current transformer winding wire help please

[Gordon W]

I am thinking of using 0.125mm wire for the secondary of a current
transformer but don't know its current rating. DAGS and wire tables give me
other specs but not current.
Any info or url would be appreciated.
Thanks
Gordon

 

[John Fields]

I am thinking of using 0.125mm wire for the secondary of a current
transformer but don't know its current rating. DAGS and wire tables give me
other specs but not current.
Any info or url would be appreciated.

 

[Genome]

I am thinking of using 0.125mm wire for the secondary of a current
transformer but don't know its current rating. DAGS and wire tables give me
other specs but not current.
Any info or url would be appreciated.
Thanks
Gordon

A guess for iron transformers is 4A/mm^2 or 4E6A/m^2 and that's for a guess
of a 30C rise above ambient based on a guess based on another guess.

Isn't it strange that if you use a word repeatedly it begins to lose its
sense?

Barry will ask you some more questions.

DNA

 

[Gordon W]

give

A guess for iron transformers is 4A/mm^2 or 4E6A/m^2 and that's for a guess
of a 30C rise above ambient based on a guess based on another guess.

Isn't it strange that if you use a word repeatedly it begins to lose its
sense?

love it. )

Barry will ask you some more questions.

DNA

Thank you Genome and John Fields for your replies, I guess I should have
given more information. Hope this is not too long and I don't show my
ignorance too much.
I've got my computer turning on/off the booster to my solar hot water system
but I don't have any feedback (closed loop to monitor).
The thought was to put a current transformer in series with the heating
element, rectify etc and feed +5V to the computer when the heater was on.
I've done an experiment with a small iron cored transformer, the bobbin is
20x10x10mm, a layer of 10A wire (the heater is 240VAC at 7.5A) well
insulated from the secondary of lots of turns (my rev counter gave up part
way through) of the 0.125mm wire.
Rectified/filtered output came out at 71V and 11mA with a 6K resistive load
and 47V and 30mA with a 1.5K load.
I can easily reduce the turns to lower the voltage but what current should I
aim for so that this 'thing' will live happily ever after ? The transformer
will be in a ventilated plastic box under the house. Ambient sometimes gets
to 35 °C.

Thanks again

Gordon

 

[Gordon W]

Thanks John, please see my reply to Genome's post for further information.

Gordon

 

[Winfield Hill]

Gordon W wrote...

Genome wrote ...

Thank you Genome and John Fields for your replies, I guess I should have
given more information. Hope this is not too long and I don't show my
ignorance too much.
I've got my computer turning on/off the booster to my solar hot water
system but I don't have any feedback (closed loop to monitor).
The thought was to put a current transformer in series with the heating
element, rectify etc and feed +5V to the computer when the heater was on.
I've done an experiment with a small iron cored transformer, the bobbin
is 20x10x10mm, a layer of 10A wire (the heater is 240VAC at 7.5A) well
insulated from the secondary of lots of turns (my rev counter gave up
part way through) of the 0.125mm wire.
Rectified/filtered output came out at 71V and 11mA with a 6K resistive
load and 47V and 30mA with a 1.5K load.
I can easily reduce the turns to lower the voltage but what current
should I aim for so that this 'thing' will live happily ever after?
The transformer will be in a ventilated plastic box under the house.
Ambient sometimes gets to 35 °C.

Whoa, making a mountain out of a mole hill. Look at those miniscule
currents you're getting on the secondary, 11mA, 30mA - you can employ
virtually any wire size you like without worry. Secondary already
wound? Fine, keep it. Generally in commercial current transformers
the primary is simply the current-carrying wire to be monitored going
through the hole in the current transformer (which may be a toroid).
In your case, that'd be one turn, which gives you higher voltages on
the secondary, for any given mandatory resistive load, but less current.
If you want to power something serious on the secondary, then add to
the primary turns and reduce that load resistor. Whatever you do, it's
not likely the secondary wire size will be chosen by anything other
than convenience.

One thing that is very important, the secondary load resistor. Keep
in mind the current transformer's primary resistance will look like
the load resistor divided by the square of the turns-ratio. You don't
want the primary to have significant voltage drop, affecting the load
you're monitoring, so keep the ratio high, and the load resistor small.
And double check, under some conditions you'll need a power resistor.

Oh, and one other thing. In experiments like this keep in the back of
your mind Jene Golovchenko's father's advice, "Tune for maximum smoke."

