## Inductor calculation

### Inductors Calculation

Here we present a practical way to calculate inductors, using parameters normally released by cores manufacturers. However it is important to know that the calculations inductors reach approximate values, depending on the own variation of parameters informed by the manufacturers and the high dependence on inductor characteristics in relation to their physical construction.

In this example we will calculate a toroidal inductor to converter application but the technique can be applied to other types of inductors.

Figure 1 present a toroidal core made of iron powder and an inductor fitted with this type of core. The iron powder core is common in applications for DC-DC voltage converters.

Figure 1

To start the calculation we need to know where the inductor will be used. We must raise the following application characteristics:

• The maximum current that will pass through the inductor;
• Maximum voltage applied to the inductor;
• Signal frequency;
• desired inductance (can be calculated);
• duty-cycle;

To build an inductor, we need to define the core, calculate the number of turns, and wire diameter to be used. These are the parameters to be derived from the calculation.

### Selecting the Core

To select the core, two characteristics are important: maximum current and work frequency. An inductor keeps the calculated features if the core doesn't saturate. As saturation depends on the magnetic field induced in the core, it is directly related to the number of turns, reluctance and maximum current.

There are several materials used in the manufacture of cores, including:

• Iron powder;
• Ferrites;
• Ferrosilicon plates;

For each material the composition may vary, causing a wide range of products. Each one is appropriate for a specific application. Iron powder, for example, is more suitable for medium and high frequency power, while ferrite cores are more suitable for lower power and higher frequencies. Ferrosilicon plates are well suited to conventional voltage transformers due to low cost and high level of saturation. But do not work for medium and high frequencies.

The important core parameters are:

• saturation field (given in Gauss or Tesla. 1 Tesla = 10,000 Gauss.)
• AL - effective inductance. Usually given in nH/turns2
• lm - Length of the magnetic circuit;
• A - core straight section;

In Figure 2 we reproduce the manufacturer specifications table of a toroidal core.

Figure 2

The table in Figure 2 reproduces the main parameters of the core.

Let's define the following variables for our calculation:

• F (A.turns *) - magnetomotriz force;
• ℜ (turns2 .henry -1 *) - Reluctance;
• AL (henry.turns -2*) - effective inductance;
• µ (henry.m -1) - magnetic permeability of the material;
• B (tesla -T) - field induced in the core;
• Ø (weber - Wb) - Magnetic flux in the core;
• A (m2) - Straight Section of the core;
• L (henry - H) - inductance;
• lm (m) - Length of the magnetic circuit;
• n (dimensionless) - Number of turns;
• i (ampere - A) - current passing through the inductor;
• V (volt - V) - Voltage applied to the inductor;

* - Ampere.turns is dimensionless. It was included in the above list only for best understanding.

Mixing of Unit Systems can not occur for the right calculation. All variables must be in the same Unit System. In the list above, units into brackets, are in the International System of Units (SI).

The summary of the formulas that apply to the inductor calculation is:

F = n. i ;       (Formula 1)

Ø = B. A;       (Formula 2)

ℜ = lm /(µ.A) ;       (formula 3)

ℜ = 1/AL;       (formula 4)

F = ℜ.Ø ;       (formula 5)

L = (n. Ø)/I ;       (formula 6)

L = AL. n 2 ;       (formula 7)

V = L. Δi/Δt ;       (formula 8)

With specifications plus the core formulas we can calculate the inductor. We will not go into details of magnetic circuits concepts involved because this article is intended to be a quick reference for inductors calculation. Details of these concepts can be found in the literature or in specific pages of eletronPi.

Here are some examples:

1) In the circuit of Figure 3, F1 is a DC source of 250V. Knowing that the key CH1 will be closed for 100uS, calculate the minimum inductance L1 so that the current in L1 reaches 500 A.

Figure3

Answer:

From formula 8:

Lmin = (VF1 . Δt) / Δi   Then:   Lmin = ( 250. 100uS) / 500 = 50 uH

That is! The minimum inductance for L1 is 50uH.

2) Knowing that we build the Figure 3 inductor using the Magmattec core model MMT052T7725 (see table in Figure 2), calculate the number of turns to achieve the desired inductance.

Answer:

From Figure 2 table:

AL = 160 nH.turns-2 = 160. 10-9 H.turns-2

From formula 7:

n2 = L / AL     Then: n2 = 50.10-6/160.10 -9

n2 = 312     Then: n = 17.7 turns

That is. We build the inductor with 18 turns.

3) The Figure 3 L1 inductor will saturate when reaching the 500A current?

Answer:

From Formula 1:

F = n. i     then: F = 9000 A.turn

The straight section area of ​​the core is given in figure 2 table: 3.38cm2. This area has to be converted to m2, which is the area unit in the International System. So A = 3,38.10-4 m2.

By formula 4:

ℜ = 1 / AL     Then: ℜ = 6,25.106 turn2. henry -1

By formula 5:

Ø = F / ℜ     Then: Ø = 1,44.10-3 Wb

Finally from formula 2:

B = Ø / A     Then: B = 1,44.10-3 / 3,38.10-4         B = 4.26 Tesla

Table in figure 2 indicates that this core saturates with 1,4T. So our inductor will saturate at the calculated conditions. The alternatives for solving the problem are:

• Choose a larger core and redo the calculations;
• Reduce the maximum current, if the project permits;
• Assemble a sandwich using inductor cores as in Figure 4;

Solution using more than one core:

Figure 4

The solution of figure 4 could be used in this example but would require three cores.