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I have a brushless DC motor with a torque constant of .0181 Nm/A. How do I calculate back emf (Vpeak/kRPM) from torque constant (Nm/A)?

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    \$\begingroup\$ For brushless DC motors Kemf = Kt. So you can multiply by the angular velocity to get the back-EMF voltage. See here and here for a full explanation of the equations. \$\endgroup\$ – embedded.kyle May 15 '14 at 20:11
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The easiest and best way to find the back-emf constant is to back-drive your motor with another motor and measure the voltage that is generated on an oscilloscope. Then measure the peak voltage of that wave form and divide that by the speed that you are back-driving the motor.

The units for torque constant (Kt) and back emf constant (Ke) are equivalent. The units for Kt are \$\dfrac{N \cdot m}{A}\$. If you expand that out to SI base units, you get \$\dfrac{N \cdot m}{A} = \dfrac{kg \cdot m^2}{A \cdot s^2}\$. Now when you write out the volt in SI base units you get \$V = \dfrac{kg \cdot m^2}{A \cdot s^3}\$. If you divide by \$\dfrac{rad}{s}\$, you end up with the same units as the units for Kt. So you get \$\dfrac{V}{\frac{rad}{sec}} = \dfrac{kg \cdot m^2}{A \cdot s^2} = \dfrac{N \cdot m}{A}\$.

So as you can see the units for Kt and Ke are equivalent. This equivalence holds between the torque constant and back-emf constant ONLY if we are talking about the "per phase" constants. The "per phase" constants are not usually what you find on a motor datasheet. On the datasheet, you are more concerned with how the overall torque relates to the current, not with the per phase relationships. If you work through the math, what this means is that \$Ke = Kt\$ (line-to-line) only if we are talking about an brushless motor with an ideal trapezoidal back-emf. For a motor with an ideal sinusoidal back-emf, the relationship is \$Ke = \dfrac{\sqrt{3}}{2}\cdot Kt\$ (line-to-line). In reality, brushless motors can't be made to have either ideal trapezoidal or ideal sinusoidal back-emfs. It's always somewhere in between. So usually if you measure the back-emf constant as I mentioned above it will be less than the Kt found on the datasheet.

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  • \$\begingroup\$ When you divide both sides by rad/s where does the "rad" go? Shouldn't that appear in the denominator of the right hand side? \$\endgroup\$ – Dsel Aug 26 '17 at 6:54
  • \$\begingroup\$ @Dsel, radians is a "dimensionless" quantity. It's actually the ratio of a length (arc length of the angle in question) to a length (the radius of the circle). So rad/s could just as easily be written as 1/s. But rad/s is a somewhat common unit for rotational velocity so I wrote it as that. \$\endgroup\$ – Eric Sep 11 '17 at 19:36

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