Maker Pro
Maker Pro

current transformer winding wire help please

R

Rich Grise

Jan 1, 1970
0
Gordon W wrote...

No, no, why don't you first try the second part of my advice, "keep
the load resistor small." That's a simple quick fix.

Maybe you can help me with a bit of brainlock here. If I put a CT, and
short the secondary, how does that reconcile with the idea of an ordinary
transformer with a shorted turn in the secondary? Is it just that the
current will be 1/N, and the voltage will be trivially small, based on
the resistance of the wire itself?

My only experience with a CT was when I built a 24VDC/120VAC, 1.2KVA
inverter from a "kit", following instructions exactly: "7 feet of
#24 teflon wire, wound on the toroid with about 7" left over for leads."
The thing was, there were two different cores in the "cores" drawer of
the parts bin, and I spent hours trying to figure out why the current
limit circuit wasn't getting activated - then finally, in exaxperation,
I wound another toroid, but with a _blue_ core, and it worked like a
charm. Apparently, the core characteristics have a _lot_ to do with it.
I've always been kind of afraid of things magnetic, but the guy I was
working for at the time designed his own ferroresonant transformers.
We'd be standing at the bench, looking at a new ferro design, and it
wasn't to spec, and he'd say stuff like, "Oh, let's just add a lam or
two" or "Hmmm - maybe use 14-lam shunts" or "Maybe a 15 uF ferro cap"
- in other words, Black Magic. ;-)

But we always got it, usually within only a few hours, and with those
exact #lams, #shunts, and cap, went right into production.

Spooky. ;-)

Thanks!
Rich
 
L

Larry Brasfield

Jan 1, 1970
0
Rich Grise said:
Maybe you can help me with a bit of brainlock here. If I put a CT, and
short the secondary, how does that reconcile with the idea of an ordinary
transformer with a shorted turn in the secondary? Is it just that the
current will be 1/N, and the voltage will be trivially small, based on
the resistance of the wire itself?

Exactly. It differs from your shorted turn only in
that the secondary current is available at the
terminals instead of looping inside the transformer.
My only experience with a CT was when I built a 24VDC/120VAC, 1.2KVA
inverter from a "kit", following instructions exactly: "7 feet of
#24 teflon wire, wound on the toroid with about 7" left over for leads."
The thing was, there were two different cores in the "cores" drawer of
the parts bin, and I spent hours trying to figure out why the current
limit circuit wasn't getting activated - then finally, in exaxperation,
I wound another toroid, but with a _blue_ core, and it worked like a
charm. Apparently, the core characteristics have a _lot_ to do with it.

One nice thing about current transformers is that, due
to the typically low voltages at the terminals, shunt
inductance plays only a small role in the response.
The downside is that magnetizing current directly
subtracts from the output current, unlike the situation
with a voltage transformer where, (approximately),
the output voltage is Ns/Np times the input voltage.
If core characteristics are affecting the output of a
current transformer very much, the core is too small
or lossy.
I've always been kind of afraid of things magnetic, but the guy I was
working for at the time designed his own ferroresonant transformers.
We'd be standing at the bench, looking at a new ferro design, and it
wasn't to spec, and he'd say stuff like, "Oh, let's just add a lam or
two" or "Hmmm - maybe use 14-lam shunts" or "Maybe a 15 uF ferro cap"
- in other words, Black Magic. ;-)

I knew a fellow like that once. When questioned,
it turned out he had well internalized the various
formulae that I still had to deal with on paper.
But we always got it, usually within only a few hours, and with those
exact #lams, #shunts, and cap, went right into production.

Spooky. ;-)

Sometimes.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Rich Grise <[email protected]>
Maybe you can help me with a bit of brainlock here. If I put a CT, and
short the secondary, how does that reconcile with the idea of an
ordinary transformer with a shorted turn in the secondary?

The transformer is just the same, but the primary is not connected
directly to a supply but **in series with the load*. It's just as if you
had an ordinary transformer in series with a lamp connected to the mains
supply and then you shorted the secondary. Let there be light!
Is it just that the current will be 1/N, and the voltage will be
trivially small, based on the resistance of the wire itself?
Yes.

My only experience with a CT was when I built a 24VDC/120VAC, 1.2KVA
inverter from a "kit", following instructions exactly: "7 feet of #24
teflon wire, wound on the toroid with about 7" left over for leads."
The thing was, there were two different cores in the "cores" drawer of
the parts bin, and I spent hours trying to figure out why the current
limit circuit wasn't getting activated - then finally, in exaxperation,
I wound another toroid, but with a _blue_ core, and it worked like a
charm. Apparently, the core characteristics have a _lot_ to do with it.

