Winding sequence, rotation direction and rotor angle: getting the three to agree

A PMSM under field-oriented control only produces smooth torque when three things agree: the physical order of the phase windings, the time order of the phase currents, and the direction in which the measured rotor angle counts. Get one of them backwards and the motor locks, growls or runs away. This article explains the rule that links them and how to recover the right mapping from an unlabelled motor on the bench.

The article assumes a three-phase PMSM under field-oriented control with an absolute position encoder on the shaft.

Fix the convention first

Direction words mean nothing until the viewpoint is fixed. Throughout this article the motor is viewed from the shaft (drive) end, and counter-clockwise (CCW) is the positive direction for angle, speed and torque. This is the mathematical convention used by most FOC derivations. Viewed from the opposite end, every "CCW" below becomes "CW", so always write the viewing end next to any direction you record.

Two angles are used. The mechanical angle θ_m is what the shaft does. The electrical angle is what the controller needs:

What "winding sequence" means

Each phase winding has a magnetic axis: the direction of the field it produces when it carries positive current. In a balanced three-phase winding the three axes sit 120° electrical apart. The winding sequence is the order in which you meet those axes as you walk around the air gap. If you meet A, then B, then C while walking CCW, the winding sequence is CCW. If you meet them in that order walking CW, it is CW.

The current sequence is the same idea in time. Currents are in A-B-C (positive) sequence when A peaks first, B peaks a third of a period later, and C a third of a period after that:

The rule: the field travels from the leading phase to the lagging phase

Put those currents into a winding whose axes are at 0°, +120° and +240° electrical and add the three pulsating fields. The backward-travelling parts cancel and what is left is a single wave of constant amplitude:

The peak of the field is on the A axis when i_a peaks, on the B axis when i_b peaks, and on the C axis when i_c peaks. So the field always moves from the axis of the phase that peaks first towards the axis of the phase that peaks next. That one sentence answers the direction question for every case:

So a stator wound A-B-C clockwise and fed A-B-C currents produces a clockwise rotating field. A synchronous rotor locks to that field, so it also turns clockwise, at a mechanical speed of ω/p. In the CCW-positive convention that is negative speed.

Which way must the rotor angle count?

FOC does not generate a rotating field on its own schedule. It reads the rotor angle and places the current vector 90° electrical ahead of the rotor d-axis (for pure q-axis current). The Park transform that does this assumes the same geometry as the figure above: θ_e is measured from the A axis, positive towards the B axis.

That gives the requirement directly. The measured electrical angle must increase when the rotor turns in the direction in which A-B-C currents rotate the field. Three conditions have to hold together:

• Direction. Turning the rotor from the A axis towards the B axis makes θ_e go up, not down.
• Scale. One mechanical revolution produces exactly p electrical revolutions.
• Zero. θ_e reads 0 when the rotor d-axis is aligned with the phase A axis.

With an absolute encoder these three conditions reduce to three stored numbers: the pole-pair count p, a direction sign s (+1 or −1) and the zero reading N₀.

Because the encoder is absolute, s and N₀ are found once and remain valid at every power-up. No alignment move is needed at start-up, but the values are only valid for that lead order and that encoder mounting. Change either and they must be found again.

An equivalent test that needs no knowledge of the winding: spin the motor by hand in the direction that makes the angle increase. The back-EMFs must then appear in A-B-C order. Back-EMF sequence and angle direction are two views of the same fact.

When all three hold, positive i_q gives positive torque, the rotor accelerates in the direction of increasing angle, and the back-EMF appears on the positive q-axis.

What happens when something is swapped

The controller sends three sinusoidal currents with a common phase angle γ:

Suppose leads B and C are swapped. The winding on the B axis now carries i_c and the winding on the C axis carries i_b. Add the three contributions along their real axes and the result is a vector of the same magnitude at angle −φ: the mirror image of the intended vector about the A axis. Swapping a different pair mirrors about the axis of whichever phase was left alone.

Under FOC with pure q-axis current the controller sets γ = θ_e + 180°, which makes i_a = −I sin(θ_e), in phase with the phase A back-EMF. The current vector is then at φ = θ_e + 90°, always a quarter turn ahead of the rotor. After the mirror the stator vector sits at −θ_e − 90°, so its angle measured from the rotor d-axis is no longer a constant 90°:

The rotor is pulled to the nearest stable point and held there. Reversing the encoder direction has the same effect, because it replaces θ_e by −θ_e in the command and again leaves a −2θ_e term. Doing both restores a constant δ, which is why two errors cancel.

This is the classic symptom: the motor jerks a few degrees, locks, and sits there drawing whatever current the loop commands. A speed loop makes it worse because its integrator winds up to the current limit.

Of the six ways to connect three leads to three terminals, three keep the positive sequence (A-B-C, B-C-A, C-A-B) and differ only by a 120° angle shift. The other three (A-C-B, C-B-A, B-A-C) reverse it. A fresh offset calibration absorbs the first kind. Only a swap or a sign change fixes the second.

