I want to simulate an BLDC motor, so I am studying corresponding mathematical model. My problem is that I am little bit confused by the torque equation.

\$e_a\$, \$e_b\$, \$e_c\$ - Back-EMF voltages (induced components of winding voltage drop)

\$i_a\$, \$i_b\$, \$i_c\$ - currents through windings

The amplitude of back-EMF voltage is proportional to angular velocity, right? So how can torque be dependent on back-EMF voltage? One would get zero torque for zero angular velocity. I know torque should be maximum at start point, but I am missing something to understand this equation..

P.S. I would get that in stable state, so I can calculate torque from back-EMF for constant speed. But how can I calculate output torque on shaft in nonstable state? Is that the equation including torque constant (\$T=K_t.I\$)?

  • \$\begingroup\$ Please edit your question to define the terms \$e_a\$, \$i_a\$, etc. \$\endgroup\$
    – TimWescott
    Commented Jul 20, 2020 at 23:34

2 Answers 2


ea, eb, ec the back EMF voltages are proportional to the actual motor speed, unlike the applied voltages which are partiallly lost driving currents ia etc through the winding resistance. Thus, ea etc are the voltages generating useful mechanical power rather than heating the windings.

Then (ea.ia + eb.ib + ec.ic) is simply the electrical power which is converted to mechanical power P.

So the overall equation is simply

Torque = Power/rotational velocity.

At zero speed, both ea etc and omega are 0 so this form of the expression is indeterminate. Instead, you can use the speed constant Kv (rpm/volt), or its radians/sec/volt equivalent ... or indeed its inverse, which is the torque constant Kt, as you posit in the question.

  • \$\begingroup\$ Ok, thank you, thats what I have been looking for. \$\endgroup\$ Commented Jul 21, 2020 at 23:33

If the \$e_a\$, \$i_a\$, etc., are the idealized voltages and currents on the armature, with magnetic and electrical losses neglected, then that equation simply follows from conservation of energy.

Let \$P_e\$ denote electrical power, and \$P_s\$ denote shaft power. Claim that there's no friction, resistance, or other losses. Then, in a universe that's not visibly changing dimensions, \$P_e + P_s = 0\$ by conservation of energy.

$$P_e = \sum_{\mathrm{all\ windings}}e_{winding}i_{winding}$$ $$P_s = T_e \omega_m$$

Do an itty bitty bit of math, to substitute \$e_a\$, \$i_a\$, etc., for my pretentious summation over all hypothetical windings, then an itty bitty bit more by dividing both sides by \$\omega_m\$, and you end up with your expression.

  • \$\begingroup\$ Can you decompose the electric power please? I know that there is a power loss on resistace of winding and the second component is power given by induced back-EMF voltage. Is there any other component? I mean, I still do not get that at the start of the motion the torque is zero according to stated equation. (ea, ab, ac are back-EMF voltages, and those are zero at zero speed -> so zero power and torque) \$\endgroup\$ Commented Jul 20, 2020 at 23:50
  • \$\begingroup\$ Before we continue, this is Stackexchange. Please edit your question with the definition of \$e_a\$, \$i_a\$, etc. Stackexchange prefers that important information is not lost in comments, but is stated in the question. While we're at it, if you could share a link, or quote more of the text around that equation, it would help us to know the original author's intent. I can assume I know, but I cannot know. You, on the other hand, actually have the text. \$\endgroup\$
    – TimWescott
    Commented Jul 21, 2020 at 1:06
  • 1
    \$\begingroup\$ @DanielŠebík - at zero speed it is indeterminate 0/0. The numerator is the power absorbed by the back-emf. It ignores resistive losses. \$\endgroup\$ Commented Jul 21, 2020 at 1:07
  • \$\begingroup\$ @Kevin White - thank you it is what I overlooked. So is there another way how you can determine torque at zero speed? \$\endgroup\$ Commented Jul 21, 2020 at 9:11
  • \$\begingroup\$ @DanielŠebík - You would normally use the torque constant of the motor that relates torque to current. \$\endgroup\$ Commented Jul 21, 2020 at 14:22

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.