Variable frequency drives

Why it matters

A variable frequency drive (VFD, also called an AC drive or inverter) turns a fixed-speed induction motor into an adjustable-speed drive. Pumps, fans, conveyors, extruders and machine-tool spindles use VFDs to match speed to the process, start gently without large inrush current, and save large amounts of energy on centrifugal loads.

Key ideas

• Why frequency. The synchronous speed of an induction motor is N_s = 120 f / P. Pole number is fixed by the winding, so the practical way to vary speed smoothly is to vary the supply frequency f.
• Why voltage must change with frequency. The air-gap flux is roughly proportional to V/f. Lowering f at full voltage would push the core into saturation and draw huge magnetising current; raising f at the same voltage weakens the flux. So below base speed a VFD keeps V/f approximately constant (constant V/f or scalar control), which keeps flux and therefore the available torque roughly constant: the constant-torque region.
• Above base speed. The output voltage cannot exceed the supply-derived maximum, so V stays at rated value while f rises. Flux falls as 1/f, available torque falls roughly as 1/f, and power stays roughly constant: the field-weakening or constant-power region.
• Low-speed boost. At low frequency the stator resistance drop is a large part of V, so drives add a small voltage boost (IR compensation) to keep starting torque.
• Power circuit (voltage-source inverter).
  1. Rectifier – usually a three-phase diode bridge converting the AC supply to DC.
  2. DC link (DC bus) – a capacitor bank (often with a choke) that smooths the DC; its voltage is about 1.35 × the supply line voltage.
  3. Inverter – six IGBTs switched by pulse-width modulation (PWM) at a carrier frequency of a few kHz to synthesise a three-phase output whose fundamental frequency and amplitude are both adjustable. The motor's inductance filters the current into a near-sinusoid.
  4. Control and protection – microcontroller/DSP running V/f or vector control, with current sensing, overcurrent, over/undervoltage and thermal protection.
• Braking. When the motor is decelerated quickly it acts as a generator and pumps energy into the DC link. A diode front end cannot return it to the mains, so the drive uses a braking chopper and resistor, or a regenerative (active front-end) rectifier.
• Vector (field-oriented) control. Splits stator current into flux-producing and torque-producing components so an induction motor can be controlled like a separately excited DC motor; it gives full torque near zero speed and fast torque response. Sensorless vector control estimates speed from voltages and currents.
• Energy saving on fans and pumps. For centrifugal machines the affinity laws give flow ∝ N, head ∝ N² and power ∝ N³. Throttling with a valve or damper wastes energy; reducing speed instead cuts power steeply.
• Side effects. Input current harmonics (use line reactors or filters), fast voltage edges (dv/dt) stressing motor insulation and causing bearing currents on long cables, electromagnetic interference, and reduced self-cooling of the motor at low speed with a shaft-mounted fan.

Formulas

N_s = 120 f / P

• N_s synchronous speed (rpm), f output frequency of the drive (Hz), P number of poles. Actual speed N = N_s(1 − s).

V / f = V_rated / f_rated (constant V/f, below base frequency)

• V phase or line voltage applied by the drive (V), f output frequency (Hz).

V_dc ≈ (3√2 / π) · V_LL ≈ 1.35 V_LL

• V_dc DC-link voltage of a three-phase diode bridge (V), V_LL supply line-line RMS voltage (V). 1.35 V_LL is the average output of the bridge on load; with a large capacitor at light load the bus rises towards the peak value √2·V_LL ≈ 1.41 V_LL.

V_LL1 = 0.612 · m · V_dc (sine PWM, linear range m ≤ 1)

• V_LL1 RMS fundamental line voltage at the inverter output (V), m modulation index.

Q₂/Q₁ = N₂/N₁, H₂/H₁ = (N₂/N₁)², P₂/P₁ = (N₂/N₁)³

• Affinity laws for centrifugal pumps and fans: Q flow (m³/s), H head (m), P shaft power (W), N speed (rpm).

