Inductor design procedure

Intent

• Turn a filter inductor’s specification (inductance, peak current, flux limit, copper-loss budget) into a core size with the core geometrical constant $K_g$.
• Carry the chosen core through the turns, the gap and the wire, and check each constraint.
• Recognize when core loss rather than saturation limits the flux density.

Theory

Four constraints

A filter inductor must meet four constraints at once [Erickson & Maksimović 2020, Ch. 11]:

1. Flux density. At the peak current the flux density stays below $B_\mathrm{max}$, which sets the fewest turns (turns for a flux limit).
2. Inductance. With those turns, the gap sets the inductance (gap length).
3. Window. The copper fills no more of the window than can be wound, the window utilization $K_u$ (window utilization).
4. Resistance. The winding’s resistance stays within the copper-loss budget $R_\mathrm{dc,max}$ (dc resistance).

The turns, the gap and the wire can each meet one constraint; the core must leave room to meet them together.

The core geometrical constant

Eliminating the turns and the wire from the flux, window and resistance constraints leaves the specification on one side and the core’s dimensions on the other [Erickson & Maksimović 2020, Ch. 11]:

$$K_g = \frac{\rho_w L^{2} I_\mathrm{pk}^{2}}{B_\mathrm{max}^{2} R_\mathrm{dc,max} K_u}$$

$K_g$

core geometrical constant

$B_\mathrm{max}$

largest flux density allowed

$I_\mathrm{pk}$

peak current

$K_u$

window utilization

$L$

inductance

$R_\mathrm{dc,max}$

largest winding resistance allowed

$\rho_w$

conductor resistivity

Core geometrical constant an inductor needs✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• linear (unsaturated) magnetic material
• flux density uniform over A_e
• a dc current spread uniformly over the conductor's cross-section

Notes

the specification's side of the four constraints (flux density, inductance, window, resistance) with the turns and the wire eliminated; any core whose own $K_g$ is at least this large can meet them, and its gap then sets the inductance. SI units, m⁵ (tables often give cm⁵: 1 m⁵ = 10¹⁰ cm⁵)

Source

Erickson & Maksimović 2020, Ch. 11 (Inductor Design: the core geometrical constant K_g); derived here

Derivation

step-by-step sympy derivation

Equation id

mag.Kg_req · equations.yaml entry

The core’s dimensions combine into the same quantity [Erickson & Maksimović 2020, Ch. 11]:

$$K_g = \frac{A_e^{2} W_A}{\mathrm{MLT}}$$

$K_g$

core geometrical constant

$A_e$

effective core area

$\mathrm{MLT}$

mean length per turn

$W_A$

winding window area

Core geometrical constant of a core✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• definition

Notes

the core's side: it grows as the fifth power of the core's linear size, so a core with twice the $K_g$ is only about 15 % larger in each dimension

Source

Erickson & Maksimović 2020, Ch. 11 (Inductor Design: the core geometrical constant K_g)

Equation id

mag.Kg_core · equations.yaml entry

A core whose own $K_g$ is at least the required value can meet the flux limit, the window and the resistance budget together; the gap then sets the inductance. The product $L I_\mathrm{pk}$ and the flux limit enter squared, the resistance budget and $K_u$ once: a higher flux limit shrinks the core most. The gap does not enter $K_g$: it is chosen after the core.

The procedure

The steps below are adapted from the $K_g$ method [Erickson & Maksimović 2020, Ch. 11]. They take the turns first, as the magnetics designer computes them. With a core of large enough $K_g$:

1. Turns. The fewest turns that keep the flux density at $B_\mathrm{max}$ (turns for a flux limit), rounded up to a whole turn. The magnetics designer takes them at the core’s smallest cross-section, where the flux density is highest.
2. Gap. The gap that gives the inductance with those turns (gap length). Fringing flux adds to the gap’s inductance, so a real gap must be somewhat longer [Dixon 2001]; for a table core, the data sheet’s gapped sets give measured values.
3. Wire. The largest copper area per turn that fits: the window times $K_u$, shared by the turns (window utilization).
4. Check. The winding’s resistance (dc resistance) then meets the budget, provided the core’s $K_g$ exceeds the requirement by enough to cover the turns added by rounding up.

When core loss limits the flux density

The procedure limits the flux density at the peak current. In a filter inductor with a small ripple the flux is mostly dc, which causes no core loss, so saturation sets the limit [Erickson & Maksimović 2020, Ch. 10]. With a large ripple, as in a DCM inductor or a flyback transformer, the ac flux amplitude (ac flux amplitude) causes core loss (Steinmetz equation) that can limit the flux before saturation does; the design must then also keep the core loss within its budget [Erickson & Maksimović 2020, Ch. 12].

