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toroidal conductivity sensor

T

timmy

Jan 1, 1970
0
hi there
i am a student and making a project on toroidal conductivity sensor to
measure conductivity.
the project is based on :that you pass a ac to toroidal core (high
permeability)
with AC through a winding. Then you place a second toroid beside that
one with another winding. The two cores are enclosed in insulation,
with a hole passing through both cores. When submersed, the first
core induces 1 turn's worth of voltage around the liquid loop that
passes through the hole. The current that voltage moves through the
liquid is sensed by the second core, acting as a current transformer.
You amplify the AC current from the second core, rectify it, and the
result represents the conductivity of the solution.
i did some calculation to find inductance:
XL=2*pi*F*L
F=8 kHz
L=4mhenry
XL=200ohms
Max voltage=3 volts
I=V/ XL
I=15ma(approx)
now please i want to know how can i start the first step to do in
practical
thanks
rohan
 
J

John Popelish

Jan 1, 1970
0
timmy said:
hi there
i am a student and making a project on toroidal conductivity sensor to
measure conductivity.
the project is based on :that you pass a ac to toroidal core (high
permeability)
with AC through a winding. Then you place a second toroid beside that
one with another winding. The two cores are enclosed in insulation,
with a hole passing through both cores. When submersed, the first
core induces 1 turn's worth of voltage around the liquid loop that
passes through the hole. The current that voltage moves through the
liquid is sensed by the second core, acting as a current transformer.
You amplify the AC current from the second core, rectify it, and the
result represents the conductivity of the solution.
i did some calculation to find inductance:
XL=2*pi*F*L
F=8 kHz
L=4mhenry
XL=200ohms
Max voltage=3 volts
I=V/ XL
I=15ma(approx)
now please i want to know how can i start the first step to do in
practical
thanks
rohan
It looks like you have done that. You have a value of inductance that
keeps the current low enough for an opamp to be able to supply it (or
nearly so). I think I would shoot for a current of about 10 ma in the
drive coil. So either a slightly higher inductance, or a higher
frequency.

Then you are ready to apply this design to some available toroidal
cores, using the manufacturer's data to determine the approximate
number of turns needed to produce the desired inductance with a given
core.

Once that is determined, you know the approximate volts applied to the
liquid loop (supply volts divided by drive inductor turns).

The current through the liquid will be that voltage divided by the
effective total liquid resistance through the hole and back around the
outside of the pair of cores. The cylinder inside the cores will
dominate this resistance, since its length to width ratio is larger
than the effective length to width ratio of the rest of the current path.

The current from the second inductor (I would use a core similar to
the first) will be the liquid current divided by the turns count on
that inductor. So you will probably want a transconductance front end
(current to voltage converter with a very low impedance input and a
high value of feedback resistance). This might be followed by a
bandpass filter to remove noise not at the excitation frequency. Then
you need an absolute value circuit and a low pass filter to find the
magnitude of that signal.

You can perform the bandpass and rectification at the same time if you
use a synchronous rectifier switched in phase with the excitation
frequency. This might consist of a -1 gain amplifier connected to the
signal and a SPDT cmos analog switch that selects either the inverted
or non inverted version of the signal to be passed through an RC low
pass filter.
 
T

timmy

Jan 1, 1970
0
hi again
thanks for ur reply
Well now i have designed a pipe like system in a loop through which the
fluid will flow.
there is a pipe on which two toroids are mounted on top, each are 4
milli henry.
i have a query is there any formulae to measure the conductivity at the
end,for example if i get a voltage at the output ,then how can it
really tell me the conductivity of the particular fluid flowing.
and also what op-amp can be the best to supply a current of around 15
ma.
thanks
rohan
 
L

legg

Jan 1, 1970
0
hi again
thanks for ur reply
Well now i have designed a pipe like system in a loop through which the
fluid will flow.
there is a pipe on which two toroids are mounted on top, each are 4
milli henry.
i have a query is there any formulae to measure the conductivity at the
end,for example if i get a voltage at the output ,then how can it
really tell me the conductivity of the particular fluid flowing.
and also what op-amp can be the best to supply a current of around 15
ma.

I think that previous advice was based on the two toroids being
immersed in the medium.

If they are outside the pipe (pipe passing through center), and there
is no electrical return between the pipe ends to produce a conduction
path, then the toroids will not couple signals through the media under
test.

RL
 
J

John Popelish

Jan 1, 1970
0
timmy said:
hi again
thanks for ur reply
Well now i have designed a pipe like system in a loop through which the
fluid will flow.
there is a pipe on which two toroids are mounted on top, each are 4
milli henry.

I cannot picture your structure from this description. The pair of
toroids must be against each other and be as surrounded by liquid as
possible, so that the liquid forms a conductive turn wrapped through
the hole, just as a wire would do.
i have a query is there any formulae to measure the conductivity at the
end,for example if i get a voltage at the output ,then how can it
really tell me the conductivity of the particular fluid flowing.

I gave you the rough approximation of the formula, last time, but you
really need some standard solutions to use as calibration points. You
might be able to find a reference on using salt water of varying
concentrations to make up standard conductivity solutions.
and also what op-amp can be the best to supply a current of around 15
ma.