 

[Genome]

love it. )

Thank you Genome and John Fields for your replies, I guess I should have
given more information. Hope this is not too long and I don't show my
ignorance too much.
I've got my computer turning on/off the booster to my solar hot water system
but I don't have any feedback (closed loop to monitor).
The thought was to put a current transformer in series with the heating
element, rectify etc and feed +5V to the computer when the heater was on.
I've done an experiment with a small iron cored transformer, the bobbin is
20x10x10mm, a layer of 10A wire (the heater is 240VAC at 7.5A) well
insulated from the secondary of lots of turns (my rev counter gave up part
way through) of the 0.125mm wire.
Rectified/filtered output came out at 71V and 11mA with a 6K resistive load
and 47V and 30mA with a 1.5K load.
I can easily reduce the turns to lower the voltage but what current should I
aim for so that this 'thing' will live happily ever after ? The transformer
will be in a ventilated plastic box under the house. Ambient sometimes gets
to 35 °C.

Thanks again

Gordon

Mr Win Hill is not wrong. I am being an idiot.

I will now proceed to Barry or demonstrate more foolishness.

Work things out arse about face.

You want sort of 5V out. Say your burden resistor is 1/4W and you run it at
1/8W so you want a 220R resistor. That gives you 23mA. 7.5A down to 23mA is
300:1 so one turn on the primary and 300 turns on the secondary.

Taking the 10mmX10mm dimensions as being ones the windings go around gives a
40mm mean length of turn. That makes the length of your secondary 12 metres.

If your wire is 0.125mm copper diametre then its resistance is about 1R6 per
metre so the resistance will be 20R. At 23mA it's going to dissipate 11mW
which, as Mr Win has pointed out, is poof.

Now, this bit might be crap.... as if the rest isn't.

On your secondary you get 6.2V, adding in some diode drops, so on your
primary you get 20mV. Treating your 50Hz as a square wave the on time is
10mS and the sum for flux density in the core is...

B = Vpri.Ton/Npri.Ae

Ae is the effective area of the core, let's say its the 10mmX10mm bit again
or 100E-6 metres^2.

Vpri = 20mV
Ton = 10mS
Npri = 1
Ae = 100E-6

Which gives B as 2 of the Teslas......

Now, I think the stuff used in 'iron' transformers saturates at about 1.5 of
the Teslas...... but Don't Panic, just worry a bit. That B value is really a
'change' in flux density and things could settle out as +/- 1 of the Teslas
and things will be OK.

Maybe.

For fucks sake, I'm making this up as I go along.

I'm sure Barry could help out here.

DNA

 

[BFoelsch]

Gordon W wrote...

Whoa, making a mountain out of a mole hill. Look at those miniscule
currents you're getting on the secondary, 11mA, 30mA - you can employ
virtually any wire size you like without worry. Secondary already
wound? Fine, keep it. Generally in commercial current transformers
the primary is simply the current-carrying wire to be monitored going
through the hole in the current transformer (which may be a toroid).
In your case, that'd be one turn, which gives you higher voltages on
the secondary, for any given mandatory resistive load, but less current.
If you want to power something serious on the secondary, then add to
the primary turns and reduce that load resistor. Whatever you do, it's
not likely the secondary wire size will be chosen by anything other
than convenience.

One thing that is very important, the secondary load resistor. Keep
in mind the current transformer's primary resistance will look like
the load resistor divided by the square of the turns-ratio. You don't
want the primary to have significant voltage drop, affecting the load
you're monitoring, so keep the ratio high, and the load resistor small.

Well, almost. The limiting factor in a current transformer is the flux
capability of the core. Generally, commercial current transformers are
proportioned so that the core saturates long before the primary voltage drop
becomes an issue. The lower the secondary circuit resistance, the smaller
the core may be. A secondary with lots of turns of small wire may have so
much resistance that it is unable to develop the secondary ampere-turns
needed to keep the core out of saturation!

Having said that, the OP may not care if the core saturates or not if all he
is doing is detecting the presence of a substantial current. However, if the
core is into saturation the secondary voltage is more difficult to
calculate. The saturated core will have a "squarish wave" change of flux
over time, and of course the secondary voltage will be the derivative of
this waveform. Too, the core heating will increase at high flux levels.

 

[Rich The Newsgroup Wacko]

A guess for iron transformers is 4A/mm^2 or 4E6A/m^2 and that's for a
guess of a 30C rise above ambient based on a guess based on another guess.

Isn't it strange that if you use a word repeatedly it begins to lose its
sense?

Barry will ask you some more questions.

Chair.

