I'm an electricians apprentice at a mine. I understand the basic operation of variable frequency drives (VFDs), rectification, filtering, and pulse width modulation. But, at the heart, all they're really doing is changing the incoming voltage into a frequency controlled replication, right?

When I'm setting up VFDs in the field though they all need the name plate information such as number of poles, full load amps etc. I can't get a good explanation of why.

Can't it just change whatever demand it gets from the motor? I know frequency is related to the number of poles in alternators but it doesn't make sense me that it would matter to a VFD using IGBTs.

Can you help me understand how the drive uses this information?

  • \$\begingroup\$ A lot of VFDs have 40 or 50 parameters that can be set.. \$\endgroup\$ Commented May 20, 2021 at 19:24
  • \$\begingroup\$ In the case of Siemens VFD's, the number can be in the thousands... but these are mostly ancillary to the physical operation of the device. \$\endgroup\$
    – rdtsc
    Commented May 20, 2021 at 19:32
  • \$\begingroup\$ The VFD doesn't have our amazing perceptory organs. So if the FLA were set to 100A on a 1/2HP (lower-voltage) motor, the VFD will ignorantly try to force 100A into it, leading to noises and vibrations we could sense, but the drive could not. It would run for a few seconds before blowing up. \$\endgroup\$
    – rdtsc
    Commented May 20, 2021 at 19:37
  • 2
    \$\begingroup\$ @rdtsc - The VFD will not force current into the load. The load takes the amount of current based upon its design parameters with the maximum current being limited by the resistance of the windings. \$\endgroup\$ Commented May 20, 2021 at 21:52
  • \$\begingroup\$ Motor current ranges up to 100:1 from full start to no load full speed. If you do not define the parameters it could be like starting a bike in 4th gear \$\endgroup\$ Commented May 20, 2021 at 22:37

3 Answers 3


If the VFD is trying to do speed control and doesn't have an external speed sensor it has to try to sense the speed electrically. But electrical speed is not mechanical speed. Running one complete electrical cycle through the motor doesn't necessarily make the motor rotate one full mechanical cycle. Often less. They are related to each other by the number of poles. Even with a speed sensor, it is best to let the VFD know how many poles there are ahead of time or else it needs to try and blindly turn one rotation to figure out the relationship which makes for jerky, weak startups.

And presumably you want the VFD protect the motor and not blow it up if the motor is overloaded so the VFD needs to know how much power the motor can handle.

You also don't want the VFD to flame up because it tried to drive the motor too hard too fast (i.e. too hard a soft start on a high inertia load).


The number of poles tells the VFD the divider between electrical and mechanical speed. For example, a two-pole motor runs a bit below 60 RPS at 60Hz, but a four-pole motor only at 30RPS.

The nominal voltage and frequency of the motor tells the VFD about the magnetization of the iron. That one must not be exceeded, because an overexcited motor is a temporary short during each AC cycle. When the VFD has to reduce the frequency to account for lower speed demand, it also has to reduce the voltage so the quotient stays constant. That's why there are special “VFD motors” — they have graciously oversized iron parts.

The nominal current needed is for telling the VFD when to detect the overcurrent situation either through excessive mechanical load or through misinformation about the maximum excitation. It's a safety measure.

  • \$\begingroup\$ I think you tried to use more popular explanation, but I would suggest the term "magnetic saturated" that not necessarily presents a short, but approximately a resistance of the windings. Of course this needs to be taken care by the VFD, but in other cases the current control is a functional and protective feature. \$\endgroup\$ Commented May 20, 2021 at 22:23
  • \$\begingroup\$ For any practical means, only having the resistance of the windings of an asynchronous machine between L and N means a short. For example, a typical winding resistance of a 350W 230V~ AC motor is 6Ω. That means it will try to draw 8kW if the inductance doesn't stop it. \$\endgroup\$
    – Janka
    Commented May 21, 2021 at 9:02
  • \$\begingroup\$ I like to not simplify the explanation that much so the reader can understand the topic. At lower frequencies than the nominal motor frequency, the V/f curve will be programmed with smaller voltage, and so 5ohms of winding resistance will have I2R losses much smaller. Working very near saturation is more volume and weight efficient, but less energy efficient. Do you agree? \$\endgroup\$ Commented May 21, 2021 at 14:11

Torque in an induction motor is proportional to slip. Rated slip is calculated by subtracting the full-load RPM on the nameplate from the synchronous RPM that is calculated from the rated frequency and the number of poles. Entering the rated current is important for proper motor protection and good performance. Rated efficiency and power factor can also be used to optimize motor performance. In addition, with the nameplate information, the VFD can perform some "tuning" tests to determine other details of the motor design. That also helps optimize the performance.

For a load like a fan or centrifugal pump, the VFD can get along without all of that, because those kinds of loads are not difficult to start and don't usually run at very low speeds. However entering that information may provide a little better overload protection and might even increase the efficiency a little. Any little bit of "tuning" that might make the motor run a couple of degrees cooler will increase the motor life.

  • \$\begingroup\$ Aren't fans easy to start? \$\endgroup\$
    – DKNguyen
    Commented May 21, 2021 at 5:48
  • \$\begingroup\$ @DKNguyen depends on the fan. Big ones have quite a lot of angular momentum and inertia \$\endgroup\$
    – Chris H
    Commented May 21, 2021 at 9:47
  • \$\begingroup\$ @DKNguyen: Yes they are easy to start for any VFD. All that is necessary is to set a low acceleration rate. Also, protections built into VFDs effectively limit the torque so that the VFD does not try to accelerate too quickly. Default settings are probably adequate most fans except large combustion draft fans, \$\endgroup\$
    – user80875
    Commented May 21, 2021 at 12:07
  • \$\begingroup\$ I don't know about fans being easy to start - i have watched teams of Siemens technicians crowded around a drive trying to tune it to start without tripping for days (and never quite getting it right). I've seen two identical fans with identical drives beside each other, one works perfectly for 20 years and the other trips twice a day. Maybe for small fans with low inertia, but ... \$\endgroup\$
    – Joel Keene
    Commented May 21, 2021 at 13:28
  • \$\begingroup\$ @JoelKeene: With a VFD, dealing with high inertia is a matter of adjusting acceleration rate or current limit. If there is a problem it could be due to the need or desire to accelerate or decelerate as quickly as the VFD-motor combination could possibly manage. On a comparative basis, fans and centrifugal pumps are less demanding than other types of loads. That does not mean that they always work perfectly. \$\endgroup\$
    – user80875
    Commented May 21, 2021 at 13:49

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