Since this current has to be supplied at large voltage swing and
significant frequency, you need an opamp with both current capability
and a fairly high gain bandwidth. If your peak current really stays
below 15 mA, then something as ordinary as an LM411 (or the LM412 if
you need duals) might be fast enough:
http://cache.national.com/ds/LF/LF411.pdf
Check out the 'current limit' and 'undistorted output swing' graphs.
 
L

legg

Jan 1, 1970
0
I cannot picture your structure from this description. The pair of
toroids must be against each other and be as surrounded by liquid as
possible, so that the liquid forms a conductive turn wrapped through
the hole, just as a wire would do.
Actually, media in a nonconductive pipe, with shorted mesh electrodes
at it's extremities could allow a more controllable conductivity
calculation. You probably wouldn't need the second toroid at all (or
even the first) if galvanic isolation weren't important.

The two-toroid method assumes insertion in volume of media much larger
than that of the toroids.

RL
 
T

Tony Williams

Jan 1, 1970
0
John Popelish said:
The current through the liquid will be that voltage divided by
the effective total liquid resistance through the hole and back
around the outside of the pair of cores. The cylinder inside
the cores will dominate this resistance, since its length to
width ratio is larger than the effective length to width ratio
of the rest of the current path.

Just using your post to pass a remark.......

I would think that it would be important to have a
guaranteed liquid seal between the two toroids.
This would be to ensure that neither toroid has an
individual single turn through it.
 
G

Glen Walpert

Jan 1, 1970
0
Just using your post to pass a remark.......

I would think that it would be important to have a
guaranteed liquid seal between the two toroids.
This would be to ensure that neither toroid has an
individual single turn through it.

It would seem possible to provide the required liquid turn with
plastic pipe, torroids on the outside, for example with 2 tees and 4
ells arranged symneterically to split the process fluid flow equally,
or just 2 tees and 2 ells jogging off to one side of the main run if
conductivity is expected to change slowly enough that the slower flow
through one side of the liquid shorted turn is not a factor. No doubt
this will provide lower sensitivity than sealed and immersed torroids
due to the higher resistance of the smaller volume and longer path of
the liquid turn, but this is unlikely to be a problem with higher
conductivity solutions IMO.
 
J

John Popelish

Jan 1, 1970
0
Tony said:
Just using your post to pass a remark.......

I would think that it would be important to have a
guaranteed liquid seal between the two toroids.
This would be to ensure that neither toroid has an
individual single turn through it.

Right. I didn't emphasize that, because I was picturing typical
commercial units, which are completely encapsulated in a single
overall covering.

http://www.emersonprocess.com/raihome/liquid/products/model_225.asp
 
J

John Popelish

Jan 1, 1970
0
Glen said:
It would seem possible to provide the required liquid turn with
plastic pipe, torroids on the outside, for example with 2 tees and 4
ells arranged symneterically to split the process fluid flow equally,
or just 2 tees and 2 ells jogging off to one side of the main run if
conductivity is expected to change slowly enough that the slower flow
through one side of the liquid shorted turn is not a factor. No doubt
this will provide lower sensitivity than sealed and immersed torroids
due to the higher resistance of the smaller volume and longer path of
the liquid turn, but this is unlikely to be a problem with higher
conductivity solutions IMO.

I have made a small volume conductivity meter by this exact method.
It suffers extreme sensitivity to bubbles, though. A bubble anywhere
in the loop changes the effective length to cross sectional area of
the conducting liquid. In the normal configuration, bubbles have
little effect unless they are in the pinch (center of the toroids).
 
R

Robert Scott

Jan 1, 1970
0
I have read all the postings in this thread, and I cannot picture the
mechanical setup. I gather that the setup ensures that the coupling
from one coil to the other is due entirely to the current induced in
the liquid and not to any mutual coupling between the coils. But how
is that possible? Can someone point me to an image or a mechanical
drawing on some web page that shows a typical setup?


-Robert Scott
Ypsilanti, Michigan
 
R

Robert Scott

Jan 1, 1970
0
Somewhere I have seen a paper on the design, and I have dissected a
probe to verify the construction. The two toroids are wound
separately and then simply stacked and encased in an insulating cover
with a hole in the middle.

I just can't figure out how that could work. The magnetic field of a
toroid is contained entirely inside the doughnut shape, unlike a
solenoid coil that projects a magnetic field way out from the coil.
Induced currents flow where there is a changing magnetic field.
Therefore the induced currents would only flow inside the doughnut of
the toriod - not through the hole in the doughnut. So how does a pair
of toroids stacked one behind the other share a loop of induced
current without also sharing a magnetic field. It seems there would
be coupling even if the conductivity of the stuff in the doughnut hole
was zero.


-Robert Scott
Ypsilanti, Michigan
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Robert Scott
I just can't figure out how that could work. The magnetic field of a
toroid is contained entirely inside the doughnut shape, unlike a
solenoid coil that projects a magnetic field way out from the coil.

Yes, so there is no appreciable coupling between the two toroids.
Induced currents flow where there is a changing magnetic field.
Therefore the induced currents would only flow inside the doughnut of
the toriod - not through the hole in the doughnut.