 

[John Perry]

Thank you Genome and John Fields for your replies, I guess I should have
given more information. Hope this is not too long and I don't show my
ignorance too much.
...
Rectified/filtered output came out at 71V and 11mA with a 6K resistive load
and 47V and 30mA with a 1.5K load.
I can easily reduce the turns to lower the voltage but what current should I
aim for so that this 'thing' will live happily ever after ? The transformer
will be in a ventilated plastic box under the house. Ambient sometimes gets
to 35 °C.
...

C'mon, guys, the detailed theory may be a bit involved, but the practice
of CT's is pretty simple!

If you have the secondary evenly wound with N turns of low-resistance
wire, and short it, the current through the secondary opposes the
current through the primary to exclude all flux from the core. So the
secondary current is 1/N the primary current, to a very good
approximation.

The farther you get from a perfect sheet of secondary current,
zero-resistance wire, and a perfect short circuit, the more flux gets
into the core, and generally the less accurate the current ratio. But
the approximation is still very good up to core saturation, insulation
breakdown, and secondary power dissipation limitation.

The only thing about the OP's CT that really concerns me is the very
thin, 0.125mm wire. Otherwise, he could just short a couple of LED's
back-to-back in series with another pair of back-to-back LED's and put
the resulting voltage into his input.

If he doesn't have enough turns, he gets too much current. 47V at 30ma
into 1.5K could make for too much current into a LED.

He's obviously using his core evenly enough to get 71V at 11ma. The 71V
is probaly limited by core saturation, causing the unequal
voltage-current ratio. The 47V at 30ma may be close enough to the
linear range of his CT that we can expect 300ma at 5v, but then he may
have winding resistance issues bringing on extra secondary voltage drop,
and maybe too much secondary power dissipation.

I'd suggest running the CT into a 10-ohm power resistor and measuring
the resulting voltage. Or better yet, directly into an ammeter. This
would tell you very nearly the actual turns ratio, and whether the
ultimate current is low enough for a LED circuit (or a Zener, or even a
specially selected low-value resistor for your needs). Run it for a
while to see if the current transformer heats up. If it stays
reasonably cool for half an hour or so, and gives a reasonable current
into the LED circuit, you may be in good shape already.

Of course, you get a free power indicator with the LED's, too .

John Perry

 

[John Perry]

Gordon W, Genome, Win Hill, etc. wrote:

If he doesn't have enough turns, he gets too much current. 47V at 30ma
into 1.5K could make for too much current into a LED.

He's obviously using his core evenly enough to get 71V at 11ma. The 71V
is probaly limited by core saturation, causing the unequal
voltage-current ratio. The 47V at 30ma may be close enough to the
linear range of his CT that we can expect 300ma at 5v, but then he may
...

Sheesh. I just realized how dumb this all was. What I should have said
was, of course, that if the 47V@30ma is near the CT's linear range, then
30ma is near the limiting secondary current for his CT and primary
current. So he may well be very close to what he needs for his use.

Of course, if his primary current changes, his secondary current will
change to maintain the flux cancellation. He needs to keep this in mind.

John Perry

 

[Fred Bloggs]

A guess for iron transformers is 4A/mm^2 or 4E6A/m^2 and that's for a guess
of a 30C rise above ambient based on a guess based on another guess.

Isn't it strange that if you use a word repeatedly it begins to lose its
sense?

Barry will ask you some more questions.

DNA

You mean Larry? Yeah- Larry should be able to add some fundamental
insight on his daffynition of current density and wire sizes- and then
try to impress us with some non-sensical and inapplicable crap he thinks
no one has ever heard of. But the fake continues to "carry on" like the
little trooper he is...you just can't crack a narcissist's universe.

 

[legg]

Mr Win Hill is not wrong. I am being an idiot.

I will now proceed to Barry or demonstrate more foolishness.

Work things out arse about face.

You want sort of 5V out. Say your burden resistor is 1/4W and you run it at
1/8W so you want a 220R resistor. That gives you 23mA. 7.5A down to 23mA is
300:1 so one turn on the primary and 300 turns on the secondary.

Taking the 10mmX10mm dimensions as being ones the windings go around gives a
40mm mean length of turn. That makes the length of your secondary 12 metres.

If your wire is 0.125mm copper diametre then its resistance is about 1R6 per
metre so the resistance will be 20R. At 23mA it's going to dissipate 11mW
which, as Mr Win has pointed out, is poof.

Now, this bit might be crap.... as if the rest isn't.

