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I have some cheap TA12-100 current transformer boards with a built in 200 Ω burden resistor rated to measure up to 5 A.

(Pictured)

I would like to set these up with an Arduino to monitor a 120 V AC outlet that will power loads up to 15 A.

I understand there will be no accuracy above the 5 A, but would it be safe to use over the rating for just detecting load presence?

The primary objective of this project is to simply measure when the outlet is powering a large load(electric saws/drills/shop tools) using the CT and subsequently close a relay to automate a dust vacuum system safely. Accuracy does not matter. However, it would be nice to output some ballpark power usage numbers just for the coolness factor.

Also, what is the best/cheapest option for my application that would allow precision measurement up to 240 V/50 A as well?

possible data sheet

Additional details:

Turn ratio 1:1000

Data sheet:

data sheet image

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  • \$\begingroup\$ What's the turns ratio? What will be the dissipation in the burden resistor at 5 A and at 15 A? What's the rating of the burden resistor? \$\endgroup\$
    – Neil_UK
    Commented Oct 24, 2022 at 6:58
  • \$\begingroup\$ 1:1000 turn ratio. I'm not sure how to answer your other questions. \$\endgroup\$
    – Andrew Podrasky
    Commented Oct 24, 2022 at 7:11
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    \$\begingroup\$ Measure how physically big the load resistor is, I can't tell from your module photo whether it's 0603 or 0402. It would have to dissipate 45 mW at 15 A, which is probably within the power rating, but a bit close to it for either size if you want long term stability. You could change the resistor for a physically bigger one to run cooler, or change it for a lower value to dissipate less power, or just go for it but not be surprised if the gain drifts a little. \$\endgroup\$
    – Neil_UK
    Commented Oct 24, 2022 at 10:03
  • \$\begingroup\$ Appears to be an 0603 resistor. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Oct 24, 2022 at 10:48
  • \$\begingroup\$ What is the appropriate resistor to use if I take the CT off of the breakout and build my own circuit? Also, if you're willing to explain how you get that dissipation number that would help me a lot. I'm new to the idea of current transformers, and I've done a lot of reading, but I don't understand how to make sure it's safe. EDIT: I now see the other responses that answer my question. I need to read this over better. Thank you. \$\endgroup\$
    – Andrew Podrasky
    Commented Oct 25, 2022 at 4:19

2 Answers 2

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The core of a current transformer will saturate if too much current is linked through that core. (Contrast this to a power transformer operated in the standard mode with a fixed ac voltage applied to it's primary.) That an excessive current can saturate the core of a transformer operating as a current transformer is due to the fact that the voltage drop seen by the primary is directly related to the current linked by the core, rather than being fixed by a constant AC voltage supply.

The current rating for a current transformer is almost always the current that the transformer can handle while giving a proportional output, rather than the current that would cause overheating. Thus, a current transformer rated for 5 amps can give a proportional output when the input is 5 amps or less.

If the input is more than the rated current, the transformer will no longer act linearly, but will generally not be damaged. When the core goes into heavy saturation, the resulting output waveform will be highly distorted. The voltage on the secondary (and consequently the current through the secondary) will actually fall during current peaks. This is because the rate of change of linked flux falls when the core is saturated.

See, for example this image from "Beyond The Knee Point: A Practical Guide To Current Transformer Saturation"

enter image description here

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  • \$\begingroup\$ -1. A CT operates exactly as a conventional power transformer until the voltage on its windings exceeds the Vs rating of the core, then it saturates. It's how you do the sums to determine this that's different. In a power transformer, it's the applied volts. In a CT, it's the voltage on the secondary. This is the secondary current times burden+secondary resistance. The latter may not be negligible, and cannot be reduced for any given CT. \$\endgroup\$
    – Neil_UK
    Commented Oct 24, 2022 at 15:05
  • \$\begingroup\$ @Neil_UK I agree that a "CT" is just a transformer and obeys the equations as all transformers. Vuyt it is operated in a different way, i.e. instead of the primary voltage being "fixed" (as in a power transformer) the primary current is "fixed". Thus at CT will saturate when current is excessive, while a power transformer will not saturate from excessive primary current, but will overheat. I'm not sure where disagree. Could you expand on your objection? \$\endgroup\$
    – Math Keeps Me Busy
    Commented Oct 24, 2022 at 15:25
  • \$\begingroup\$ expand on my objection - it's the winding voltage, not the primary current, that determines whether the CT will saturate or not. Without a burden, it takes a very low primary current. With a short circuit, a large current is needed. With a finite burden, somewhere in between. A wide range of primary currents, but the same secondary voltage, required for saturation. You have to know what the burden is before you calculate the primary saturation current. The winding voltage saturation (as long as compted with burden+sec resistance) is independent of the burden resistor. \$\endgroup\$
    – Neil_UK
    Commented Oct 24, 2022 at 16:38
  • \$\begingroup\$ @Neil_UK If the burden resistance is fixed, as it is in many cases, and as it appears to be in this case, then the magnetomotive force in the core will be proportional to the primary current, and thus, there is a specific primary current level that will cause saturation. I agree that if the burden resistance is varied, the primary current needed to cause saturation will vary. Is there some specific language in my answer that you object to? \$\endgroup\$
    – Math Keeps Me Busy
    Commented Oct 24, 2022 at 16:51
  • \$\begingroup\$ Yes, I object to the first sentence, as it refers to the primary current, not the net current, creating saturation. The second sentence makes an incorrect contrast between power and CT transformers. The third sentence links the primary voltage drop to the primary current, when it's the burden resistor that sets this ratio. 3 out of 3 in the first paragraph alone. Now bearing in mind the OP asks about safety, and has already dismissed accuracy, what you have said seems to be also missing his point. \$\endgroup\$
    – Neil_UK
    Commented Oct 24, 2022 at 19:26
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With 15 A input, the output will be 15 mA, and the 200 burden resistor will dissipate about 45 mW. It looks pretty small, but will probably handle this current. You could try it and see how hot it gets after about 5 minutes.

The specs say 7 amps maximum.

(edit) I designed a small primary injection circuit breaker test set (PI-250) using a similar CT rated for 200 amps (TE1200), and I found it to be linear within about 1% from 2 to 1000 A with a 1 ohm burden resistor, and from 0.5 to 200 A with a 10 ohm resistor. Test results:

Current transformer test results

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    \$\begingroup\$ Thank you, I will try some tests and maybe try replacing the burden resistor with a much larger dissipation resistor for safety factor? This circuit is also never really going to be energized for more than a few minutes at a time, I'm inclined to believe it's very low risk. \$\endgroup\$
    – Andrew Podrasky
    Commented Oct 25, 2022 at 4:31

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