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# Re: [femm] SRM Inductance plot at constant current

- To: femm@yahoogroups.com
- Subject: Re: [femm] SRM Inductance plot at constant current
- From: David Meeker <dmeeker@ieee.org>
- Date: Tue, 06 May 2003 10:20:05 -0400
- In-reply-to: <006c01c313d3$0f32f1d0$66eb5e18@bradsgateway>
- References: <006c01c313d3$0f32f1d0$66eb5e18@bradsgateway>
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`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@ieee.org
    www: http://femm.berlios.de/dmeeker

- Follow-Ups:
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    - From: "Brad Frustaglio" <fuzz@wildrail.net>
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    - From: "Brad Frustaglio" <fuzz@wildrail.net>

- References:
  - SRM Inductance plot at constant current
    - From: "Brad Frustaglio" <fuzz@wildrail.net>

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