I have a DC motor with a torque constant of 1.05 mNm/A. How do I calculate back emf constant from torque constant (mNm/A)?

I have information that is in the maxon catalog.(DCX 10 S metal brush motors)

enter image description here

  • 2
    \$\begingroup\$ There is an equation in the wikipedia page for motor constants that relates Kv and Kt. You just have to solve it for Kv. en.wikipedia.org/wiki/Motor_constants \$\endgroup\$
    – user57037
    May 25, 2016 at 5:36
  • \$\begingroup\$ And in SI units, solving it is particularly simple : Kv = 1/Kt. So back EMF = speed/Kt. (speed is in radians/second, of course) \$\endgroup\$
    – user16324
    May 25, 2016 at 11:06
  • \$\begingroup\$ @BrianDrummond Can you expand your equation Kv=1/Kt with SI units, I want to know how Nm/A turns into V/rpm, thanks. \$\endgroup\$ May 25, 2016 at 12:13
  • \$\begingroup\$ @Marko : It doesn't. Like I said, in SI units, speed isn't in rpm but radians/second. Simply rearrange the equalities in my other comment, you'll see speed/voltage (Kv) = current/torque (1/Kt). It basically follows from conservation of energy (naturally, ignoring losses). \$\endgroup\$
    – user16324
    May 25, 2016 at 12:21
  • 1
    \$\begingroup\$ You get it, you just don't see you get it yet... :-) You have enormous practical experience with motors so this is worth doing... Multiply both sides with V * Nm, and you get ...Nm * rad/s = A * V. See it yet? Just in case you don't, that says Power = Power. \$\endgroup\$
    – user16324
    May 25, 2016 at 12:29

4 Answers 4


OK I'm going to turn the comments into an answer since the question raises a surprisingly subtle point, and one that is most easily grasped if you consistently use the SI system rather than traditional units.

Because a motor translates electrical power into mechanical power (and vice-versa in generator mode) it must obey the conservation of energy.

So (ignoring friction, resistive and other losses) power in = power out.

Or, voltage * current = rotational speed * torque.

Rearranging, Voltage/Speed = Torque/Current.

Torque/Current (Nm/A) is known as the torque constant Kt.

Speed/Voltage (rad/s/volt) is known as the speed constant Kv (commonly seen is RPM/V but here expressed in SI units.

So, given the torque constant for a motor, the speed constant is also known, and presumably its inverse is known as the back EMF constant in some circles (though I haven't personally ever seen that).

EDIT : Following Gregory Kornblum's comment : who says it's the same power? The principle of conservation of energy.

Now clearly this is the simplest, most ideal situation - as I said above, ignoring all losses. You can define anything any way you like, but the most generally useful approach is to start with the simplest ideal situation, then separately account for energy losses until you have a satisfactory model for your purpose.


First of all you should have done a bit of research before asking this question, you also havent mentioned what sites did you go to before asking and also why other similar questions could not answer your problem.

Now as mkeith said the wikipedia page on Motor constants answers your question. Wikipedia Also you haven't specified which kind of motor it is. If its a brushless motor then $$Kemf=Kt$$ or the back emf is equal to the torque. The different types of motor will always have a specific ratio between EMF and the Torque. This EE.SE answer might also help you : Stack Exchange answer This is also a good website for learning about such calculations Micromo.com If you still have a doubt comment and I'll try to clear it.


you cannot make the assumption that \$K_t \equiv K_e\$ for a couple of reasons.

  1. the definition of Kt & Ke
  2. where Kt and Ke are defined

now #1 is easy enough to manage. Kt is Nm/A. Peak of the AC (or quasi if BLDC) will give the torque product. The equivalent for Ke is V/\$\omega\$ peak line-line for mechanical velocity.

An ideal motor, with a stator-pack that DOES NOT saturate. Ke and Kt (for the above statement) are interchangeable & if you wanted rms phase voltage a simple factor is all that is needed.

However, there is no such thing as an ideal motor & this is where the main difference comes into play.

\$K_t\$ is determined at PEAK current.

\$K_e\$ is defined as the OPEN-CIRCUIT voltage.

