I am not a physicist, but my current project drives me to some physics-related computations, hence seeking help.

I have some ultrasonic transducers, 5938D-25LBPZT, for which very limited information is available.

Transducer Diagram Transducer Photo

In order to define the parameters for an ultrasonic booster and amplification horn for this device, I need to compute the displacement amplitude of its radiating surface. The procedure I am following, and the base data (or where data is unavailable, my assumptions) are thus:

Operating power: $50\ \mathrm W$ RMS (the transducer is rated for $60\ \mathrm W$)
Emitting horn material: stainless steel SS316 (best-guess, unfortunately, as SS316 is typical in other such transducers)
Emitting face diameter: $5.865\ \mathrm{cm}$ (measured)
Frequency at resonance: $f=24\,989\ \mathrm{Hz}$ (measured)
Best-case power transmission: $90\ \%$ (in forward direction, according to various sources)


Density of SS316: $\rho=7.8\ \mathrm{g/cm^3}$
Velocity of longitudinal sound propagation: $C=5.8\ \mathrm{mm/μs}$ (from this reference, approx)
Should the axial velocity of sound be considered instead for this calculation?

Calculations so far:

Face area: $27.016\ \mathrm{cm^2}$ (ignoring area lost to bolt-hole)
Sound intensity at face: $I = 90\ \% \times \text{power}/\text{area} = 1.666\ \mathrm{W/cm^2}$
Displacement amplitude $A$:
$$A=\sqrt{\frac{2I}{(\rho C)\omega^2}}$$ where $\omega = 2 \times \pi \times f$
Thus, $A = 172.8296514\ \mathrm{nm}$

My questions are:

  1. I am not sure I have managed to convert everything to consistent SI units, so I need help confirming that I've got that part right
  2. If anyone has experience in such power ultrasonic piezo transducers, a confirmation that my results are more or less within the realms of plausibility would really help.
  3. Is there some significantly incorrect assumption or constant value in the foregoing?
  4. Should I be using some online resource that trivially calculates all this from the rated power and other inputs, instead of struggling with it manually?

Update: Adding diagrams to show the excitation direction and materials involved.

Ultrasonic Stack
(please ignore Diamond insert)

This is the final arrangement of the Ultrasonic Stack. The booster and horn in the diagram above will both be SS316. In order to compute dimensions for these two parts for resonance, I need the displacement amplitude of the front face of the steel "radiation head" in the transducer (first diagram here), which is to be bolted to the booster. The bolt-hole in the radiation head can be seen in the photograph above.

Another diagram, for a similar "Ultrasonic Stack", with the Langevin transducer itself split up. This diagram has a cylindrical radiation head, as opposed to the cone section shape in the transducers I need to compute for.

Ultrasonic Stack
(please ignore Wedge)

(Last two diagrams are from articles on ScienceDirect.com)


1 Answer 1

  1. Your calculations appear to be correct. I believe the density in your equation corresponds to the density of the material through which your sound wave is propagating. Your usage of steel indicates that you are sending sound through a steel plate (not air). If this is intentional, your calculations are correct.
    You should be using the axial sound speed if your transducer is oscillating perpendicular to the transmission direction. This would apply, for example, if you had a transducer and a receiver, with their transmission axes oriented parallel to one another, both placed on the same side of a long plate. If, on the other hand, the transducer is oscillating parallel to the transmission direction (e.g., at the end of a steel rod, transmitting along the length of the rod), then the longitudinal sound speed is the appropriate speed. (If you intend to transmit through air, then shear waves are not supported. Moreover, in the case of air, the equation you used is not wholly applicable, because the emitting horn is a finite body, rather than a quasi-infinite medium as assumed by the equation.)

  2. The value you have computed is reasonable. There are a number of (pay-wall blocked) academic articles discussing transducer amplitudes. These amplitudes range between minimum detectable values of a bit over 1 angstrom (0.1 nm) to rupture inducing amplitudes of ~5 micron (5000 nm). Of course, these values depend heavily on the driving frequency, power, and geometry of the transducer. However, you should have confidence in the values you have computed. I would add that, since the uncertainty in your material density and sound speed is somewhat large, so too will be the uncertainty in your amplitude. To more accurately reflect this uncertainty (using only significant figures, rather than a rigorously computed uncertainty), your amplitude is about $1.7\times 10^{-5}$cm.

Edit in response to comment: I understand better now what it is you are doing. The equation you have used above is for computing the amplitude of sound waves propagating through a uniform medium of known density and rigidity (equivalently, known density and sound speed). In the geometry you have laid out, you would want the longitudinal sound speed, because the driver axis is parallel to the horn, which is the transmitting body. However, I must qualify that the use of a small booster and horn invalidates the above result. The result you have computed is the amplitude of oscillations that would be driven, given a known power flux and frequency in an infinite material. This kind of analysis would be reasonably well suited to sending signals through girders in bridges, and finding defects if anomalies in the expected result are found. In the arrangement you are describing, the very point of the booster is to defeat this response by creating a non-uniform distribution of mass thereby modifying the amplitude of the transducer. If you are trying to compute the necessary booster and sonotrode (horn) that you will need, you need to consult the manufacturers specifications as to how they define the driver amplitude. It appears to me that boosters are usually designed around two parameters -- the gain, and the frequency. The transducer, on the other hand, primarily specifies frequency and power. The amplitude will depend a great deal on the rest of the system, i.e., the mass of the horn, and whatever the wedge may be in contact with. While trying to find out more on the subject, I believe I found an answer to your question number 4, although I'm not sure how much use it will be to you. Suffice to say, these types of problems are handled through finite element analysis, not the simple and elegant, but very limited equation you have used above.

  • $\begingroup$ Thank you for this response. Yes, the medium of propagation is steel, please see the new diagrams and explanation I have added to the question. About axial v/s longitudinal speed, I don't really know, as some computations I have found (e.g. for ultrasonic boosters) use axial, others use longitudinal. Transmission is straight down the line from the electrode down the radiation head through the booster to the horn tip. For now, my computation needs are for the radiation head surface. $\endgroup$ Feb 13, 2013 at 14:38
  • $\begingroup$ @AnindoGhosh: See updated answer. $\endgroup$
    – KDN
    Feb 13, 2013 at 18:50
  • $\begingroup$ That very sonotrode calculator (and it's more advanced downloadable version) was what triggered my question: That tool requires the unloaded displacement amplitude of the radiation head, in order to compute sonotrode dimensions. $\endgroup$ Feb 13, 2013 at 18:54
  • $\begingroup$ If they really mean unloaded, then I would (naively) assume they mean the displacement in air. It looks like their calculator assumes an input amplitude, rather than an input power. In this case, you might consider specifying the amplitude you would like to achieve and then perform the calculation. If you are planning on using a booster, this will change the amplitude and increase the power draw of the transducer. The amplitude you computed might be a good starting place, but this is a really a question for an engineer. $\endgroup$
    – KDN
    Feb 13, 2013 at 19:06
  • $\begingroup$ Thank you for the insights you have already provided. The reason I'm in the middle of this is that the engineering firm involved delivered non-fucntional dimensions, and the client figured that an electronics engineer is pretty much the same as any other, and there I was :-) $\endgroup$ Feb 13, 2013 at 19:17

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