[SOLVED] Cooling ultrasonic transducer

[engineer1000]

I am running a 50W ultrasonic transducer. I have a 1KV sine wave 153KHz across the transducer with a current of approx 0.05A giving me 50W.

For this particular application I need to run the transducer loose, i.e not bonded to a tank.

The effect of this is the transducer gets very hot very quickly. I have found that if the transducer is house in a a small box and hit with compress air (25PSI) I can keep it cool enough.
I want to replace the compressed air with a fan. Can I get one with the same cooling effect? I was thinking of a hair drying with the heater element removed.

 

[DineshSL]

Also you can consider about fan and air flow guiding systems used to cool in PC processors (and processors in graphic-accelerator cards).

Thanks

 

[chuckey]

You need to make a heat sink that fits tightly around the transducer to increase the heat dissipating area. if its big enough, no fan will be required.
Frank

 

[jeffrey samuel]

the heat sink must fit exactly as there is a possibility of malfunction if the sink is smaller

 

[engineer1000]

Thanks guys,
Can any one explain by the transducer is getting hot ? Someone said it was to do with impedance matching but I don't think that this is true

 

[chuckey]

The transducer is getting hot because not all the power it gets is turned into audio power. If you are very clever you can acoustically match the physical aperture of the transducer to the physical properties of the air. Consider if you try and paddle a boat with a tea spoon as a paddle or a one metre square sheet of metal. With either one you would get hot but the boat would not go very fast!!
Frank

 

[DineshSL]

If it generate lot of heat, that is wasting energy. Check the applied voltage and current are according to specification.

 

[jeffrey samuel]

the power given to the transducer is shed out as heat that is the cause of the heat

1 reason is impedance mismatch many reasons are also responsible like high current, power operation is varied

 

[FvM]

As mentioned in your previous thread, we can assume that the power dissipation is mostly caused by mechanical lossed in the material and roughly proportional to transmitted ultrasonic power. In addition, there may be a certain amount of dielectric losses in detuned operation.

Thermal resistance in forced air cooling is as matter of surface area and flow velocity. It sounds unlikely to achieve the cooling effect of compressed with a fan.

 

[engineer1000]

If I try impedance matching the transformer with the transducer . The transducer is approx 20ohms ( set my manufacture) and set the turns ratio of the transformer to match. When a transducer is connected across it my output voltagfe goes from approx200V to 1000V , this seems to make the impednace matching calcs invalid.
So imedance matching doesn't seem to help.
Can someone please confirm my thoughts ?

 

[FvM]

20 ohm would be in transducer series resonance. But you should clarify if it's specified without additional reactive elements connected to the transducer.

 

[engineer1000]

At resonance I think the impedance should be mostly resistive .
The equations Vp/Vs = n = (ZP/ZS)square root.
if I get the impedance of my gen and I know the transucer is 20ohms. This doesn't work at resonance as the voltage across the transducer is amplifed to 1000v + by the transducer so the equatuions don't seem to work.
I'm trying to impendance match to reduce the heat on the loose transducer

 

[FvM]

The high voltage suggests that you're operating the transducer near parallel resonance, but not necessarily excactly at Fp. In so far I don't understand which equations you're referring to.

 

[engineer1000]

https://www.electronicsarea.com/impedance_matching_transformer.asp

From this link on impedance matching

 

[FvM]

O.K., I misunderstood the meaning of indices p and s I think,the problem is that the transducer apparently isn't operated at the series resonance frequency.

 

[engineer1000]

I think resonant frequency can be found by looking that the max current and the max amplitude of the sine wave when connected to the transducer . I have this frequency, the manufacurer of the ceramic rings has said they would expect about 1kv across the transdcuer at resonace.
I have measured the output voltage of several gens when connected to a transducer and they are all about 1kV. Now given this, I am struggling to work out how to match the impedance of the gen to the transucer.

 

[chuckey]

"I am running a 50W ultrasonic transducer. I have a 1KV sine wave 153KHz across the transducer with a current of approx 0.05A giving me 50W." from your first post, no you have not, you have a VA of 50W, that is current times voltage. This is not real power. Real power is only dissipated in resistors, later on you say that the resistive impedance of the transducer is 20 ohms, this means that the voltage drop across the resistive part is about .05 X 20 = 1V and so the real power is 1V X .05 = .05W. All the other voltage that you measure is across the series capacitance. This gives a series impedance of ~1kV/.05 = 20K ohms. To make this thing work properly you have to add in series an inductive impedance of 20K ohms. When you used your transformer, its leakage inductance made up for this, so the circuit started to resonate (good) and the voltages rose (good) and so you should have brought your driving power down, else you will over dissipate the transducer. Measuring voltages around a circuit that is somewhere near resonance, will only tell you that much, NOT how much real power is involved.
Frank
Frank

 

[D.A.(Tony)Stewart]

What is your schematic with transducer PN.

Driving with square wave or sine?

 

[FvM]

Measuring the impedance locus (complex impedance versus frequency) is the only way to clarify things.

See also this previous discussion: https://www.edaboard.com/threads/213634/

 

[manha7]

P = V*I*cos(theta), theta is a angle difference between V and I. Angle difference is induced from a transducer Impedance(C0). Impedance matching is removing C0 by L(Transformer). Matching circuit is equal to a L-C resonance circuit. This thema is equal to Power factor.