P = T·ω

• P power (W), T torque (N·m), ω speed (rad/s). In the constant-torque region P rises in proportion to speed.

Worked examples

Example 1 (standard). A 400 V, 50 Hz, 4-pole induction motor is run from a VFD with constant V/f at 30 Hz. Its slip speed at rated torque is 40 rpm and, under constant V/f, stays about the same at any frequency. Find the drive output voltage, the synchronous speed and the speed at rated torque.

1. V = V_rated × f / f_rated = 400 × 30/50 = 240 V.
2. N_s = 120 f / P = 120 × 30/4 = 900 rpm.
3. Speed at rated torque N = N_s − slip speed = 900 − 40 = 860 rpm.

Answer: 240 V, N_s = 900 rpm, N ≈ 860 rpm.

Example 2 (GATE level). A 30 kW centrifugal fan runs at full speed for 4000 h a year. The required airflow drops to 80 % of maximum. Compare (a) the shaft power if speed is reduced with a VFD, and (b) the annual energy saving relative to full-speed running. Also find the DC-link voltage of the VFD on a 400 V supply and the maximum linear-range fundamental output line voltage with sine PWM.

1. Flow ∝ N, so speed ratio N₂/N₁ = 0.8.
2. Power ratio P₂/P₁ = (0.8)³ = 0.512, so P₂ = 0.512 × 30 = 15.36 kW.
3. Saving ΔP = 30 − 15.36 = 14.64 kW; annual energy saved = 14.64 × 4000 = 58 560 kWh (ignoring drive and motor efficiency changes). A damper would cut the power far less, because the fan still runs at full speed.
4. V_dc ≈ 1.35 × 400 = 540.2 V.
5. V_LL1,max = 0.612 × 1 × 540.2 = 330.8 V. This is below 400 V, which is why practical drives use third-harmonic injection or space-vector PWM (raising the limit by about 15 %) or slight over-modulation to deliver full rated voltage.

Answer: 15.36 kW, about 58 600 kWh per year saved; V_dc ≈ 540 V; V_LL1,max ≈ 331 V.

Common mistakes

• Changing frequency without changing voltage. Low f at full V saturates the motor; constant V/f is needed below base speed.
• Using f = P N / 120 with the actual rotor speed and calling it the supply frequency; that relation uses synchronous speed. An induction motor needs slightly more than 50 Hz to run at 1500 rpm on load.
• Assuming torque stays constant above base frequency; in the field-weakening region it falls roughly as 1/f.
• Applying the cube law to constant-torque loads such as conveyors, where power falls only in proportion to speed.
• Forgetting braking energy: fast deceleration of a high-inertia load trips the drive on DC-bus overvoltage unless a braking resistor or regenerative front end is fitted.
• Assuming a shaft-mounted fan cools the motor adequately at very low speed under full torque.

For GATE ME

Expect numericals on synchronous speed at a given drive frequency, constant V/f voltage settings, speed under load with constant slip speed, power at constant torque versus speed, and fan/pump energy savings using the affinity laws. Conceptual MCQs test the stages of a VFD, why V/f is held constant, constant-torque versus constant-power regions, and the purpose of PWM and braking resistors. Practise the cube-law saving calculations quickly.

Quick check

1. A 2-pole motor is fed at 40 Hz from a VFD. What is the synchronous speed?
2. A 415 V, 50 Hz motor on constant V/f is run at 20 Hz. What voltage should the drive apply?
3. A pump's speed is cut to 70 %. What fraction of the original power does it draw?
4. Name the three stages of a voltage-source VFD power circuit.
5. Why is a braking resistor fitted to some VFDs?

Answers: 1. 2400 rpm. 2. 166 V. 3. 0.343 (about 34 %). 4. Rectifier, DC link, inverter. 5. To dissipate energy returned to the DC link when the motor regenerates during deceleration.