I'm seeing a 90 degree phase shift with a current transformer, I don't get it. The setup is pretty simple.


simulate this circuit – Schematic created using CircuitLab

The lamp is 100 watts. The current transformer is a junk drawer toroid, was used as a line filter, with ten turns. The lamp is one turn.

enter image description here

But here is what is confusing.

enter image description here

You can see the turn on at the start, switched on with a zero crossing triac. yellow is (1)120AC and red is (2)CT_OUTPUT

Because it is a purely resistive load I would expect to see the output of the current transformer to follow the voltage. But it is as close to 90 degrees out of phase as it gets.

What am I missing?


On one of those line filter toroids with some 50 turns and a .33 ohm burden resistor, this is what it looks like. And good enough for what I'm up to. Thanks all for keeping me looking in the right places.

enter image description here

  • \$\begingroup\$ How are you driving the primary of your CT? You haven't really explained that. \$\endgroup\$ – Andy aka Jun 6 '18 at 10:28
  • \$\begingroup\$ Hi @Andy aka The single white wire through the toroid is the light bulb circuit. I say the lamp is one turn and have a photo. What else do I need to add? \$\endgroup\$ – lakeweb Jun 6 '18 at 15:39

What you should do when using a current transformer unless you want odd results is: -

  • Make sure that the secondary burden resistor value is low enough so that its impedance (when "seen" from the primary side) is much, much lower than the magnetization inductive reactance.

If you don't do this (i.e. you make the burden too high) then the voltage developed across the primary is largely 90 degrees out of phase with the primary current. This is what you are seeing because the 5 ohm is too big.

This is NOT how CTs are operated when measuring current - the primary inductance may be around 100 uH and, at 50 Hz this has an impedance of 0.03 ohms. When the burden is "seen" from the perspective of the primary, the value is reduced by the turns ratio squared to be in parallel with the inductive reactance.

So if the burden is 1 ohm and the turns ratio is 100:1, when viewed from the primary, the burden will look like 0.1 milli ohms and this takes about 300 times more current than the magnetization reactance (30 milli ohm).

That is how CTs are meant to work and this is why they don't introduce a 90 degrees phase shift. With ten turns and 5 ohms, the reflected impedance to the primary is 50 milliohms and likely to be of the same order (or bigger than) as the magnetization reactance from a single turn through the core.

Thus, in this example most of the current entering the primary is taken through the magnetization inductance. With a ferrite core (as per the picture), the magnetization inductance is probably no more than 10 uH and so the reactance is therefore 3 milliohms i.e. massively lower than the 50 milliohm of the burden seen at the primary.

Bottom line - use a CT into a low value resistor burden or get results that seem odd. A 10:1 ratio requires a burden resistor that is a fraction of an ohm at most.

  • \$\begingroup\$ Hi Andy, Yes. That is what I came up with yesterday, my update. With about 50 turns and .33 ohms the junk ferrite core works fine at an amp. I just need to see if it saturates at the current I plan to use it at. Thanks. \$\endgroup\$ – lakeweb Jun 6 '18 at 16:11
  • \$\begingroup\$ Yes, saturation current could be a problem - 1 ampere turn is the magneto motive force and if you divide that by the mean length of the core you get "H". Then using the permability of the core (maybe not known) you can look at its BH curve and determine if it is close to saturation. Or just suck it and see. Nothing much to lose. \$\endgroup\$ – Andy aka Jun 6 '18 at 16:16
  • \$\begingroup\$ Hi Andy, The verdict is in. With a bathroom heater and a variac, it follows my fluke clamp-on, in phase, beautifully! Went to 12 amps. It will work for my application fine. \$\endgroup\$ – lakeweb Jun 6 '18 at 17:33
  • \$\begingroup\$ @lakeweb that's a great result and you would never get that without frying the core (saturation) if you didn't have 0.33 ohms converted to 123 micro ohms by the turns ratio! \$\endgroup\$ – Andy aka Jun 6 '18 at 18:00


The voltage induced in the secondary winding is proportional to the change of magnetic flux in the core. The magnetic flux is directly proportional to the current through the primary.

U ~ dI/dt

As an experiment, you can use a DC supply and turn it on and off to observe the effects.

  • \$\begingroup\$ Hi Oskar,What you say makes perfect sense to me. In fact, I'm at a loss as to why current transformers are advertised to have proportional input and output current. \$\endgroup\$ – lakeweb Jun 5 '18 at 23:32
  • 1
    \$\begingroup\$ No this is not correct for a perfect (or real) transformer, as is obvious if you think about how ordinary transformers are used. Either the currents are proportional or the voltages are proportional. What you are not accounting for is that the flux is fully coupled and the secondary current also makes a field counteracting the primary. (I think). What you are talking about are loosely coupled fields. \$\endgroup\$ – Henry Crun Jun 6 '18 at 1:39
  • 1
    \$\begingroup\$ Isn't the whole point here to have a minimal load on the secondary? The current in the secondary should be very small (after considering the turns ratio) and will create a very weak counteracting flux. \$\endgroup\$ – Oskar Skog Jun 6 '18 at 13:42

To be a current transformer assumes that it acts as a transformer, at your frequencies and impedances:

  • there must be tight magnetic coupling at the frequency
  • the unloaded internal impedance of the windings is much greater than the external impedances, so that the external resistances dominate.

In this case the impedance of 10 turns at 50Hz must be much greater than 5 ohms.

Your junk draw toroid is probably a ferrite mix made for filtering RF noise. i.e it is an RF ferrite mix.

Perhaps it is designed to have Z=500ohms at 1MHz

Then it will only have X <= 500*(50/1E6) = 25milliohm at 50Hz

We can consider this a perfect transformer, with an inductor L1 across it. In this case, the dominant effect is the 25mohm X of L1 not the 5ohm R.

Being L, the current and voltage are at 90 degrees.

Being very low, the voltage is much much lower than what you were expecting from an R of 5 ohms.


simulate this circuit – Schematic created using CircuitLab

CTs are wound on an iron metal core not ferrite. The mu of ferrite is too low.

  • 1
    \$\begingroup\$ Hi Henry, thanks. So I broke open one of those old school wall transformers. It had a steel laminated core and tried again. Same thing. And the output is all current. Not until I shunt R1 down to the resistance of the winding does the voltage drop. And same voltage without R1 as 5 ohms. \$\endgroup\$ – lakeweb Jun 5 '18 at 22:21
  • \$\begingroup\$ Do you mean that you used the secondary of the transformer (low voltage,thick wire) side connected to the 5ohm R, and made a single turn primary on the 120V side? \$\endgroup\$ – Henry Crun Jun 5 '18 at 23:00
  • \$\begingroup\$ Hi Henry, no. I used my die grinder with a cut off wheel and completely removed the original winding. Then wound it like my ferrite try. \$\endgroup\$ – lakeweb Jun 5 '18 at 23:25
  • \$\begingroup\$ What do you mean by "all current"? You should have 600mV rms across the 5ohms, and 120mA rms current if you put an ammeter across the 5ohms \$\endgroup\$ – Henry Crun Jun 5 '18 at 23:30
  • \$\begingroup\$ Hi Henry, I mean until the burden resistor gets to fractions of an ohm, the output stays the same. And update.. I took my fluke clamp-on and looked. Open circuit and there is that 90 degree shift. shunt it with a couple of ohms and there is no phase shift. I'm going to see how many secondary turns I can get on that laminated core... \$\endgroup\$ – lakeweb Jun 5 '18 at 23:51

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