Worked example

An example filter inductor’s specification and the core geometrical constant it needs:

K_g inductor example

Symbol	Value	How it is obtained
$\rho_{20}$ resistivity at 20 °C	17.2 nΩ·m	given
$\alpha_{20}$ temperature coefficient at 20 °C	0.00393 1/K	given
$T_w$ winding temperature	100 °C	given
$L$ inductance	100 µH	given
$I_\mathrm{pk}$ peak current	5 A	given
$I_\mathrm{rms}$ rms current	4.5 A	given
$B_\mathrm{max}$ largest flux density allowed	300 mT	given
$R_\mathrm{dc,max}$ largest winding resistance allowed	30 mΩ	given
$K_u$ window utilization	0.4	given
$\rho_w$ conductor resistivity	22.7 nΩ·m	Resistivity of a conductor at its temperature
$K_g$ core geometrical constant	0.0525 cm⁵	Core geometrical constant an inductor needs
${P}_{\mathrm{cu}}$ power	608 mW	Conduction loss in a resistance

It needs a $K_g$ of 0.0525 cm⁵. At its rms current, the resistance budget of 30 mΩ allows a copper loss of 608 mW. The two cores of the magnetics designer’s table, with the values their data sheets give:

K_g inductor example: The table cores against the K_g the example needs ($K_g$ = 0.0525 cm⁵)

Core (data sheet)	$A_e$	$W_A$	$\mathrm{MLT}$	$K_g$	Kg over the need
E 25/13/7 (EF 25), N87	52.5 mm²	61 mm²	50 mm	0.0336 cm⁵	0.641 (too small)
ETD 29/16/10, N87	76 mm²	97 mm²	52.8 mm	0.106 cm⁵	2.02 (meets it)

The E 25/13/7 core is too small. The ETD 29/16/10 core has about twice the $K_g$ needed, and the magnetics designer carries out the rest of the procedure on it:

Inductor on ETD 29 example

Symbol	Value	How it is obtained
$L$ inductance	100 µH	given
$I_{L,\mathrm{pk}}$ peak inductor current	5 A	given
$I_{L,\mathrm{rms}}$ rms inductor current	4.5 A	given
$\Delta i_L$ inductor ripple, half peak-to-peak	500 mA	given
$f_s$ switching frequency	100 kHz	given
$B_\mathrm{max}$ largest flux density allowed	300 mT	given
$T_w$ winding temperature	100 °C	given
$b_w$ winding breadth	19 mm	given
$K_{u,\mathrm{max}}$ largest window utilization that can be wound	0.4	given
$d_w$ bare wire diameter	0.8 mm	given
$d_o$ wire diameter over its insulation	0.86 mm	given
$k_s$ strands in parallel	3	given
$M_\ell$ layers of the winding (Dowell's M)	4	given
$A_e$ effective core area	76 mm²	TDK Electronics, ETD 29/16/10
$A_\mathrm{min}$ smallest cross-section of the core	71 mm²	TDK Electronics, ETD 29/16/10
$l_e$ effective magnetic path length	70.4 mm	TDK Electronics, ETD 29/16/10
$A_{L0}$ inductance factor of the ungapped set	2.2 µH	TDK Electronics, ETD 29/16/10
$W_A$ winding window area	97 mm²	TDK Electronics, ETD 29/16/10
$\mathrm{MLT}$ mean length per turn	52.8 mm	TDK Electronics, ETD 29/16/10
$N_\mathrm{min}$ fewest turns for Bmax, at Amin	23.47	Fewest turns that keep the peak flux density at a limit
$N$ number of turns	24	rounded up
$A_L$ inductance factor the design needs	173.6 nH	Inductance from the inductance factor
$\mu_e$ effective permeability of the ungapped set	1622	Inductance factor of a gapped core
$l_g$ gap, without fringing	0.5067 mm	Air-gap length that gives an inductance with N turns
$B_\mathrm{pk}$ peak flux density (at Amin)	293.4 mT	Peak flux density from inductance and peak current
$B_\mathrm{ac}$ ac flux-density amplitude	27.41 mT	AC flux-density amplitude from the current ripple
$\rho_w$ conductor resistivity (at Tw)	22.66 nΩ·m	Resistivity of a conductor at its temperature
$\delta$ skin depth (at fs)	0.2396 mm	Skin depth of a conductor
$A_w$ bare conductor area of one turn	1.508 mm²	Copper area of one turn of round strands
$R_\mathrm{dc}$ winding dc resistance	19.04 mΩ	DC resistance of a winding
$\eta$ porosity of a layer	0.6717	Porosity of a layer of round conductors
$\varphi$ layer thickness over skin depth	2.425	Dowell's layer thickness for round conductors
$F_R$ ac resistance factor (at fs)	26.69	AC resistance factor of a foil winding (Dowell)
$P_\mathrm{dc}$ loss at the dc resistance	385.6 mW	Conduction loss in a resistance
$K_u$ window utilization	0.3731	Window utilization of a winding

The turns are rounded up at the smallest cross-section. Three strands in four layers fill less of the window than $K_u$ allows, and the winding’s dc resistance stays below the budget. Dowell’s factor at $f_s$ is large for this winding, but only the ripple flows at $f_s$, and its rms value is a small fraction of the current here (see winding loss).

