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'[EE]: How does one measure power in an ultrasonic '
2001\02\08@142940 by Chris Eddy

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For the purposes of running experiments, I have the need to apply a high
voltage 40KHz wave to an ultrasonic transducer.  The waveform could be
sine or square (will both drive the ultrasonic)?  My next challenge is
to measure in real time the total power consumed by the transducer in
real use.  I assume I must measure volts, amps, and multiply (taking
phase into account).  Ultimately, I want to convert to Joules (1J= 1W *
1S).  I would very much like to get 12 bit or better final results.

I could sample and do math (but 80+KHz sounds like a monster), or I
could use one of those power meter watt calculation chips, say from
Analog Devices.  I do not care which way, as long as I get the highest
precision one could ask within reason.

Any old ultrasonic hands out there??

Thanks
Chris~

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2001\02\09@063526 by mike

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On Thu, 8 Feb 2001 12:25:31 -0500, you wrote:

>For the purposes of running experiments, I have the need to apply a high
>voltage 40KHz wave to an ultrasonic transducer.  The waveform could be
>sine or square (will both drive the ultrasonic)?
> My next challenge is
>to measure in real time the total power consumed by the transducer in
>real use.  I assume I must measure volts, amps, and multiply (taking
>phase into account).  Ultimately, I want to convert to Joules (1J= 1W *
>1S).  I would very much like to get 12 bit or better final results.
>
>I could sample and do math (but 80+KHz sounds like a monster), or I
>could use one of those power meter watt calculation chips, say from
>Analog Devices.  I do not care which way, as long as I get the highest
>precision one could ask within reason.
>
>Any old ultrasonic hands out there??
>
>Thanks
>Chris~
Why do you want to measure it? Could you just measure the DC power into the driver - this would be a
lot easier. If you want to optimise efficiency, measure DC in and sound out
(voltage across a receiver fixed very close to the transmitter (to
avoid acoustic effects)).
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2001\02\09@094416 by Chris Eddy

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Mike Harrison wrote:

> Why do you want to measure it?
> Could you just measure the DC power into the driver - this would be a
> lot easier.
> If you want to optimise efficiency, measure DC in and sound out
> (voltage across a receiver fixed very close to the transmitter (to
> avoid acoustic effects)).
>

Mike;

This project involves some physics experiments, and they want to measure the
energy being driven into the experiment.  I understand your point, and would
do that under other circumstances.

In the mean while, I discovered in reading that if I take the RMS of the
volts and current, multiply them together, then multiply by the cos of the
phase angle, I will have real power.  Multiply by sin phase and I get
imaginary power (vars).  I think that I can use two RMS to DC converters, and
also use a comparator on the sine signals to generate a pulse train where the
width of the pulse is phase.  Mash it all into a PIC, and I can calculate
power.

Thanks
Chris~

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2001\02\09@131847 by Olin Lathrop

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> In the mean while, I discovered in reading that if I take the RMS of the
> volts and current, multiply them together, then multiply by the cos of the
> phase angle, I will have real power.

True, but this assumes that both the voltage and current waveforms are
sinusoids.


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Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, olinspamKILLspamembedinc.com, http://www.embedinc.com

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2001\02\09@145444 by Chris Eddy

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Yes, both sine waves.  I intend to filter the driving signal to clean up the
40KHz signal into a pure sine wave.. The next question is, will an ultrasonic
transducer respond to 40KHz sine just as well as square wave??

Chris~

Olin Lathrop wrote:

> > In the mean while, I discovered in reading that if I take the RMS of the
> > volts and current, multiply them together, then multiply by the cos of the
> > phase angle, I will have real power.
>
> True, but this assumes that both the voltage and current waveforms are
> sinusoids.

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2001\02\09@183258 by Olin Lathrop

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> Yes, both sine waves.  I intend to filter the driving signal to clean up
the
> 40KHz signal into a pure sine wave.. The next question is, will an
ultrasonic
> transducer respond to 40KHz sine just as well as square wave??

How "well" it responds has to do with the transducer and the desired effect.
A square wave contains much of its power in the fundamental, but also has
significant harmonics, with the first harmonic being at 3 times the
fundamental.  If the transducer can't "see" 120KHz, then it shouldn't
matter.  Even if it can respond to 120KHz, it still won't matter if the rest
of the system doesn't care if 120KHz signals are present.


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Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, olinspamspam_OUTembedinc.com, http://www.embedinc.com

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2001\02\10@035331 by Peter L. Peres

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I think that getting 12 bits out of it is not realistic for something that
is not lab equipment and individually adjusted. I think that you can
calibrate the output power vs. drive voltage for each transducer and
amplifier and get away with just measuring input power to the amplifier
during use. The calibration curve should correct for any changes in
efficiency with varying drive, temperature, etc. Aging may also play a
role in efficiency. I think that this is much simpler than measuring
active power on a 80 kHz wave that can be very much out of phase when the
transducer is not coupled to the medium.

