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I am looking at a current sense transformer, PE-51687NL (Digikey). I must be missing something in understanding the datasheet.

Digikey says the inductance is 20mH. I assume that this is the "single-turn" primary inductance of the magnetic core.

The datasheet says the secondary has 100 turns and the minimum secondary inductance is 2mH.

I know that the turns ratio is related to the inductance ratio: L1/L2 = (n1/n2)^2.

No matter how I twist the numbers I can't get the datasheet to make sense! Would someone help me make sense of this datasheet?

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    \$\begingroup\$ 20 mH, or even 2 mH, is an enormous inductance to be the primary of a current transformer. A CT's primary inductance is usually very low. I'm guessing whoever put the info into digikey's site made a mistake. \$\endgroup\$
    – Hearth
    Nov 13 '21 at 19:42
  • \$\begingroup\$ The schematic for PE-51687NL is the figure 2A. So, what is needed is only the value of the termination resistor which is generally a low value as 20 Ohm ... notes 4 & 5 : RT=100 Ohm max which give -> 1V / A (20 A max -> 20V ! ) Just verify the bandwidth of this wiring. You just have to pass a wire through the CT and you have then a 100:1 "transformer". When using the CT for lower current, pass more turns through the CT. Sensitivity will then be increased ... but max current must be lower ! \$\endgroup\$
    – Antonio51
    Nov 13 '21 at 21:23
  • \$\begingroup\$ For comparison see this D1870L or CS1100L and note 2 : coilcraft.com/getmedia/5cabc2b0-54ea-4f3d-a76a-5681314bc042/… \$\endgroup\$
    – Antonio51
    Nov 13 '21 at 21:29
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It's a 100:1, through-hole current transformer (CT).

There are errors in the Digikey description.

enter image description here

Here's the datasheet.

enter image description here

The 100 turn secondary,with an inductance of 2.0 mH, is to be terminated with a 100 Ω burden.

With 20A through the 1 turn primary, the secondary current would be (1/100) * 20 = 0.2 A and the secondary voltage 0.2 * 100 = 20 V.

enter image description here

In other words, the CT scale factor would be 1 secondary Volt per 1 primary Ampere.

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  • \$\begingroup\$ I appreciate this response because it helped to confirm my understanding of how transformers work. I also realized that the datasheet is misprinted! PE-51687NL is in fact a 20mH secondary. \$\endgroup\$
    – aosborne
    Nov 14 '21 at 21:32
  • \$\begingroup\$ Hi aosborne, Many thanks for the information. I am pleased to know that my answer was useful. \$\endgroup\$
    – vu2nan
    Nov 15 '21 at 3:21

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