What happens when you press a piezo disk

[socratidion]

Or rather: I know that if I push on a piezo buzzer disk in a particular way, the voltage reading on my oscilloscope (reading the difference between the two attached wires) jumps up, then falls again, even though I'm still pressing. Nothing I can do to stop it going down to zero. Then when I release the pressure, the voltage drops below zero, and then bobs back up to zero again. So why, exactly, does that happen? I've got some theories, but I can't find any kind of confirmation (information on the subject is either way too technical for me, or simple but off the point, e.g. it talks about short pushes, impacts, and vibration; nothing about continuous pressure). The best I can come up with is: under pressure, the disk spews out a finite amount of charge. When that charge is gone, the voltage drops. When the pressure is released, the piezo wants the charge back again, so the voltage drops below ground to suck the charge in. Is that broadly right? Where is the charge going when I press? Where is it sucked from when I release? Does my oscilloscope probe play any role, e.g. closing the circuit, making available a path to ground?

Now say I attach a capacitor, between the positive and ground wires. Do I expect anything different to happen? What difference might it make that the piezo voltage change is reflected in the capacitor (not just disappearing into thin air as it did before when there was an open circuit)? By observation, using 47nF with a small buzzer disk, things seem to happen a little more slowly, more smoothly. Again I wonder if my probe is clouding the issue by draining off charge.

[ataradov]

So why, exactly, does that happen?

That's the essence of piezo effect. I'm sure Wikipedia will have excellent artcle on this.

EDIT: It sure does! https://en.wikipedia.org/wiki/Piezoelectricity

[nowlan]

You should look at nike+ipod for fun.
http://electronics.howstuffworks.com/gadgets/fitness/nike-ipod1.htm

[uncle_bob]

Hi

Quick answer - piezo devices convert electric field to strain (or displacement if you prefer). They are bi-directional devices. They are just as happy to do the reverse (convert strain to field).

Since they are just converting the strain (pressure) to a voltage on the plates (a capacitor) there is essentially zero current available. Once you drain the charge with your scope probe, no more voltage. A small charged capacitor on your probe would do the same thing.

Bob

[rs20]

socratidion,

Your intuition is spot on -- the oscilloscope provides a path to ground (1 megaohm, but still). The voltage cannot stay high indefinitely while you have the disk pressed down, because you're only putting a finite amount of energy in when you depress the disk. A non-zero voltage for an indefinite time is an infinite energy! Finite energy input cannot yield infinite energy output!

So yes, it's as if there's a capacitor in series with the piezo element. The 1 megaohm resistor discharges that cap, or as an alternative viewpoint, the signal is AC-coupled so DC doesn't survive.

[vk6zgo]

The piezo device doesn't know the difference between when you push the diaphragm,& when you release it.
Both actions deform the piezo material & generate a voltage.
To quote my first Year Electronics Lecturer:

"Quartz,Tourmaline & Rochelle Salts exhibit the characteristic known as Piezoelectricity"----------& so on

I must have been impressed to remember the introduction verbatim after 50-odd (& some of them were very odd )years.

[djacobow]

If you tap it three times, you get your wish.

<ducks>

[socratidion]

Thank you for your replies.
The wiki article was of course my first recourse: I simply didn't understand the maths, and for the rest, could not see how the information there had a bearing on my specific question. (I've got a lot of respect for Wikipedia's technical material, actually, and I've often found it to be more accurate than printed sources, but also tend to find that it only makes sense if you already know what it is saying).

There must be some current available when you press the disk, no? Just a very small amount. I mean, as long as there is a voltage, there must be a charge, and if there's a charge, there must be the possibility of a current.

During the time I am pressing the disk down, I am continuing to expend energy (sometimes I increase the force, to see if I can stop the voltage from dropping again). So I don't see why the voltage should not stay at the high level. It's not that I'm expecting infinite energy: I expect it to stop sometime, after I let go.

OK, so if it's my probe that's causing the minuscule charge to drain off, I tried testing that: I didn't touch the disk with my probe until a few seconds after I had started pressing the disk. The voltage reading was near zero, so either the charge had leaked off some other way (e.g. into the breadboard) or the theory was wrong.

Concerning my final question (what happens when I make a circuit by connecting a capacitor to the terminals of the disk?), I should add that I chose 47nF for the capacitor because it was just over twice the capacitance I measured from the disk itself, with my multimeter: 18nF. My mind is still unsettled by the question. If the voltage rises on the piezo, some charge must flow into the capacitor, so the cap voltage will also rise; but as the charge flows out of the piezo, the piezo voltage must fall. I guess they'll meet in the middle, and the capacitor will always lag slightly behind (but there's no resistor in the circuit, so it must appear almost instantaneous): and the difference in capacitances might explain the 'slowing down' effect I observed? Also does the relative size of the two capacitances impose a ceiling on the voltage produced by the piezo?

Please note, my questions are very specifically about continued pressure on a disk. I have spent three days trying to get general information from 'introductions' and 'how-it-works' type sources. Eventually you just have to ask someone, don't you?

[ataradov]

During the time I am pressing the disk down, I am continuing to expend energy

You are applying force, but you don't move anything after a certain point, so you are not making work in a physical sense of this word.

I expect it to stop sometime, after I let go.

I can make it infinite, just jam it into a vice

charge had leaked off some other way

It did. Piezo ceramic material has its own resistance, so your energy went into heating it up just a little bit.

but there's no resistor in the circuit

With the levels of currents resulting from your press, everything is a significant resistance. Even resistance of the air is not infinite.

