Current transformer

What?

What?

John

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

Let me help you.... Here's the Jive Translation:

Would've been supa' fine if ah' had knode some current transfo'ma' isn't some very baaaad transfo'mer. Ah be baaad... Instead uh two terminals, it acts mo'e likes some sin'le terminal, wid yo' poo' burden resisto' caught in de middle. What it is, Mama! Ah sheeit, such be basic research...

Cheers

All transformers are current transformers, see the Faraday's lows of mutual induction. In practice it can be configured in current, voltage or power mode depending on application.

Mathew Orman

His current transformer is out of date...

News to Faraday, my transformer doesn't transform. At least, not very well. At least, not within the first, about a microsecond.

Simple example: go buy a bog standard Triad CT206, it rings like a bell at a rather low frequency (a MHz or so)!

In fact, I'm pretty sure it's acting as a toroidial resonator, which isn't much heard of, helical resonators are more common. Same idea, but with four quadrants.

Tim

Your using it as an air core transformer. Didn't Telsa invent something similar ;)

Cheers

It sure feels like it, but I'm pretty sure the core is still mu_r >> 100 at whatever resonant frequencies these things are doing it at. I'd call it a ferrite-loaded helico-toroidial resonator, or something ungainly like that.

Since L is large, that means F is small (~MHz), and the impedance is high -- sadly, the Q is also high, so the impedance (ESR, since it's a series resonant equivalent) at the feedpoint (i.e. burden R) is very low.

It's very interesting to rotate the "primary" around the core and watch the nodes in the standing wave. I should see about making an air-cored version, driving it with RF and putting it in a low pressure neon environment or something.

Tim

Huh???

On a sunny day (Wed, 30 Nov 2011 20:40:20 -0600) it happened "Tim Williams" wrote in :

I appreciate your deep knowledge of the subject, however maybe I should not read about it, as it makes something that I always perceived as extremely simple, sooooo complicated.

Current transformer: current_in / turns_ratio = current_out. (for a low load resistor or short). current_out x load_resistor = voltage_out.

Of course there are some details, but those can usually be disregarded as the errors caused by those are very small. NORMALLY.

One lesson: Do Not Leave Out The Load Resistor, as then Ze Foltage Kan Beacome Ferry High

On a sunny day (Wed, 30 Nov 2011 20:26:34 -0500) it happened "Martin Riddle" wrote in :

LOL

You seem to have failed to take into acount the parallel capacitances of the windings. No electronic component is pure - in the sense of presenting only resistive, capacitative or inductive impedance - and inductors/transformers are more imperfect than most.

-- Bill Sloman, Nijmegen

od

ll,

maybe he connected the wrong kind of 0 terminal capacitor?

NT

Some people don't understand that the "burden" resistor value has a maximum value before a current transformer goes crappy on you. ...Jim Thompson

We don't have a lot of 1-terminal parts on our schematics. A few, usually.

John

Also, typical current transformers are designed for low-frequency performance. If the burden resistor is high, and there is a lot of HF voltage on the primary conductor, it will capacitively couple to the secondary. I have heard of people making the secondary winding with coaxial cable, with the shield connected only at the one end, of course, to get rid of this capacitive effect. Tim mentions MHz in his original post, so that could be the problem.

Jon

Indeed! Back in my GenRad days, when I was designing switchers (before there were switcher-specific chips :-), I was fond on running my current transformers into essentially a virtual ground. ...Jim Thompson

Test pins, antennas, Van de Graaf generators, cattle prods,....

Cheers

Phil Hobbs

Indeed, effective parallel parasitic capacitance is a valuable concept. Sadly, it's just that, a concept -- the *actual* capacitance from end to end of, say, a solenoidal coil (i.e., as more advanced modelers call it, a helical resonator) is dramatically smaller than the turn-to-turn capacitance.

Consider, if instead of a helix, you had a stack of rings. It's the same basic structure, except skewed by a turn, so the turns aren't turns, they're loops. Now ground ALL the rings, except for just the two ends. What is the capacitance between those two rings?

The capacitance will not only be small due to distance, but almost entirely shielded by the turns inbetween them. When you unground them, all the intervening turns have their own capacitance, but it still doesn't even act as an ideal capacitive divider, because there is finite propagation delay along the structure (i.e., the speed of light) and because the interspersed turns have a comparable loading all their own (self capacitance to free space as well as "mutual capacitance" to adjecent turns).

The same is true of the toroid, with the added boundary condition that the magnetic field must be equal at both ends -- in the helical resonator, they can be equal or opposite, allowing (N + 1) / 2 wave resonances; toroids only allow integer N. Speaking of which, it stands to reason that the bandwidth of this resonance should correspond to the evenness of the winding; if the leads are not at exactly 0 and 360 degrees (give or take the external reactance between them), the wave can be skewed by that many degrees, across the unwound portion of the core.

Tim