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Current Transformers (CT)s work like any other transformer: through Faraday's law of induction, the changing magnetic flux (from the AC current flowing through the conductor) induces a proportional voltage on the secondary coil. In this case, the primary is a single winding: the conductor through a core. Typically, the CT's are used as a measurement device, and the secondary is "burdened" with a known resistance value. So, reading the voltage across the resistor gives an accurate representation of the current you're measuring.

I see CTs as sources for "energy harvesting", and have even experimented with them myself, but my question is: does a CT technically load down the original circuit?

In other words: if I have an AC circuit drawing 20A and clamp on a Current Transformer and use the CT as part of an "energy harvesting" scheme to eventually power a 10mA LED, microcontroller, or what have you... where does that 10mA come from?

Is it 10mA more being drawn from the mains, on top of the 20A? Or is the 10mA coming from an otherwise non-utilized magnetic field being converted to energy I can take advantage of (truly energy harvesting)?

Edit:

So, to sum up, a CT is simply adding a series impedance, which actually reduces the current. This actually reduces the power consumed.

schematic

simulate this circuit – Schematic created using CircuitLab

So, per the circuit above, if we were to add a CT with N = 500 turns, and put a 2 ohm burden resistor on it, the resistance (in series) placed on the conductor would be 2 * (1 / 500)^2 = 8 uOhm

Power used by the whole circuit is reduced when adding a CT. Before adding the CT, the power consumed was Vs^2 / Load. After adding the CT, the power consumed is Vs^2 / (Load + 8uOhm)

Please correct me if you feel I'm wrong

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    \$\begingroup\$ Are you asking if they generate energy out of nothing? \$\endgroup\$ – PlasmaHH Apr 16 '18 at 9:31
  • \$\begingroup\$ The energy is coming from the mains. Not mA for mA, but mW for mW. \$\endgroup\$ – Dampmaskin Apr 16 '18 at 9:34
  • \$\begingroup\$ While I know that it is not free energy, I am not sure the actual source of the energy. Is the "downstream" circuit deprived of the 10ma? \$\endgroup\$ – InfiniteZoom Apr 16 '18 at 9:34
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If your load (aka burden) is 10 mA RMS and your primary current is 20 amps RMS then you have a turns ratio of 1:2000. If the load voltage is (say) 5 volts RMS then the load power is 50 mW.

That 50 mW is taken from the primary. What sort of impedance takes 50 mW from 20 amps? V = 50 mW / 20 amps = 2.5 mV RMS. Not much of a volt drop. The effective impedance seen in the primary winding is 2.5 mV / 20 amps = 125 micro ohms.

Is it 10mA more being drawn from the mains, on top of the 20A?

What you will see is a slight reduction in AC load voltage (2.5 mV) and this naturally results in a slight decrease in the 20 amps taken by the load (assuming it is a linear load).

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  • \$\begingroup\$ Ok, so is this a true statement: before adding the current transformer, the circuit was drawing 20A, let's call that that 4800W at 240V. After adding the current transformer, the circuit now uses more energy, about 50mW more. So, the circuit is now using 4800W (actually a little less because that load won't draw 20A anymore do to slight voltage drop) + 50mW \$\endgroup\$ – InfiniteZoom Apr 16 '18 at 18:07
  • \$\begingroup\$ No, after adding the current transformer the AC load (kettle, appliance etc.) is using slightly less electricity because it has only 230.9975 volts across it. The CT and its load (aka burden) steal 2.5 mV. The CT adds a series resistance of 125 micro ohms. \$\endgroup\$ – Andy aka Apr 16 '18 at 19:14
  • \$\begingroup\$ Ok, but the 2.5mV came from a calculation using 20A, but since the (assuming resistive/linear 12ohms) load would draw 20A from a 240V supply, and now is provided 239.9975V, it will be now be 19.9998A, and now we're iterating... Is this only able to be solved using diff eq / a PSPICE? I realize it's negligible, but I'm very interested in accounting for every last nW \$\endgroup\$ – InfiniteZoom Apr 18 '18 at 2:00
  • \$\begingroup\$ No, an iteration isn't needed. If the load is constant at 12 ohms then the total load is 12 ohms + 125 micro ohms and current from a 240 volt supply is 19.9997916688 amps. Pointless to worry about I'm sure. @InfiniteZoom \$\endgroup\$ – Andy aka Apr 18 '18 at 7:49
  • \$\begingroup\$ Thank you. You've helped me realize that the series impedance is the same no matter what the current is. A burden load is hooked up to the secondary of the CT. That burden load is transformed to the primary, by the square of the turns ratio. This transformed burden impedance is just sitting there, added in series to the remaining circuit, waiting for "normal" current to go through it. Nothing spooky or iterative about it! \$\endgroup\$ – InfiniteZoom Apr 20 '18 at 2:07
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Yes it appears as a small series voltage drop in the wire.

The current through the wire is unchanged (20A).

So say you put a diode across a 500turn CT.

The current will be 20A/500=40mA.

The diode voltage is 0.7V.

The burden voltage will be 0.7/500 = 1.4mV, or 28mW

The mains is 240V. Your load only gets (240V-1.4mV) or 1.4mV*20A= 28mW less power

In theory the secondary voltage of the CT will be whatever it takes to force 40mA to flow. So you could get 5V or 50V or 500V (and should not leave it unterminated lest you discover 240V*500). You should have a voltage limiting element there (diodes, zeners, leds)

In practice small CT's can't get very high voltages and this can be an obstacle to using them for powering circuits directly. (and this is why there is a bag of small 500T CT's in my workshop)

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  • \$\begingroup\$ Thanks for your answer. Can you clarify how it is that the original load can still draw the same 20A. Let's assume it is a purely resistive load. This upstream CT now presents itself as a voltage drop, theoretically reducing the mains voltage across the original load by that amount. But we used 20A in our calculation of the CT! I know this is isn't a chicken egg situation, but I'm having a hard time wrapping my head around it. \$\endgroup\$ – InfiniteZoom Apr 16 '18 at 18:05
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    \$\begingroup\$ Ok, you got me. The current through the load is reduced because the voltage drops, by 240V/1.4mV=116uA. But I'm an engineer, so 2+2 is close enough to 4. \$\endgroup\$ – Henry Crun Apr 16 '18 at 21:34

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