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'[EE]: Having difficulty interfacing a current sens'
2001\07\29@070840 by Sanjay Punjab

picon face
I have an application where I need to monitor and
digitize the current waveform output of an audio power
amplifier into a subwoofer (speaker).
Previously, I used a hall-effect based sensor, but for
cost purposes (consumer product), it is impractical.
So instead, I have decided to use a 1 milli-ohm
current sense resistor. The problem now is creating an
electrical interface to convert the instantaneous
current (0 - 50 amps) into a proportional
instantaneous voltage (0-2.5v). Since the audio output
is not referrenced to ground, some type of
differential circuit must be used. But although the
voltage across the resistor will always be small, the
voltage at each of the 2 resistor taps, with respect
to ground can swing as high as much as +/-100 volts.
Unfortunately most instrumentation op-amps that can
handle such a high common-mode input voltage, are also
quite expensive. In addition, a textbook differential
op-amp circuit doesn't provide the needed accuracy,
even with .1% resistors, since any component
tollerance is multiplied by the gain of the circuit
(50). I though it would be easy finding an interface
circuit that would solve my problem, especially since
almost every DMM must use something similar for DC
current measurement. But I have had no luck. I would
appreciate some advice, even better a schematic.
Thanks


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2001\07\29@075702 by Spehro Pefhany

picon face
At 04:07 AM 7/29/01 -0700, you wrote:
>I have an application where I need to monitor and
>digitize the current waveform output of an audio power
>amplifier into a subwoofer (speaker).
>Previously, I used a hall-effect based sensor, but for
>cost purposes (consumer product), it is impractical.
>So instead, I have decided to use a 1 milli-ohm
>current sense resistor.

Nothing like making things difficult...

>The problem now is creating an
>electrical interface to convert the instantaneous
>current (0 - 50 amps) into a proportional
>instantaneous voltage (0-2.5v). Since the audio output
>is not referrenced to ground, some type of
>differential circuit must be used. But although the
>voltage across the resistor will always be small, the
>voltage at each of the 2 resistor taps, with respect
>to ground can swing as high as much as +/-100 volts.

You've created an extrememly difficult problem for
yourself. While this is possible (for example, isolate
the circuitry to the front end, build it inside of a
Faraday cage that is connected to one side of the
resistor and use transformers and/or optocouplers to
transfer power and signal across the barrier), it
may not be the best way to tackle this.

{Quote hidden}

Most of them float the front end (the whole meter on
cheap ones) at the signal voltage. Ones that don't tend
to have the box within a box arrangement if they have
decent AC performance (as does my nice 6 digit DMM).
You can handle the higher voltage by dividing down the
voltage, but as you have found, the resistor ratios
(and parasitic capacitance ratios..) have to be within
very narrow limits. If you want accuracy of +/-500mA,
and you divide the inputs down to +/-10V, that is
an accuracy of 50 microvolts DC (you could AC-couple
it to get rid of this), and 5ppm in the
resistor ratios (good luck). The B-B IN148 is designed
for similar applications and is probably close to
state-of-art for monolithic designs
www-s.ti.com/sc/psheets/sbos123/sbos123.pdf
but CMRR is only 70dB min, and you need more like
106dB just to get a lousy half an amp of accuracy.
Other commercial inamps have the CMRR but not the
input range, and are costly ($6/1K), and you'd have
to add some fancy (eg. $$ Caddock) resistor sets and
prolly some HF compensation to get the input range
and CMRR.

>But I have had no luck. I would
>appreciate some advice, even better a schematic.

Maybe you could use a CT  (limited high frequency
performance due to the typical tape-wound core) or an
active current balancing scheme using a ferrite CT
and a feedback winding. If you only need to work
up to a few hundred Hz (subwoofer), then the CT
may be the easiest way. Only a couple of $USD.

Best regards,
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2001\07\29@081939 by Bob Ammerman

picon face
I would probably set up a galvanically isolated subsystem that 'rode' on the
speaker waveform. It would convert the analog value to digital and then
transmit it over an isolation barrier (optoisolator) to the rest of the
instrument.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

{Original Message removed}

2001\07\29@090953 by Spehro Pefhany

picon face
At 07:58 AM 7/29/01 -0400, you wrote:

P.S. One thing implicit in my previous posting, but I'll state it
explicitly is that 0.5A produces a heck of a lot of sound compared
since hearing has a logarithmic response. If your interest is in
audible sound level rather than volts and amps, you'd have to make the
subsystem that much better (orders of magnitude better).

