I am struggling to wrap my head around this. I understand that a transformer is not going to pass DC, and that every current transformer (CT) has some pass band. But what properties drive the high pass behavior, and how can I estimate what this corner frequency would be? I understand that it has something to do with the inductance of the primary, core material, etc. But if I know my turns ratio and core material is there some rule of thumb to estimate where the low frequency cutoff is? I read this article on CT design but I am still trying to grasp this. Perhaps if someone could draw an equivalent circuit for me showing whatever parasitic RLC components are in play.
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1\$\begingroup\$ This doesn't answer the question well either but it appears to be core saturation that causes the minimum core frequency: electronics.stackexchange.com/questions/100968/… \$\endgroup\$– hortaAug 6, 2014 at 19:24
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\$\begingroup\$ Hmm, I did a quick search before posting this but didn't see that post. Thanks for linking it. I will review. \$\endgroup\$– NickAug 6, 2014 at 19:26
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\$\begingroup\$ Another couple links that're related but not answering your specific question: electronics.stackexchange.com/questions/117281/… electronics.stackexchange.com/questions/83072/… It sounds like it's an experimentally found thing if it's not on a manufacturer's datasheet. Otherwise you have to work through the equations for core saturation and then determine at what frequency you'll saturate. \$\endgroup\$– hortaAug 6, 2014 at 19:27
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\$\begingroup\$ My issue is I have no data sheet. I've been told that thing was basically wound by some guy in his garage. I know the turns ratio and the materials but don't have any good experimental data. I was trying to see if I could ballpark this to try explain some other stuff going on in my system w/o having to actually go and measure it just yet! But if I am reading your links right it sounds like my burden resistance might be a factor too. That makes things interesting because that's not fixed. \$\endgroup\$– NickAug 6, 2014 at 19:37
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\$\begingroup\$ Even if you found the correct equations for your calculations, I wouldn't be surprised if they're an order of magnitude off. That's just the way the real world works. Why is your burden resistor changing? \$\endgroup\$– hortaAug 6, 2014 at 19:42
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You can arrive at an estimate by first calculating the inductance of the secondary. This may be on your data sheet, or it can be calculated from the number of turns, turn area, and core material. From there, the high pass response will depend on the input impedance to the amplifier you are using to buffer / amplify the sense signal off the secondary or the termination resistance (whichever is lower... probably the termination resistance). Specifically, it should be fl=Rterm/(2*pi*Lsecondary).
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\$\begingroup\$ So by termination resistance do you mean the effective parallel burden resistance? As I said in a comment above, I just have a fixed shunt and then some dividers to adjust the range. There is no amplifier - the the arbitrarily high resistance of the input to an ADC is the only other parallel resistance in the circuit. The dividers also do some biasing. \$\endgroup\$– NickAug 6, 2014 at 20:24
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\$\begingroup\$ Yeah, I think what I mean by termination resistance and what you mean by burden resistance are the same thing... \$\endgroup\$ Aug 6, 2014 at 20:25
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2\$\begingroup\$ ... or just get an LCR meter and measure it! \$\endgroup\$ Aug 6, 2014 at 20:28
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\$\begingroup\$ @Dave That's probably the right idea. I don't have an LCR handy right now. But I probably need to order a cheap handheld one. I have 1000 turns in my secondary, a 3cm radius, and a 0.5cm coil radius. The core material should have a permeability of around 4,000. Therefore my secondary's inductance should be around 2.01H. My termination resistence can be as low as 3Ohm. So my LF cuttoff should be around 0.2Hz! That seems really low - I imagine the current would need to be through the roof to magnetize the core. \$\endgroup\$– NickAug 6, 2014 at 20:51
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\$\begingroup\$ Yeah, that seems unrealistic. You will be able to sense the AC voltage on the primary side above that frequency, but of course that voltage will be very low at low frequencies. Specifically it will be Iprimary*2*pifLprimary... \$\endgroup\$ Aug 6, 2014 at 21:27