M4 · Design a CCM buck inductor

Goal

• Turn an inductor’s specification into the core geometrical constant $K_g$ it needs, and choose a core.
• Finish the design in the magnetics designer: the fewest turns for the flux density, the gap for the inductance, and a winding that fits the window within its resistance budget.
• Find what fails first when the core is too small, and what it would take to make it work.

Before you start

• Read the design procedure and the magnetics designer.
• Know where the specification comes from: the inductance and the peak current from the converter (the buck and the converter designer), and $\Delta i$ as half the peak-to-peak ripple.

Steps

1. The specification

The example inductor has $L$ = 100 µH, a peak current of 5 A and an rms current of 4.5 A. It may reach $B_\mathrm{max}$ = 300 mT, its winding may fill $K_u$ = 0.4 of the window, and its winding resistance budget is 30 mΩ.

2. The K_g it needs

Work out the $K_g$ the specification needs (the core geometrical constant needed), with the wire’s resistivity at the winding’s temperature (resistivity at temperature), and the copper loss the resistance budget allows at the rms current (conduction loss). Enter both below.

3. The core

Compare your $K_g$ with the table cores in the design procedure’s worked example and take the smallest one that meets it. Note by how much it does.

4. Turns, gap and window

Open the magnetics designer on that core:

▶ Open in the magnetics designer: Inductor on ETD 29 example
Preset from the example mag-kg

Check the fewest turns by hand (fewest turns for B_max), at the smallest cross-section $A_\mathrm{min}$. Then read the gap for the inductance at those turns (gap length), how much of the window the winding fills, and its dc resistance against the budget.

5. The smaller core

Switch the core to E 25/13/7. Find the first thing that fails: the turns the flux density needs, the window, or the resistance. Then change one thing, the wire, the budget or $B_\mathrm{max}$, and say what the change costs.

Acceptance criteria

• The core geometrical constant $K_g$ the example inductor needs.
  Your value (cm⁵)Check
  Show the example's value

  0.05246 cm⁵ (from kg-inductor: K_g inductor example)

• The copper loss its resistance budget allows at the rms current.
  Your value (W)Check
  Show the example's value

  607.5 mW (from kg-inductor: K_g inductor example)

• The smallest table core that meets the $K_g$, and by how much.
• In the designer, the fewest turns for $B_\mathrm{max}$ at $A_\mathrm{min}$, checked by hand.
• The gap for the inductance at those turns, a winding within $K_u$ of the window, and its resistance within the budget.
• On the E 25/13/7 core, what fails first, and what one change would fix it at what cost.

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Gotchas

• $B_\mathrm{max}$ is not the material’s saturation flux density. It lies below it at the core’s hottest, with a margin, as the magnetics designer’s page explains.
• The turns follow from the flux density at the smallest cross-section, where it is highest, not at the effective area $A_e$.
• The gap estimate ignores fringing. A real gap must be longer for the same inductance, which is why data sheets give the $A_L$ of gapped sets.
• $\Delta i$ in the designer is half the peak-to-peak ripple. Entering the full swing doubles the flux swing, and the core loss it predicts grows more than that, as a power of the swing (core loss per volume).

Go deeper

• Book chapter Inductor Design (Ch. 11) — R. W. Erickson, D. Maksimović · retrieved 2026-09-24
  The constraints a filter inductor must meet, the core geometrical constant K_g that turns them into a core size, and a step-by-step design procedure.
• Course Magnetics for Power Electronic Converters — University of Colorado Boulder (Coursera) · retrieved 2026-09-23
  Magnetic circuits, core and winding losses, and inductor and transformer design.

Quiz

1. You allow twice the flux density $B_\mathrm{max}$, all else unchanged. The $K_g$ an inductor needs:

   1. Halves
   2. Falls to a quarter
   3. Doubles
   4. Stays the same
   Answer and explanation

   Correct answer: B The $K_g$ needed is inversely proportional to the square of $B_\mathrm{max}$.

2. You halve the winding's resistance budget, to halve its copper loss. The $K_g$ needed:

   1. Halves
   2. Doubles
   3. Falls to a quarter
   4. Stays the same
   Answer and explanation

   Correct answer: B The $K_g$ needed is inversely proportional to the resistance allowed: less resistance takes more copper and a larger window.

3. Why does the magnetics designer compute the fewest turns at the core's smallest cross-section $A_\mathrm{min}$?

   1. The flux density is highest there, so that is where the core saturates first
   2. The winding sits there
   3. The gap is there
   4. Data sheets give no other area
   Answer and explanation

   Correct answer: A The same flux passes through every cross-section, so its density is highest where the area is smallest.

4. The fewest turns come out as 17.2. How many do you wind?

   1. 17: fewer turns, less copper
   2. 18: rounding up keeps the flux density below $B_\mathrm{max}$
   3. 17.2, as a partial turn
   4. Any number between 15 and 20
   Answer and explanation

   Correct answer: B More turns lower the flux density at the same peak current. Rounding down would let it exceed $B_\mathrm{max}$.

5. The gap estimate ignores fringing. Compared with it, the gap that really gives the inductance is:

   1. Shorter
   2. Longer
   3. The same
   4. Zero
   Answer and explanation

   Correct answer: B Around the gap the flux spreads beyond the core's area, so the real gap conducts more flux than the estimate assumes and must be longer.

6. In a CCM filter inductor with a small ripple, how does the flux swing $B_\mathrm{ac}$ compare with the peak flux density?

   1. It is a small part of it
   2. It equals it
   3. It is twice it
   4. It is zero
   Answer and explanation

   Correct answer: A The flux follows the current. Only the ripple, a small part of the current, swings, so the core loss is small while the peak is set by the dc current.