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Current per Unit of Area in a Transformer or inductor turn.

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Patrick

Mon Nov 23 2015, 05:55PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

heres the pic of the TSC 50-all "flat line" the one for my planars.

60 C = 140 F @ 52 mw/cc

20 C = 68 F @ 60 mw/cc

for a 40 degree C change, we see a 14% difference.

Sulaiman

Mon Nov 23 2015, 06:57PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

If you are going for maximum power throughput using a power ferrite
it would be good to operate near to the temperature at which core losses are a minimum
but it would be even better to operate at a slightly lower temperature where the loss vs. temperature curve is still negative
to avoid thermal runaway.

DerAlbi

Mon Nov 23 2015, 11:11PM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

Wolfram, i love you.

Patrick

Tue Nov 24 2015, 01:56AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Sulaiman wrote ...

If you are going for maximum power throughput using a power ferrite
it would be good to operate near to the temperature at which core losses are a minimum
but it would be even better to operate at a slightly lower temperature where the loss vs. temperature curve is still negative
to avoid thermal runaway.

do you mean where the graph is still negative in slope.

Sulaiman

Tue Nov 24 2015, 07:46AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

yes.
theoretically, if you operate at the temp. where core loss is minimum all is good
but a little extra heat would cause increased core losses so even more heating.......

Since you are aiming for maximum utilisation of your transformers
a thermal sensor/alarm/switch may be a simple addition ?

Patrick

Sun Nov 29 2015, 07:01PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

ok so ive confirmed with a micrometer my copper foil is 0.004 inches thick, and 0.2 inches wide. Which mean 0.004 x 0.2 = 0.0008 square inches of cross section.
as seen here:

this equates to 1020 circular mils. I'm estimating that 10 amps continuous can go through my copper section. (this presumes not much current drop from end to end)

When I unroll ATX power supply transformers, I see about 800 circular mills that seems at the center of the insulated bobbin capable of 13-15 amps. (current rating of there product going into a main board)

radiotech

Sun Nov 29 2015, 08:06PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

The failure of iron core inductors lies with the insulation, assuming the windings are not
forcibly ejected by magnetic or mechanical forces.

So the operating temperature considerations can lie within the class of insulation chosen.
All well and good, but copper does two things to destroy the coatings. It reacts with almost
all practical varnishes, in a way aluminum can not. It also expands and contracts from one extreme
to another between hot and cold.

Given that iron transformer cores vibrate, and motor rotors spin and try and cast out
the copper, at high speeds, the windings are also dipped in more varnish to glue them in
place.

All this contributes to what you find in a burned out winding. Insulation failure.

The bottom line is this. When you have counted up the pennies in energy efficiency, you find
that windings operating at higher loading, save much more money at the expense of shorter
service life.

Just another degree of freedom in the design tree.

Patrick

Sun Nov 29 2015, 08:10PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

It will be ferrite core copper foil embedded in epoxy.

Patrick

Mon Nov 30 2015, 06:34PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Let me attack this problem more logically . . . 20 AWG seems similar in surface area to my foil conductor. 20 AWG has 10.15 ohms per 1000 feet. so 10.15 ohms / 1000 ft = 0.01015 per foot. now for the important part. like a resistor, calculating just normal DC wattage. (though an over simplification) 0.01015 ohms X 15 amps = 0.152 volts across end to end of the 1ft wire. 0.152 X 15 = 2.3 watts.

I can live with that.

DerAlbi

Mon Nov 30 2015, 10:51PM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

Do you have an air gap in your transformer core? A foil conductor would be extremely bad then.

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