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'How can I know the cable type ??????'
1998\12\29@122645 by  Hi, how are you ? :

I  have a doubt:

I«m trying to know How can I calculate the cable type (AWG or somethig
like it) if a know how much current will flow throught it (some 40mA).

I need to know it because want to build a bobbin.

Any formulas ????

thanks.  >        I«m trying to know How can I calculate the cable type (AWG or somethig
>like it) if a know how much current will flow throught it (some 40mA).
>
>        I need to know it because want to build a bobbin.

If I understand correctly, you want to wind a coil or solenoid. This is
more complicated than just the max current capability of the wire.

First, the wire in the center of the coil will heat more than the wire near
the ends.
If there are multiple layers, then the inner layers will heat more as well.
The insulation's melting temperature is usually the limiting factor in an
air-core coil.
At 40mA, you're not likely to have any problems unless you have a LOT of turns.

If your coil has any core material other than air, (Iron, ferrite) then yu
need to also look at how much magnetic flux will be produced, (product of
amps and turns) and make sure this is below the saturation limit of the core.  > >        I«m trying to know How can I calculate the cable type (AWG or
> somethig
> >like it) if a know how much current will flow throught it (some
> 40mA).
> >
> >        I need to know it because want to build a bobbin.
>
> If I understand correctly, you want to wind a coil or solenoid. This
> is
> more complicated than just the max current capability of the wire.
>
[snip]

> At 40mA, you're not likely to have any problems unless you have a LOT
> of turns.
>
>
all good points. I'd like to add that skin effect also is
often a limiting factor, especially at high frequencies.
To get good efficiency out of the coil, you want to balance
core and copper losses. As stated above, it can definitely
get more complicated than just the current-carrying capacity
of the wire.

that having been said, a good starting point is to use 500
circular mils per ampere (of RMS current), where a circular
mil is the the area of a circle 1 mil in diameter.
(area in square inches would be (1/4)*PI*E-6).

if the 40mA is the peak current of a square wave, you need to
convert to rms(Irms = Ipeak*sqrt(Ton/T).

Then multiply Irms*500 to get the number of circular mils.

For example, assume Irms = 40mA, you'd need
0.04 * 500 = 20 circular mils. so you need a wire
with a diameter of sqrt(20) = 4.5 circular mils.
That's between #37AWG and #36AWG. use #36AWG.

(#36AWG has a diameter of 5 mils. That's an
area of 25 circular mils, or 19.63E-6 sq. in)

hope this helps.  At 01:54 PM 12/30/98 -0500, Eisermann, Phil wrote...
>
>        Then multiply Irms*500 to get the number of circular mils.
>
>        For example, assume Irms = 40mA, you'd need
>        0.04 * 500 = 20 circular mils. so you need a wire
>        with a diameter of sqrt(20) = 4.5 circular mils.
>        That's between #37AWG and #36AWG. use #36AWG.
>
>        (#36AWG has a diameter of 5 mils. That's an
>        area of 25 circular mils, or 19.63E-6 sq. in)

Because there is a potential error here of > 10%, I thought I would point it out
:

Circular area is PI*r^2, so #36 is 3.1416 * 2.5 mils ^2 = 19.63 mils^2.

For 20 mils^2, the diameter would be 2 * sqrt(20 mils^2 /PI) = 5.05 mils

Somehow, you got the in^2 correct, but the mils^2 wrong!?

Mike  > At 01:54 PM 12/30/98 -0500, Eisermann, Phil wrote...
> >
> >        Then multiply Irms*500 to get the number of circular mils.
> >
> >        For example, assume Irms = 40mA, you'd need
> >        0.04 * 500 = 20 circular mils. so you need a wire
> >        with a diameter of sqrt(20) = 4.5 circular mils.
> >        That's between #37AWG and #36AWG. use #36AWG.
> >
> >        (#36AWG has a diameter of 5 mils. That's an
> >        area of 25 circular mils, or 19.63E-6 sq. in)
>
> Because there is a potential error here of > 10%, I thought I would
> point it out:
>
> Circular area is PI*r^2, so #36 is 3.1416 * 2.5 mils ^2 = 19.63
> mils^2.
>
i will agree that the area of a circle 5 mils in
diameter is 19.63 square microinches.

> For 20 mils^2, the diameter would be 2 * sqrt(20 mils^2 /PI) = 5.05
> mils
>
no, because the area *IS*NOT* 20 square mils,
but 20 circular mils. There is a difference
between area in square inches and circular mils.
I figured out the required diameter to get 20
circular mils, then calculated the area in
square inches.

if you use a diameter of 5.05 mils, you would
get 25.5 circular mils. That's equal to
~20 square microinches, as you pointed out.
In the previous example, a diameter of 5 mils
gives you 25 circular mils, which is equal to
~19.63 square microinches.

is anyone confused yet?  At 03:02 PM 12/30/98 -0500, Eisermann, Phil wrote...
>
>> Circular area is PI*r^2, so #36 is 3.1416 * 2.5 mils ^2 = 19.63
>> mils^2.
>>
>        i will agree that the area of a circle 5 mils in
>        diameter is 19.63 square microinches.
>
>> For 20 mils^2, the diameter would be 2 * sqrt(20 mils^2 /PI) = 5.05
>> mils
>>
>        no, because the area *IS*NOT* 20 square mils,
>        but 20 circular mils. There is a difference
>        between area in square inches and circular mils.
>        I figured out the required diameter to get 20
>        circular mils, then calculated the area in
>        square inches.
...
>        is anyone confused yet?

I think I understand the logic - if a "circular mil" is the area of a circle 1 m
il in diameter, then the area of that circle is PI * r^2, which is .784 square m
ils, hence the conversion factor.