Just consider one toroid for now. The fluid encircles the core, passing
through the hole in the middle, and thus forms a single turn secondary
winding, into which a voltage is induced by a current in the toroid. The
resulting current depends on the resistance (and hence the conductivity)
of the fluid 'turn'.

Now add the second toroid, through which the fluid 'turn' also threads.
The current in the fluid induces a voltage in the winding on the second
toroid. Note that this would occur even if the fluid circuit were very
long, so that the two toroids were so far apart that no direct magnetic
coupling could occur. The toroids could even be oriented at
right-angles.
So how does a pair of toroids stacked one behind the other share a
loop of induced current without also sharing a magnetic field.

The sharing of the current loop is simply a matter of the physical
arrangement, just like three adjacent links of a chain, and does not
require or imply a sharing of magnetic field.
It seems there would be coupling even if the conductivity of the
stuff in the doughnut hole was zero.

Not if the toroids are perfectly uniformly wound, so that there really
IS no external field. I suspect that in practice some cancellation of
residual coupling is necessary, by bleeding a compensating current from
one winding to the other so as to get zero output voltage with no fluid
present.
 
T

Tony Williams

Jan 1, 1970
0
I would think that it would be important to have a
guaranteed liquid seal between the two toroids.
This would be to ensure that neither toroid has an
individual single turn through it.
[/QUOTE]
Right. I didn't emphasize that, because I was picturing typical
commercial units, which are completely encapsulated in a single
overall covering.

http://www.emersonprocess.com/raihome/liquid/products/model_225.asp

Ah yes, thank you. That construction also provides a defined
conductor diameter through the toroids. I also had not
appreciated the need to keep the sensor assembly well away
from conductive metalwork.... obvious when pointed out.
 
T

timmy

Jan 1, 1970
0
hi there
i have made a pipe with tees .like in a loop and two toroids mounted on
top close enough,
i just passed a 8khz frequency to one of the toroid using the signal
generartor and i get some output on the oscilloscope ,and i get some
voltage on channel 1 and channel 2 in milli volts.So how can one find
conductivity using voltage ,is there any formula to use
thanks
rohan
 
J

John Popelish

Jan 1, 1970
0
timmy said:
hi there
i have made a pipe with tees .like in a loop and two toroids mounted on
top close enough,

I get no mental picture from these words.
i just passed a 8khz frequency to one of the toroid using the signal
generartor and i get some output on the oscilloscope ,and i get some
voltage on channel 1 and channel 2 in milli volts.So how can one find
conductivity using voltage ,is there any formula to use

Short circuit the second toroid. The conductivity is proportional to
the current through the short. You will have to calibrate the
constant of proportionality with a solution of known conductivity.
 
J

John Popelish

Jan 1, 1970
0
timmy said:
hi
i have attached the picture of the pipe on which the two toroids are
mounted.
please see
thanks
rohan http://rapidshare.de/files/4274743/toroid1.doc.html
Thank you. It looks quite similar to a sample loop conductivity meter
I have tested. This arrangement should provide a complete sample
conductivity loop if you can be sure the pipes stay filled and do not
trap bubbles of any significant size, especially in the smaller
diameter sections.

You need to amplify the signal from the receiving toroid by using an
inverting opamp configuration with no resistor in the input leg. Just
the toroid shorted from signal common to - input, and a feedback
resistor to set the gain. This provides an output voltage that is
proportional to inductor current.

If you need more than 1 gain stage, additional stages can be AC
coupled (capacitor in series) to block DC errors, since you are
interested in only the excitation frequency. You can read
conductivity with an ordinary digital AC meter (with a correction
factor to convert the voltage reading to conductivity). or you can
rectify the signal to DC and read the conductivity with any kind of DC
volt meter appropriate for the voltage, to convert the signal to a
number with an A/D converter and get that value into a computer
(perhaps with something like one of these:
http://www.dataq.com/products/startkit/di194rs.htm )
 
J

John Popelish

Jan 1, 1970
0
timmy said:
hi
i have attached the picture of the pipe on which the two toroids are
mounted.
please see
thanks
rohan http://rapidshare.de/files/4274743/toroid1.doc.html

One other comment: With this sort of arrangement, there is no need to
have the two toroids right against each other. If it helps to
eliminate any cross talk between them (any output signal when the
sample loop is full of air, for example) you can place them anywhere
around the sample loop. for instance, one may be on the top
horizontal pipe and one on the bottom horizontal pipe. They need to
be closely stacked only for the direct immersion application as is
common with the commercial probes. The important thing is that one
core induce current in the loop, and the other core measure that loop
current.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that John Popelish <[email protected]>
It looks quite similar to a sample loop conductivity meter I have
tested.

I would move the toroids further apart, so as to reduce any stray direct
coupling, unless the conductivity is very low. A Faraday shield between
them would be good, too. A sheet of, preferably, copper 0.5 mm thick or
more, with a hole for the pipe, connected by a narrow slit to the edge
so as not to make a shorted turn. Make it at least twice the outside
diameter of the toroids.
 
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