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Current transformer

T

Tim Williams

Jan 1, 1970
0
Would've been nice if I had known a current transformer isn't a very good
transformer. Instead of two terminals, it acts more like a single
terminal, with your poor burden resistor caught in the middle. Ah well,
such is basic research...

Tim
 
C

Cydrome Leader

Jan 1, 1970
0
Tim Williams said:
Would've been nice if I had known a current transformer isn't a very good
transformer. Instead of two terminals, it acts more like a single
terminal, with your poor burden resistor caught in the middle. Ah well,
such is basic research...

What?
 
M

Martin Riddle

Jan 1, 1970
0
John Larkin said:
What?

John

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
 
A

admformeto

Jan 1, 1970
0
Tim Williams said:
Would've been nice if I had known a current transformer isn't a very good
transformer. Instead of two terminals, it acts more like a single
terminal, with your poor burden resistor caught in the middle. Ah well,
such is basic research...

Tim

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

http://www.faster-than-light.us/
 
T

Tim Williams

Jan 1, 1970
0
admformeto said:
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.

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
 
M

Martin Riddle

Jan 1, 1970
0
Tim Williams said:
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
 
T

Tim Williams

Jan 1, 1970
0
Martin Riddle said:
Your using it as an air core transformer. Didn't Telsa invent something
similar ;)

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
 
T

Tim Williams

Jan 1, 1970
0
Bill Sloman said:
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.

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
 
T

Tim Williams

Jan 1, 1970
0
Actually Jim, that's exactly the problem.

A high impedance load will work perfectly.

Note the comment about series resonance...

You've been spending too much time on ICs where the transmission lines are
approximately the resistivity of nichrome and the inductivity of mud ;-)

Tim

P.S. Tonight I also enjoy "cooking" with wine ;-)
 
J

John S

Jan 1, 1970
0
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

Or you could just measure it.
 
O

Oppie

Jan 1, 1970
0
Martin Riddle said:
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...

Consider yourself bitch slapped ( ;-P
 
J

Joerg

Jan 1, 1970
0
John said:
We don't have a lot of 1-terminal parts on our schematics. A few,
usually.

There are a lot of parts that can transform themselves from
multi-terminal parts into 1-terminal parts. Usually after some smoke has
wafted off. In really bad cases they can turn into a 0-terminal part.
Usually after a lot of smoke has wafted away, or after a loud bang.
 
J

josephkk

Jan 1, 1970
0
Would've been nice if I had known a current transformer isn't a very good
transformer. Instead of two terminals, it acts more like a single
terminal, with your poor burden resistor caught in the middle. Ah well,
such is basic research...

Tim

Say what?

?-)
 
T

Tim Williams

Jan 1, 1970
0
Bill Sloman said:
Once you've got deep enough into high frequencies that the propagation
delay along the solenoid matters, you are obviously into a different
ball-park, but I was a bit surprised how useful the simple lumped
component model was in my application.

Oh, if you've got the right model and the system is reasonable (possibly,
it must be non-dispersive), it even works all the way up to the first
resonance. This is true of 1/4 and 1/2 wave resonators, in coax /
stripline form, and still reasonable for microstrip (which is dispersive).
Probably because the infinnitessimal circuit is a scaled down model of the
same layout. Take a chunk of coax and it looks like series L and parallel
C on almost any scale (limited by diameter, when higher modes take over);
Z_o = sqrt(L/C) and F_res = 1 / (2*pi*sqrt(L*C)) are always true.

I don't think it applies for helical and thin toroidial resonators,
because the calculated effective parallel capacitance and inductance vary
with frequency (going to extreme values at resonance, of course); and this
is going to be a consequence of the infinnitessimal model, which involves
mutual inductances and chains of capacitors, producing dispersive modes
that are difficult to model in the same way.

Tim
 
P

Phil Allison

Jan 1, 1970
0
"vkj"
Why? I dont understand why one cant simply use the open ckt voltage on
the
secondary as a measure of the primary current.

** For the simple reason that a "transformer" always converts voltages
according to the turns ratio on the windings. When you feed an iron cored
transformer prmary with a current, the resulting voltage drop is undefined
and very non-linear.
If you consider the ct. transformer as a coupled inductor, then
the actual calculations are quite simple.


** An inductor could be used to measure AC current (at a single frequency)
as there is a simple ratio between current and voltage drop ( long as the
inductive reactance is much bigger than the resistance). A secondary winding
would enlarge the voltage to be measured - but that is no longer a *current
transformer* at all and loses most of the advantages.

The original idea of a current transformer was to allow safe and simple
measurement of high AC supply currents with a standard moving iron ammeter.
No rectifiers are needed, no high voltages involved and there is virtually
no effect on the circuit being measured.


.... Phil
 
P

P E Schoen

Jan 1, 1970
0
"DJ Delorie" wrote in message
Because a current transformer is normally in the 1000:1 range.
Do you really want to try to measure 120,000 volts?

Actually, the voltage would never get anywhere near that high, although it
could get high enough to damage the insulation. For one thing, a 1000:1
ratio is a 5000:5 CT (usually), which is fairly rare. Most systems run
currents of 100 to 500 amps or so. If you do run current through a CT with
an open secondary, it will just function as an inductor, and it will soon
saturate at 60 Hz so the primary voltage drop will be only several volts.
Thus the secondary will probably produce only several hundred to a thousand
volts or so. However, this is often beyond the voltage rating of the control
wiring and could cause breakdown of insulation, arcing, and other damage.
And much higher spikes will be created when the power is switched on and
off. This could be especially bad if the CT is in a PWM motor control line
where the current may be switched at 20 kHz or so.
(also: this is why it's important to hook up the load resistor
*before* you put the current transformer on the primary wire)

Most CTs are shipped with a shorting jumper or spring clip on the secondary.
And some are protected by a pair of high current anti-parallel diodes. But
sometimes you may need more than 0.6V peak to drive the load, and since CTs
sometimes operate at 2x to 5x nominal rating, that can be a lot of power to
dissipate in the diodes.

Paul
 
S

Sylvia Else

Jan 1, 1970
0
Actually Jim, that's exactly the problem.

A high impedance load will work perfectly.

Only as long as the primary current doesn't drive the core into saturation.

Sylvia.
 
P

Phil Allison

Jan 1, 1970
0
"vkjerkoff"
Sorry, I dont think thats true at all.


** Then you are know nothing, fucking idiot.

FFS go test one !!!!!!!!!!!!!

Then PISS off to hell.
 
J

Jasen Betts

Jan 1, 1970
0
Why? I dont understand why one cant simply use the open ckt voltage on the
secondary as a measure of the primary current. Heres my argument:
For sinusoidals therefore, the open circuit is proportional to the prim.
current. If you consider the ct. transformer as a coupled inductor, then
the actual calculations are quite simple.

works fine with linear loads which draw sinusoidal current,
doesn't work for phase control, bridge rectifiers feeding capacitors,
or other active loads.
 
J

josephkk

Jan 1, 1970
0
Because a current transformer is normally in the 1000:1 range. Do you
really want to try to measure 120,000 volts?

(also: this is why it's important to hook up the load resistor *before*
you put the current transformer on the primary wire)

That is not all, maybe not even the most important. The flux in the core
when operated properly in the CT mode is nearly zero, and if the secondary
is open that is NOT true, and the core could easily saturate. BTW the
relevant IEEE standards make a distinction between metering and
protection/relaying CT and their performance properties in overload.

?-)
 
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