Stepper, BLDC and Reluctance Motors

The variable-reluctance motor, from the phase-and-tooth formula:

The same answer from the pole geometry, which is the more physical route. Each excitation aligns the nearest rotor tooth with the newly energised stator pole, and the rotor therefore moves by the difference between the stator and rotor pole pitches:

This is the vernier action that gives every stepper its fine step: two coarse pitches, 60° and 45°, differencing to a fine one of 15°. If \(N_s = N_r\) the step angle would be zero and the motor could not index at all.

The hybrid motor:

The standard industrial stepper. Its 50 rotor teeth are split between two axially displaced cups, offset by half a tooth pitch, with a permanent magnet between them — the arrangement that makes 200 steps possible from a machine with only two windings.

Resolution in the three drive modes:

A warning about the last two rows. Micro-stepping divides the commanded position by 16, but it does not divide the error. A typical hybrid stepper is accurate to about ±5 % of one full step — roughly ±5.4 arc-minutes — and micro-stepping cannot improve on that, because the error comes from tooth-pitch imperfections and from detent torque, not from the drive. Micro-stepping buys smoothness and quiet running; it does not buy accuracy.

Why the variable-reluctance motor is coarse is simply that it has six rotor teeth against fifty. It is cheap, has no magnet, and has no detent torque when de-energised — sometimes an advantage, sometimes not. The hybrid motor's magnet gives it detent torque, a much larger holding torque per unit volume, and the fine step that made it the standard.

Derive the speed relation rather than quoting it. Each pulse advances the shaft by \(\beta\) degrees, so \(f\) pulses per second advance it by \(\beta f\) degrees per second:

At 1000 pulses per second:

Ten thousand steps a second is at the top of what a hybrid stepper will do; the winding inductance limits how fast the current can be established in each phase, and torque falls away steeply beyond a few kilohertz.

Pulse counts are pure division:

Both are whole numbers, which is not an accident of the arithmetic but a requirement of the machine: a stepper can only stop at a multiple of \(\beta\). A commanded move of 40° would be 22.2 steps — unreachable, and the controller would round it to 22 steps (39.6°) and carry the 0.4° forward.

Time for the long move, ignoring the acceleration and deceleration ramps:

The variable-reluctance motor at the same pulse rate turns more than eight times faster, because each of its steps is more than eight times larger:

The last VR entry is fantasy — no small motor reaches 25 000 rev/min — and that is the point. Fine resolution and high speed pull in opposite directions, and the pulse rate is the same for both.

What is exact here and what is not. The position is exact: 5100 pulses is 25.5 revolutions to the last arc-minute, provided not one step is lost. The speed and the time are exact only while the motor keeps up. Command a rate the motor cannot follow, or accelerate too fast, and steps are missed silently — there is no feedback to notice. That single vulnerability is why steppers are used for light, predictable loads and abandoned in favour of servos everywhere else.

Prove the equality from energy conservation. The electrical power converted is the product of the back e.m.f. and the current; the mechanical power produced is torque times angular speed. In an ideal conversion these are the same power:

Check the units: \(1\ \text{V·s/rad} = 1\ \text{W·s/(A·rad)} = 1\ \text{J/(A·rad)} = 1\ \text{N·m/A}\). The same physical quantity, read in two directions.

In the units data sheets prefer:

Loaded to 1.2 N·m, the current follows first because torque is set by current alone:

Power and efficiency:

The two end points of the characteristic:

A straight line from no load to stall, exactly as for a separately excited d.c. machine — which is what a BLDC is, with the commutator's job given to the transistors.

Comment on the stall current. 120 A is six times the working current and would dissipate \(120^2\times0.4 = 5.76\) kW in a winding designed to lose 160 W. Nothing in the machine prevents it: at standstill there is no back e.m.f., and only the 0.4 Ω stands between the battery and the winding. Every BLDC drive therefore carries a current limit in the controller, and it is that limit, not the motor, that sets the usable stall torque. The 7.2 N·m in the table is a mathematical extrapolation and not an operating point.

Convert the inductance slope to radians, because the torque formula is in SI units and a degree is not:

Equivalently 3.467 mH per degree. Forgetting this conversion is the single commonest error in switched-reluctance problems and gives an answer 57 times too large.

The instantaneous torque while the phase conducts:

Note what is not in this expression: no flux linkage with a rotor winding, no slip, no excitation. The rotor is a lump of shaped steel, and the torque exists only because moving it changes the inductance the stator sees.