Very much so.
I've always been kind of afraid of things magnetic, but the guy I was
working for at the time designed his own ferroresonant transformers.
We'd be standing at the bench, looking at a new ferro design, and it
wasn't to spec, and he'd say stuff like, "Oh, let's just add a lam or
two" or "Hmmm - maybe use 14-lam shunts" or "Maybe a 15 uF ferro cap" -
in other words, Black Magic. ;-)

Ferroresonant design IS pretty much an art. Everything is non-linear and
the published characteristics of the core material are only a rough
guide. Ordinary linear 50-60 Hz transformers are no big deal, but it's
far easier after you have designed the first 50 or so and kept all the
data in a Black Book. That eliminates a lot of tedious arithmetic, with
the errors that is prone to.
 
G

Gordon W

Jan 1, 1970
0
John Perry said:
But, remember, Win, that 320mv was with a 1.5K resistor -- far too much
for any reasonable current transformer circuit, and hundreds of times
the normal burden resistance.

He wants to drive an led (optocoupler) from a primary of 7.5A. To do
this directly would require 7500/20=325 turns ratio! It simply isn't
worthwhile.

What he has, when he puts in a reasonable burden resistance, does
exactly what he needs, although he needs to dissipate 0.12W in his
28-ohm shunt resistor.

Yes, he could use an active circuit to reduce the burden impedance to
near zero, and have practically no power in the burden circuit, and
almost zero volts on the primary. But then he'd have to add power
supply, amplifier circuit, ...

Is it really worthwhile for his present project? Now he knows how to do
the job right. Save the purism for the future!

Gordon, unless you want to ice your cake even more, don't bother. You
already have a cool-running transformer that will do everything you
wanted to do, and you've learned enough to do the job with confidence if
you should need to do it again.

John Perry

Thanks John, yes I will do it again, the next project is a 2.6kW
air-conditioner and then :eek:)
It's great being retired but I wish I'd stayed awake in class back then.

Thanks again for your help

Gordon
 
G

Gordon W

Jan 1, 1970
0
Winfield Hill said:
Gordon W wrote...

No, no, why don't you first try the second part of my advice, "keep
the load resistor small." That's a simple quick fix.

Thanks again Win for your reply. What I said above was an effort to combine
advice from two replies and be brief at the same time. Not a good idea.
The site that legg gave me showed a diagram with 1 pri turn and 500 sec
turns and a 68 ohm resistor. My CT has 10 or 12 pri turns and nothing like
5000 sec turns. Far from ideal.

The 1.5k resistor I started with gave a 320mA pri current. After your advice
I used the 10 ohm which reduced the pri current to 196mA. With rectification
etc the 22 ohm I finished up with gave 20mA for the LED and 215mA pri
current.
You said 320mA is way too high so I think 215 or even 196 is still too high.

So the rewind thinking. Keep the ratio high (I'll start with 1:500) keep the
resistor low (I'll try 10 to 20 ohms) keep the pri current low (hopefully it
will be ) :eek:)
As John Perry says, my CT works (not really ideal but it works) so I won't
bin it but I would like to do better and after all the good advice I've been
given the next one should be.

Thanks again for your help

Gordon
 
F

Fred Bloggs

Jan 1, 1970
0
Larry said:
Exactly. It differs from your shorted turn only in
that the secondary current is available at the
terminals instead of looping inside the transformer.

Exactly? As usual you're such a dumbsh_t you didn't even read what the
other idiot said...

One nice thing about current transformers is that, due
to the typically low voltages at the terminals, shunt
inductance plays only a small role in the response.
The downside is that magnetizing current directly
subtracts from the output current, unlike the situation
with a voltage transformer where, (approximately),
the output voltage is Ns/Np times the input voltage.

That's because they were designed that way, idiot. Error bandwidth and
step-down ratio- they do not just appear.
If core characteristics are affecting the output of a
current transformer very much, the core is too small
or lossy.

Nah- more likely that his magnetizing inductance was way too small,
shunting *all* the load current around the reflected secondary winding
resistance + burden resistor.

I knew a fellow like that once. When questioned,
it turned out he had well internalized the various
formulae that I still had to deal with on paper.

Ahh- couldn't put him down then...
 
L

Larry Brasfield

Jan 1, 1970
0
[Derf transform applied.]

Fred Bloggs said:
Larry Brasfield wrote:

In my post, I quoted Rich Grise asking:
If I put a CT, and short the secondary, how does that reconcile
with the idea of an ordinary transformer with a shorted turn in the
secondary? Is it just that the current will be 1/N, and the voltage
will be trivially small, based on the resistance of the wire itself?
and answered as Fred quoted:
Exactly. It differs from your shorted turn only in
that the secondary current is available at the
terminals instead of looping inside the transformer.

Exactly? [derf] you didn't even read what the other [derf] said...

I read Rich's proposition and agreed with what he
was trying to say. So, yes, I meant "Exactly". Of
course, an electronical genius could see that there
are myriads of trivial distinctions to be made which
show that Rich didn't quite get it but the electronical
genius does. I consider such games to be stupid.
One nice thing about current transformers is that, due
to the typically low voltages at the terminals, shunt
inductance plays only a small role in the response.
The downside is that magnetizing current directly
subtracts from the output current, unlike the situation
with a voltage transformer where, (approximately),
the output voltage is Ns/Np times the input voltage.