The other way is to measure actual power output using an external sensor.

I am not an ultrasonics old hand, though.

Peter

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2001\02\10@055241 by Scott Stephens

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On Thu, 8 Feb 2001 12:25:31 -0500, you wrote:

>For the purposes of running experiments, I have the need to apply a high
>voltage 40KHz wave to an ultrasonic transducer.  The waveform could be
>sine or square (will both drive the ultrasonic)?

Yes, but hopefully you have a good acoustic conjugate impedance match at
your frequency of interest, or you will have, depending on power (<1Watt) a
broken transducer or coupling. So higher harmonics may not couple well, and
result in heating of the adhesive (silver epoxy?). Should be able to fix
that with a cap and inductor filter. I wonder if you could wrap a bifilar
(Ruthroff), resonant directional coupler for that frequency, and measure it
the way you would an RF signal?


>I could sample and do math (but 80+KHz sounds like a monster), or I
>could use one of those power meter watt calculation chips, say from
>Analog Devices.  I do not care which way, as long as I get the highest
>precision one could ask within reason.

Doesn't Microchip have an app note for a watt-hour meter, that does cosine
and power-factor conversion? As long as the signal is sinusoidal and
repetitive, it should be adaptable.
.
Scott

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2001\02\10@100447 by mike

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On Fri, 9 Feb 2001 18:01:41 -0500, you wrote:

>> Yes, both sine waves.  I intend to filter the driving signal to clean up
>the
>> 40KHz signal into a pure sine wave.. The next question is, will an
>ultrasonic
>> transducer respond to 40KHz sine just as well as square wave??
>
>How "well" it responds has to do with the transducer and the desired effect.
>A square wave contains much of its power in the fundamental, but also has
>significant harmonics, with the first harmonic being at 3 times the
>fundamental.  If the transducer can't "see" 120KHz, then it shouldn't
>matter.  Even if it can respond to 120KHz, it still won't matter if the rest
>of the system doesn't care if 120KHz signals are present.
Most piezos are basically capacitors, and driving them is basically
about charging and discharging that capacitance, so harmonic content
is not usually an issue - the capacitance and mechanical inertia takes
care of it (this may be less the case for low-mass ultrasonic units).
Max output usually happens if you charge and discharge the piezo
capacitance as quickly as possible, i.e. with a square wave. The
current waveform will tend to look pretty spiky, as it would driving a
pure capacitance. Square wave drive also means you lose less power in
the driver.  
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2001\02\10@100500 by mike

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On Fri, 9 Feb 2001 12:56:14 -0500, you wrote:

>Yes, both sine waves.  I intend to filter the driving signal to clean up the
>40KHz signal into a pure sine wave.. The next question is, will an ultrasonic
>transducer respond to 40KHz sine just as well as square wave??
No - you will get less output. US transducers and piezo discs are
essentially capacitive, but also have strong mechanical resonance
effects, which combine to produce some strange electrical
characteristics. Using an efficient squarewave driver stage, and
measuring the DC into it will be by far the easiest way to measure
transducer input power (which of course will often have a somewhat
nonlenear correlation to output power)
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2001\02\10@154657 by Peter L. Peres

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>In the mean while, I discovered in reading that if I take the RMS of the
>volts and current, multiply them together, then multiply by the cos of the
>phase angle, I will have real power.  Multiply by sin phase and I get

Or multiply the current and the voltage information together in an analog
multiplier and measure the RMS of the result without caring for phase.
This is the standard power meter approach btw.

Peter

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2001\02\10@164526 by Spehro Pefhany

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At 12:11 PM 2/10/01 +0200, you wrote:

>Or multiply the current and the voltage information together in an analog
>multiplier and measure the RMS of the result without caring for phase.
>This is the standard power meter approach btw.

You just need to measure the average of (low pass filter) the
instantaneous current * voltage, that *is* the power.

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2001\02\11@111638 by 772-3129

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> Or multiply the current and the voltage information together in an analog
> multiplier and measure the RMS of the result without caring for phase.

You would take the average, not RMS, of the instantaneous voltage times
current.


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Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, RemoveMEolinEraseMEspamEraseMEembedinc.com, http://www.embedinc.com

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2001\02\12@023740 by Vasile Surducan

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On Fri, 9 Feb 2001, Chris Eddy wrote:

> This project involves some physics experiments, and they want to measure the
> energy being driven into the experiment.  I understand your point, and would
> do that under other circumstances.
>
> In the mean while, I discovered in reading that if I take the RMS of the
> volts and current, multiply them together, then multiply by the cos of the
> phase angle, I will have real power.  Multiply by sin phase and I get

 No. You'll have the input power which is totally different from real
mechanical output power. Think what's happend if you switch only a little
the input frequency. Input power will be the same. Output power could be
zero.
You need a transducer. Usual a weight or pressure measuring sensor.
Vasile

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