Current Transformer to 0-10 Volts

Yes.

Yes.

Be careful. An open secondary on a CT is a nasty thing. Take no chances - ask an expert if you are unsure.

The 5 ohm resistor may be to big. You need to match or be less than the CT secondary "Burden" rating. Typically that value is less than 1 ohm. The voltage developed across that resistance is then amplified by a wide band fixed gain amp., thus the signal conditioner. As Homer pointed out, do not run the secondary into an open circuit or at high impedance. It will then act as a PT possibly creating thousands of volts, doing all kinds of nasty things!

Also keep in mind, the secondary waveform will be the actual current waveform, AC. If you need RMS, or DC output you further need to refine that in your signal conditioner.

It's all AC and variable waveshapes must be considered. You'll need an additional power supply and some electronics to rectify, smooth and amplify the signal to 0-10Vdc. E.g. 100:1 CT to a 1ohm resistor, then to an AD737 true RMS converter, then to a x50 opamp, then to the PLC. john

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Well, I'm not a CT guy, but I'd think that the output current would look a lot like the input current, like in any transformer.

I'd use a big resistor - 2A at 10V is 20 watts! =:-O

Good Luck! Rich

Note that 10V at 2A is 20 watts you'll be dissipating in the load resistor! I'd suggest that you go to a much higher turns ratio.

100:1 (500:5) might be reasonable. Then the output is 0.2A. You could use an even higher ratio. If the transformers are easy to get, I'd look at even 1000:1, though you may have some trouble finding those. If you used your original 10:1 ratio, the ten volt drop would reflect back as a 1 volt drop in your line, which is way more than you need to allow. A ten volt output at 1000:1 reflects only ten millivolts drop along the monitored line--actually somewhat more because of less than perfect coupling, but still not a lot. A ten volt output at 1000:1 or even 100:1 is also much more likely to be a reasonable burden for the transformer. At 1000:1, a ten volt output with 20A in the primary is 20mA secondary current and 200mW dissipation.

You'll need to rectify the output; an op-amp precision rectifier is appropriate. Then you need to convert the output to whatever the PLC wants. The precision recitifer can easily be made to put out 0-10 volts, even if the input is only 1 volt instead of 10, but then you need power to run the op amps. If you _know_ that your circuit will always have some minimum current in it and the ratio between min and max isn't too great, you could probably arrange to run the op amps on the current transformer output, but that's not a wonderful idea from the standpoint of precision, since the amplifier power will appear as additional transformer load. A better idea would be to use the 4-20mA loops, especially if these sensors will be some distance from the PLC. You can find example 4-20mA circuits in op amp manufacturers' data sheets. I believe that Linear Technology is one good source for such circuits. Try their ap notes, too.

Too bad you didn't have this need a year or so ago. Marlin P. Jones had some nice split-core 4-20mA AC current transducers with a jumper for, um 10A, 20A and 50A full scale as I recall, for about $10 each. But even if you have to pay $100 or more each for them, you'll probably be better off buying them. I have a feeling from the last sentence of your posting that you'll be in trouble trying to build them.

Cheers, Tom

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You need a module from LEM. They're current transformers with built in signal conditioners (often self powered) giving a PLC analog compatible signal - 0-10, 0-5 or 4-20mA

HTH

Tim

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And 20 Watts!

there's not (m)any CTs with that sort of output. a 20VA 60Hz transformer weighs a few kilograms...

typicaly curent transformers produce milliwatts.

aim for the sort of current you can feed to an op-amp eg: a 1000:1 transformer into a virtual earth or a small resistor.

then you probably want rectify and average it (if you want RMS that'll add complexity)

it sure does.

Bye. Jasen

You can get CTs with 100 mA rated output. There are also high ratio PCB mounted (or not) CTs with primaries from 10 to 200 amps, and output up to

10 VRMS. Digikey has them TE1020-ND for $7.80, or you can get a wide range of CTs from CR Magnetics:

They also have transducers that might do what you need.

Paul

snip

LTC1967 and it is simple, and cheap

martin

You can get the Triad CSE187L current Xfmr from Mouser for $2.47 ea in qty of

1. formatting link

It has 1:500 turns ratio, current range of 0.1A - 30A. More details at

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On Feb 6, 11:43 am, "Tom" wrote:

If measuring the average absolute value of the waveform is acceptable (that is, if you don't need "true RMS"), and if the input resistance of the PLC's 0-10V input is high (say >100kohms), there's an easy way to do this, and the accuracy will be quite good, even down to currents under an amp, for 20A full scale.

You need to get current transformers (CTs) that can output a fairly high voltage--at least 12 volts RMS. Expect these will have a high turns ratio. For example, the 2000:1 CR8350-2000 from CR Magnetics (thanks, Martin, for the link) should do the job. The catalog page lists the effective turns ratio, Te, as 2037:1. We'll be detecting the average of the absolute value of the waveform. Assuming it's a sine, the average is about 0.900 times the peak (2*sqrt(2)/pi). We'll put a resistive load on the CT, such that 20A RMS sinewave in the monitored wire results in 10.0V average magnitude across the resistor: 10V*2037/(20A*0.900) = 1131 ohms. Actually, we'll make it a bit more, and add a way to trim it down to the right value to calibrate it: say 1.21k ohms 1%, paralleled with (10k fixed in series with a 20k trimmer). We won't connect that load directly to the CT output, but rather through a bridge rectifier. There is some advantage to using Schottky diodes, but it's minor, and you can just as well use 1n4148 or similar silicon diodes. The resistive load goes across the bridge output. Since the CT looks like a current source, the diode voltage drop is unimportant, so long as it's not so large the CT can't put out the voltage. Now we just need to average the output; what's across the load is the absolute value of a (nominally) sine wave. We could just put a capacitor across the load, but that requires the CT output to slew quite a few volts rapidly; better to isolate the capacitor with at least a resistor, and even better, an inductor. For the very low DC current involved, you can use the winding of a small audio transformer to get high inductance, tens of henries for a nominal "20k ohm" winding. Then use something like a

47uF cap at the output to smooth out the voltage. You may want to add a little resistance in series with the inductance to reach critical damping; then the response of the circuit will be reasonably "crisp"-- much more so than if you just use a resistor instead of the inductor.

After construction, calibrate it: put a known 20A through the monitored wire, read the voltage on the PLC, adjust the trimpot so it reads 10.00V. Now reduce the current to 10A, and make sure the PLC reads 5.00V. Check the linearity further by putting in, say, 1A, and checking for 0.500V from the PLC. If it's not as linear as you'd like, consider applying a calibration in the PLC. But with the rectifier connected to the CT and the load to the rectifier output, you've removed a significant source of nonlinearity.

Cheers, Tom