Can we harvest enough energy (4-5 watts of power) from a custom made Rogowski coil?

The primary conductor is 33kV and has an average current of 10A. I'm aware of the induced voltage and the mutual coupling is weak because of the air core. There should be no other power source except a rechargeable battery or a supercapacitor for initial state of harvesting, if required. This question might be not crystal clear, so I will go thoroughly through all your answers and clarify.

  • \$\begingroup\$ For a Rogowski that is long enough, yes. However, you might find it needs to be very, very long. Choose a volume for your coil. Compute the inductance of the primary in that volume. You can now compute the available power from 10A rms flowing in the primary. Scale it up to meet your 5 watts. If you need to measure as well as power, you might be better with two coils, one Rogowski for measurement, one with a ferrous core for power. \$\endgroup\$
    – Neil_UK
    Dec 19, 2020 at 10:49
  • \$\begingroup\$ Unfortunately, no matter the voltage on the conductor, it's only the current that matters for magnetic coupling. And the current isn't much! \$\endgroup\$
    – Hearth
    Dec 19, 2020 at 16:12
  • \$\begingroup\$ @Neil_UK I get the point now. I would like to go with your suggestion. When we use materials like ferrous or silicon steel it gets saturated easily. So we will have to consider keeping the secondary ends short circuited (or under the saturation point voltage).Could you tell how can we harvest enough with these changed conditions. I have thought of using an Op amp to provide a virtual ground. I'm not sure how to harvest energy from it after that arrangement. \$\endgroup\$ Dec 19, 2020 at 17:26
  • \$\begingroup\$ Saturation is not necessairly a problem, especially with a single turn primary. Too busy to do actual numbers now, I might manage to do it later. Even at 33 kV, do you have the option to loop several turns of it into a core? Do you have the option to pass it through a toroid, or are both ends captive so you can only clamp something round it? (a current transformer is easy to make with an existing toroidal transformer by threading the primary cable through the hole). What's the diameter of the cable over the insulation? \$\endgroup\$
    – Neil_UK
    Dec 19, 2020 at 17:30
  • \$\begingroup\$ @Neil_UK Unfortunately there won't be a possibility of looping the main conductor through the core. I'm considering a situation where the primary conductor is sent through the core one time.Cable won't be thick more than like 2 inches. If you can tell me an idea of arrangement to do more efficient harvesting with a core included I would study them and ask if there are still questions. \$\endgroup\$ Dec 19, 2020 at 17:49

1 Answer 1


In order to harvest 5 W from a conductor carrying 10 A, we need to generate a voltage drop on the conductor of at least 0.5 V. As the primary is a single turn, that means the core must be able to generate 0.5 V per turn.

If we assume a sinusoidal current, at 50 Hz, the core must support a total rms flux of volts/2πf = 0.5/314 = 1.6 mWeber. The flux in Webers is the product of core area, and flux density in Tesla.

If we assume a 1 T rms flux density (a nice round number, and the practical maximum for ordinary transformer steel), the core area therefore needs to be 1.6m m2, or about 40 mm x 40 mm.

Can we actually achieve a 1 T flux density in the core? The cable is said to be 2" in diameter, so has a circumference of about 160 mm. The core will have to sit clear of the cable, so let's estimate 250 mm for the magnetic length of the core. The cable carries 10 Arms, so that generates an H field of 40 A/m in a length of 250 mm. The B field will be Hμ0μr, so the required μr is given by B/Hμ0 = 1/40*4π10-7 = 20k.

20k relative permeability! That's if you want to keep the core area down to 40 mm by 40 mm. You can get that permeability with mu-metal or permalloy, but not the saturation level. Easily available transformer core steel will only deliver μr in the 2k to 5k range, so you'll need significantly more area than 40x40 mm, 40x400 mm if you use 2k steel.

These large numbers are due to restricting the primary to a single turn, of large diameter, while trying to harvest watts.

Once you have a core large enough to intercept the power you need, then you can design the secondary to suit the end voltage you want to produce. It may help to resonate the transformer inductance with a shunt capacitor on the secondary, to save having to make the core even larger to allow for the effect of the finite primary inductance on the power throughput.

During one part of my career, I worked with one of the well known UK energy suppliers, on a clamp-on self-powered energy meter. The maximum primary diameter was 10 mm, so the core length was nearer to 80 mm than 250 mm. The core was about 7 mm x 20 mm. The target harvest was a few mW, not a few watts. The electronics was designed with μA-level opamps and MCU, which then spent most of the time asleep.

If you can engineer your required power down to 5 mW rather than 5 W, you may have a more practical design.

  • \$\begingroup\$ @Andyaka would you sanity check these numbers for me. \$\endgroup\$
    – Neil_UK
    Dec 20, 2020 at 5:54
  • \$\begingroup\$ I've just seen this request. My only thought is that if the primary current is passing (mainly) through the magnetization inductance, then that current is unredeemable in the secondary. You want the 0.5 volts to be produced almost entirely by the projected burden resistor and that means a projection of 0.05 ohms from the burden to primary. And, that means you want the mag impedance to be at least ten times more than this i.e. 0.5 ohms so that the majority of the current flows in the referred burden resistor. Hope this helps and happy new year. Virtually any modest CT is incapable of this. \$\endgroup\$
    – Andy aka
    Jan 25, 2021 at 11:17
  • \$\begingroup\$ @Andyaka When I worked at the well known energy supplier, we harvested energy to run a current monitor from the CT (mW). However, the inductance was resonated out with a C. I've not suggested this to the OP as I'm having difficulty quantifying the size of the improvement in the harvesting. Whether it was just an improvement in voltage output, or whether the power could be improved as well. \$\endgroup\$
    – Neil_UK
    Jan 25, 2021 at 11:59
  • \$\begingroup\$ I can't see a problem with this regards the feasibility of it working but the devil will be in the detail. \$\endgroup\$
    – Andy aka
    Jan 25, 2021 at 12:45

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