Try it

▶ Open in the equation explorer: Core geometrical constant an inductor needs
Preset from the example kg-inductor

Sweep $B_\mathrm{max}$: the required $K_g$ is inversely proportional to its square, so doubling it divides the $K_g$ needed by four. Then halve $R_\mathrm{dc,max}$: halving the copper loss doubles the $K_g$ needed.

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

In the magnetics designer, switch the core to E 25/13/7. It needs more turns, and the same wire no longer fits the window: the designer warns. Wire thin enough to fit has a resistance above the example’s budget.

Bench exercise

Wind the design on a gapped core set. Measure its inductance with an LCR meter at a frequency well below its self-resonance, and its dc resistance with four wires (see Measuring magnetics). Then find the current at which it saturates with the pulse test described there, and compare it with the design’s peak current.

Gotchas

• $K_g$ leaves out core loss. It covers the flux limit, the window and the copper loss. With a large flux swing, check the core loss as well [Erickson & Maksimović 2020, Ch. 12].
• Round up, then check again. Rounding the turns up, or taking them at a smaller cross-section, adds turns and resistance: a core with barely the required $K_g$ can then miss the budget.
• The gap estimate ignores fringing. Use a data sheet’s gapped set, or expect a real gap somewhat longer than the estimate [Dixon 2001].

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.
• Book Magnetics Design Handbook (SLUP132) — L. H. Dixon, Texas Instruments (Unitrode) · retrieved 2026-09-23
  Lloyd Dixon's magnetics design handbook, reproduced by TI from the 2001 Unitrode edition. The ti.com copy requires a free myTI login.
• 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. Which three constraints does the core geometrical constant $K_g$ combine?

   1. The flux limit, the gap and the window
   2. The inductance, the gap and the core loss
   3. Saturation, core loss and temperature
   4. The flux limit, the window and the winding resistance
   Answer and explanation

   Correct answer: D Eliminating the turns and the wire from the flux-density, window and resistance constraints leaves $K_g$; the gap is chosen afterwards and sets the inductance.

2. With everything else fixed, $B_\mathrm{max}$ is raised from 0.2 T to 0.3 T. By what factor does the required $K_g$ change?

   1. 0.67: it falls
   2. 1.5: it rises
   3. 0.44: it falls
   4. 2.25: it rises
   Answer and explanation

   Correct answer: C The required $K_g$ is inversely proportional to the square of $B_\mathrm{max}$, and 0.2/0.3 squared is 0.44.

3. The allowed copper loss is doubled at the same rms current. What happens to the required $K_g$?

   1. It halves
   2. It doubles
   3. It falls to a quarter
   4. It is unchanged
   Answer and explanation

   Correct answer: A Twice the loss at the same current allows twice the winding resistance, which enters $K_g$ once, in the denominator.

4. A core has twice the $K_g$ of another core of the same shape. How much larger is it in each dimension?

   1. About 26 %
   2. About 41 %
   3. Twice
   4. About 15 %
   Answer and explanation

   Correct answer: D $K_g$ grows as the fifth power of a core's size, and the fifth root of 2 is 1.15.

5. Why check the winding resistance again after rounding the turns up?

   1. More turns share the window, so each has less copper, and the winding is longer: the resistance rises as the square of the turns
   2. Rounding up lowers the flux density below $B_\mathrm{max}$
   3. Rounding up changes the core's $K_g$
   4. The gap no longer sets the inductance
   Answer and explanation

   Correct answer: A The core's $K_g$ covers the budget at the exact number of turns; rounding up or taking the turns at a smaller cross-section adds resistance that only a margin in $K_g$ covers.

6. In which part is core loss, rather than saturation, most likely to limit the flux density?

   1. A filter inductor with a small ripple
   2. An inductor carrying only dc
   3. A flyback transformer or a DCM inductor, whose flux swings widely every period
   4. A choke at mains frequency
   Answer and explanation

   Correct answer: C Core loss comes from the ac flux. A filter inductor with a small ripple has little of it, while a large ripple swings the flux widely every period.