Best regards,
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2001\07\29@091954 by lizana

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face
Hi piclisters.

Take a look at LEM products.

http://www.lem.com


Regards.

Luis


{Quote hidden}

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2001\07\29@095810 by lizana

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Hi piclisters.



Sanjay: take a look at LEM products.

http://www.lem.com


Regards.

Luis

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2001\07\29@101514 by Olin Lathrop

face picon face
> I have an application where I need to monitor and
> digitize the current waveform output of an audio power
> amplifier into a subwoofer (speaker).
> Previously, I used a hall-effect based sensor, but for
> cost purposes (consumer product), it is impractical.
> So instead, I have decided to use a 1 milli-ohm
> current sense resistor. The problem now is creating an
> electrical interface to convert the instantaneous
> current (0 - 50 amps) into a proportional
> instantaneous voltage (0-2.5v). Since the audio output
> is not referrenced to ground, some type of
> differential circuit must be used. But although the
> voltage across the resistor will always be small, the
> voltage at each of the 2 resistor taps, with respect
> to ground can swing as high as much as +/-100 volts.

Forget the ground referenced diff amp.  You need too much common mode
rejection for that.  You are trying to measure 50mV on a 100V signal.  Thats
20 * Log10(100V/50mv) = 66dB common mode rejection just to get the noise
signal the same size as the signal of interest.  Let's say you want at least
1 part in 100 accuracy (a little less than 7 bits), then you need another
40dB of common mode rejection for a total of 106dB.  Clearly you need to
attack this in a different way, and this doesn't even get into the very high
common mode range required of the diff amp.  I can think of two approaches
off the top of my head:

One way is to float a regular amp with one end of the current sense
resistor.  This would require a separate small power transformer to make the
floating supply voltages.  This floating section would digitize the signal
and pass back the result to the rest of the world via opto isolators.

Another way which I like a little better for this problem is to use a small
transformer which you might construct yourself.  The primary would only be a
turn or two of thick wire, and wouldn't add significant inductance in series
with the speakers.  The secondary can be more turns of a fine guage wire
because there will never be current thru it.  Your ground referenced circuit
measures the open circuit voltage produced by the secondary.  You will want
an electrical ground shield between the primary and everything else to
eliminate capacitive pickup problems.  The output will actually be
proportional to the current and its frequency, but a single pole filter
should be able to compensate for this over the frequency range of interest.
Most of the tricky part will be making it work at the low end of the
frequency range.


********************************************************************
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(978) 742-9014, TakeThisOuTolinEraseMEspamspam_OUTembedinc.com, http://www.embedinc.com

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2001\07\29@120924 by Spehro Pefhany

picon face
At 10:08 AM 7/29/01 -0400, you wrote:

> The output will actually be
>proportional to the current and its frequency

Actually a current transformer (CT), if you terminate it in a low enough
resistance will be independent of the frequency over a wide range.

----   ||(----------------x-----------> input
1t    ||) 500t         [Rl]
----   ||(------0v        |
                         0V

The single-turn primary is just a loop through the core (say a tape-wound
toroidal core). With a 500t secondary and a 10 ohm load resistor you
will have a 1V output for 50A input. By putting a back to back diodes
in there with the 10.0 Ohm in series with one to ground, you will get
a DC output of 1V at the cost of a slightly larger core.

To CT
(other lead to 0V)

-x-----|<|----- 0V
|
x-----|>|----x---------> low pass filter, 0-1V for 0-50A in the primary.
             |
            10.0
             |
            0V

The core material is determined by the upper frequency, the core size by
the lowest frequency of interest, required output burden, and the
required accuracy. Probably a 1" O.D. tape wound toroid would do it.

Best regards,

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2001\07\29@122121 by Roman Black

flavicon
face
Hi Sanjay, sorry I don't know where you were trained,
but I was taught that to get good reliable measurements
of "active" signals you need to allow 2% to 5% of the
total for the measurement equipment. You mention a
sense resistor of 1 milliohm, this is ridiculous. I have
a good milliohm meter, and any electrical connnection will
give you a few milliohms resistance. Your sense resistor
will have two of these connections.

