Motor DC-Motor Brushless-DC-Motor – Why Are My BLDC Motor Constants Kt and Ke Different?

brushless-dc-motordc motormotorrms

I'm new to working with motors. Reading up on BLDC theory, it is repeatedly said that in SI units, the Ke and Kt will be numerically equal. I downloaded the motor spec excel sheet of the T1101 motor by MTI. After choosing a particular winding and input voltage, it generates its motor data. As expected, the Ke = (30/pi/Kv). However, the Kt value displayed on the data sheet is different from the Ke value displayed. I'm confused as to the discrepancy. This is true both for peak and RMS categories.

Which brings me to my second point of confusion. Should I be using peak or RMS voltage/current in calculating motor performance? As a mechanical engineer, I'm trying to get a certain torque and RPM for some purpose. I look at a formula that relates torque and RPM to current, no load current, torque constant, voltage, etc. In these formulae, do I input the RMS or peak for voltages and currents?

I very much appreciate the help.

Best Answer

Please keep in mind how each are defined

\$K_e\$ is defined as the PEAK line voltage per mechanical rotor velocity with a fundamental equation \$K_e = \frac{V_{pk,ll}}{\omega_m}\$

\$K_t\$ is defined as the PEAK torque per phase current with a fundamental equation \$K_t = \frac{T}{A}\$

The units of \$K_e = \frac{V \cdot s}{rad} = V\cdot s\$ (since radians are unitless

The units of \$K_t = \frac{N\cdot m}{A} = \frac{J}{\frac{C}{s}} = \frac{J\cdot s}{C} = V\cdot s \$

\$K_e\$ and \$K_t\$ have exactly the same units and in the ideal case ( no mechanical drag, no magnetic saturation) they are comparable. I say comparable not equal because there is the \$\sqrt{3}\$ factor due to one being line-line and the other being phase.

In practice... \$K_t\$ is defined at rated current and as such there is magnetic saturation resulting in \$ K_t < \sqrt(3) \cdot K_e \$ (how "less than" depends on saturation)

From your comment

The spec values on the Excel sheet are what I am confused about. The Kt is 0.55 N-m/Arms and the Ke is 0.318 Vrms/rad/s. I was under the impression that the numerical value for both would be the same, given that they are expressed in SI units.

\$ K_t < \sqrt(3) \cdot K_e \$
0.55 < \$\sqrt{3} \cdot\$ 0.318
0.55 < 0.5508

Thus your electrical machine datasheet is in alignment