Filter Inductor Design

Design Calculations for Filter Inductor

Design Calculations for filter inductor of ferrite core will be explained here. These inductors are common and essential part of power electronic converter circuits. Main objective of inductor in converter circuits is to eliminate unnecessary harmonics in circuit to get smooth output. As in power electronic devices switching is very high so ferrite core material is used for filter inductor.

Calculations are performed for inductor design so that required inductance can be achieved, core of inductor is not saturated while operation and copper loss (winding resistance) is within permissible limit.

These calculations are performed on basis of Effective Measure of core area (A_g).  First we will discuss parameters and Constraints required for Designing inductor in this article.

Design Constraints:

In figure 1 circuit diagram of inductor is shown (core and winding). Sufficient air gap is in core so that core is not saturated when there is peak current (actual current plus ripple current).

Circuit Diagram of Reluctances in Inductor Core
Figure 1: Circuit Diagram of Reluctances in Inductor Core

Let, L_g be the air gap length, \mu _c be the permeability of core and A_c be the cross sectional area of core. Then, reluctances are given as below:

 R_g = \frac{L_g}{\mu _0 A_c}

R_c = \frac{L_c}{\mu _c A_c}

So, applying loop analysis in above circuit diagram in figure 1, we get

 n I = \phi (R_g + R_c)

As, air gap reluctance is much greater than that of core reluctance so above equation can be approximated as:

n I \approx \phi R_g ----------------- (1)

Our objective is to obtain appropriate inductance without saturating the core so, Following are constraints for designing inductor.

1.   Maximum Flux Density:

For maximum current, there will be maximum flux and at that flux, core must not be saturated.

As, we know know that \phi = BA so putting this in equation (1) we got:

nI_{max} =B_{max} \frac{L_g}{\mu_0}

This is first constraint that must satisfy while designing inductor. where, number of turns of coil (n) and length of air gap is unknown at moment.

2. Inductance:

Inductance must be obtained using:

L=\frac{n^2}{R_g} = \frac{\mu_0 A_c n^2}{L_g}

To this moment, n, core area and air gap length are unknown.

3. Winding Area:

Winding area is limited so winding must bound within the window of core. Let A_w be the area of wire to be used for winding. If there is n number of turns, then total area required for winding is n\times A_w.

Another factor of Window utilization factor (K_u) also involved in this. Let W_a be total available window area then for winding, (K_u \times W_a) area will be available. K_u have value between 0 and 1.

So, another constraint for inductor design is that available core area must be greater or equal to winding total area. i.e.

 K_u \times W_a \geq n \times A_w

4. Winding Resistance:

As we know that, Resistance of a conductor is given by:

R= p \frac{L_b}{A_w}

As winding has n number of turns and with each one complete layer of winding, length is changed so we use mean length per turn to define length of copper wire. is Resistivity and is the property of material and for copper the resitivity is 1.724 \times 10^{-6} \Omega -cm.

L_b = n (MLT)

so, a constraint is then,

R=p \frac{n \times MLT} {A_w}

5. Geometrical constant (Kg):

Some of the parameters (number of turns, gap length and winding area) were unknown in above constraints, substituting values of unknown in terms of known parameters we got following relation:

 \frac{A_c^2 W_A}{MLT} \geq \frac{p L^2 I^2}{B_{max}^2 R K_u} -------------------(2)

The relation on left side is core geometrical constant. For required maximum flux density, current and core size, this constant have standard values. So this becomes a constraint that this constant must be greater or equal to relation on right hand side. This actually summarize above constraints.

Conclusion:

Equation (2) shows that core size increases by increasing inductance (L) or maximum required current (I). For greater flux density, core size is reduced. So, for larger inductance either core area is increased which is result of increase in core material. Larger inductance can also be achieved by more copper wire (number of turns) which will increase window area.

So, there is trade off between these two parameters such that equation (2) is satisfied. i.e. geometric constraints are fulfilled to achieve objective.

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Asad Ullah

I am MSc Electrical Scholar under a fellowship program. Working with Modeling and Simulation software related to my field are my activities in leisure.

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