Driving an Ultrasound Transducer

[Mono]

Hello!
Thanks to the help provided in this thread I was able to produce the pulsed ultrasound signal that I needed. (1.5 MHz sinewave, pulsed at 1kHz, 20% duty cycle, 21 V p-p)
The general design ended being as follows: a signal generator that produces the sinewave, a 555 circuit that controls a 4066 which switches the sinewave and finally a LM318 that amplifies the signal to 21 V peak to peak.

I was getting the correct signal by measuring with the oscilloscope without the transducer load. As I connected the transducer, the circuit worked for about 30s and the shut off and some smoke appeared. It seems no big damage was caused.

What happened? What I have read leads me to believe that I need a transducer driving circuit (disclaimer: I am a total electronic noob). I thought the LM318 handeled the "power"part by amplifying the signal, but it seems that this is a current issue, right?

So my question is: do you have any design advice for an ultrasound transducer driver for this kind of signal?

BTW, on the LM318 datasheet on page 9 it shows a circuit for "isolating large capacitive loads". Will that suffice or do I need other completely different driver circuit?

 

[davenn]

ok
what sort of transducer were you putting that signal into ? -- do you have a link to its data sheet ?
maybe either a) it couldnt handle 1.5MHz or b) it couldnt handle 21Vp-p or c) combination of both

a circuit layout/schematic of your project would help us guys to help you

going by the SMA connectors in that DDS sig gene that your link provided it looks to be possibly a RF sig generator

anyway lets see your cct diag
and a complete description on what you are trying to achieve so we all dont have to play 20 questions

cheers
Dave

 

[(*steve*)]

Where did the smoke come from?

Smoke from electronics rarely causes "no big damage", it's usually pretty catastrophic for the devices that smoke.

I have a no smoking policy and eject any components that refuse to comply.

 

[Mono]

Dave:
Here is a basic layout:

It worked perfectly until I connected the transducer.
Also, the transducer is capable of operating at 1.5 MHz and 21 V p-p (one of the few things I knew from the beginning of the project)

Steve:
It is really that "I'm hoping" for no big damage.
I have to go to the lab to remeasure everything (no oscilloscope at home).
It seems the smoke came from the signal generator, but I have to really take it to the lab to know for sure.

 

[(*steve*)]

The signal generator is one place I would have expected smoke *not* to have come from...

 

[davenn]

ok 2 questions to answer

again .... what is the transducer do you have a link to its data sheet?

and as steve asked where did the smoke issue from ?
I would almost guess the Op-amp
any device that issues smoke is usually terminal


cheers
Dave

 

[BobK]

Voltage is not the only thing that determines the power output. The LM318 can only output about 20ma of current. At 21V p-p (about 15V rms) this corresponds to an impedance of 600 Ohms for the transducer. If it's impedance is less than that you the opamp is being overloaded.

Bob

 

[Mono]

Hi guys. Thanks for your help so far!

It actually WAS the 318 with the smoke. I have the signal generator mounted just over the zone the 318 was, so the smoke came from below. I also thought it could be the generator because it turned off when the smoke appeared, but luckily it was the 318.

Dave, I can't find any datasheet for the transducer. It has no marks in it and it was given to me in the lab for this project. I think I should measure it's impedance before designing a driver. Will try it tomorrow at the lab.

 

[(*steve*)]

The piezo device will have a very high resistance, however it has a significant capacitance and the majority of the current will be due to charging and discharging that current several million times each second.

To achieve that voltage swing, you may need a driver capable of several amps.

 

[BobK]

As Steve said, the resistance is not useful. You need to measure the impedance at 1.5MHz. You can do this by connecting your signal generatator thorugh a resistor and then measuring the voltage across the transducer. The resistor and the impedance of the transducer then form a voltage divider for the AC signal, and measuring the voltage across the transducer will allow you to calculate it's impedance at 1.5MHz.

Use a resistor that is big enough that it does not lower the output of the signal generator much. Probably 1K would work.

So, the equation would be:

Vt = Vin * (Zt / (Zt + R)

Where:
Vt = Voltage across transducer (p-p)
Vin = Voltage output by signal generator (p-p)
Zt = Impedance of transducer
R = series resistor

Solving for Zt you get:

Zt = R / ( Vin / Vt - 1)

Bob

 

[(*steve*)]

Vt = R / ( Vin / Vt - 1)

Vin / Zt?

edit: or is that solving for Zt, so Zt = ...

 

[BobK]

Yep, should be Zt =

Thanks for catching that, I have fixed the original post.

Bob