I’m (trying to) building a ultrasonic cleaner for cleaning PCBs.
I have 2 ultrasonic 100W 28khz PZT transducers connected in a parallel fashion.
link: http://postimg.org/image/aksjbdtuv/
C7,C8,C9,C10,R5,R6,L4,L5 form the equal vent of the transducers.
The resonance frequency of the transducers seems to be f = 28659.9hz.
Therefor the circuit should operate at this frequency.
To compensate for the capacitive of the transducer I have added a inductor L3.
According to the math the impendence of the transducers with the compensation coil would be Z=3.445 ohm.
So the secondary of the transformer (L1,L2) should be 3,445 ohms?
Then the inductance should be 19.13085395µh.
Please ignore the primary halve of the circuit it is not calculated, the values do not make sense.
Can anyone confirm I’m on the right track here?
I’m in a learning phase so please be patient. 😀
I have 2 ultrasonic 100W 28khz PZT transducers connected in a parallel fashion.
An externally hosted image should be here but it was not working when we last tested it.
link: http://postimg.org/image/aksjbdtuv/
C7,C8,C9,C10,R5,R6,L4,L5 form the equal vent of the transducers.
The resonance frequency of the transducers seems to be f = 28659.9hz.
Therefor the circuit should operate at this frequency.
To compensate for the capacitive of the transducer I have added a inductor L3.
According to the math the impendence of the transducers with the compensation coil would be Z=3.445 ohm.
So the secondary of the transformer (L1,L2) should be 3,445 ohms?
Then the inductance should be 19.13085395µh.
Please ignore the primary halve of the circuit it is not calculated, the values do not make sense.
Can anyone confirm I’m on the right track here?
I’m in a learning phase so please be patient. 😀
Last edited:
I got some time to calculate the transformer, and this is what i got so far.
http://postimg.org/image/tw9y7pth9/
The current is much lower as suspected, is this because of the high Q on the secondary?
An externally hosted image should be here but it was not working when we last tested it.
http://postimg.org/image/tw9y7pth9/
The current is much lower as suspected, is this because of the high Q on the secondary?
Your pics are utterly unusable, and I don't see what you mean by "vent", but anyway you should operate the transducers at the conditions stated by the manufacturer. At the resonance frequency, the transducer has to be resistive by definition, there should be no need to add somethingC7,C8,C9,C10,R5,R6,L4,L5 form the equal vent of the transducers.
The resonance frequency of the transducers seems to be f = 28659.9hz.
Therefor the circuit should operate at this frequency.
To compensate for the capacitive of the transducer I have added a inductor L3.
Certainly not, the transformer needs to have a large enough magnetizing inductance to make it negligible, and probably more importantly, it should have a large enough v*s product to operate at the intended voltage/frequency.According to the math the impendence of the transducers with the compensation coil would be Z=3.445 ohm.
So the secondary of the transformer (L1,L2) should be 3,445 ohms?
Then the inductance should be 19.13085395µh.
Thank you for your kind reply, with "Vent" I actually mend "equivalent".
I'm sorry for the bad quality of the pictures I will re-upload them.
These transducers are always capacitive if it is in resonance, because the equivalent circuit of a transducer shows a capacitor in parallel with the resonance circuit. That’s why an inductor is added to get the phase shift close to 0.
About the transformer: I’m quite new to this array, see picture 2 of the calculated values i have found.
I'm sorry for the bad quality of the pictures I will re-upload them.
These transducers are always capacitive if it is in resonance, because the equivalent circuit of a transducer shows a capacitor in parallel with the resonance circuit. That’s why an inductor is added to get the phase shift close to 0.
About the transformer: I’m quite new to this array, see picture 2 of the calculated values i have found.
The effect of the parallel capacitance will normally be tiny, and will be compensated by a small shift in the resonance frequency, you don't need to add explicit compensation components.These transducers are always capacitive if it is in resonance, because the equivalent circuit of a transducer shows a capacitor in parallel with the resonance circuit. That’s why an inductor is added to get the phase shift close to 0.
If you want to drive them at a different frequency, then it could be important.
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