ultrasonic Ringing Phenomenon

[engineer1000]

I am feeding a ceramic ultrasonic transducer with a 300V pk pk sine wave via series inductor or choke . When the transducer hit resonance the voltage across the transducer rises to approx 1000V which is called ringing.
Can someone explain this Phenomenon and why?

 

[chuckey]

No its not called ringing, ringing occurs only on the edges of a fast changing step waveform, like a square wave when sine wave damped oscillations can be seen. What you are seeing is the actual resonance of the series C L and R circuit.
Basically the reactance of the C opposes the reactance of the L, leaving the the voltage across the R equal to the input voltage. I could do the calculations for you but I am too lazy, but they go like this:- If the R = 300 ohms, then at resonance, the current is 1A (I = V/R). But because of the reactance of the C and L being 1000 ohms, with a current of 1A there will be 1 X 1000 volts dropped across each one BUT WITH OPPOSITE SIGNS. Now you can't measure the volt drop across the C because its inside the transducer, you measure the voltage across the L & R. This is why you get such a high voltage. - nothing to worry about!!
Frank

 

[engineer1000]

Thank you for your replay. Could you just explain where you got the reactance of C and l (1000ohms). A thought at resonance Xl AND Xc canceled each other out leaving the real part R

 

[chuckey]

The capacitive reactance is part of the transducer. YOU put in the series L.
Frank

 

[engineer1000]

The capacitive reactance is part of the transducer. YOU put in the series L.
Frank

Sorry That didn't really answer my question I have a series L in line with my transducer, but at resonance don't they cancel out leaving R?

 

[chuckey]

Yes that is what you want, this means that all the voltage you send to the transducer is doing useful work also that transformers work. If your signal source does not like a 20 ohm load, you can utilise your "L" as part of a PI network to magically transform your 20 ohms into just about any resistive impedance, so as to match your signal source.
Frank

 

[engineer1000]

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Thanks Frank, Could you expand on the PI network idea?