toroidal conductivity sensor

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.

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

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 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.

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:

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

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

In article , John Popelish wrote: [snip]

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.

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).

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 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

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.

So you don't think that adding a second winding on the powered toroid would produce an output voltage via transformer action?

The liquid is a shorted one turn winding wrapped around the powered toroid. But that loop also passes through the second toroid, acting as a primary of a second transformer the winding on the second toroid is the secondary of that Transformer. Schematically:

--- || ----- || --- )|| | | ||( )|| | | ||( )||( )||( )|| | | ||( )|| | | ||(

--- || ----- || ---

The middle loop is the shorted liquid turn including the resistance of the liquid.

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

I read in sci.electronics.design that Robert Scott wrote (in ) about 'toroidal conductivity sensor', on Sat, 20 Aug 2005:

Yes, so there is no appreciable coupling between the two toroids.

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.

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.

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.

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.

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

I get no mental picture from these words.

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.

hi i have attached the picture of the pipe on which the two toroids are mounted. please see thanks rohan

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: