0
\$\begingroup\$

I want to design the impedance matching circuit for an underwater acoustic power amplifier. Because the power amplifier should work in a range of frequencies, the output impedance changes due to the frequency variation (i.e. I think I need a wideband impedance matching). Thus, the common matching methods such as transformers and filters are not suitable for my purpose. I just know the input and output impedances, nothing more.

How do I solve this problem? Would an audio transformer work? If not, what should I use instead?

\$\endgroup\$
  • \$\begingroup\$ What is the transducer you are driving? In general if you can adjust the impedance of your amplifier then make it as low as possible. Impedance matching (for max power.) is usually only needed when you can't change the amp impedance, such as the 50 ohms used at RF. (A quote) "If you can only vary the load impedance (the source impedance is fixed), the best power transfer occurs when the load impedance matches the source impedance. If you can vary the source impedance, its best impedance is ZERO!" \$\endgroup\$ – George Herold Nov 10 '14 at 17:41
  • \$\begingroup\$ Are you saying that the transducer impedance changes with frequency? Please be more specific: What range of frequencies, what power levels, what are the impedances involved, and how much error tolerance you can allow. \$\endgroup\$ – Dave Tweed Nov 10 '14 at 19:18
  • \$\begingroup\$ The transducer I am driving is a piezo one. \$\endgroup\$ – farshid Nov 11 '14 at 16:49
1
\$\begingroup\$

If the transducer is a piezo, it will exhibit strong impedance variation with frequency. There will be no one fixed optimum source impedance other than zero (or as close to zero as you can get). Also, if it is a piezo, you won't be varying the frequency very much. The best drive happens when excited at the resonant frequency. Because of the relatively high impedance, you may need a significantly high drive voltage. Typically a conventional audio power amplifier is not a good driver because they are designed to drive a low impedance (usually 4 ohms to 32 ohms).

Some piezo transducers are resonant above audio. But, if it is for underwater use, the frequency can be as low as 2KHz. You have to know the resonant frequency! If it is for echo-ranging, then transient response is critical.

This is not a very clear response, but it is very significant challenge to design a "generic piezo driver". Much more specific design guidelines can be provided if you can specify the piezo resonant frequency, the impedance at the resonant frequency, and the desired output power.

Jim Wagner Oregon Research Electronics

\$\endgroup\$
0
\$\begingroup\$

You would be doing very well to get more than about an octave or so of useful transmit power bandwidth out of a piezo or piezo composite transducer.

The usual approach is to use a tuned transformer right behind the element to provide both voltage step up and to tune out most of the fixed capacitance, then back this up with an L or T network (with series resistor to kill the Q and limit the misbehavior at the band edges).

The general approach to power amplifiers is to design them to drive a low impedance (which is potentially highly reactive, SOA is a real issue) to a hundred V or so, and then to design a match for whatever transducer you are using to suit. I think that if I was trying for an amp that could adapt to different transducers automatically, it would include an impedance bridge and relay switched L network at the output.

2KHz would be a rather large tonpills or free flooded ring element which is a somewhat specialized thing often used in arrays for applications like sub bottom profiling, most civilian transducers are resonant a decade or so higher then that.

Here is what a fairly typical omnidirectional broadband transmit element looks like Neptune sonar D11-BB note the two spikes on the admittance plot that pretty much define the edges of the usable bandwidth, note also that the broadband devices usually have a thinner piezo ceramic and so are less able to cope with large voltages.

Regards, Dan.

\$\endgroup\$

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.