Induction Motor Drives: V/f and Vector Control

Because flux is proportional to \( V/f \), so the two changes move it in opposite directions.

• Reducing \( V \) at fixed \( f \) reduces the flux. The machine is under-fluxed: torque capability falls as \( V^2 \), and it may stall, but nothing is damaged. This is the mechanism of Chapter 21's stator-voltage control.
• Reducing \( f \) at fixed \( V \) increases the flux, in inverse proportion. Halve the frequency and the flux doubles.

And the iron will not allow it. A machine is designed with its rated flux density close to the knee of the B–H curve — typically 1.5–1.7 T for silicon steel, where saturation begins around 1.8–2.0 T. Asking for twice the flux does not give twice the flux; it gives a modest increase in flux and an enormous increase in magnetising current, because the core has left the linear region entirely.

What actually happens in the first seconds:

1. Magnetising current rises by a factor of five to twenty.
2. Stator copper loss rises as the square of that current.
3. Core loss rises steeply as well, because the flux density is higher.
4. The drive either trips on overcurrent — the usual outcome — or, on a supply with no protection, the stator insulation fails thermally.

The one-line rule: voltage down is safe and weak; frequency down alone is unsafe. That asymmetry is the entire reason V/f control exists.

Starting torque rises, breakdown torque is unchanged, efficiency falls.

Starting torque — up. The peak of the curve sits at \( s_{max} = R_2'/\sqrt{R_{th}^2+(X_{th}+X_2')^2} \). Increasing \( R_2' \) increases \( s_{max} \), sliding the whole curve towards higher slip. Choose \( R_2' \) so that \( s_{max} = 1 \) and the machine develops its full breakdown torque at standstill — the largest starting torque it is capable of.

Breakdown torque — unchanged. \( R_2' \) does not appear in \( T_{max} \) at all. This is the single most useful fact about the curve: rotor resistance relocates the peak without changing its height.

Efficiency — down, and by a predictable amount. The air-gap power divides as

so the rotor circuit dissipates a fraction \( s \) of the air-gap power regardless of where that resistance physically sits. Running at 30% slip means throwing away 30% of the air-gap power as heat — some in the rotor bars, most in the external resistors.

Which is why this is a starting method and not a speed-control method. It is entirely reasonable to accept large losses for the few seconds of a start, and entirely unreasonable to accept them continuously. Wound-rotor speed control by resistance was used for decades on crane and mill drives precisely because nothing better existed; an inverter now does the same job at 97% efficiency.

And the modern echo of the idea: deep-bar and double-cage rotors achieve the same effect without slip rings, using the skin effect. At standstill the rotor frequency equals the supply frequency, current crowds into the top of the bar and the effective resistance is high; at running slip the rotor frequency is 1–2 Hz, the current fills the bar, and the resistance is low. One rotor, both characteristics.

The machine is under-fluxed at the starting frequency, so it cannot produce the torque the loaded conveyor needs, and the drive supplies more and more current trying.

The mechanism. Flux depends on \( E/f \), not \( V_1/f \), and \( E = V_1 - I_1(R_1+jX_1) \). At 3 Hz on a 400 V, 50 Hz machine, constant V/f gives 24 V line — and the \( I_1R_1 \) drop at starting current may be 20 V of it. Almost nothing is left to magnetise the machine.

Why it starts empty and not loaded: the empty conveyor needs little torque, so it draws little current, so the \( I_1R_1 \) drop is small and the flux is adequate. Loading it raises the current, which raises the drop, which reduces the flux — exactly when more torque is needed. The fault is self-aggravating.

The correct fix, in order:

1. Increase the boost towards the rated \( I_1R_1 \) drop — typically 3–6% of rated voltage for a small machine.
2. Better, enable automatic boost / IR compensation if the drive has it, so the offset tracks the actual current.
3. Extend the acceleration ramp. Much of the starting current goes into accelerating the inertia, not into the steady-state load.
4. Check the current limit setting — many drives ship with 110% and a conveyor start legitimately needs 150%.

Now the over-correction, which is the part people miss. Set the boost high enough for a loaded start and the machine is over-fluxed whenever it runs light at low speed:

• The iron saturates, and magnetising current rises far more than proportionally.
• Losses rise while output does not, so efficiency at low speed collapses.
• The motor gets hot at low speed — where its own shaft-mounted fan is barely working — and hums audibly.

The recognisable symptoms of the two faults are opposites: too little boost trips on overcurrent under load; too much boost runs hot and noisy on no load. If you see both, the answer is not a compromise value but automatic boost.

Because both the flux and the leakage reactance work against you, and each contributes a factor of \( f \).

