Re: [femm] Transformer Forward curent and magnetizing current

[David Meeker <dmeeker@xxxxxxxx>]

jire602000 wrote:
>   Hello ,
> everybody .
>
> I want to simulate a transformer with one primary and one secondary 
> winding .They have the same number of turns ; ratio = 1 .
> I want to see what happen with different current  loads which I call 
> the forward current . 
>
> I assume using the in circuit property to fix the current in each 
> windings because the forward current in a transformer is  at phase 0 
> and the magnetizing current is at phase PI/2 ,  j*Lmagnetizing
> *
>
>
> primary current =  forward current  -  j * magnetizing * current/2
> secondary current = - forward * current - j * magnetizing * current/2
>
> I am OK  ? Let me know .
>
> Any example to help ?
>
> Ciao.
Because of flux leakage, eddy currents, etc., you don't know exactly
what the relationship is between voltages and currents.  One of the
objectives of doing a finite element analysis is to resolve this sort
of effect.  Probably the best thing to do is to "go back to the
basics."  

We could write, for the primary,

(d flux1)/(dt) + R1 i1=v1

and for the secondary,

(d flux2)/(dt) + R2 i2=v2

where flux1 and flux2 are the flux linkages of the primary and
secondary, respectively.  We could assume that R1 and R2 are the DC
resistances of the coil that you'd calculate analytically (not exactly
true if skin effects are important).

If  we assume that the materials are linear, we can assume that the
flux linkage has the form:

flux1 = L1 i1 + M i2
flux2 = M  i2 + L2 i2

We don't know the parameters L1, L2, and M, but we could identify them
by doing  three finite element runs, each at the frequency at which the
transformer will run.

In the first run, the primary would carry 1 amp.  L1 is then the result
of the A.J integral taken over coil 1.

In the second run, the secondary would carry 1 amp, L2 is the result of
the A.J integral taken over coil 2.

In the third run, both the primary and secondary would carry 1 amp. 
The result of the A.J integral taken over both coils is L1 + L2 + 2M,
from which we can deduce M, knowing L1 and L2. 

We can then use the equations:

j*omega*(L1 i1 + M i2) + R1 i1=v1
j*omega*(M  i1 + L2 i2) + R2 i2=v2

to solve for the currents that we'd use to simulate any particular set
of applied terminal voltages.

Note that L1,L2, and M can be complex-valued, e.g. if your core
material is laminated or hysteretic.  The imaginary parts are
associated with losses--when you multiply by j*omega, the imaginary
parts become real and look like resistance.

If nonlinear materials are used, things are somewhat more of a pain. 
You'd have to iterate to find values of current that satisfy:

(d flux1)/(dt) + R1 i1=v1
(d flux2)/(dt) + R2 i2=v2

where flux linkage for a coil would be obtained by integrating A.J over
the coil and dividing by the current carried by the coil.

Proximity effects can be an issue in transformer design, depending on
the application.  Some notes on including these effects are at
http://femm.berlios.de/proxapprox.zip

Dave.
-- 
David Meeker
dmeeker@xxxxxxxx
http://femm.berlios.de/dmeeker