That's because they were designed that way [derf]. Error bandwidth and step-down ratio- they do not just appear.

An amusing comment in a thread about considerations
in the design of a current transformer. I assure you, it
is possible to misdesign a current transformer such that
shunt inductance is too small. In fact, I know a fellow
who pretends to know transformers who asserts so.
Nah- more likely that his magnetizing inductance was way too small, shunting *all* the load current around the reflected secondary
winding resistance + burden resistor.

Really? Whatever makes you claim *all*? From such
a self-supposed smart guy, that's quite a blunder. With
more careful reading, from Rich's having "spent hours
trying to figure out why the current limit circuit wasn't
getting activated", you could logically deduce only that
he got less output than he expected from the device.
Ahh- couldn't put him down then...

Unlike you, I don't get my jollies that way.
 
F

Fred Bloggs

Jan 1, 1970
0
Larry said:
I read --- proposition and agreed with what he was trying to say.
So, yes, I meant "Exactly". Of course, an electronical genius could
see that there are myriads of trivial distinctions to be made which
show that --- didn't quite get it but the electronical genius does.
I consider such games to be stupid.

Really? That's all you do is play games. The question and your response
are stupid- typical pseudo intellectual fare musing over descriptive
aspects of some fundamental property you don't come close to understanding.

An amusing comment in a thread about considerations in the design of
a current transformer. I assure you, it is possible to misdesign a
current transformer such that shunt inductance is too small.

Not really- and your assurances are worthless around here- you know
absolutely nothing and your experience level is zero. So take your
pretentious little viewpoint and go shove it.

In fact, I know a fellow who pretends to know transformers who
asserts so.

Who cares...
Really? Whatever makes you claim *all*? From such a self-supposed
smart guy, that's quite a blunder. With more careful reading, from
---- having "spent hours trying to figure out why the current limit
circuit wasn't getting activated", you could logically deduce only
that he got less output than he expected from the device.

Not really, retard- no telling to what extreme the idiot went to trying
get enough signal- a big jump in AL going from an RF iron powder to
ferrite pot core it seems. ---- is another worthless p.o.s. like
yourself- uneducated riffraff who poses as engineer on NG under a
variety of names, and here we find out he was always a half-assed
piddler with little to no skill set.
Unlike you, I don't get my jollies that way.
Oh yes you do- and there is ample proof of that- until you get your butt
kicked, then you switch horses. Everyone knows quite well what you are...
 
L

Larry Brasfield

Jan 1, 1970
0
[Derf transform applied.]

Fred Bloggs said:
Larry Brasfield wrote: ....
[derf]

....
[derf]
....
[derf] is [derf] like
yourself- [derf] who poses as engineer on NG under a
variety of names

For the record, I have been employed for decades as an
electronics design engineer, specializing in analog design,
and I have never posted under any name other than the
one my parents gave me at birth. Anybody who claims
otherwise is a liar, as "Fred Bloggs", or whoever posts
under that fictitious name, repeatedly proves himself to be.

[derf]
 
F

Fred Bloggs

Jan 1, 1970
0
Larry said:
For the record, I have been employed for decades as an
electronics design engineer, specializing in analog design,
and I have never posted under any name other than the
one my parents gave me at birth.

Nah- that's more of your bs- we have your number- no way in hell have
you *ever* been employed as an "analog designer"- you are a muddle
headed jackass and incompetent. We have caught you fabricating stories
and posting erroneous information on far too many occasions- you are a
worthless and ignorant idiot- wrong too consistently to be anything
other than a complete poser.
 
L

legg

Jan 1, 1970
0
The 1.5k resistor I started with gave a 320mA pri current. After your advice
I used the 10 ohm which reduced the pri current to 196mA. With rectification
etc the 22 ohm I finished up with gave 20mA for the LED and 215mA pri
current.
You said 320mA is way too high so I think 215 or even 196 is still too high.

So the rewind thinking. Keep the ratio high (I'll start with 1:500) keep the
resistor low (I'll try 10 to 20 ohms) keep the pri current low (hopefully it
will be ) :eek:)
As John Perry says, my CT works (not really ideal but it works) so I won't
bin it but I would like to do better and after all the good advice I've been
given the next one should be.

If the turns ratio remains constant, varying the sensor's secondary
load resistor should not affect either the primary or secondary
currents, while the sensor remains linear.

Nonlinearities produced by saturation (with large sensing resistors
and voltage burden) would be expected to reduce the current reported
by the sensor, as the output waveform truncates prematurely each time
it reaches saturation. This is the opposite of the trend that you
report.

Only the main power circuit load, that appears in series with the
sensor's low-turns-count primary winding, is expected to be able to
dominate current flow. A current transformer is supposed to be a
comparatively low power sensing device that affects the power circuit
being monitored as little as possible.

The greater the voltage burden placed on the power circuit by the
sensor's primary, the less it looks like a current transformer, the
more it looks like another voltage transformer in series with the main
load.

RL
 
Top