I am building a TIG welder, and already have all the full bridge converter working nicely. To close the loop and regulate the arc current I have a 0.3mOhm shunt, and I need to amplify the voltage on it.
The distance between the shunt and the circuitry is about 30 cm.

My first approach was to just run the two sense wires to the control board, and there use INA240 differential amplifier. There seems to be so much interference induced into the measurement, that instead of a sawtooth signal representing the output inductor current I get something that is twice the frequency, so I suspect high common mode noise overloads the amplifier. The inductor current was measured with a current probe so I am sure the amplifier output does not match the input at all.

So I am thinking about moving the amplifier next to the shunt. Now, how would you power this to get the best possible result? I am leaning towards using an isolated power supply just for the amp, so that the common mode swings at the amp side don´t disturb its supply voltage.
Second thing is how to get the signal back to the control board? I am considering converting the signal to differential and sending that back, or using current output instead of voltage output.
Or maybe a linear optocoupler circuit?

 

You probably have a Grounding issue.

You need to insure that You have a "Star"-Grounding arrangement, FOR EVERYTHING.
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The distance between the shunt and the circuitry is about 30 cm.

which circuitry? the distance between shunt and the INA240 should be more like 3mm, not 300mm. layout and short sense leads are critical

 

There is also an issue of common mode voltages. The shunt voltage is a few millivolts but they are both at some much higher voltage of the arc circuit. THAT voltage then appears at both inputs to he instrument amplifier. That voltage is called the "common mode voltage" For that an isolated inputs amplifier is the simple way to go.

 

Based on the rality of the large common mode voltage it makes sense to consider using an isolation amplifier, which is an amplifier that is not referenced ,to external circuits, so that within limits the common voltage has no effect on the output. That would reduce,much of the noise and leave you with the average arc current signal and the arc voltage noise.

 

First of all sorry for ghosting the thread, I got fever and it took me some time to recover.

which circuitry? the distance between shunt and the INA240 should be more like 3mm, not 300mm. layout and short sense leads are critical

I think this is the first thing to try. However the shunt itself is about 100mm long so the distances won´t be in millimeters anyway.

After the shunt voltage gets amplified on the other side, I am still thinking what way should I transmit the signal to the receiving end at the control board.
In the picture is the simplest way.
Second option would be converting it to differential and then recovering it.
Third option which I don´t know if would be better interference wise, would be V to I conversion and making a current loop output, reading on a shunt at the control board.
Isolation amp is probably the last resort, but I have made those before and it is not too complicated.

 

I have some nice amplifier/shunt modules that us an older technology Burr Brown isolation amplifier that can provide quite a bit of gain,like fifty millivolts boosted to five volts, with a few hundred volts of isolation. So there are devices that will do both isolate and amplify. And certainly they are less expensive than that BB module. That particular module is good because the power supply it has is isolated from both the input side and the output side.

 

Since the distance is 30cm, it is best to use a digital communication solution.

 

Have you considered a hall-effect sensor?

 

Really, A shielded twisted pair, with the shield tied to the common at the amplifier end should be adequate. Going to a digital format at the shunt will require a shielded enclosure and shielded wiring for the power and and the digital signal wiring.. So there is a trade-off in complexity and it is not so simple.

 

what is the form factor of the shunt? i would try designing PCB that fits right on its sense terminals.

 

That is the way the isolation amplifier boards that I have fit: screwed down to the instrument terminals of the shunt hardware. I offer a suggestion for the board design, which is to identify the terminal connections as to +and -, for both power and signal.

 

that is what i have seen others do too, all electronics is right on the PCB that sits on the shunt. then the 30cm cable has amplified signal, rather than something in mV

 

There seems to be so much interference induced into the measurement,
that instead of a sawtooth signal representing the output inductor current
I get something that is twice the frequency,
so I suspect high common mode noise overloads the amplifier.

I think you need non-inductive current shunt, like this:
https://powertekuk.com/coaxial_shunt.htm

 

I have been presuming that the 0.003 ohm shunt is of the more conventional form, with two fair sized brass blocks a a strip of material between them, usually spaced 2 to 3 CM apart, on a common insulation block. About as non-inductive as it can get. But of course, I have not seen the shunts used for over 1000 amps or at radio frequencies, they may be quite different. But this application is for much lower frequencies and lower current. So it does not seem that special form factors would be required.

 

Second option would be converting it to differential and then recovering it.

That would be a better noise free solution, and you could use a piece of mic cable and phantom power it with the B+ you need. So that even helps you there moving to a shielded cable.
easiest way is to run +/- 15 on the balanced signal wires, Then pick them off at the other end to drive a balanced line driver.

 

With a folded return wire the inductance cancels out. A non-inductive current shunt can be made with Litz magnet wire folded using the ends with a null area loop using some twists in the wire.

The spectrum of welding extends beyond the visible range, so the skin effects are important.

I have measured welder (diffusion bonder) currents over 10 kA in diffusion bonding Zirc-shims in Monel steel tubes where the current must increase as the weld travels around the two tubes being bonded and the resistance progressively reduces between the large solid copper wheel electrodes.
- I used the massive copper welder legs to create a current sense with 10 mV drop at 10kA on 4Voltage arc voltages and used RG58 coax to sense the voltage at 90 deg. right angles which is critical. The problem is the wires do not have huge CMRR unlike an INA >> 100 dB at f. You need shielded twist pairs and may need double shielded wires using a local earth ground. Wires and shunts may have inductance about 10 nH/cm +/-50% depending on l/d ratio.

Consider more shielding and/or clam-shell Baluns (low and high mu ferrite) .
This will help raise the CM impedance so that a loaded Vdiff added to the input of the INA will improve the Vin signal CMRR before it enters the INA. The wires would have to be matched in impedance, and length to 1 part in 10^5 to achieve 100 dB CMRR so this is why the INA is not enough.

I suggest a CM Pi filter (line filter design) for sensing the arc current.

 

That would be a better noise free solution, and you could use a piece of mic cable and phantom power it with the B+ you need. So that even helps you there moving to a shielded cable.
easiest way is to run +/- 15 on the balanced signal wires, Then pick them off at the other end to drive a balanced line driver.

There is no way that the voltage signal from the shunt is not differential. OR did they mean the amplified signal?? It is not clear. And the way to transport a differential signal is with a shielded pair of conductors. If external signals may be strong, then a twisted shielded pair.

 

There is no way that the voltage signal from the shunt is not differential. OR did they mean the amplified signal?? It is not clear. And the way to transport a differential signal is with a shielded pair of conductors. If external signals may be strong, then a twisted shielded pair.

Balanced line driver can always accept an unbalanced input.
But since the OP isn't converting it into a signal, a balanced driver will work with separate power wires. But I would still use a piece of shielded cable. And of course an instrumentation amp like an INA851 would work too.

 

Exactly right about the two conductor shielded cable. But I am not so very certain about the line driver because I am thinking they are not high precision devices. And at least some line drivers are for digital applications, not linear at all.