My suggestion for good accuracy is to sacrifice 5% of
your total power to the subwoofer load and use that
to measure the current waveform, then of course allow
for that 5% in your calcs. Don't ever expect accurate
results from a 1 milliohm resistor!!
:o)
-Roman



> {Original Message removed}

2001\07\29@124020 by Spehro Pefhany

picon face
At 02:21 AM 7/30/01 +1000, you wrote:
> You mention a
>sense resistor of 1 milliohm, this is ridiculous. I have
>a good milliohm meter, and any electrical connnection will
>give you a few milliohms resistance. Your sense resistor
>will have two of these connections.

You can do Kelvin connections. I've done 15,000 ADC digital meter
systems, and you sure don't want to put more than a few tens of
micro-ohms in there. 10uOhms => 2,250 W loss. Accuracy better
than 1% (system).

Best regards,

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2001\07\29@125106 by Roman Black

flavicon
face
Spehro Pefhany wrote:
>
> At 02:21 AM 7/30/01 +1000, you wrote:
> > You mention a
> >sense resistor of 1 milliohm, this is ridiculous. I have
> >a good milliohm meter, and any electrical connnection will
> >give you a few milliohms resistance. Your sense resistor
> >will have two of these connections.
>
> You can do Kelvin connections. I've done 15,000 ADC digital meter
> systems, and you sure don't want to put more than a few tens of
> micro-ohms in there. 10uOhms => 2,250 W loss. Accuracy better
> than 1% (system).


Hi Spehro, you've completely lost me here and
of course I want to learn why! ;o)
How and why did you go about this?? And also would
you recommend a 1mohm sense system? This seems
insane to me. I often do medium/high power DC motor
work and appreciate any info you can share here...
:o)
-Roman

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2001\07\29@131017 by Mike Kendall

flavicon
face
Who cares where Sanjay was trained?  I've  met some real morons in my day
that hold college degrees from reputable colleges.
Aren't we talking about the sensitivity of the circuit used to make the
measurement and the power dissipation/efficiency of any impedance
transformation device used.  A traditional analog meter uses 10kohms/1volt.
This why the older SAMS circuits gave voltage reading points with the "load"
of the meter taken into account.  A modern digital multimeter (DMM) has a
sensitivity of 10megaohms/1volt, I've even seen one with 20mohms/volt once.
Another classic example is RF current meters in transmitters.  Certainly an
RF current meter inserted in a transmission line path will not require 5% of
the radiated power.  If this was so, a 1.5kw transmitter would lose 25watts
to the meter and it would require a small heatsink. If the analog meter
movement was replaced with a digital readout that had an active power
supply, would it suddenly require 5% of the radiated power?  I've never seen
a heat dissipating RF current meter in my life.     The earlier idea of
winding a transformer seems to have much more merit to  me than to give up
5% of the power.
> Hi Sanjay, sorry I don't know where you were trained,
{Quote hidden}

> > {Original Message removed}

2001\07\29@132220 by Spehro Pefhany

picon face
At 02:50 AM 7/30/01 +1000, you wrote:

>How and why did you go about this??

Kelvin connections are just 4 -wire resistors, so the end
resistance doesn't matter. Imagine you took a bar of low
tempco metal and drilled 4 holes into it, like this:

    <--------- R ------------->
 A  B                         C  D
----------------------------------
| o  o                         o  o |
----------------------------------

On the outer two (A & D) put some hefty bolts in there to connect to
copper bus bars, on the inner two you stake a couple of pins in there
and solder to the bar and to wires. Want to trim the resistor?
Get your file and thin the middle down a bit. You use B & C as
the output. Typically 50, 60 or 100mV would be the full-scale
output voltage. Notice that if there is a bit of extra resistance
in connection A, say, it won't affect the measurement, there are
effectively 3 resistors, and only the middle one matters for
the measurement. Ra-b and Rc-d can vary a bit. Rb-c is the
critical one, but there are no high-current connections to it,
very little current flows through wires to B & C.

You don't usually make these, of course, there are companies that
specialize in just this sort of thing.

>And also would you recommend a 1mohm sense system? This seems
>insane to me.

I don't!, but mostly because it's floating on top of a +/-100V
signal. This means a very high CMRR is required, and it is
AC, so you may have to worry about a pF here and there.
Were it to be grounded or close to grounded and DC,
no problem, mon! I just wanted to point out (perhaps for
other applications) that it's not impossible.

> I often do medium/high power DC motor
>work and appreciate any info you can share here...

Probably useful for that kind of work, particularly
for low voltage. A 2 HP 24V motor would be ~63 A, so
0.001 Ohm would be reasonable (4W dissipation, 63mV
signal). Low side sensing is easier but you can high-
side sense it and level shift it with a current source.

Best regards,
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2001\07\29@132427 by David VanHorn

flavicon
face
>
> > > > I have an application where I need to monitor and
> > > > digitize the current waveform output of an audio power
> > > > amplifier into a subwoofer (speaker).


Cheat :)
Drive the speaker with a voltage to current converter.
I built a stereo amp some years back, using the LM-12 in this configuration.
It always mystified me why we drive a device that converts current into
sound pressure, with a voltage waveform.
The current amp applies a very interesting waveform to the load, but the
sound is at least the same, and probably somewhat better, though my ears
are too old to hear it. (Youth IS wasted on the young!)

Seriously, the current transformer sounds like the most practical approach.
Alternatively, you could do a couple turns around a ferrite disc, and make
that hall sensor much more sensitive, and maybe less costly too.


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2001\07\29@144506 by Chris Carr

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Resend.....to the list this time, sorry Luis

{Quote hidden}

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2001\07\29@180639 by Olin Lathrop

face picon face
> > The output will actually be
> >proportional to the current and its frequency
>
> Actually a current transformer (CT), if you terminate it in a low enough
> resistance will be independent of the frequency over a wide range.

All true, but I wasn't talking about a current transformer.  I specifically
mentioned measuring the open circuit voltage of the secondary.  In a current
transformer, you essentially measure the short circuit current of the
secondary.  Both will work here, but I thought the open circuit method would
present a slightly more desirable load to the circuit under test and would
have a better frequency response when used with the single pole low pass
filter.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
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2001\07\29@184325 by Olin Lathrop

face picon face
> Hi Sanjay, sorry I don't know where you were trained,
> but I was taught that to get good reliable measurements
> of "active" signals you need to allow 2% to 5% of the
> total for the measurement equipment.

I don't believe this, care to justify it?  Frankly this sounds like one of
those conclusions reached in a particular case where the answer was
remembered but the conditions governing the particular case were forgotten.

Suppose I showed you a wire and asked you to design a circuit to measure the
current through it up to +-100A with 8 bits accuracy and minimal
intrusiveness, and I wanted the measurement circuitry isolated with the
current information also passed back thru some sort of isolation.  Other
than the dielectric strenght of the voltage isolation, would the circuit
change any if I told you the wire could be up to 10V from ground?  What
about 100V, or 1KV?  Would you take 2-5% of the voltage in each case?  Of
course not since the voltage doesn't really enter into the design except for
the isolation strength.  You'd use the minimum sense resistor so that under
worst case conditions you still have a good 8 bits, which has nothing at all
to do with the voltage on the wire.

> You mention a
> sense resistor of 1 milliohm, this is ridiculous.

I wouldn't be so fast to say that.  With 1mohm you get +-100mV max signal.
That is enough to maintain a signal to noise ratio of 50dB (8 bits) after
amplification to a convenient level for an A/D, like 0-5V.  That only
requires a gain of 25.  It would be tough to maintain this accuracy for DC
due to offset voltages, but the problem specifically said audio going to a
speaker.  In that case things can be AC coupled with rolloff around 10Hz or
so, thereby eliminating offset voltages as a problem.

> I have
> a good milliohm meter, and any electrical connnection will
> give you a few milliohms resistance. Your sense resistor
> will have two of these connections.

True, but there won't be any current folowing thru the sense leads, or what =89=tle current there is will still be proportional to the current you are
trying to measure.  I agree that the "resistor" is just a length of wire,
preferably one that needs to be there anyway.  And at 100A you need to be
extra squeaky clean about good connections.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
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2001\07\29@184335 by Olin Lathrop

face picon face
> It always mystified me why we drive a device that converts current into
> sound pressure, with a voltage waveform.

I thought that too a long time ago -- until I tried it.  It didn't sound as
good.  Then I did some measurements and found the frequency response was
worse than when driven with a fixed voltage.  Then the obvious finally
became obvious to me as to why this is.

