---
A guess for iron transformers is 4A/mm^2 or 4E6A/m^2 and that's for a guess
love it.
)given more information. Hope this is not too long and I don't show my
Thanks John, please see my reply to Genome's post for further information.
Whoa, making a mountain out of a mole hill. Look at those miniscule
Mr Win Hill is not wrong. I am being an idiot.Genome <ilike_spam@yahoo.co.uk> wrote in message
news:Ojhae.1420$p06.691@newsfe3-gui.ntli.net...
"Gordon W" <welch@hypermax.net.au> wrote in message
news:1114209775.518036@nntp0.iseek.com.au...
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?
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
Well, almost. The limiting factor in a current transformer is the flux
Chair.
C'mon, guys, the detailed theory may be a bit involved, but the practice
.Sheesh. I just realized how dumb this all was. What I should have said
You mean Larry? Yeah- Larry should be able to add some fundamental
Thank you to everyone, you've been very helpful and I appreciate yourGordon W, Genome, Win Hill, etc. wrote:
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
...
A guess for iron transformers is 4A/mm^2 or 4E6A/m^2 and that's for a
...
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
)Gordon W, Genome, Win Hill, etc. wrote:
...John Perry wrote:
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
...
After reading (many times) what you all have said I did as John Perry
suggested. I used a fan heater as a load on the primary (9.5A) and a 100 ohm
5W resistor on the secondary and ran it for an hour. I used the 100 ohm
because it gave a good sine wave output without distortion.
The drop across the primary was 320mV, the voltage on the secondary 11V with
a current of 108mA.
After an hour I could hold my finger on the transformer core without a worry
but of course I couldn't do that with the resistor. The windings remained
cool to the touch.
OK, Gordon,
I did it again with the 100 and a 10 ohm resistor and there is a very slightGordon W wrote:
Gordon W, Genome, Win Hill, etc. wrote:
...John Perry wrote:
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
...
After reading (many times) what you all have said I did as John Perry
suggested. I used a fan heater as a load on the primary (9.5A) and a 100
ohm
5W resistor on the secondary and ran it for an hour. I used the 100 ohm
because it gave a good sine wave output without distortion.
The drop across the primary was 320mV, the voltage on the secondary 11V
with
a current of 108mA.
After an hour I could hold my finger on the transformer core without a
worry
but of course I couldn't do that with the resistor. The windings
remained
cool to the touch.
OK, Gordon,
Think of the secondary as an AC current source. No matter what
impedance you put across its output, it will put out 1/88 the primary
current, up to the linear limit of the core. A 10-ohm resistor will
dissipate 1/10th the power of your 100-ohm resistor, and your primary
voltage will be 1/10 its present value. Likewise, your secondary
voltage will be 1.1V, 1/10 what it was with the 100-ohm resistor.
Now, consider that you'll likely need 20ma to get reliable signal from
your optocoupler. You need to shunt 90'ish ma from the 1.8'ish volts
the optocoupler's LED will drop: 1.8V/.09A=20 ohm in parallel with your
optocoupler. Add your rectifier bridge (or don't, unless you really
want a DC on-off signal), and you're done.
Of course, I'd really like to see your measurements with a 10-ohm
resistor on the secondary, so we can be really sure you have the turns
ratio nailed down. The closer you are to saturating, the less
confidence we can have in your measurements. And you've already shown
that the CT can handle the expected current, so things only get better
from here on.
Yes I will have to do it again as the real deal pulls 7.5A and I'm hoping
True. In crude AC current monitoring, a double limit flux swing can be
Thank you for your advice and the url, both very helpful
My advice was, "keep the ratio high, and the load resistor small,"
Thanks Win, I used a WAG to wind the transformer and as legg says I need to
But, remember, Win, that 320mv was with a 1.5K resistor -- far too much
already have a cool-running transformer that will do everything you