Current Induced in Wire-Wound Inductorexamples|products|Murata Software Co., Ltd.

Example27 Current Induced in Wire-Wound Inductor

General

  • The current induced in a wire-wound Inductor is analyzed.
     

  • The current distribution and the resistance are solved.
     

  • This will be an axisymmetric analysis.
     

  • Unless specified in the list below, the default conditions will be applied.
     

 

Analysis Space

Item

Settings

Analysis Space

Axisymmetric

Model unit

mm

 

Analysis Conditions

Item

Settings

Solvers

Magnetic Field Analysis [Gauss]

Analysis Type

Harmonic analysis

Options

Deselect Inductance Calculation.

 

The frequency is swept from 1MHz to 10MHz in 1MHz interval.

Tab

Setting Item

Settings

Harmonic analysis

Frequency

Minimum: 1×10^6[Hz]

Maximum: 10×10^6[Hz]

Step

Linear step: Division number: 9

Model

The axisymmetric analysis is performed. For a model which is symmetric around Z axis,

the simulation is done for the X-Z section only. It is virtually 2D analysis.

The calculation is much faster than 3D analysis. The axisymmetric analysis is well suited for this exercise.

 

 

A core is placed in center, and the coil is wound on the core.

The coil is formed by 16 turns in Z direction and 4 layers in X direction. It is 64 turns in total.

The right half section of the model is analyzed.

 

 

Body Attributes and Materials

Body Number/Type

Body Attribute Name

Material Name

Mesh Size

0-63/Sheet

65/Sheet

Body_Attribute_001 to 064

CORE

008_Cu *

CORE

0.01

 

* Available from the Material DB

 

Set the current as follows.

Body Attribute Name

Tab

Settings

Body_Attribute_001 to 064

Current

Waveform: AC

Current: 1[A]

Turns: 1[Turns]

Direction: +Y direction

Boundary Conditions

No setting.

Results

The following are solved.

1) Magnetic flux lines

2) Current density distribution

3) Frequency characteristics of the inductor’s resistance

4) Inductance

 

1) Magnetic flux lines

The figure below shows the magnetic flux lines at 10MHz.

 

2) Current density distribution

The figure below shows the current distribution. The current flows uniformly at 1MHz. At higher frequencies, the current flows unevenly especially in the outer windings.

 

3) Frequency characteristics of the inductor’s resistance

The chart below is the inductor’s resistance. It is higher at higher frequency.

  • How to calculate the resistance
    Resistance R can be calculated from the loss P [W] displayed on the output window:
    R = 2*P, where P = 1/2*R*I*I and I = 1[A] peak.

 

4) Inductance

The inductance can be calculated as follows. The magnetic energy Em, the current I and the inductance L satisfy

Em = 1/2*L*I*I As Em = 1.563e-5[J] is obtained at I = 1[A] through the simulation,

L = 31.26[uH]