On your secondary you get 6.2V, adding in some diode drops, so on your
primary you get 20mV. Treating your 50Hz as a square wave the on time is
10mS and the sum for flux density in the core is...

B = Vpri.Ton/Npri.Ae

Ae is the effective area of the core, let's say its the 10mmX10mm bit again
or 100E-6 metres^2.

Vpri = 20mV
Ton = 10mS
Npri = 1
Ae = 100E-6

Which gives B as 2 of the Teslas......

Now, I think the stuff used in 'iron' transformers saturates at about 1.5 of
the Teslas...... but Don't Panic, just worry a bit. That B value is really a
'change' in flux density and things could settle out as +/- 1 of the Teslas
and things will be OK.

True. In crude AC current monitoring, a double limit flux swing can be
tolerated before saturation (due to excessive detection winding
volt-seconds) clips the output signal.

The OP might be advised to check out physical dimensions of varieties
of commercial 60Hz current transformers for comparison. These can be
surprisingly small, with rated volt-second burden vs accuracy listed.

http://www.coilcraft.com/pwrsense.cfm

In his application, he obviously just needs to reduce the sensor
resistance value so that the sensed output voltage is within his
required range. The ideal current transformer load is a short circuit.

If he wants to modify turns to do this, he should reduce the primary
turns. The larger the turns ratio of a current transformer, the
smaller the shorted output current will be, and the less affect the
sensing resistor or rectifier will have on the operation of the
circuit.

RL

 

[Gordon W]

True. In crude AC current monitoring, a double limit flux swing can be
tolerated before saturation (due to excessive detection winding
volt-seconds) clips the output signal.

The OP might be advised to check out physical dimensions of varieties
of commercial 60Hz current transformers for comparison. These can be
surprisingly small, with rated volt-second burden vs accuracy listed.

http://www.coilcraft.com/pwrsense.cfm

In his application, he obviously just needs to reduce the sensor
resistance value so that the sensed output voltage is within his
required range. The ideal current transformer load is a short circuit.

If he wants to modify turns to do this, he should reduce the primary
turns. The larger the turns ratio of a current transformer, the
smaller the shorted output current will be, and the less affect the
sensing resistor or rectifier will have on the operation of the
circuit.

RL

Thank you for your advice and the url, both very helpful

Gordon

 

[Winfield Hill]

Gordon W wrote...

Thank you for your advice and the url, both very helpful

My advice was, "keep the ratio high, and the load resistor small,"
IIIRC, didn't you say something about a 320mV drop on the primary?
That's huge! Unless you have a specific goal to get X volts and
X mA to power your load circuit, you should reduce the load resistor
and get the primary voltage down. For ideal operation that is...

 

[Gordon W]

Gordon W wrote...

My advice was, "keep the ratio high, and the load resistor small,"
IIIRC, didn't you say something about a 320mV drop on the primary?
That's huge! Unless you have a specific goal to get X volts and
X mA to power your load circuit, you should reduce the load resistor
and get the primary voltage down. For ideal operation that is...

Thanks Win, I used a WAG to wind the transformer and as legg says I need to
reduce the primary turns and as you say keep the ratio high so it's back to
a rewind and after all the good advice I've received I hope to get it right
this time.

Thanks again

Gordon

 

[John Perry]

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!

Thanks Win, I used a WAG to wind the transformer and as legg says I need to
reduce the primary turns and as you say keep the ratio high so it's back to
a rewind and after all the good advice I've received I hope to get it right
this time.

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

 

[Winfield Hill]

Gordon W wrote...

Winfield Hill wrote ...

Thanks Win, I used a WAG to wind the transformer and as legg says I
need to reduce the primary turns and as you say keep the ratio high
so it's back to a rewind and after all the good advice I've received
I hope to get it right this time.

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.

 

[Winfield Hill]

John Perry wrote...

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.

Yes, that's right, and although I wrote, "keep the ratio high, and
the load resistor small," Gordon saw only the first part available
to him and felt he had to rewind his transformer, whereas simply
using a smaller load resistor would drive the primary voltage down
to a reasonable value, and allow using a low-wattage load resistor.

 

[Rich Grise]

John Perry wrote...

Yes, that's right, and although I wrote, "keep the ratio high, and
the load resistor small," Gordon saw only the first part available
to him and felt he had to rewind his transformer, whereas simply
using a smaller load resistor would drive the primary voltage down
to a reasonable value, and allow using a low-wattage load resistor.

For what it's worth, I've learned more about current transformers than
I knew I had room for in my head, from following this thread.

Thanks to all of you!
Rich