I am building a boost converter 12VDC-24VDC with output current of \$I_{out}=250\:\mathrm{mA}\$ and input current of around \$I_{in}=600\:\mathrm{mA}\$ and need to design an inductor for it. I've calculated that the inductance has to be \$L=125\:\mathrm{\mu H}\$, the switching frequency is 400kHz.

I have a ferrite core with 5mm diameter and 20mm length (I can cut it if necessary) and permeabillity of 12. I am also using 32 AWG wire (because of skin effect) which has a diameter of 0.2mm and \$R=538.3\:\Omega\$ for 1000m.

The equation of cylindrical coil is \$L=\frac{N^2\cdot\mu_o\cdot\mu\cdot A}{l}\$ where \$N\$ is number of turns, \$\mu_o\$ is permeabillity of free space \$\mu\$ is relative permeabillity of the core, \$A\$ is cross section area of coil and \$l\$ is length of the coil.

There are dozens of ways one can design an indutor based of number of turns and the length of coil, but increasing the length also increases the resistance of the inductor.

So what is the maximum resistance the inductor can have under these circumstances, how can I calculate it? So I know later on what length to take, and from this the number of turns.


I've calculated the inductance based on the equations found on this website, page 40.

Input ripple current: \$\Delta I=0.2*0.6=0.12A\$

Duty cycle: \$D=\frac{24-12}{24}=0.5\$

Inductance: \$L=\frac{12*0.5}{0.12*400000}=125\mu H\$

  • \$\begingroup\$ The maximum resistance is probably limited by the threshold where something is damaged by heat. Hopefully you aren't anywhere near that: the ideal resistance is zero since any resistance represents a loss. \$\endgroup\$ – Phil Frost Jan 10 '15 at 13:53
  • \$\begingroup\$ Keep in mind you will also have hysteresis losses, which look like resistance. \$\endgroup\$ – Phil Frost Jan 10 '15 at 13:54
  • \$\begingroup\$ @PhilFrost What is the maximum wattage/temperature such a wire can handle? Is there a way to calculate this? \$\endgroup\$ – Golaž Jan 10 '15 at 14:30
  • \$\begingroup\$ Yes, but again, that's the wrong way to approach this problem. If you are anywhere near overheating the inductor you've built a very inefficient converter. \$\endgroup\$ – Phil Frost Jan 10 '15 at 19:00

Skin effect does not mean that you have to use thin wire, it only means that you don't get all the benefit of using thicker wire.

Your calculated inductance sounds way too high for 400kHz operation (maybe an order of magnitude or so). I suggest you double-check the calculation.

You can get lower DC resistance in two ways- use thicker wire (but you only benefit with the diameter of the wire rather than the diameter squared above the skin depth) or you can use multiple thin wires in parallel (ideally in a Litz arrangement, but a simple slow twist is much better than nothing).

If your inductor is high resistance you'll get I^2R losses from the wire (effectively series resistance) on top of the core losses (which look like a frequency-dependent parallel resistance) and your DC-DC efficiency will suffer.

  • \$\begingroup\$ Please see edit. \$\endgroup\$ – Golaž Jan 11 '15 at 11:17
  • \$\begingroup\$ Your calculation looks fine, but you could choose a higher input ripple current and use a smaller inductance, down to maybe half. \$\endgroup\$ – Spehro Pefhany Jan 11 '15 at 14:13
  • \$\begingroup\$ Ok, will do. The question how to design one is still valid tho. \$\endgroup\$ – Golaž Jan 11 '15 at 14:17
  • \$\begingroup\$ I think the problem is underconstrained. You should consider efficiency, which will place an upper limit on the resistance. \$\endgroup\$ – Spehro Pefhany Jan 11 '15 at 15:14
  • \$\begingroup\$ I was thinking of 90% efficency. So approximately 10% would get lost by I^2R, and from this I could calculate resistance, right? \$\endgroup\$ – Golaž Jan 11 '15 at 15:56

If you know the diameter of the core and the inductor is close-wound on it, then the length per turn is $$ Lt = \pi \times (D +d),\ $$ where:

\$ Lt\$ is the length of the turn,

\$ \pi \$ is about 3.14,

\$ D \$ is the diameter of the core, and

\$ d \$ is the diameter of the wire.

The length of the wire used to wind the core will then be:

$$ Lw = n \times Lt $$

where \$ n \$ is the number of turns,

and the resistance of the winding will be the resistance of the wire per unit length times the length of the wire in the coil.

Generally, though, you'd be primarily concerned with the resistance of the coil in terms of its effect on the Q of the coil, and you'd measure out the length of the wire needed to wind the coil, regardless of its resistance, plus a little extra, just in case.

  • \$\begingroup\$ Can you provide more details about this Q factor? \$\endgroup\$ – Golaž Jan 10 '15 at 14:29
  • \$\begingroup\$ Sure. A quick and dirty description is that "Q" is generally considered to stand for "Quality", and is the ratio of reactance to resistance. That is, if at some frequency the coil exhibits a reactance of 100 ohms while the resistance through it equals 1 ohm, it's said to have a Q of 100. A much more comprehensive explanation is here. \$\endgroup\$ – EM Fields Jan 10 '15 at 14:56
  • \$\begingroup\$ I am sorry my english isnt that good, what do you mean by " and you'd measure out the length of the wire needed to wind the coil" \$\endgroup\$ – Golaž Jan 11 '15 at 11:19
  • \$\begingroup\$ Instead of measuring the resistance to determine the length of the wire, you'd measure its length physically. Inches, meters... \$\endgroup\$ – EM Fields Jan 11 '15 at 15:12

Resistive losses are reduced by increasing the wire diameter of your inductor wire.

Therefore, choose the thickest wire that still allows you to achieve your needed inductance.

Using Litz wire will further reduce your losses.

By the way, the absolute must-have book on inductance is Grover, but it's a very tough read.


The maximum resistance is probably limited by temperature. A higher resistance will mean higher losses in the core, which means more heating. If the core becomes too hot its magnetic properties will change and you will have a bad time.

The losses will be approximated by Joule heating:

$$ P = I^2 R $$

How this affects temperature will depend on the thermal resistance of your inductor. Without a datasheet it would have to be determined experimentally.

However, you will run into other problems well before you hit this absolute maximum. Resistance means losses, and losses mean inefficiency. You probably want your converter to be reasonably efficient. If the converter is any good at all, losses will be low enough that temperature is not a concern. Otherwise, what you have built is a heater, not a boost converter.

  • \$\begingroup\$ You provided usefull information but if going by the maximum resistance isnt the right approach to design an indcutor, what is? \$\endgroup\$ – Golaž Jan 11 '15 at 11:53

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