Re: [femm] SRM Inductance plot at constant current

[David Meeker <dmeeker@xxxxxxxx>]

Brad Frustaglio wrote:

> Hi All,
>
>
>
> I have been using FEMM for a few years now. Many thanks to Dave, 
> excellent program.
>
>
>
> I have question on evaluating the A-J integral and then calculating 
> the self inductance from the result.
>
>
>
> I am modeling a 4 phase SRM with 8 stator poles and 6 rotor poles. I 
> am attempting to extract the torque profile and inductance profile 
> with one phase excited at constant current as the rotor moves through 
> one stroke. The torque magnitude and shape looks reasonable. However I 
> am having trouble understanding how to calculate the inductance 
> correctly. The profile shape looks as expected I am just concerned on 
> the magnitude of inductance. My method is:
>
>
>
> 1. Highlight the blocks with the current flowing ( In my case there
> are 2 current blocks per pole (one on each side of the stator
> pole) for a total of 4 current blocks per phase.
> 2. Evaluate the A-J integral at each respective rotor position
> 3. The self inductance as defined is int(A-J)dV /i^2) where i is
> the coil current
>
>
>
> I think I am running into trouble on the coil current term. What is 
> the correct coil current to use to evaluate the inductance correctly?
>
>
>
> The value of one current block or the sum of all current blocks. Also 
> this is actually amp-turns. Not the actual current flowing in the wire 
> itself. Right?
>
>
>
> For instance: the current in one block in the model is defined as 
> 11.15 MA/m^2. The coil area is 2.6903 x 10^-5 m^2. So the magnitude of 
> current in the coil block is 300 A-T. Is this the current I use for 
> the evaluation of self inductance using the A-J integral?
>
>
>
> Regards,
>
>
>
> Brad Frustaglio
The /i/ in the equation actually /is /the actual current flowing in the 
wire itself. Like in your example, say that your coil is made of 60 
turns of wire. Since you have 300 Amp*Turns, the current in the wire 
would be 5 Amps. The 5 Amps is then the value that you'd use as /i/ in 
the equation. There's also an example that might be relevant to you at 
http://femm.berlios.de/induct1/induct1.htm

One last thought--switched reluctance motors are typically run in a 
highly saturated condition. The quantity that you are computing with 
the A.J integral is (flux linkage/current), but it doesn't have the same 
relationship to stored energy as inductance in a linear problem. 
Furthermore, this "nonlinear inductance" appears in a subtly different 
way in any electric circuit equations that you might write. For example,
the electric circuit equation that applies to your case is:
D(flux linkage)/Dt + R i = v
where the Dt is meant to represent the /total/ derivative with respect 
to time. If inductance, L, is not a function of current, this 
simplifies to the usual:
L di/dt + R i = v
However, if L is a function of i, we'd get:
(L(i) + i*dL/di)*di/dt + R i = v

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