I am trying to measure the output power for BLDC controller, in order to calculate the efficiency of the controller. I came to know that two watt-meter method is one of the method used to calculate the three phase AC power.

enter image description here

But instead of using two wattmeters, i have.....

  • Two clamp meters (AC/DC clamp meter) for measuring Ir and IY.

  • Two multimeters (AC/DC) for measuring VRB and VYB.

  • My input power is DC 48 volt. P = V × I will give me the input power.

  • But to calculate the output power has complexity.

W1 = IR × VRB and W2 = IY × VYB

W1 + W2 gives the output power.

For voltage measurement VRB and VYB:

My bldc controller output is trapezoidal waveform with variable output frequency, Vrb and Vyb can be calculated by using ac multi meter. Because ac voltage does not depends on the Frequency, whatever the Output frequency the Vrms will be the same i guess. Correct me friends if i am wrong.

For current measurement Ir and Iy : Since the AC current must not be same for different frequency at all time. How to calculate the AC current at my output frequency say (eg: 100 Hz).

Edited... Sorry i forgot to say something, I am driving the bldc controller with bldc motor in a load setup.

  • \$\begingroup\$ I've fixed some of the random and missing capitalisation and shown how to write subscripts (<sub>...</sub>) but ran out of energy to finish them all. \$\endgroup\$ – Transistor Aug 25 '20 at 14:28

To measure power in the situation shown, the instrumentation system must be able to determine the instantaneous values of voltage and current, multiply current X voltage at corresponding instants and integrate the results over the period of the waveform. You can not measure the power using the instrumentation described.


You can always figure out real power by averaging the instantaneous power of an element (V instantaneous * Iinstantaneous).

If you only have RMS meters then you have to sit down with your theoretical waveform and calculate it out real power analytically, then RMS analytically, and then compare the two to know the appropriate fudge factor to apply to your RMS numbers so that you can adjust the numbers from your meter. But phase and waveform shape cannot change from your theoretical or it will throw things off. That is to say, there is no real way to do it with just RMS meters since they do not retain the phase information.

  • \$\begingroup\$ Then what is the way to find the efficiency of the bldc controller. \$\endgroup\$ – Bud Aug 25 '20 at 16:46
  • \$\begingroup\$ @Bud You need equipment suitable for the task. The other way is to apply a known load to the motor or measure the torque, and measure its RPM to measure the mechanical output power directly. \$\endgroup\$ – DKNguyen Aug 25 '20 at 17:48
  • \$\begingroup\$ ALthough, I suppose that if you just use a Y-resistive load instead of a motor there will be no phase shifts. In that case, you could measure the output power of the BLDC controller with RMS meters but you would have to rig it so it actually runs without an actual motor there (fake commutation signals). That will get you some gauge on efficiency, but since there will be no flyback since no inductance then it won't be exactly the same as if a motor were there. Certainly it won't let you find any inefficiencies due to commutation. \$\endgroup\$ – DKNguyen Aug 25 '20 at 17:52
  • \$\begingroup\$ What equipment and method are required to calculate or find the bldc controller efficiency. \$\endgroup\$ – Bud Aug 25 '20 at 18:21
  • \$\begingroup\$ @Bud You either need a three-phase power meter (which is probably too expensive and specific), or an oscilloscope with one voltage probe and one current probe if you assume balanced phases. Preferably you would have three voltage probes and three current probes and a 6 channel scope, but that's asking for a lot. Instead of current probe, you could use a current sense resistor and differential probe. You would use the multiplication and averaging function of the scope to calculate average instantaneous power \$\endgroup\$ – DKNguyen Aug 25 '20 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.