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Current transformer waveform severe distortion

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mrinalmani

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Hi
I've wound a simple current transformer with 1:100 turn ratio using a small toroidal nanocrystalline core.
It more or less works OK, but there are two issues.
1. The current aligns to voltage at every cycle end but goes goes slightly out of phase during falling slope of the voltage. (Please refer to attached figure)
2. The secondary current drops suddenly if the value of primary current exceeds about 5A RMS. This causes severe distortion.
In my measurement setup, there's an auto-trasnformer connected to a 25Ohm resistive load and the primary current is varied by varying the transformer knob. The oscilloscope probes are connected to:
(a) the voltage across 25 Ohm load
(b) the voltage across burden resistor which is 1 Ohm. (Secondary winding resistance is 2 Ohm, so total burden = 3 Ohm)
Yellow is primary current and blue is secondary current.
Capture1.JPG
Capture2.JPG

Any one with experience in current transformers please help.
Thank you
 

FvM

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It's core saturation, transformer maximal ∫Vdt exceeded. You should be able to calculate the saturation point according to given core data. Nanocrystalline core has very sharp saturation characteristic, as seen in the waveform. As the burden is mostly formed by the secondary winding resistance, you need to rewind the transformer with more secondary turns or thicker wire.
 

mrinalmani

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Thanks for the reply.
But since this is a current transformer, ∫Vdt concept doesn't quite fit here. Or am I missing something?
The current is already dictated by the primary current. If the core were to saturate it should rather saturate at the top of sinusoid. But the photo looks as if it actually saturated when ∫Vdt exceeded a limit because the current drops off when ∫Vdt is is high and not when current is high. I don't understand.
I did some other tests.
Instead of 1:100, i used 1:50 configuration and the drop happened almost at half the current as compared to 1:100. (Same "external " burden but internal would have reduced to half as turns were removed)
Then I used 1:200 and the drop happened at twice the current compared to 1:100.
 

FvM

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But since this is a current transformer, ∫Vdt concept doesn't quite fit here. Or am I missing something?
B ~ ∫Vdt applies to any kind of transformer with magnetic core.

If the core were to saturate it should rather saturate at the top of sinusoid.
Why? Integral of sine wave is cosine waveform, it's maximum occurs at sine zero crossing.

Then I used 1:200 and the drop happened at twice the current compared to 1:100.
Yes, as expectable.
 

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Easy peasy

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try 5 ohms of burden resistor - this will lower the V.mS and hence the peak B excursion and put you closer to being inside the BH loop - accurate mains CT's use only about 100mT of a possible 1.2T swing ... ( for accuracy )
 

KlausST

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Hi,

try 5 ohms of burden resistor - this will lower the V.mS
Currently he uses 1 Ohms burden, why should a 5 Ohms burden lower Vms?

I'd say: the lower the value the better.
In ideal case a negative burden (to almost compensate internal resistance) with external Opamp circuit could be useful.

Klaus
 

FvM

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25 ohm is the primary load, I think.

OP needs to understand basic transformer law so that he can calculate useful CT operation range and expectable saturation limits.
 

mrinalmani

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I figured out the solution. Thanks everyone, thanks FvM.

- If the transformer was ideal then with secondary shorted out (0 burden) there would be no magnetizing current and net flux in core would be zero.
- As we increase burden, it will act in parallel to the magnetizing inductance and current will divide in ratio of impedance between Xm and R. What I forgot was that Xm >> R and therefore current through Xm is 90 degree out of phase with line current through R. And therefore if the core had to saturate, it will saturate when line current is near zero axis because that's when Xm is near peak. And this is what is exactly happening. I was expecting it to happen at peak of line current.

The calculations are as follows:
Primary inductance = 30uH
Xm = 30u*314 = 10mOhm
Primary referred R = 3/10000 = 0.3mOhm
SO we have Xm = 3mOhm and R = 0.3mOhm
B = LI/NA
N = 1, and A = 7mm2. So flux density per ampere line current would be
(30u/7u)*0.03 = 170mT. ""(0.03 factor because Xm = 10mOhm and R = 0.3mOhm so roughly 3% line current flows through Xm and the rest 97% through R.)
Now the core saturated when peak line current was nearly 8A. This would mean B of 170m*8 = 1.36T. Which seems OK. (Of course the saturation will not happen at the instant when line current was at peak, but the peak flux would be 1.36T)

Thanks again everyone
 

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