Why looking at the stator is not enough

On a distributed winding you can often read the sequence off the sample. Walking around the bore, the phase belts appear in the order A+, C−, B+, A−, C+, B−. The direction in which you read that order is the direction of the A-B-C sequence.

On a fractional-slot concentrated winding the same reasoning can give the wrong answer. What matters is the electrical position of each tooth, which is p times its mechanical position. Take three teeth A, B, C placed CCW at 0°, 120° and 240° mechanical:

The same thing happens between 12-slot/10-pole and 12-slot/14-pole machines. They can share an identical stator, but the 10-pole rotor rides the 5th MMF harmonic and the 14-pole rotor rides the 7th, and those two harmonics travel in opposite directions. Same winding, same currents, opposite rotation. This is why a measurement is more trustworthy than an inspection.

Finding the right mapping on an unlabelled motor

Either method below gives the terminal labels, the angle direction and the angle offset together. Mark the shaft so its direction is easy to see, and note the viewing end.

Method 1: back-EMF, no drive needed

1. Disconnect the motor from the drive. Connect three equal resistors (a few kΩ) from the terminals to a common point to create an artificial neutral, and put a scope channel on each terminal with respect to that point.
2. Spin the shaft steadily in the direction you want to call positive (CCW from the shaft end here). A hand drill works well.
3. Pick any terminal and call it A. The terminal whose voltage peaks 120° later is B. The one that peaks 240° later is C. These labels are now correct for that direction by construction.
4. Watch the position sensor during the same spin. The count must increase. If it decreases, set the direction sign s to −1.
5. Check the scale: the number of electrical cycles per mechanical turn is the pole-pair count p.
6. Find the zero. With the rotor turning in the positive direction, the phase A back-EMF is e_a = −ωψ sin(θ_e). So θ_e = 0 at the falling zero crossing of e_a, and e_a peaks at θ_e = −90°. Capture the encoder reading at that crossing: it is N₀.

If you can only measure line-to-line, the sequence test is unchanged (v_ab, then v_bc, then v_ca), and v_ab leads e_a by 30°.

Method 2: DC alignment steps

A DC current vector pulls the rotor d-axis onto itself. Stepping the vector in a known electrical direction and watching the shaft reveals everything. Use a current-limited supply or the drive itself at roughly 10 to 30% of rated current, with the shaft unloaded.

1. Label the terminals A, B, C arbitrarily and apply step 1. The rotor snaps to the A axis. Record the encoder reading: this is N₀.
2. Apply steps 2 to 6 in order. Each one should move the shaft by 60°/p mechanical in the same direction.
3. If the shaft steps CCW, your labels are a CCW-positive A-B-C. If it steps CW, swap the labels on any two terminals and repeat from step 1.
4. The encoder reading must increase at every step. If it decreases, set the direction sign s to −1.
5. Count the steps in one full mechanical turn. The count equals 6p, which confirms the pole-pair number.

Friction and cogging torque stop the rotor slightly short of the true axis. For a better offset, approach the zero position once from each side and average, and repeat over several pole pairs. The automatic commissioning routine in most FOC drives is this same procedure with a slowly rotating vector: it compares the commanded electrical angle with p times the encoder angle, and a slope of +1 means sequence and direction agree while −1 means one of them is reversed.

Worked example: 12-slot/10-pole motor, 12-bit absolute encoder

The numbers below are illustrative. The motor has p = 5 and the encoder gives 4096 counts per revolution, so one electrical revolution is 4096 / 5 = 819.2 counts and one 60° step is 136.5 counts, or 12° of shaft rotation.

First pass. The three leads are unmarked. Call them X, Y, Z and try A = X, B = Y, C = Z. Run the six alignment steps of Method 2:

The shaft moves 12° per step, but clockwise from the shaft end, and the count falls. So this labelling is a CW-positive A-B-C, which is not the convention we chose. Note that count and shaft agree with each other: the count falls when the shaft turns CW, so it will rise when the shaft turns CCW. The encoder direction is already right.

Second pass. Swap two labels: A = X, B = Z, C = Y. Repeat:

Now the shaft steps CCW and the count rises by about 136.5 per step. The last reading wraps past 4095: 3310 + 819.2 = 4129.2, which the encoder reports as 33. That is the A axis of the next pole pair. Continuing all the way round takes 30 steps, confirming 6p = 30 and p = 5.

One count is 0.44° electrical on this motor, so the encoder resolution is not the limit. Cogging is: repeat the 0° step on a few pole pairs, approaching from both sides, and average the implied N₀. Finish with a small positive i_q command and confirm that the shaft turns CCW with the count rising.

Commissioning checklist

• Viewing end and positive direction are written down.
• Rotor spun in the positive direction gives back-EMF in A-B-C order.
• Measured angle increases in the same direction, with p electrical turns per mechanical turn.
• Angle reads zero with the rotor aligned to phase A.
• Each current sensor is mapped to the phase it actually measures, with the right polarity.
• A small positive i_q turns the shaft in the positive direction, and i_d stays near zero as speed rises.