1
\$\begingroup\$

If I have an AC line powering some device, I can put a CT clamp around either the hot or neutral wire, and measure some low ac voltage and current. I understand this is because the AC line creates a magnetic field due to the right hand rule, and I am essentially harvesting the power with a transformer. I have two questions:

  1. Does a CT clamp add more load to the AC line, or is it that the AC line would create that magnetic field no matter what, and I am just harvesting it?

  2. How can I get more total power out of a CT clamp? I am not able to loop the input around multiple times, but would like more power. Will more secondary windings do this? widen the ferrite core? I'm not sure.

Any help appreciated.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ CTs convert current to voltage signal for later processing, it's a type of sensor. A sensor must not interfere so much with the quantity needed to be measure. Why do you want more power from them? \$\endgroup\$
    – Long Pham
    Sep 8, 2021 at 22:16
  • 1
    \$\begingroup\$ I want to use it as a current transformer to power a small device that will clamp to an ac line \$\endgroup\$ Sep 8, 2021 at 22:47
  • 1
    \$\begingroup\$ @LukedukeAnimations Look up the author, Gaikwad. At least two papers are easily found on this topic: "Evaluation of dimensional effect on electromagnetic energy harvesting", which is directly available for free at that link, and also a year later in 2019, "Electromagnetic Energy Harvesting to Power the Micro-power Temperature Sensor", which isn't so freely available on the web. \$\endgroup\$
    – jonk
    Sep 8, 2021 at 23:55

2 Answers 2

1
\$\begingroup\$

Question 1 has answers in my question Primary voltage drop in current transformer on this site.

Question 2: CTs have a VA rating just as VTs do. You calculate the maximum burden (resistance) using the VA rating and the rated current. For example, a 5 A, 3.75 VA CT can have a maximum output voltage of 3.75 / 5 = 0.75 V. That means that the maximum output burden (at the rated primary current) will be R = V / I = 0.75 / 5 = 0.15 Ω. Remember that the voltage will fall with falling primary current.

Tip:

  • Voltage transformers like open circuits (no load) and don't like short-circuits (overload).
  • Current transformers like short circuits (no load) and don't like open circuits.
\$\endgroup\$
3
  • \$\begingroup\$ So is there any rule to calculate how much load the CT is adding to the line? \$\endgroup\$ Sep 9, 2021 at 2:13
  • \$\begingroup\$ Yes. Power drawn from line = power drawn from CT / efficiency. Efficiency will be close to 1 for a decent CT. I have revised the second paragraph due to late night calculation error. \$\endgroup\$
    – Transistor
    Sep 9, 2021 at 6:53
  • \$\begingroup\$ I have the simulation here: tinyurl.com/yfstrbw4 . When I increase resistance on the secondary, I can draw more from the primary. Even when I simulate the CT identically, the simulation shows that increasing resistance on secondary would increase voltage and current in the secondary. IRL, when I increase resistance, voltage drop across the resistor increases to a max of what it would be as an open circuit, in my case (5v). As close as I can measure current stays the same. Is this because the CT is weakly coupled bc few primary turns? or what is going on here? \$\endgroup\$ Sep 14, 2021 at 0:23
1
\$\begingroup\$

A CT might be able to draw 100 mW of linear power from a few kW power source drawing 10A so it's not efficient nor free energy unless tapped into someone else's line.

A novel method I read about storing extra energy is using a nano crystalline CT with very high mu that would normally saturated 3/4 of the time is counteracted with a control coil to prevent saturation but still stored more energy than put back might boost up to 46%. Their CT with a lot of effort (!) was able to store almost 1/2 W. But the impedance must be matched to get MPT (max power transfer with 50% thru)

The limitation is the thermal resistance of thousands of windings to create 1/2 W with matched impedance from N^2 and the voltage available into a linear load, not LEDs. Plus the ability to added a second winding to counteract the saturation effects that kill the inductance and ability to store energy. But the linear load gets an irregular pulse response and not a sine wave. That's the challenge.

\$\endgroup\$
1
  • \$\begingroup\$ Any links to this novel method or things I can google? \$\endgroup\$ Sep 9, 2021 at 2:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.