- It has to do with turns on the same core. If a different core, then you

would be parallelling inductors. Basic equation is

L, Henrys = 0.4 * pi * N^2 * Ac * 10e-8 / lg

N = number of turns

Ac = core cross-section in cm^2

lg = gap length for cut cores and laminates, or equivalent for

powder cores

From the above equation, it is irrelevant as to how many parallel

windings. It is based on core geometry and turns. As for flux and

saturation, that is a different equation that is based on material and

current.

Derek Koonce

DDK Interactive Consulting Services

On 3/19/2013 10:08 PM, sstandfast wrote:

>

> As I am sitting here rewinding a transformer for the 3rd time in two

> days (The first time I ran out of window area before I finished

> winding. The second time I nicked the insulation on the magnet wire

> and developed a short between the primary and secondary.), I began to

> contemplate magnetics in general and started to ponder something that

> I can't quite figure out yet. If you wind an inductor with stranded

> wire (or multiple parallel solid wires) why is it that you do not end

> up with N inductors in parallel, where N is the number of strands used

> to wind the part? E.g. Say I wind 10 turns on a core with an Al value

> of 100mH/1000T^2, I would expect to get an inductor of 10uH. Now say I

> needed to use 2 strands of #28AWG wire to carry the current load, why

> is it that I don't have two 10uH inductors in parallel or 5uH when

> it's all said and done? I'm sure it has something to do with the flux

> coupling in the core but right now I don't see it. Perhaps when my

> mind is not dulled by counting turns. Speaking of which was that #74

> or #75?...Crap...1...2...3...

>

> Shawn

>

>

[Non-text portions of this message have been removed] - Interesting question!

The key insight is probably this: The strength of a magnetic field around a

wire is proportional to the total current through the wire. It doesn't

matter whether the wire consists of a single strand (carrying, say 1A) or

multiple strands (each carrying 1/n A).

A changing magnetic field induces a "back EMF" in the wire which is

proportional to the rate of change and which exactly balances the applied

voltage across the wire and limits the rate at which the current rises in

the wire. The back EMF is per unit length of the wire (e.g. 1V per cm).

Thus, again it doesn't matter whether the wire consists of a single or

multiple strands.

[Non-text portions of this message have been removed] - On 3/20/2013 12:08 AM, sstandfast wrote:
>

The flux linking a coil is proportional to N * I. Where N is the number

> As I am sitting here rewinding a transformer for the 3rd time in two

> days (The first time I ran out of window area before I finished

> winding. The second time I nicked the insulation on the magnet wire

> and developed a short between the primary and secondary.), I began to

> contemplate magnetics in general and started to ponder something that

> I can't quite figure out yet. If you wind an inductor with stranded

> wire (or multiple parallel solid wires) why is it that you do not end

> up with N inductors in parallel, where N is the number of strands used

> to wind the part? E.g. Say I wind 10 turns on a core with an Al value

> of 100mH/1000T^2, I would expect to get an inductor of 10uH. Now say I

> needed to use 2 strands of #28AWG wire to carry the current load, why

> is it that I don't have two 10uH inductors in parallel or 5uH when

> it's all said and done? I'm sure it has something to do with the flux

> coupling in the core but right now I don't see it. Perhaps when my

> mind is not dulled by counting turns. Speaking of which was that #74

> or #75?...Crap...1...2...3...

>

> Shawn

>

> _

of turns and I is the current. You will get the same flux if you use

one winding with 100 turns and 1 amp as you will with 2 windings of 100

turns with !/2 amp in each winding.

The other Howard

[Non-text portions of this message have been removed] - Right, that's what I get for trying to think after 10:00 pm. When you parallel the strands, they each divide the total current by N (assuming each strand has equal impedance) thus the total amp-turns on the core remains constant. (Provided you wrap each strand with the same number of turns.) Since the core geometry is constant then the effective cross-sectional area (Ae) remains constant. That means that the total flux (B dot Ae) is constant and is equal to the sum of the flux contributions of each strand. That means that the integral volt-seconds remains the same and that means the slope of the E*dt vs I curve is constant and thus the inductance is constant. To represent it mathematically:

H = Ns*Is/Le; where Ns = # turns per strand, Is = Current in an individual strand, and Le = effective magnetic path length

Assuming that we have X number of parallel strands of equal resistance wound on the core and the total inductor current (Itot) remains constant then:

Is = Itot / X

That means that the individual flux density of each strand is:

Bs = u*H = u*((Ns*Is)/Le) = u*((Ns*Itot)/(X*Le))

where u = u0*ur and u0 is the absolute permeability of free space and ur is the relative permeability of the core material.

Thus the total flux in the core is:

Phi = Sum(A Dot Bs(i), i = 1, i=X) which equals:

Phi = X*Ae*Bs

Using Faraday's law, the integral volt-seconds is equal to the number of turns times the total flux or:

V*dt = Ns*Phi

Substituting what we know into this equation:

V*dt = Ns*X*Ae*Bs = Ns*Ae*X*u*((Ns*Itot)/(X*Le))

Simplifying the above yields:

V*dt = u*Ae*Itot*Ns^2/Le

We also know that L equals the slope of the line of the V*dt vs I curve so dividing both sides by Itot gives:

V*dt/Itot = L = u*Ae*Ns^2/Le which is the standard formula for inductance. You can also see that this equation is independent of X thus proving that you can wind as many parallel turns on a core as you want but you will always get the same inductance value. Another interesting thing would be to try and figure out what happens when you wind an inductor with multiple strands and unequal numbers of turns between them. I would think that the transformer action that occurs would attempt to balance the current through the windings so that all had an equal number of Amp-Turns. I.E. the winding with the most turns would carry the least current and all the other windings would carry a proportionally larger current. I predict that the total inductance will be the same in that case too, just with unequal current sharing between the parallel windings. I won't go into a derivation of those equations but it would be a fun exercise.

Thanks to everyone who helped get the light bulb to turn on. Like I said, I shouldn't be thinking late at night because it makes me question the simplest of truths.

Shawn

--- In Electronics_101@yahoogroups.com, Jan Kok <jan.kok.5y@...> wrote:

>

> Interesting question!

>

> The key insight is probably this: The strength of a magnetic field around a

> wire is proportional to the total current through the wire. It doesn't

> matter whether the wire consists of a single strand (carrying, say 1A) or

> multiple strands (each carrying 1/n A).

>

> A changing magnetic field induces a "back EMF" in the wire which is

> proportional to the rate of change and which exactly balances the applied

> voltage across the wire and limits the rate at which the current rises in

> the wire. The back EMF is per unit length of the wire (e.g. 1V per cm).

> Thus, again it doesn't matter whether the wire consists of a single or

> multiple strands.

>

>

> [Non-text portions of this message have been removed]

>