recently I've needed to buy a brushless motor, but because I couldn't find any motor with wanted characteristics, I've decided to make my own. I know that it's very precise job, but even if something goes wrong, I will enjoy just trying to construct motor. As perfectionist, first I must plan everything (if you aren't intrested in what I've already done, I suggest skipping to the end - there is my problem and question :P).

First I've tried to make a formula for calculating the number of turns on one coil (\$n\$). Maximal rotational speed in BLDC motor is the speed at which moving magnets induce voltage equal to negative voltage across the coil.

Faraday's law of induction: \$\varepsilon = -\frac{\Delta \phi}{\Delta t}\$

I have \$n\$ turns of wire, so: $$\varepsilon = -n \frac{\Delta \phi}{\Delta t}$$

In my case \$\Delta t\$ is equal to time, in which next magnet "appear" next to given coil: $$\Delta t = \frac{1}{K_{V} U m}$$ where:
\$K_{V}\$ - in my case it will be rotates per second per volt - not per minute
\$U\$ - voltage across one phase
\$m\$ - number of magnets in motor

\$\Delta \phi\$ is equal to change of magnetic flux during \$\Delta t\$. It means doubled magnetic flux of single magnet (\$\phi\$): $$\Delta \phi = 2 \phi$$

That means: $$\varepsilon = -\frac{2 \phi n}{\frac{1}{K_{V} U m}} = -2 \phi n K_{V} U m = - U_{c}$$ where:
\$U_{c} = \frac{U}{N}\$ - voltage across one coil
\$N\$ - number of coils per phase

Final effect: $$n = \frac{1}{2 \phi K_{V} N m}$$

And at this moment I need your help. When I'm trying to estimate how many turns coil should have, I need to know magnetic flux of single magnet (in my case neodymium one). In the Internet you can find tables containing properties of the materials the magnets are made of, but there isn't magnetic flux (what is understandable).

My question is: is it possible (if yes, how?) to calculate magnetic flux of magnet with given dimensions, knowing remanence, coercivity and "energy density" (the product of B and H) of material the magnet is made of?

Sorry for all my English mistakes and thanks for answers.

  • 1
    \$\begingroup\$ Good luck with that.... \$\endgroup\$
    – Trevor_G
    Commented Mar 7, 2017 at 22:57
  • 1
    \$\begingroup\$ yes, it's possible. Calculating motors isn't exactly a new field in electrical engineering (Siemens himself says hello). In practice, everyone building motors has (potentially pretty advanced) finite element simulators that make the analytic solution actually usable for anything that doesn't consist of "perfectly round, perfectly even conducting across their diameter, constant temperature, perfectly spaced" coils, magnetics and gaps. \$\endgroup\$ Commented Mar 7, 2017 at 23:29

1 Answer 1


That link you provided, 'tables' gives nice curves of B and H for their various materials, along with 'load lines' for the demagnetising force.

You can use an equivalent of 'ohms law' to find the field. Think of the magnet as a battery, a source of H. The terminal current will be B, as B rises, the H will fall as a result of its internal 'impedance' (reluctance). It is H that forces a field through materials between the magnet faces.

To a first approximation, all of the H will be 'dropped' across your high reluctance air-gap. In comparison, the iron pole pieces have negligible reluctance.

The curves already have a load line drawn on them. The flux and H field will be at the intersection of the load line, and the magnet loop. I suspect the load line labels are normalised to the magnet length, check with that supplier, I'm not going to devote the time to investigation. You have to make sure you understand what normalisation has been used with any set of graphs. If that's the case, and for the example of N35 material (just the first one on the list), you see the 1.0 line, which means an airgap of length equal to the magnet length, it intersects the curve at 1.1T (20C line). If you halve the airgap (2.0 line), you get only 1.15T, if you double the airgap (0.5 line), you drop to 0.6T.

Obviously the reluctance of the iron pole pieces will lower the field very slightly, and the reluctance of any inadvertent airgaps between magnet and pole pieces will lower the field a lot.

Also obviously, there is such a range of motors available, that getting one of the next size up from your 'unique' specifications will be far cheaper, faster and more efficient than rolling your own. However, as a learning and machining exercise, go for it!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.