True, the force pushing the center of the speaker cone is directly
proportional to current, not voltage.  However, you hear the result of the
displacment of the whole speaker cone, not the force applied to it.  The
mechanical system has relative resonances and dead spots a various
frequencies.  When a resonance occurs, the speaker's impedance goes up, and
the inverse happens at a dead spot.  With constant current drive, when the
impedance goes up, more power is delivered, thereby exaggerating these bumps
in the frequency response.  With a constant voltage drive, the power goes
down as the impedance goes up, which compensates a bit for the bumps.

Sometimes the road less traveled is less traveled for a reason.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, .....olinspam_OUTspamembedinc.com, http://www.embedinc.com

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2001\07\30@020825 by Vasile Surducan

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On Sun, 29 Jul 2001, Olin Lathrop wrote:

> One way is to float a regular amp with one end of the current sense
> resistor.  This would require a separate small power transformer to make the
> floating supply voltages.  This floating section would digitize the signal
> and pass back the result to the rest of the world via opto isolators.

Agree ! But you don't need to digitize the signal, there are some simple
methods of using a pair of optocouplers and one or two operational
amplifiers (both in the same package ) so what you've got is a floated
analogic signal proportinal with the input current, ready to be measured
with an AD tehnique ( sigma-delta, comparator or standard ad ) from
microcontroller ( if you have one...)
But the second method Olin proposed is better and is well known in
switching power supplies world. ( using a curent sense transformer )
Vasile

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2001\07\30@103709 by Mike Mansheim

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face
> True, but there won't be any current folowing thru the sense leads,
> or what =89=tle current there is...

just curious, is '=89=tle' jargon of some sort, or just an e-mail bit
scramble?

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2001\07\30@110145 by Roman Black

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face
Olin Lathrop wrote:
>
> > Hi Sanjay, sorry I don't know where you were trained,
> > but I was taught that to get good reliable measurements
> > of "active" signals you need to allow 2% to 5% of the
> > total for the measurement equipment.
>
> I don't believe this, care to justify it?  Frankly this sounds like one of
> those conclusions reached in a particular case where the answer was
> remembered but the conditions governing the particular case were forgotten.


Yes I must apologise to Sanjay and the rest of
the list! I hastily read his post and thought it
was for testing of a 100v speaker system, ie
transformer driven, and I just snapped at the idea
of using a 1 milliohm resistor to measure this.
I'll make sure to read slower in future, i've
been doing some real long hours lately...
Sorry! :o)
-Roman

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2001\07\30@110152 by David VanHorn

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face
At 06:33 PM 7/29/01 -0400, Olin Lathrop wrote:
> > It always mystified me why we drive a device that converts current into
> > sound pressure, with a voltage waveform.
>
>I thought that too a long time ago -- until I tried it.  It didn't sound as
>good.  Then I did some measurements and found the frequency response was
>worse than when driven with a fixed voltage.  Then the obvious finally
>became obvious to me as to why this is.
>
>True, the force pushing the center of the speaker cone is directly
>proportional to current, not voltage.  However, you hear the result of the
>displacment of the whole speaker cone, not the force applied to it.  The
>mechanical system has relative resonances and dead spots a various
>frequencies.  When a resonance occurs, the speaker's impedance goes up, and
>the inverse happens at a dead spot.  With constant current drive, when the
>impedance goes up, more power is delivered, thereby exaggerating these bumps
>in the frequency response.  With a constant voltage drive, the power goes
>down as the impedance goes up, which compensates a bit for the bumps.
>
>Sometimes the road less traveled is less traveled for a reason.

Interesting.
I must have lucked out with my speakers.

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2001\07\30@110157 by Douglas Butler

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face
For the low audio band I would recommend using a current transformer.

If accuracy is a problem, you can use the three winding technique.  The
current to be measured goes through a primary with few turns.  A
cancelling current goes through a second primary with lots of turns,
therefore little current.  Both primaries have the same amps*turns in
opposite directions so flux in the core is zero.  Keeping the flux at
zero removes core nonlinearities from affecting the system.

The secondary has lots of turns and is an error signal to sense when the
primaries are out of balance.  The secondary drives an op amp integrator
which drives the cancelling primary.  A shunt resistor in the ground
side of the cancelling primary provides your voltage output.

Sherpa Doug

> {Original Message removed}

2001\07\30@112040 by Roman Black

flavicon
face
David VanHorn wrote:
{Quote hidden}

I thought about doing closed-loop speakers, ie
using a CD player laser or similar to measure
cone travel, then using a high speed digital
PWM to move the speaker cone to the precise
position. More like a high speed voice coil servo
than an analog speaker.