Start from the high-speed approximation. Above base speed the reactances dominate the resistances, so

Now count the frequency dependence of each term with \( V \) held at its ceiling:

• \( V_{th}^2 \) — constant. The inverter cannot give more.
• \( \omega_s \) — proportional to \( f \).
• \( X_{th}+X_2' = 2\pi f(L_{th}+L_2') \) — also proportional to \( f \).

Two factors of \( f \) in the denominator, none in the numerator: \( T_{max}\propto 1/f^2 \).

The physical reading. Flux falls as \( 1/f \), which alone would give \( T \propto 1/f \). But the leakage reactance also rises with \( f \), which limits the current that a given voltage can drive into the rotor circuit — a second, independent \( 1/f \). The machine loses torque capability twice over.

The consequence for drive sizing. Constant power needs \( T \propto 1/f \). Since the ceiling falls faster, the two must cross:

Beyond it, the drive is limited by pull-out rather than by heating, and torque must be derated as \( 1/f^2 \). A machine with a 2:1 breakdown ratio gives a much narrower constant-power range than one with 3:1, which is a specification worth checking before promising a wide-range drive.

Contrast with the DC machine of Chapter 25, where the constant-power region is bounded by commutation and by the risk of residual-flux runaway, not by a pull-out torque. The two machines run out of field weakening for entirely different reasons.

The reference frame is being oriented — rotated so that one of its axes lies exactly along the rotor flux vector.

What that buys, in two steps:

1. The flux becomes one-dimensional. By construction \( \lambda_{qr}=0 \) and \( \lambda_{dr}=\lambda_r=L_mi_d \). The flux now depends on one current component and one only.
2. Torque becomes a product of two independent things.
   Result

   \[ T = \frac{3}{2}\frac{P}{2}\frac{L_m}{L_r}\lambda_r i_q \]
   which is \( T = k\Phi I_a \) with \( i_d \) as the field current and \( i_q \) as the armature current.

Now the PI question, which is the more interesting half. A PI controller drives steady-state error to zero only for a constant reference — its integrator has infinite gain at DC and finite gain everywhere else. Ask it to track a 50 Hz sinusoid and it will always lag in phase and fall short in amplitude, because at 50 Hz its gain is no longer infinite.

In the rotating frame there is no sinusoid. A balanced three-phase current at 50 Hz, viewed from a frame rotating at 50 Hz, is two constants. The PI controller sees a DC reference, applies infinite DC gain, and achieves exactly zero steady-state error. The same controller that could not track the current in stationary coordinates tracks it perfectly here.

Two practical points that follow:

• This is why FOC uses ordinary PI controllers rather than anything exotic. The transformation does the hard work; the controllers can be textbook.
• The alternative — staying in stationary coordinates and using resonant controllers tuned to the fundamental — exists and is used in grid converters, but it needs a new resonant term for every frequency of interest, whereas the rotating frame handles the fundamental once and for all.

The cost, stated honestly: everything depends on knowing \( \theta \). An angle error of \( \delta \) means \( i_q \) is not purely torque-producing, and the decoupling degrades roughly as \( \cos\delta \) for torque and \( \sin\delta \) for the unwanted flux disturbance. Ten degrees is tolerable; forty degrees is not control at all.

Because a hoist needs full torque at zero speed, and V/f control produces none.

The three requirements a hoist imposes, and how V/f fails each:

• Holding torque at standstill. At \( f=0 \) there is no rotating field, so no rotor EMF, so no rotor current, so no torque — regardless of applied voltage. The load would descend.
• Controlled torque on the way up and down. V/f commands frequency and lets the load choose the slip, so torque is an outcome. A hoist must limit torque to protect the rope and the gearbox.
• Four-quadrant operation. Lowering a load is generating: the motor is driven by the descending mass and returns power to the DC link. V/f drives can do this, but the transition through zero speed is exactly where V/f is weakest.

What to specify:

1. Closed-loop field-oriented control with an encoder. This is not optional for a hoist — sensorless vector loses orientation below 1–2 Hz, which is precisely the region a hoist lives in during load pick-up and set-down.
2. A regenerative front end or a braking resistor. Lowering a 30 kW load returns roughly 30 kW to the DC link, which will trip the drive on overvoltage in under a second otherwise. An active front end recovers it to the mains; a braking chopper and resistor dissipate it, which is cheaper and very common.
3. A mechanical brake, interlocked with the drive. Electronic holding torque disappears the instant power is lost. Safety standards require a mechanical brake, and the drive must be configured to apply torque before the brake releases and to hold torque until it is re-applied — the “brake sequencing” parameters that every hoist drive provides.
4. Torque proving. Before releasing the brake, the drive verifies it can produce the commanded torque. If it cannot — an open phase, a failed encoder — the brake stays on.

The general lesson for drive selection: ask what happens at zero speed and what happens when the load drives the motor. Those two questions separate the applications a V/f drive can serve from the ones it cannot, far more reliably than power rating or speed range.