2
\$\begingroup\$

I am looking to build a circuit capable of driving a 600W piezoceramic transducer with an impedance of 200Ω at a frequency of 50kHz. Since it is being used in a sonar application only short pulses(around 400uS) will be output around 2-3 times a second.

I am looking to drive it from a 12V source which will need to be stepped up to around 300V. Initially I was thinking on building an amplifier circuit, inputting a sine wave and using the amplifier to drive a transformer.

However I was wondering. Would it be possible to drive step-up transformer with a square wave, simplifying the circuit greatly, and use a band pass filter on the output in order to convert it into a sine to be fed into the transducer?

Any help would be much appreciated. Thanks.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Band pass filter on 600W output? Hm. Think where the "filtered" power will go. why not just use an off-the-shelf 12V power inverter? \$\endgroup\$
    – Eugene Sh.
    May 12 '17 at 17:30
  • \$\begingroup\$ I see, into the filtering components and not the transducer? \$\endgroup\$
    – smith1993
    May 12 '17 at 17:36
  • \$\begingroup\$ Assuming your transducer is operating at resonance, the transducer itself is a bandpass filter. Your average power is less than 1/4-W. Seems entirely feasible to use a step up transformer. \$\endgroup\$ May 12 '17 at 17:55
  • 1
    \$\begingroup\$ Usually you first scale the voltage up then just transmit with the square wave. You use the transducer itself to filter out the harmonics. \$\endgroup\$ May 12 '17 at 17:59
  • \$\begingroup\$ Your short pulse contains only 20 cycles of 50 kHz. So you need a fairly wide-band step-up method. I'd agree with Spehro, but a resonant load (a piezo) on that transformer will result in currents in the primary winding to be other-than-square, if you intend on square-wave drive. \$\endgroup\$
    – glen_geek
    May 12 '17 at 18:16
2
\$\begingroup\$

I think you need 346V^2/200ohms=600W

Z at 600W is 200 ohms at series resonance? if valid or V is just rounded down, ok.

then at 12V the impedance ratio is \$(346V/12V)^2=831\$ so the driver sees 300/831=0.36 Ohms

For 1% loss or 6W the transistor ESR (RdsOn or rCE) must be 1% or 3.6 mOhms.

Next what frequency? and waveform? square wave? then you need twice the peak voltage to get peak-peak voltage at 50% duty cycle for 326Vrms if you spec is valid and thus RdsOn.

The transformer DCR primary winding must be similar to RdsOn or less for same losses and you probably need Litz wire for low inductance and use a half bridge driver with +12V on centre tap. as well low ESL wiring from driver is necessary.

I'll add more details as you show more specs. on my assumptions. Do you have a datasheet?

Since this is just peak power, what is the duty cycle and thus average power? 400us/333ms x600W = 0.72W.

You need a well balanced transmission line to prevent pulse ringing and thus false echos from long decay times , which limits your short range.

  • so the Q must be low 0.7 like in a good RF amplifier or matched impedance at 50% efficiency . Power loss is not a problem but high Q resonance at a 50kHz frequency with >10 harmonics from a high turns ratio transformer is harder to doand conflicts with shutoff fast response times, than a DC-DC convertor with >50 us response time to loop bandwidth.

    Thus I would step up DCDC supply to say get an equivalent load of 8 Ohms or a turns ratio of 5 and much higher cheaper RdsOn MOSFETs supply low power for lower current driver to get 600W peak.

  • then consider a design for class E audio with 50kHz

Thus with a low turns ratio of 5 is a smarter idea to get 1us decay time on the 400 us pulsed 50kHz carrier, lower current driver and easier to manage impedance ratios.

just an idea.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Thanks for your informative reply. The transducer in question is: link Using a DC-DC converter to boost the initial voltage would actually be beneficial to the design since it would allow me a wider range of input voltages and was something I was thinking on using and have built a few before. So with a turns ratio of 5 I would need around 70 volts to get an equivalent load of 8 ohms? \$\endgroup\$
    – smith1993
    May 12 '17 at 20:08
  • 1
    \$\begingroup\$ Here is a lot more information on the transducer link \$\endgroup\$
    – smith1993
    May 12 '17 at 20:50
  • 1
    \$\begingroup\$ I suppose I should also state that a specific pulse length isn't all that important. I still need to experiment with this once I am able to receive return echos. I have seen circuits where a transducer is connected directly to the secondary of a transformer and the primary side has been fed with a single pulse. This led to a decaying 50kHz sine wave which would be suitable. \$\endgroup\$
    – smith1993
    May 12 '17 at 22:48
  • 1
    \$\begingroup\$ link \$\endgroup\$
    – smith1993
    May 13 '17 at 0:01
2
\$\begingroup\$

The usual approach (I used to do that stuff for a living) for a narrowband driver is to use a tuned transformer (Wind the secondary to have an inductance that resonates with the transducer fixed capacitance) and go for a sane sort of step up ratio.

This is then typically driven by a third or fourth order matching network, generally with some resistance in the series arm (They tend to go very low Z near the band edges), input power is generally a H bridge. It helps to have a smallish boost (or flyback) taking your 12V up to something more sane and charging a honking great capacitor, a couple of hundred volts at the input to the H bridge stomps all over doing it with 12V.

I did a 1kVA (About 500W of real power @ 10-20KHz) design off a 12V rail, the tuned transformer had a 1.5 turn primary, an air gap measured in mm and a magnetising current of ~100A! It worked well, but my god was it painful, output was hitting 800V RMS.

\$\endgroup\$
3
  • \$\begingroup\$ @ Dan Mills why the air gap? \$\endgroup\$ May 13 '17 at 3:23
  • 1
    \$\begingroup\$ Core saturation, with that much magnetizing current the core needed a huge gap to avoid saturating the ferrite. IIRC the transformer was an ETD49, with a whole load of parallel wound litz as the primary wound over a few hundred turns on the secondary. \$\endgroup\$
    – Dan Mills
    May 13 '17 at 13:22
  • \$\begingroup\$ What sort of matching network should I be looking at? \$\endgroup\$
    – smith1993
    May 13 '17 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.