The stroke geometry. An 8/6 machine has \(m = 8/2 = 4\) phases and \(N_r = 6\) rotor poles, so

The stroke angle equals the inductance-rise interval, so as one phase reaches alignment the next begins its rise. The torque is therefore continuous and the average equals the instantaneous value, 9.93 N·m.

Shaft power at 1500 rev/min:

Check it by energy, which is an entirely independent route. The work done in one stroke is the integral of the torque through the stroke:

Two consequences of the \(i^2\). First, the torque does not depend on the direction of the current, so each phase needs only a unidirectional switch — an asymmetric half-bridge of two transistors and two diodes, not a full bridge. Second, torque is proportional to the square of the current, exactly as in the series motor of Set 45, which gives the same generous overload torque and the same requirement for current limiting.

Where the model is optimistic. Holding the current perfectly constant over the whole 15° requires a supply voltage large enough to force it in against a rising inductance, and to extract it again before the inductance starts to fall. Neither is possible at high speed, so the real torque falls short of 9.93 N·m as the speed rises, and the current waveform becomes a rounded pulse rather than a rectangle. The figure computed here is the low-speed, current-controlled limit.

Start from the two-reaction equations of Set 37 with the excitation removed. There is no field winding, so \(E_f = 0\) and the terminal voltage is supported entirely by the two reactance drops. Taking the rotor direct axis as reference and \(\delta\) as the angle of \(\mathbf V\) from the quadrature axis:

The power is the sum of the two axis products, and the saliency is what makes it non-zero:

If \(X_d = X_q\) the bracket vanishes and the machine makes no torque at all. Saliency is not a refinement here; it is the entire mechanism.

Substitute the numbers:

Working one load angle in full, at \(\delta = 20^\circ\):

The four operating points:

The torque peaks at 45° but the power factor peaks at 30°. The two optima do not coincide, so a reluctance drive is normally worked short of maximum torque.

The best power factor, in closed form. Maximising \(\cos\phi\) over \(\delta\) gives the classical result

Exactly the table's entry at 30°. It depends only on the saliency ratio \(X_d/X_q = 3\) — not on the voltage, the frequency or the size of the machine.

Why the saliency ratio is everything. A ratio of 3 caps the power factor at 0.50 and the machine needs 43 A to deliver 13.3 kW where a comparable induction motor would need about 25 A. Modern axially-laminated and flux-barrier rotors reach ratios of 8 to 10, giving \(\cos\phi_{max}\) of 0.78 to 0.82, and it is that improvement alone — not any change in the underlying theory, which is Set 37's — that has brought the synchronous reluctance motor back into commercial use as a rare-earth-free alternative to the permanent-magnet machine.

Total inertia first, the coupling being rigid:

The restoring torque near a detent is a spring. Displace the rotor by \(\theta\) mechanical radians and the electrical displacement is \(N_r\theta\), so

Why that number matters. A stepper is almost undamped. Drive it at a stepping rate near 87 steps per second and each pulse arrives in phase with the rotor's own oscillation; the swings build up, the rotor overshoots by more than a step, and the motor stalls or runs backwards. Every stepper data sheet shows this low-frequency resonance dip, and every controller either accelerates quickly through it, adds mechanical damping, or uses micro-stepping to spread the impulse.

The start–stop rate. To start without losing a step, the motor must complete one full step within one pulse interval, starting from rest. Take the available accelerating torque as the holding torque less friction:

This is an estimate and should be quoted as one. It assumes the full holding torque is available throughout the step, whereas the true torque follows \(T_h\sin(N_r\theta)\) and averages about \(2/\pi\) of the peak; it also ignores the winding's electrical time constant. Both make the real start–stop rate lower — typically 200 to 250 steps per second for a motor of this size — so the figure is an upper bound.

Above the start–stop rate the motor must be ramped. With acceleration and deceleration profiles a hybrid stepper of this size will slew at several thousand steps per second, but it can neither start nor stop there. The 282 pulses/s figure is the boundary between "command it and go" and "ramp it".

The index move, at 2000 pulses/s — well above the start–stop rate, so a ramp is assumed and the figures below are the flat-out part of the move:

The leadscrew. One revolution advances the nut by the pitch, so

Half-stepping halves the resolution figure and halves the feed rate at the same pulse rate — and, as Problem 1 noted, does nothing whatever for the accuracy, which is limited by the leadscrew's own pitch error and backlash long before it is limited by the motor.