If you happen to have a "lazy motor" that is inefficiently using the stator pack and ONLY operating in the linear region of the B-H curve then yes... \$K_t \approx K_e\$ but that is a very, very poorly design motor.

The optimal point of a motor design is around hte knee and as such \$K_t != K_e\$. Its close but not 1:1. There is no magic "frig factor" to convert between \$K_t\$ and \$K_e\$ because it is dependent on the magnetic design & operating point.

If you do not have access to the magnetic design work & are not provided with \$K_e\$, the only guaranteed way is to backdrive the motor and determine the open-circuit voltage for a given rotor velocity... Or accept the deviation

The majority of the time \$K_t\$ is more useful, not only for torque calculation but also for operational BackEMF as when the motor is loaded, at speed, the core is becoming saturated and naturally \$K_e\$ drifts towards \$K_t\$.


These two constants Kt and Kv are not related, knowing torque constant won't help you to calulate the generated voltage. There is just one way, spin the motor and measure voltage and speed, make a table and write a linear function.

EDIT, with enlightenment of Brian Drummond:
\$V\cdot I=I\cdot K_T[\dfrac{Nm}{A}]\cdot V\cdot\frac{1}{K_V[\dfrac{V}{rpm}]} \cdot{\frac{2\pi[rad]}{60[s]}} \$

\$K_V[\dfrac{V}{rpm}]=\dfrac{2\pi}{60}\cdot K_T[\dfrac{Nm}{A}]=\dfrac{2\pi}{60}\cdot K_e[\dfrac{V\cdot s}{rad}]\$
\$K_V[\dfrac{V}{krpm}]=\dfrac{2000\pi}{60}\cdot K_T[\dfrac{Nm}{A}]=\dfrac{2000\pi}{60}\cdot K_e[\dfrac{V\cdot s}{rad}]\$

\$K_V[\dfrac{rpm}{V}]=\dfrac{60}{2\pi}\cdot \dfrac{1}{K_T[\dfrac{Nm}{A}]}=\dfrac{60}{2\pi}\cdot \dfrac{1}{K_e[\dfrac{V\cdot s}{rad}]}\$
\$K_V[\dfrac{krpm}{V}]=\dfrac{60000}{2\pi}\cdot \dfrac{1}{K_T[\dfrac{Nm}{A}]}\$

Also it can be written that 3 phase PMSM is little different:
\$P=M\omega = \sqrt{3}\cdot V \cdot I\$
\$K_V[\dfrac{V}{krpm}]=\dfrac{2000\pi}{\sqrt{3}\cdot 60}\cdot K_T[\dfrac{Nm}{A}]\$

These are data of 3phase PMSM:


\$K_T=0.835\$ calculating \$K_E\$ for 3ph PMSM gives result 50.5 V/krpm, which is approximately the declared 53.0 V/krpm.

  • \$\begingroup\$ The torque constant and back-emf constant are numerically equal for a DC motor, this comes from the motor and generator equations F=BIl and E=Blv. \$\endgroup\$
    – Chu
    May 25, 2016 at 11:04
  • \$\begingroup\$ Sorry, Kv and Kt are intimatey related, since voltage * current = speed * torque = power. \$\endgroup\$
    – user16324
    May 25, 2016 at 11:07
  • \$\begingroup\$ @Chu Can you write an equation, how Kt and Kv are related? Also it shall be noted that OP has only Kt, no power, speed,...etc. \$\endgroup\$ May 25, 2016 at 12:16
  • \$\begingroup\$ \$\small F=Bli\$ and Torque, \$\small T=Fr=Blri\$, where \$\small r\$ is radius of armature. Torque constant, \$\small K_T=\frac{T}{i}=Blr\:NmA^{-1}\$. \$ \small E=Blv\$, where \$\small v=\omega r\: ms^{-1}\$ is velocity of armature, and \$\small \omega\$ is angular velocity. Hence \$\small E=Blr\omega\$, and back-emf constant is \$\small K_V=\frac{E}{\omega}=Blr=K_T\$. So the two constants are numerically equal, but different units. I didn't downvote, btw. \$\endgroup\$
    – Chu
    May 25, 2016 at 13:09
  • \$\begingroup\$ @MarkoBuršič, see above comment \$\endgroup\$
    – Chu
    May 25, 2016 at 13:18

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.