Has anyone else done this or know anyone working
on it??
-Roman

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2001\07\30@112636 by Douglas Butler

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> >
> > > > > I have an application where I need to monitor and
> > > > > digitize the current waveform output of an audio power
> > > > > amplifier into a subwoofer (speaker).
>
>
> Cheat :)
> Drive the speaker with a voltage to current converter.
> I built a stereo amp some years back, using the LM-12 in this
> configuration.
> It always mystified me why we drive a device that converts
> current into
> sound pressure, with a voltage waveform.
> The current amp applies a very interesting waveform to the
> load, but the
> sound is at least the same, and probably somewhat better,
> though my ears
> are too old to hear it. (Youth IS wasted on the young!)
>
About 1980 I worked with some HiFi audio professors at college.  I asked
about current drive of speakers and was told that it had been tried,
they had also tried using displacement feedback from the speaker cone.
But neither of those techniques produced the right "sound".  There are a
lot of (seemingly) irrational things that go into high end audio, and I
put it down as another of them ;-)

I work in sonar now.

Sherpa Doug

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2001\07\30@114202 by David VanHorn

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face
>
>I thought about doing closed-loop speakers, ie
>using a CD player laser or similar to measure
>cone travel, then using a high speed digital
>PWM to move the speaker cone to the precise
>position. More like a high speed voice coil servo
>than an analog speaker.
>
>Has anyone else done this or know anyone working
>on it??
>-Roman


Someone marketed something like this, but I can't recall the details.

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2001\07\30@115444 by Alan B. Pearce

face picon face
>Has anyone else done this or know anyone working
>on it??
>-Roman


I cannot remember who did it, but I recall someone doing it with a piezo
feedback element attached to the voice coil.

On the other hand my work project is to attempt to emulate something very
similar consisting of a piston to pump helium gas in a cooler. The piston
moves about 4 to 5 mm at max movement, and has an LVDT attached to provide
positional feedback. The whole thing also gets bogged down in drying to
minimise the vibration of the mechanical unit to stop the satellite it will
be on from vibrating and spoiling the nice pretty pictures the camera will
be taking.

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2001\07\30@120449 by Dal Wheeler

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I've seen a few servo driven subwoofer projects.  They're using a number of
different position sensors, but I haven't actually seen a laser detection
system.

Most of the successful ones use a linear actuator with a position sensor in
the drive.  I've seen a project with accelerometors glued to the voice
oil.  --Not sure about the success of that one.

My yamaha computer speaker system has a servo driven sub, although I'm not
sure about the measuring method.  Maybe I'll hafta take 'em apart.

Lots of projects come up under search for "servo subwoofer".  I haven't seen
anything to do higher frequencies --yet.  --Laser positioning might allow
that.
-Dal

{Original Message removed}

2001\07\30@150318 by Chris Carr

flavicon
face
Roman Black wrote:
> David VanHorn wrote:
> >
> > At 06:33 PM 7/29/01 -0400, Olin Lathrop wrote:
> > > > It always mystified me why we drive a device that converts current
into
> > > > sound pressure, with a voltage waveform.
> > >
> > >I thought that too a long time ago -- until I tried it.  It didn't
sound as
> > >good.  Then I did some measurements and found the frequency response
was
> > >worse than when driven with a fixed voltage.  Then the obvious finally
> > >became obvious to me as to why this is.
> > >
> > >True, the force pushing the center of the speaker cone is directly
> > >proportional to current, not voltage.  However, you hear the result of
the
> > >displacment of the whole speaker cone, not the force applied to it.
The
> > >mechanical system has relative resonances and dead spots a various
> > >frequencies.  When a resonance occurs, the speaker's impedance goes up,
and
> > >the inverse happens at a dead spot.  With constant current drive, when
the
> > >impedance goes up, more power is delivered, thereby exaggerating these
bumps
> > >in the frequency response.  With a constant voltage drive, the power
goes
{Quote hidden}

If I remember correctly Philips did a system with specially constructed
speakers with 2 voice coils, one doing the driving in the normal way and the
other used to provide motional feedback. This was (is?) also a technique
used in sub-woofers in High End Car Audio Systems (JVC, I believe). It
seemed to be fairly effective in moving the small finite air mass within a
vehicle.