But in case of piezo materials, electrons are generated while variable force is applied. It is like a spinning generator, it only works while you spin it.

[rs20]

During the time I am pressing the disk down, I am continuing to expend energy (sometimes I increase the force, to see if I can stop the voltage from dropping again). So I don't see why the voltage should not stay at the high level. It's not that I'm expecting infinite energy: I expect it to stop sometime, after I let go.

It's not terribly important, since you can understand it anyway you like -- but I must correct you on these points.

-- Firstly, if you a pressing down with a fixed force on the disk, you are not continuously delivering mechanical power to the disk for as long as you press down. This is because you are equivalent to a heavy mass sitting on the disk. If one could continuously extract power from the weight of a stationary heavy mass on a piezo disk, then worlds energy problems would be solved! Power equals force multiplied by velocity; velocity equals zero in this case.
-- Secondly, I was portraying a hypothetical where you never let go. If you never let go, and if we assume that the voltage stays high, then we conclude that infinite energy is delivered. But this is impossible (since finite energy was put in), so one of our assumptions must be wrong: the assumption that the voltage stays high.

[Brumby]

During the time I am pressing the disk down, I am continuing to expend energy

During the deformation of the piezo device, you are doing work which gets converted into electrical energy.  Once you have stopped, no more work is being performed (even though you might be maintaining the force), so no more electrical energy is created.

Your muscles may be expending energy to keep your finger in place, but you could remove you finger and simply put a weight pressing down with the same force.  Logic will tell you that you cannot expect energy out of a system that is static like this.

As the piezo is a capacitive device, the electrical energy appears as a voltage across the capacitor and a load - your meter - will close a circuit and current will flow, discharging the electrical energy.  ** Internal resistance will also drain charge from one side to the other.

In this state, there is still some potential energy stored in the piezo device in the form of the mechanical deformation - like a leaf spring.  This is kept in place by the external force of your finger.

When you remove your finger, the stored mechanical energy is allowed to act (and is, thus, doing work), creating another quantity of electrical energy to be generated as the piezo returns to its rest state.


The piezo device will only produce electrical energy during the process of deforming or relaxing.  Stop the movement and energy production stops as well.

** Edit: Added reference to internal resistance

[Ian.M]

For the purposes of circuit analysis, you can treat your disk as a current source in parallel with a capacitor and a high value resistor.   

You've measured the capacitance as 18nF.

The current source's current is proportional to the rate of change of applied pressure.  Constant pressure = no current.  Increasing and decreasing pressure produce currents in opposite directions.

The shunt resistance is very high - for a Maplin YU87U piezoceramic disk transducer, it was higher than the 40Meg my meter could measure to. However if you are touching the disk, YOU are a much lower leakage resistance and the output rapidly decays (After a step change in pressure, it will follow the usual exponential RC discharge curve). To minimize this, it is essential to cover the disk with a low leakage insulator e.g. polypropylene or PTFE sheet and press on that.   

Added shunt capacitance will increase the RC time constant and decrease the output voltage in proportion to the total capacitance.

Interfacing to a piezo sensor can be quite tricky due to the combination of high output voltage and extremely high input impedance needed to get an acceptable LF response.  You have to design your circuit to cope with mechanical shocks generating enough voltage to blow your buffer amp, which requires input clamping to protect it.

[helius]

There must be some current available when you press the disk, no? Just a very small amount. I mean, 1: as long as there is a voltage, there must be a charge, and 2: if there's a charge, there must be the possibility of a current.

Interestingly, both arms of the syllogism here are false. 1, because voltage is simply the difference in the electric field, and electric fields do not only come from charges. They also come from moving magnetic fields. A spinning bicycle wheel with magnets on it, fully grounded and equipotential, induces voltages in its surrounding space. An ideal voltmeter could detect those voltages, but real meters are not ideal: they do not have infinite input impedance. Their finite impedance is like a parasitic resistance that draws charge out of the DUT. 2, because charges in insulators do not flow. The dielectric inside a capacitor has charges that orient themselves against the polarity on the plates; but these dielectric charges can never flow anywhere. Far from being insignificant, they are the whole reason the capacitor works in the first place, and give rise to the fascinating phenomenon of dielectric absorption.

[alsetalokin4017]

For your amusement:

[socratidion]

Thank you, everyone. That certainly addresses my question, and gives me a lot to chew on besides.

[TimFox]

For the purposes of circuit analysis, you can treat your disk as a current source in parallel with a capacitor and a high value resistor.   

You've measured the capacitance as 18nF.

The current source's current is proportional to the rate of change of applied pressure.  Constant pressure = no current.  Increasing and decreasing pressure produce currents in opposite directions.

The shunt resistance is very high - for a Maplin YU87U piezoceramic disk transducer, it was higher than the 40Meg my meter could measure to. However if you are touching the disk, YOU are a much lower leakage resistance and the output rapidly decays (After a step change in pressure, it will follow the usual exponential RC discharge curve). To minimize this, it is essential to cover the disk with a low leakage insulator e.g. polypropylene or PTFE sheet and press on that.   

Added shunt capacitance will increase the RC time constant and decrease the output voltage in proportion to the total capacitance.

Interfacing to a piezo sensor can be quite tricky due to the combination of high output voltage and extremely high input impedance needed to get an acceptable LF response.  You have to design your circuit to cope with mechanical shocks generating enough voltage to blow your buffer amp, which requires input clamping to protect it.

It may be easier to treat the device as a voltage source in series with the 18 nF capacitance, instead of a current source.  The piezo effect converts the applied force to a voltage.