Analysis and examples of power conversion by variable-frequency drives.
ElectricalElectrical power is defined as the following:
Power (P) = √3 x Voltage (V) x Current (I) x Power Factor (PF)
In an ideal VFD, the following would hold true:
Powerin = Powerout
But because a VFD has inefficiencies and requires a small amount of power consumption to power the brains of the drive, the input power will be slightly greater than the output power. For this, we will assume that this extra power draw is negligible.
With these two equations, we can then define the relationship between the VFD’s input and output:
Vin x Iin x PFin = Vout x Iout x PFout
Taking these equations into account, let’s use a 100-hp motor as an example with the following properties:
Power = 100 hp
Speed = 1,785 rpm
Voltage = 460 V
FLA = 115 A
Power factor = 0.86
Assume that the motor is running at 60 Hz on a VFD, drawing a no-load current of 40 A on the output of the VFD. With this, one would assume that the input current would also be the same, 40 A. However, using an ammeter on the drive’s input, a person is reading nearly zero amps! How is this possible? Is the drive creating power somehow? The answer simply is no, the drive is not creating power. The power factor causes this “discrepancy” in current When a motor is running at no load, the motor’s power factor can be assumed to be zero, not 0.86 86 as stated on the nameplate. The reason the power factor isn’t at 0.86 is because this is the motor power factor at full load. Alternatively, mechanical (friction) and electrical (resistive) losses in the motor prevent the power factor from being zero when running no load, but we’ll assume these losses to be zero just like we did for the VFD. Therefore, you would have the following:
Pout = 460V x 40A x 0
Pout = 0
Because the output power is zero, the input power also will be zero. With a fixed input voltage, the two variables would be current and power factor. Because current is needed for a power factor to exist, both current and power factor are zero, which means the low input current reading is indeed correct.
This explains why the input current to the VFD is so low when the motor is operating under no load conditions. But what about under load? The same concept still applies when the motor is under load. For example, assume the same motor is now operating at half speed, 30 Hz and producing full motor rated torque and drawing the motor’s full-load amps (FLA). This means that the electrical power that the motor is drawing is:
Pout = √3 x 230V x 115A x 0.86 = 39.4kW
Because the VFD is a power converter, this means that the input current is (assuming a 0.89 input power factor from a 3% line impedance):
Iin = (39.4kW) / (√3 x 460V x 0.89) = 55.6A
Because the VFD is operating at half speed and under full load, the input current is less than half of the output current.
|Power (kW)||Voltage (V)||Current (A)||Frequency (Hz)||Power Factor|
In this example, the input current is less than half of the output, a result of having a higher power factor on the input side.
The difference in power factor between the input and output side of the motor is what makes it possible to have a higher output current than input current. Assuming the motor is now running at full load and using the same power factor values, your input current now becomes:
Iin = (√3 x 460V x 115A x 0.86) / (√3 x 460V x 0.89) = 111A
which is 4 A lower than the output current.
|PFin < PFout||Iin < Iout|
|PFin = PFout||Iin = Iout|
|PFin > PFout||Iin > Iout|
MechanicalThe current a VFD draws on the input side also can be related to the mechanical power a motor is delivering. The basic relationship for motor power is:
Powermechanical ∝ Speed x Torque
This means that if the motor is operating at half the speed and producing full torque, the motor is outputting half of its rated power. Consequently, if the motor is running at full speed and producing half torque, the motor is also outputting half of its rated power.\
Because of motor losses, the power relationship between the electrical power going into the motor and the mechanical power is:
PowerElectrical = (Powermechanical) / (EfficiencyMotor)
Revisiting the above example, if the motor is operating at 30 Hz, half the motor’s rated speed and producing full torque, then the mechanical power being produced is 50 hp. Assuming that the motor is 95% efficient, the electrical power that’s required is:
PowerElectrical = (50HP x 0.746) / 0.95 = 39.3kW
which means that the current on the input side of the VFD will be approximately 55 A. This same current will also hold true even if the motor is operating at full speed and producing half torque.
Table 1. Operating points for examples*
|Motor speed (rpm)||Motor torque (ft-lb)||Motor power (hp)||Input power (kW)||Input current (A)|
|*Motor is rated for 100 hp with a rated torque of 295 ft-lb.|
Ultimately, a VFD is merely a power conversion device that converts the fixed voltage and frequency of incoming power to a variable voltage and frequency output to provide the variable speed capabilities for which it was designed. Keep in mind the variables associated with electrical power (voltage, current and power factor) and their relationships when comparing the VFD’s input to its output. This also will hold true when using the motor’s mechanical power (speed and torque) to determine the amount of input power/current to the VFD. Taking all the variables into consideration, one can be pleasantly surprised to find the input current lower than the output current