Bouncing a laser beam off a small mirror glued to a speaker cone should work
for bass speakers. I have used that technique with a rotating mirrors to
project an oscilloscope type display on a wall.

Regards

Chris Carr

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2001\07\30@150654 by Olin Lathrop

face picon face
> > True, but there won't be any current folowing thru the sense leads,
> > or what =89=tle current there is...
>
> just curious, is '=89=tle' jargon of some sort, or just an e-mail bit
> scramble?

Email bit scramble.  What I typed was "or what little current there is".



********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, @spam@olinspam_OUTspam.....embedinc.com, http://www.embedinc.com

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2001\07\30@155816 by D. Schouten

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> > Has anyone else done this or know anyone working
> > on it??
>
>
> If I remember correctly Philips did a system with specially
constructed
> speakers with 2 voice coils, one doing the driving in the normal way
and the
> other used to provide motional feedback. This was (is?) also a
technique
> used in sub-woofers in High End Car Audio Systems (JVC, I believe).
It
> seemed to be fairly effective in moving the small finite air mass
within a
> vehicle.

Kenwood also did something like that, called 'Sigma drive' to hugely
increase the damping factor of the system. They used four wires for
each speaker connection, two large diameter conductors for the
amplified audiosignal, and two smaller wires to measure the back emf
from the single voice coil. Although technically the concept worked
great, they have stopped using it while there was too much amplifier
damage in the market due to confused customers wiring the system
wrongly. A Dutch (:-)) pro audio manufacturer named Stage Accompany is
still using such an approach with very good results.

Daniel...

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2001\07\30@200437 by Kelly Kohls

picon face
Have a look at http:\\http://www.servodrive.com.  These are driven (at least the
ones I'm familiar with) by servo motors.

Kelly Kohls, N5TLE
Dallas, Texas
Homepage: http://www.qsl.net/n5tle/
There's never enough time to do it right, but plenty of time to do it over.

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2001\07\30@200904 by Spehro Pefhany

picon face
At 10:41 AM 7/30/01 -0500, you wrote:

>Someone marketed something like this, but I can't recall the details.

Philips, in the late 1970's or early 80's, IIRC.

Best regards,

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2001\07\31@134154 by Peter L. Peres

picon face
imho a Hall current sensor+amplifier for DC use will span the audio band
if its amplifier is modified a little. Its output should be proportional
to instantaneous current.

Peter

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2001\07\31@134205 by Peter L. Peres

picon face
Roman, imho look into coil feedback instead. A second coil is added (or
exists) to the existing one and used as instantaneous speed feedback in
the amplifier. This is analog and it's pretty good at suppressing speaker
self-resonance and other artifacts. Someone was making speakers like this
I think. You can also use a capacitive sensor between the metalized cone
center and an extra electrode you provide (coin, disc etc).

Physically sensing position (optical, etc) makes little sense since the
transmission medium (air) does not transmit 'dc' and the static sensor
will have a poorer linearity than the voice coil (which is 0.1% or better
usually). You are mostly interested in motions that were not oredered by
the voice coil, and getting rid of them, so the speed is more interesting
than the exact position of the cone.

The output from the speed sensor is added directly to the normal feedback
loop of the amplifier. It is VERY easy to set up an oscillator like this
(when having 180 degrees phase through mechanical resonances etc). Allot a
lot of time for tinkering and debugging.

Peter

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2001\07\31@134213 by Peter L. Peres

picon face
> On the other hand my work project is to attempt to emulate something
> very similar consisting of a piston to pump helium gas in a cooler.
> The piston moves about 4 to 5 mm at max movement, and has an LVDT
> attached to provide positional feedback. The whole thing also gets
> bogged down in drying to minimise the vibration of the mechanical unit
> to stop the satellite it will be on from vibrating and spoiling the
> nice pretty pictures the camera will be taking.

Two identical opposed pistons and valving machined in the cylinder walls
as slits ? (see boxer motors and Jumo 205 WW2 era aircraft diesel engine).

Peter

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'[EE]: Having difficulty interfacing a current sens'
2001\08\01@065007 by Alan B. Pearce
face picon face
>imho a Hall current sensor+amplifier for DC use will span the audio band
>if its amplifier is modified a little. Its output should be proportional

I would agree with this. You are rapidly finding that measuring the current
with a resistor is a cheap way to do it, but the electronics that follows is
not cheap or easy. A commercial clamp on hall effect probe will soon become
a LOT cheaper and easier.

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