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So I've got a -12v +12v ADC input limit and I've got a 200 turns ratio current transformer. My input Max current is 30A which is stepped down to 150mA. How do I determine the right burden resistor so I get a -+12V input for my ADC, also do I need a regulator and filtering for this circuit.

Cheers

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  • \$\begingroup\$ You should not go higher than the ohms specified in the data sheet or you won't get accurate results. What does the data sheet say? \$\endgroup\$
    – Andy aka
    May 12, 2015 at 7:41
  • \$\begingroup\$ Here's the datasheet eltime.co.uk/resources/product/datasheet_66.pdf \$\endgroup\$
    – Joh b
    May 14, 2015 at 9:28
  • \$\begingroup\$ If you read the data sheet, you'll see that for the range it covers, the max burden resistor is 25 ohms and it may be smaller because you haven't stated which model. The math is easy so where are you having problems? \$\endgroup\$
    – Andy aka
    May 14, 2015 at 18:55

2 Answers 2

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Ohms law applies as long as the transformer is operated well enoug away from saturation.

At 150 mA to get 12V you need R = V/I = 12/0.15 = 80 Ohms. 12V is getting moderately high for some CTs (current transformers) - you should check the specification sheet to see if there is a stated upper limit.

If max current is 30A (RMS rather than peak) then the peak current will be 1.414 (= sqrt(2)) times as high for a sinewave and you need to reduce your burden resistor accordingly if you want 12V max. ie Rb = 80/1.414 = 56.7 Ohms. Using somewhat less than the calculated values will give a proportionately lower voltage and allow you some safety margin for waveforms that are non sinusoidal for whatever reason. If your waveforms have been badly distorted by equipment such as switching power supplies or reactive loads or loads with poor power factors you may wish to use a significantly lower burden resistor to accommodate semi-artbitrary waveforms within your ADC's input range.

If you are feeding this signal to an ADC to measure the waveform voltages as the current changing then you do not want to add anything that distorts the waveform significantly - which is what a filter does by definition. The same applies to a regulator. If you want to only measure peak values or mean values or some other related measurement then you need to provide more information.

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  • \$\begingroup\$ If you are only looking for the RMS value of the current then you may be able to use a AC to RMS converter circuit or IC and measure the steadystate DC level at a much lower sampling rate. With just a current transformer and burden you will have to sample at least twice as fast as the input frequency (quite a bit more to maintain any kind of accuracy) to have repeatable results. beat frequencies between your line and sampling may result in strange results too. \$\endgroup\$
    – KalleMP
    May 12, 2015 at 8:38
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    \$\begingroup\$ @KalleMP You didn't specify it, but even using an AC to RMS converter, you still absolutely MUST have a burden resistor on a CT. Without one, the CT can develop dangerously high voltages, and even break down internally. \$\endgroup\$
    – R Drast
    May 12, 2015 at 10:42
  • \$\begingroup\$ @RDrast Quite right. A burden is needed for any CT exactly for the reasons you mention and the fact that it would almost certainly destroy the RMS circuit without, choice of value might be more flexible though. My thoughts were on his needs to filter a line frequency input waveform when he might not even want the wave shape (valuable info if doing power factor measurements or pulse shape monitoring though). \$\endgroup\$
    – KalleMP
    May 29, 2015 at 5:31
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Just wanted to add as an answer, since it wasn't covered in the original answer that for any commercially available current transformer, it should have an IEC rating of VA, and be marked so, or for IEEE transformers should have a "B-xx" rating.

For the IEC rated transformers, you have to select a burden resistor that is equal to or less than the VA rating to maintain accuracy. For IEEE transformers, you must select a burden resistor (technically, impedance) that is equal to or less than the 'xx' number. A B-2.0 rated device can handle up to two ohms burden load.

Probably about 95% of the small devices out there, with a primary rating of anywhere from 5 to 5000 amps are designed to handle a two ohm burden load, but that isn't a guarantee.

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  • \$\begingroup\$ Interesting that there is some consensus on a 2 Ohm burden. Just goes to show that some of us are armchair instrumentation folks with limited personal experience. \$\endgroup\$
    – KalleMP
    May 29, 2015 at 5:33
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    \$\begingroup\$ It isn't just a consensus, but virtually all current transformers are designed around a 2 Ohm burden, and have been for decades. The reasons vary, but one that holds a lot of weight goes back to the analog days, where you could select just about any CT to feed an analog meter movement. Generic meters were labeled 0-100%, but could easily be customized, so you stock only one meter, 30 different CT's, and all was golden. \$\endgroup\$
    – R Drast
    May 29, 2015 at 10:38

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