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I need a current transformer for an induction heating project. After searching Digi-Key for a transformer meeting my specifications (turns ratio, current, mounting type, etc.), the cheapest model I found was this. Unfortunately it specifies a frequency range of 50 kHz to 500 kHz, whereas my induction heater works at frequencies as low as 20 kHz. I simply couldn't find any other model that meets my specs (especially with regards to mounting) on Digi-Key.

Now, in principle I would avoid using a part outside spec, but due to lack of options, I've been trying to find out what sets this lower limit of 50 kHz for the frequency of this particular part, hoping that some particularity of my application would mean I could indeed use it at lower frequencies.

One of the answers to this question (by Spehro Pefhany) mentions the \$L/R\$ constant for the secondary. For my application, I selected a burden resistor of 3.3Ω, which along with the stated inductance of 22.4 mH, works out to a 6.8 ms time constant, therefore a 23.4 Hz cutoff frequency. Even if it were necessary to take the CT's secondary resistance of 2.9Ω into account, the cutoff frequency would approximately double. Either way, the attenuation at 20 kHz should be negligible. This doesn't seem to be the limiting factor.

The datasheet notes that the peak flux density should remain below 2,000 Gauss, and an equation is given to compute this:

$$ B_{PK} = \frac{8 \times V_{REF} \times d \times 10^5}{N \times f} $$

Where \$V_{REF}\$ is the voltage applied to the burden resistor, \$d\$ is the duty cycle, \$N\$ is the turns ratio, and \$f\$ is the frequency in kHz. Plugging in all values for my application, I get \$B_{PK} < 1000\$ Gauss, worst case (including some added margin). So this doesn't seem to be the limiting factor either.

Therefore I ask: does anyone have a plausible theory why this current transformer shouldn't work for 20 kHz signals?

By the way: should I be worried about the fact that my current, although switched at 20+ kHz, has a 120 Hz component (due to the mains frequency)? My gut feeling is that it doesn't matter, but I'd thought I'd ask nevertheless.

Edit: as per a commenter's request, this is the datasheet for the CT. Regarding the waveform, it can be approximately described as a 20+ kHz sine wave multiplied by a 60 Hz full-wave rectified sine wave. I've sketched the waveform in the figure below, however I used a 2.4 kHz waveform in place of the 20 kHz waveform, because with the full 20 kHz waveform the plot became unreadable.

Waveform

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  • \$\begingroup\$ I'm puzzled as well. You've got to keep the secondary load, reflected though turns squared, much less than the primary inductance. Even if you went for a 50ohm burden resistor, say to match an oscilloscope, that's 1.25mohm at the primary, compared to its impedance of at least 175mjohm at 50k, and 70mjohm at 20kHz. \$\endgroup\$
    – Neil_UK
    Commented Oct 30, 2015 at 21:15
  • \$\begingroup\$ The 120 Hz component of current could be an actual big whopping show stopper and you left it to the end to mention this. I think you need to state exactly what the current is. Don't do a quick retort - think about it a while and state what it is worst case. You also need to provide a link to the transformer's data sheet (.pdf). If it aint got one then find one that has. \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2015 at 21:29
  • \$\begingroup\$ @user44635, what's a mjohm? \$\endgroup\$
    – Transistor
    Commented Oct 30, 2015 at 22:39
  • \$\begingroup\$ The data sheet is poor and doesn't appear to give the saturation levels. Are you saying (as your picture imples) that there is no 120 Hz component? \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2015 at 23:25
  • \$\begingroup\$ There is a 120 Hz component. The envelope of the fast waveform shown on the picture has a period of 8.33 ms. The time axis on the graph spans 18 ms (although the final ~1.3 ms of the drawing is blank). \$\endgroup\$
    – swineone
    Commented Oct 30, 2015 at 23:26

1 Answer 1

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There isn't a 120 Hz component -- it is multiplied by the 20 kHz, thus giving sidebands at 20k + 120 and 20k -120.

The CT's limit is core saturation; at lower frequencies, you can't run as high a current through it -- basically limited proportional to frequency.

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  • \$\begingroup\$ Of course you're right, no 120 Hz component -- not sure what the heck I was thinking. But the real question here is: will this core really saturate? In my question I pointed out I'm nowhere near the limits of peak flux density, which as I understand is just different wording for saturation. Wouldn't that mean I'm still operating the part within spec, even if below the indicated minimum frequency? \$\endgroup\$
    – swineone
    Commented Nov 3, 2015 at 3:31
  • \$\begingroup\$ From your equation above, as f reduces, VREF should reduce also to keep B below the limit. I assume that the output is not AC coupled ? That would also limit the lower frequency. Is there any way you can calibrate it ? You can use a standard signal generator (50 ohm output), and just wind 10-100 turns through the CT. 1 V in 50 ohm is 20 mA -- 100 turns will 'look' like 2A to it. Check with a 'scope what you are delivering to it. \$\endgroup\$
    – jp314
    Commented Nov 8, 2015 at 5:20

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