cables to 1 termination resistor of 1 ohm . Now when the pulses arrive at the termination resistor the polarity is reversed so it is not 1 but 2 volts across the termination resistor. 2 volt across a 1 ohm resistor is 4 watts. My cost to put a 1 volt pulse in the cables was 1 watt each so how could I have 4 watts of heat where the two cables meet at the 1 ohm termination resistor? 2 watts went in and 4 watts came out. This has to be wrong. The cable is 10 light seconds long so the pulse generator can not see the termination resistor for the 1 second period. It only sees the reactive properties of the cable , Z = 1. Two 1 watt pulses end up being a 4 watt pulse where the cables meet at the 1 ohm termination resistor.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 08, 2018, 04:01:37 pm
content: Simple: two cables in parallel is a Thevenin source of 1/2 ohm. Terminating that with 1 ohm leads to a reflection. Sooner or later, that energy comes back to the supplies (or just rings down in the cables, some minutes later).Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: February 09, 2018, 06:32:44 am
content: Quote from: T3sl4co1l on February 08, 2018, 04:01:37 pmYes from a resistance / reactance point. A pulse propagating down a cable will see the resistor as 1 ohm therefore match the cable 1 ohm Z impedance. . It will also see the continuation of the other cable connected to the same resistor as a 1 ohm Z impedance. 1 ohm resistor and 1 ohm Z load so effectively 1/2 ohm as you said so a 50 percent reflection that will return to the source. However I the trouble maker sent a pulse down the other cable timing it just right so that it would meet your pulse where the resistor is. You are welcome. Now what happens? Resistance , reactance plus information it self . the pulse that is additionally complicated by identical information pulse A and pulse B that is propagating in opposite direction meeting at the resistor at the same time. Too many variables on the table. Let us forget the resistor for now and deal with reactance and information only.Reactance is 1 ohm Z but when pulse A runs head on into pulse B they will see each other as a infinitely high impedance as information A is identical to information B , 1 second positive pulse. This should lead to a 100 percent non inverted reflection A and B off each other returning them back to where they came from. A problem here is you can not test to prove this. Reflection A and B are identical so you can not say it is A returning or simply B arriving from the other pulse sourceAdding a little marker such as a bump on top of one of the pulses does not help either as different information is not reflected only same information.Invert one of the pulses Hmmm. That one does not work either as a reflection from opposite information is seen as 0 ohm Z therefore a inverted reflection that would be identical to the pulse from the other source.This is a no win. Did the pulses go through each other or did they reflect off each other. I all else fails return to Newtonian physics and repeat the experiment. One slinky spring with a pulse added at both ends. This one does not work either as the slinky pulses meeting in the middle could be considered going through each other or reflecting off each other. It is subjective at the end of the dayFor what it is worth I am running with pulses reflecting off each other for identical information that is propagating in opposite directions. Forgot about the 1 ohm resistor. Problem for another day.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 09, 2018, 01:12:57 pm
content: Hardly "too many variables". The system is linear and obeys superposition!Regardless of pulse polarity and timing, each pulse fires down one line, reflects off the mismatch, and continues on down both lines.To solve for the waveforms, you draw a lattice diagram:https://www.eeweb.com/app-notes/transmission-line-theory-reflections-and-lattice-diagramsEach transmission and reflection event (a wavefront hitting a port) is a vertex on the diagram, and the edges are labeled with the amplitude of that wavefront. You sum up all the wavefronts hitting and exiting each port, and the waveform is read off along the axis.That's how you make it simpleTim

author: John Heath
author_rank: Frequent Contributor
timestamp: February 09, 2018, 11:16:57 pm
content: Quote from: T3sl4co1l on February 09, 2018, 01:12:57 pmHmm. Your link leads to this link.https://www.eeweb.com/app-notes/transmission-line-theory-reflections-and-lattice-diagramsThis link leads back to your link? It an endless loop. Doe this mean money lattice needs to move from my wallet to their hands before there is a lattice diagram in my future?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 09, 2018, 11:20:48 pm
content: Looks like this...Tim