# BLDC Motor Constants: Why is my Kt and Ke different?

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.

• How different are the values from the spec? Oct 6, 2018 at 2:54
• @BlairFonville 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. Oct 6, 2018 at 5:49

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)

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

• I think both these answers are good, illustrating different aspects of the problem. I lean towards JonRB's answer (a) as sqrt(3) is suspiciously close to the discrepancy and (b) the alternative would be an unusually inefficient BLDC motor (though plausible for a toy brushed motor) Oct 6, 2018 at 12:12
• The sqrt(3) is valid for 3phase, star connected machines. I can't find the derivation in the machines book I have at home. It is however PEAK so I suspect when the OP wrote rms there was a mistake
– user16222
Oct 6, 2018 at 14:29
• Thank you everyone for your help. Although as I said, since I'm new to motor theory, this has made me more confused then before, especially regarding what peak and RMS really means and what it represents. Here's an image link to the manufacturer's Excel spreadsheet: ibb.co/dwiBgz I am trying to reconcile my motor theory equations with this, while sitting behind a computer and not actually doing measurements (oscilloscope, etc.) The equations I am trying to reference to are here: ibb.co/iMiVPK and here: ibb.co/cwgBWz Oct 6, 2018 at 17:16
• Hopefully the Excel image clears things up. I do not where to use the RMS values and where to use PEAK values when I'm trying to apply them in the equations I linked. Again, the help is greatly appreciated. For context, I have a propeller whose characteristics are determined (torque reqd vs RPM) and batteries that are already selected. This may be the wrong order in how to select a motor, but it is the situation I find myself in. Oct 6, 2018 at 17:22
• @JonRB Additional note, if you like at the Excel image I sent, the sqrt(3) thing doesn't work for the peak values. I wonder if it was just a coincidence that it worked out in one of the cases? Oct 6, 2018 at 19:08

Kv == Kt in an ideal lossless motor, when defined consistently.

You should use the average (not rms) when computing torque and speed, because these are linear, that is they depend on the first power of the variable.

You should use rms current when computing heating of the windings, because this depends on the current squared, the second power.

It's inappropriate to use Kv or Kt in an expression with rms values. Changes in the waveform will change the measured value.

DVMs generally give one of two measurement types for AC waveforms, cheap ones give you 'average rectified scaled as sinewave rms', and expensive ones that claim it give you 'true rms'. Either placed in the phase line of a BLDC motor will approximate the true average current, but the error will be in double figures of percent. You would expect this sort of level of error from ignoring ESC losses, friction, air viscosity, winding resistance, so it's probably adequate.

If you want better accuracy than this, then you could use an oscilloscope and compute it from the waveform. It's probably better to just over-design by a bit and measure the torque on the final system.

The design process usually goes like this. (1) use Kv to make sure your battery voltage is high enough for your maximum speed. (2) make sure your power supply will deliver enough power to meet the motor output power plus losses, as the motor will draw whatever current it needs from the ESC to meet the torque, which will draw whatever it needs from the PSU to meet the current.

Unless you actually need to control the torque directly, there's usually little need to use Kt in your calculations.

Why would the motor supplier publish an equation using Kt and rms measurements? Perhaps if they assume a waveform (popular ESC type) then they are factoring-in the expected scaling error?

• Thank you for your help, Neil. I have linked some clarifying info in the comment replying to the other answer. To be clear, I am doing this "design" from a computer without the ability to analyze actual signals, etc. Oct 6, 2018 at 17:23