I have not seen "circular" units used before, is this common practice when deali
ng with wire gauges?

Mike

Mike - "de Bug" W016451 MI License TDI BUG
Remote Window Controller: http://members.tripod.com/~remotewindow/  At 15:02 12/30/98 -0500, Eisermann, Phil wrote:
>        is anyone confused yet?

there is definitely an advantage in using all those measures like square
inches, circular mils, fractional inches, mils, AWG: you have an insider
code nobody understands how you get to your results... plain mm and square
mm would be too easy to figure out for the not initiated :-)

ge  [snip circular mil stuff]

> I have not seen "circular" units used before, is this common practice
> when dealing with wire gauges?
>
hard for me to say if it is "common".  In terms of a mfg.
specifying it in catalogs... probably not. Not in any of
the ones i have, at any rate. I run into it every now
and then. Pressman uses it in his book on SMPS. I have
seen it in some sort of training manual for electricians
from somewhere on the internet that gives wire data,
which includes circular mils. It's probably about as common
as "slugs" for MechE's. rare, but you still see it every
now and then. At least, that's my impression.

anyone else that's heard of (or uses!) circular mils
care to join in?

it is a nice convenience to use. Since circular mils are
already in units of "diameter circles" (not sure how to say
this properly), it makes it easy to calculate required wire
size. if i need 20 circular mils, i just take the square root
to get the diameter. If you're blessed not to have to work
in the english system, you could do the same thing with
millimeters, to get circular millimeters *grin*  But in the
metric system, i think it is more common to use amps per
square centimeter (eg there is no equivalent to
"circular mils" Not that i know of, anyway).  Ricardo

Lookup a Standard Wire Guage table.
It will tell you reistance per 1000 feet in circular mils.
It will also tell you the maximum Amperages per gauge
IE like 12G is 20 Amps... and 16G is 7 Amps...

Eric Borcherding  You might take a look at the (US) National Electrical Code, Tables
310-16 thru -19.  Basically, in the power world the limitations on a
conductor's amperage are based on max allowable voltage drop, and
maximum allowable temperature to limit damage to the insulation. 2-3
percent is usually the max used in design.  Code allows a max of 5% from
electrical service entrance entrance to final load. Multiple NEC tables
are provided due to the different insulation types available and
environmental installation, such as conduit, direct earth burial or
open-air, all of which affect the cooling rate.  But luckily your only
concern is likely voltage drop, and at milliamp amperage is likely nil.
For example, per Table 8, Conductor Properties (NEC 1996) copper
conductors have the following DC resistance per thousand feet.  From
this you can calculate your voltage drop based on amp draw:

#18 AWG (1620 Circ. Mills)  -  7 strands -  8.45 ohms per k feet
#16 AWG (2580 Circ. Mills)  -  7 strands -  5.29 ohms per k feet
#14 AWG )4110 Circ. Mills)  -  7 strands -  3.26 ohms per k feet

I guess my second life as an "electrical engineer" after switching from
"electronic engineer" may be of *some* help here after all <grin>.

"Eisermann, Phil" wrote:
{Quote hidden}  Eisermann, Phil wrote:
{Quote hidden}

I am.
According to my limited knowledge, current only flows on
the skin of a conductor because a potential difference
cannot exist inside a conductor (like a Faraday cage).
Because we work with less-than-perfect conductors, some
current will also flow a little deeper into the conductor,
but the heavy current flows on the surface.

This is why a ribbon can conduct more than a rod. Following
this potentially flawed reasoning, I would expect the
circumference of a conductor to be of more significance than
the diameter.

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|--------------------------------------------------|  Eisermann, Phil wrote:

>         it is a nice convenience to use. Since circular mils are
>         already in units of "diameter circles" (not sure how to say
>         this properly), it makes it easy to calculate required wire
>         size.

It seems to me the practicality here is that windings are essentially
a rectangular array, by which I mean that the turns are side-by side and
on top of one another.  The space between the circle and the square is
entirely wasted, so no point counting it!
--
Cheers,
Paul B.

'How can I know the cable type ??????'
1999\01\05@140904 by  [snip]

> I am.
> According to my limited knowledge, current only flows on
> the skin of a conductor because a potential difference
> cannot exist inside a conductor (like a Faraday cage).
> Because we work with less-than-perfect conductors, some
> current will also flow a little deeper into the conductor,
> but the heavy current flows on the surface.
>
> This is why a ribbon can conduct more than a rod. Following
> this potentially flawed reasoning, I would expect the
> circumference of a conductor to be of more significance than
> the diameter.
>
yes, but we are talking about wires here,
and wires are round. until you asked, i've
never thought about it. Here's my guess
as to why its done this way: The skin depth
is related to frequency, and perhaps for the
vast majority of appliations, it doesn't
have enough of an effect for it to matter.
Those applications where it does matter
may be so few (in terms of where it does
matter) that it is not economically attractive
for manufacturers to make a different type.

for what it's worth, larger transformer
applications do, in fact, use "ribbons"
of copper instead of wires. Litz wire is
another alternative.  Hello,

>         for what it's worth, larger transformer
>         applications do, in fact, use "ribbons"
>         of copper instead of wires. Litz wire is
>         another alternative.

I'd like to add that nearly all transformers over 15W in volume consumer

that the tape windings have a secret advantage over wire: An electrostatic
shield can be obtained by putting a single winding of tape over and under
the winding to be isolated and by grounding one end of each 1-turn
winding, and leaving the other unconnected. This is a very effective
shield and also reduces RFI when used as the outermost winding on a SMPSU
transformer.

Also, winding tape is often cheaper and easier than wire, plus heat
conduction is better and, and, and...

Peter

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