I want to design a driver for PKT40A from ThorLabs. The spec sheet indicates "Power (Typical, Instantaneous @ 3 kV)" of 700W, but the "Impedance at Resonant Frequency" of <12 ohms. By the Ohm's law driving 3kV through 12 ohm load will generate power of 750 kilowatts, which is much greater than indicated 700 watts. So I'm clearly missing something. There's also a graph in the spec that indicates impedance of <1ohms at the resonant frequency, which gives ridiculous 9MW when driving sinusoid with 3kV amplitude and 20kHz frequency through it. Maybe the indicated power is that of the produced sound waves and not that of electrical consumption? But that makes it terribly inefficient. The spec also indicates "Drive Voltage Range" of 0-5 kV. Does that mean driving 5kV will generate more than 700 watts? How does the amplitude of the AC driving the transducer relate to it's power output in general and how does one derive safe max values for operation from the specs?

  • \$\begingroup\$ @periblepsis The source impedance should be low. See my answer. \$\endgroup\$
    – RussellH
    Commented Jan 15 at 0:35

2 Answers 2


Looking at the data sheet, it calls for a DC bias voltage that may vary from 0 to 5kV. The actual driving voltage at 20 kHz will be a lot less (with an impedance of say, 6 ohms, the voltage would be about 65 VRMS for 700 watts. Presumably, the recommended bias voltage is 3 kV. No driving circuitry is shown but there would have to be provisions to separate the high voltage DC bias from the AC drive signal. This might be as simple as an inductor. You might consider contacting the manufacturer and asking about suitable drive circuits

  • \$\begingroup\$ How do we get value of 65 VRMS from the 700 W @ 3 kV? \$\endgroup\$ Commented Jan 14 at 23:48
  • \$\begingroup\$ "actual driving voltage at 20 kHz will be a lot less ": The total voltage, ac+dc, must be less than 5000V \$\endgroup\$
    – RussellH
    Commented Jan 15 at 0:37
  • \$\begingroup\$ @beardeadclown From the data sheet, it appears the impedance is 0.8+j0 ohms at 20 kHz. Thus, for 700 W you would need 23.7 Vrms at 20 kHz on top of a DC bias. When you put a load on the transducer, the impedance will change. It would be wise to measure the loaded impedance. The DC bias, as recommended in the data sheet, should be under 3 kV to prevent arcing of the wiring. \$\endgroup\$
    – qrk
    Commented Jan 15 at 1:46
  • \$\begingroup\$ I see, so at 21.4 kHz we would need 60 Vrms, hence the wave generator should support amplitudes of 30 - 90 V. I don't however understand where the need for biasing comes from? After all, "Drive Voltage Range" is 0 - 5 kV. \$\endgroup\$ Commented Jan 15 at 2:03

The 12Ω resistor is not the load. The chart showing the resonances ia measured with no load. Here the load means either a mechanical or thermal load that the transducer is used for.

Applying transducers includes the dynamics of the non-electrical side as well.

A crude lumped sum model example for an ultrasonic heater involes the relationsips between thermal power P (heat flow in watts) and temperature T in kelvins.

Thermal resistance is: $$R_T=\frac{T}{P}$$ and Thermal capacitance is fined as: $$P=C_T\frac{dT}{dt}$$

There is a conversion (hopefully) constant K that converts the electrical signals to/from the thermal signals.

Like a transformer the transducer "secondary", the thermal side, is reflected to the electrical side, apearing as a load shown in series with the internal resistance of the transducer, as shown in the diagram.

For a mechanical agitator, the mechanical dynamics would also create a load reflected to the electrical side.

Your challenge is to model these dynamical systems to effect your design,


simulate this circuit – Schematic created using CircuitLab

Be aware that the conversion constant may not be the same for \$C_T\$ and \$R_T\$. I am not sufficiently practiced to provide an acurrate thermal or mecanical model with the ralationship to the electrical system. Perhaps others can add some information in this regard.

The main purpose of this answer is to show why the 12Ω resistor is not the load but the internal resistance of the transducer. This is a parasitic value that should be kept as small as possible. At 20KHz it is 800mΩ.

Do not operate at high voltages with no mechanical load. Even the thermal heating will provide an effective mechanical load. Applying voltage causes bending of the ceramic. Applying a high voltage at no load could overbend, cracking the transducer.

Don't allow the voltage to be negative. The dc bias is required to prevent the sinusoidal voltage from driving the input negative.

  • \$\begingroup\$ My main takeaway is that the amplitude can be as high as 5kV? In that case how does one calculate power and current at different amplitudes to choose wire gauges and component ratings accordingly? \$\endgroup\$ Commented Jan 15 at 0:43
  • \$\begingroup\$ You need to know the mechanical load on the transducer (force times velocity is power). For a heating system there will be another conversion through friction from heat. This will depend on the equipment thath the transducer is attached to, and how it is attached. This has to come from the application, not the datasheet. Since the application is not included, this is as far as I can go. Since I have little experience with these this is as far as I can go. You need to study how the mechanical/thermal systems convert to power and current to determine voltage and wire gauge. \$\endgroup\$
    – RussellH
    Commented Jan 15 at 1:07
  • \$\begingroup\$ Since this is converting electricity to sound waves I don't think the concept of mechanical load is applicable. \$\endgroup\$ Commented Jan 15 at 1:14
  • \$\begingroup\$ Sound waves are mechanical vibrations. The sonic environment may provide a different conversion constant, by the electrical equivalent diagram is the same. \$\endgroup\$
    – RussellH
    Commented Jan 15 at 1:19

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