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
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);
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;
- 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.
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
Ø = B. A;       (Formula 2)
ℜ = lm /(µ.A) ;
ℜ = 1/AL;
F = ℜ.Ø ;       (formula
L = (n. Ø)/I ;       (formula
L = AL. n 2 ;
V = L. Δi/Δt ;
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.
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.
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
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?
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
- 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:
The solution of figure 4 could be used in this example but would require three cores.