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I use a current transformer for pulsed current measurement (with DC offset). It seems to me the core stores some energy after pulse disappear. The transformer works at about 30khz a it has 2 turns on primary and about 300 on secondary. The secondary is loaded with 100 ohm resistor with 300mV voltage across.

Is there some way how to reset the core so the output signal will not be shifted? I am thinking about disconnecting the secondary from 100 ohm right after pulse and quick discharge to higher impedance.

schematic

simulate this circuit – Schematic created using CircuitLab

Top is primary current, bottom is 100ohm resistor voltage: enter image description here

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  • \$\begingroup\$ Show the circuit and provide details the current in the primary. Show the CT output waveform too. \$\endgroup\$
    – Andy aka
    Jun 10 at 13:51
  • \$\begingroup\$ @Andyaka I added the waveforms a and schematic. \$\endgroup\$ Jun 10 at 13:59
  • \$\begingroup\$ What's the purpose of this measurement anyway? Could you reduce the repetition rate? \$\endgroup\$
    – Hearth
    Jun 10 at 17:04
  • \$\begingroup\$ @Hearth I am doing a self-oscillated resonant converter so rectified "sine" is getting from LC tank. There are two CT making a feedback for controlling the switch. The repetition ratio in approx. 50% according load, etc... It's not much powerful but in boosted LC tank flows a relatively huge current. \$\endgroup\$ Jun 10 at 17:42
  • \$\begingroup\$ @MichalPodmanický if we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. In other words you should be accepting an answer to this question. \$\endgroup\$
    – Andy aka
    Sep 10 at 12:46

5 Answers 5

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The first thing you have to realize is this: -

enter image description here

So, if you want to recover the waveform (so that it looks like the input waveform) you have to process it with an op-amp clamp that keeps the most negative voltage at 0 volts. Or use an MCU to process the signal.

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The easiest way to counter a DC offset on a current transformer is to add an additional "reset" winding alongside the primary and apply a DC current through that, calculated (and/or trimmed) to oppose the DC offset.

schematic

simulate this circuit – Schematic created using CircuitLab

This will cancel the DC component of the flux in the core, and allow you to sense larger currents without saturating the core. The current in the reset winding needs to be \$I_{rst}=I_{pri,DC}·\frac{N_{pri}}{N_{rst}}\$; use a multi-turn reset winding if your primary current's DC component is large.

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  • \$\begingroup\$ Interesting. The DC current is applied all the time or just after the pulse? If always i think the resetting DC current can be replaced with permanent magnet attached to the core. \$\endgroup\$ Jun 10 at 14:58
  • \$\begingroup\$ @MichalPodmanický At all times, while the circuit is in use. There's no need to keep it on when the circuit's not being used. And I wouldn't replace it with a permanent magnet; that would be difficult to tune to the appropriate flux (and you can't really insert it directly in line with the magnetic path anyway, because current transformers almost universally use toroid cores). \$\endgroup\$
    – Hearth
    Jun 10 at 15:10
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When the source is switched and unidirectional, the burden must also be switched and unidirectional.

An example application from my archives:

30V boost converter and AC output Source: my website, https://www.seventransistorlabs.com/Images/Mag_Amp_PSU.png

The bottom half of this can be ignored for present purposes.

Notice the CT primary (1T) is in series with the switching MOSFET, so the source is switched and unidirectional. The secondary (150T) has a large burden resistor across it (1k), which determines the peak reset voltage; then a diode into the nominal (small value) burden resistor.

Mind that this doesn't show phasing for some reason, which should be set in the obvious direction (transistor on = FR102 on).

As long as duty cycle is less than some maximum (the time required for the CT to reset into the 1k, several L/R time constants), the reading will remain zero-based.

Notice that another CT and diode could be wired-OR into the burden resistor. If you needed continuous current monitoring (say for an average current mode control), a second CT could be placed in series with the SBL2040, the secondary wired in exactly the same way (with its own reset resistor and diode). Pulse width would then have to be limited at the top and bottom end (i.e. it only reads correctly for, say, 5 to 95% PWM). I would generally recommend a Hall effect current sensor over two CTs, as it's less limited, but it is an option sometimes.

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  • \$\begingroup\$ Seems you do it similar what I mention in question and so let the core reset though 1k (high impedance). Originally i was thinking about disconnecting the secondary by transistor. Now I see it is possible with diode only also. Cool stuff. \$\endgroup\$ Jun 11 at 19:41
  • \$\begingroup\$ Notice that this is the general solution to the general question. You added in a comment that you're making a resonant converter, which permits a simplification that requires no diodes at all. But as you didn't make that part of the question, it is not answered as such here. \$\endgroup\$ Jun 11 at 20:34
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To reset a magnetic item (in this case, the current transformer core) takes an AC induction that slowly decreases. One might accomplish this with a switched incandescent light bulb and thermistor current limiter. This connects to a winding on the current transformer, which is out of circuit until the reset button is pressed.

schematic

simulate this circuit – Schematic created using CircuitLab

The PTC thermistor starts cold, at low resistance, but warms up in a few seconds to drop the AC applied to the reset winding; the meter winding is also on the same core, so it is a transformer (i.e. you don't want the reset winding to exceed normal range of the metering circuit; select lamp or other dummy load, and thermistor, accordingly).

The pushbutton switches the lamp circuit ON, drawing initial current according to the lamp and thermistor's cold resistance, then that current diminishes (as the thermistor resistance rises with thermistor temperature). After a few seconds, release the pushbutton, and allow a few minutes of cool-down time before resetting again.

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  • \$\begingroup\$ Is this a deguassing circuit? I'm not sure that's what they want; ferrite cores have low remenance and shouldn't need degaussing unless grossly overloaded. \$\endgroup\$
    – Hearth
    Jun 10 at 14:42
  • \$\begingroup\$ @Hearth Yes, it's to degauss a core that has taken on magnetization due to DC bias; most 'current transformer' items in my experience have laminated iron cores. \$\endgroup\$
    – Whit3rd
    Jun 10 at 14:59
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    \$\begingroup\$ Perhaps my experience with CTs is not the norm, then; I have mostly used Pearson CTs for near-metrology-grade measurements. Those are made with ferrite (or perhaps metal powder?) cores. \$\endgroup\$
    – Hearth
    Jun 10 at 15:13
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  1. Turn off the power: Before working on the CT, make sure to turn off the power supply and ensure your safety.

  2. Disconnect the secondary circuit: If the CT has a secondary winding or conductor loop, disconnect it or short-circuit the secondary terminals.

  3. Remove primary current: If the CT is connected to a live circuit, disconnect the primary current flowing through it.

  4. Demagnetize the core: Pass an alternating current through the secondary winding in the opposite direction of the primary winding. This creates an opposing magnetic field that demagnetizes the core.

  5. Gradually reduce the current: Start with a high magnitude of alternating current and gradually decrease it over time. This allows the magnetic flux in the core to decrease to zero.

  6. Remove the current: Once the current is completely removed, the opposing magnetic field dissipates, and the core is demagnetized.

  7. Reconnect and reassemble: Reconnect any disconnected components and reassemble the CT as necessary.

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  • \$\begingroup\$ The OP appears to be saying that this is a problem over a time of about 500 microseconds. It might be difficult to perform this suggested procedure on that sort of timescale ;) \$\endgroup\$ Jun 11 at 17:24

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