🧲 Inductance, Ke, Kt & Pole-Pairs

Before field-oriented control can do anything clever, it needs to know the machine it is driving. That knowledge is not a vibe, it is a small set of measured constants: two inductances, one electrical coupling constant that wears two hats, the number of pole-pairs, and the shape of the cogging. Get them in the right order and characterized correctly, and the controller’s math just works. Get one wrong and the loop fights the motor instead of driving it. This lesson is about what those constants are, how you measure each one, and why two of them are secretly the same number.

By the end, you can

1. Explain what Ld and Lq are and how a current-step or LCR-meter test measures them
2. Measure the back-EMF constant Ke by spinning the motor open-circuit and reading line voltage versus speed
3. Show that in consistent SI units Kt = Ke, and use Kt to convert a torque demand into a coil current
4. Characterize pole-pairs and cogging, and relate electrical angle to mechanical angle
5. Calculate the coil current the robot-hand finger motor needs for a given grip force

Intuition first

Think of the motor as a translator that converts between two languages: volts-and-amps on one side, radians-and-newton-metres on the other. A good translator is consistent. If it turns a sentence one way, it turns the answer back the same way. The motor does exactly this. Drive current in and you get torque out at a fixed exchange rate. Spin the shaft by hand and you get voltage out at a fixed exchange rate. Those two exchange rates are not two unrelated facts you have to look up separately. They are the same coupling seen from opposite ends, which is why, in honest SI units, one number does both jobs.

The inductances are a different kind of fact. They describe how hard it is to change the current in the windings, not how much torque a steady current makes. Current cannot jump in an inductor, so the inductance sets how fast your controller can push current where it wants it. A motor with a salient (shaped) rotor is even more interesting: it presents a different inductance depending on which way the rotor magnets are pointed relative to the current you are pushing. That directional difference is the whole reason the d-q frame exists, and it is what makes some motors produce extra torque for free.

So three ideas, in order: the inductances govern how fast you can steer current, the coupling constant governs how that current becomes torque (and how speed becomes voltage), and the pole-pairs plus cogging tell you how the electrical world maps onto the physical angle of the shaft. Characterize them in that order and the controller in the next lesson falls right out.

Ld and Lq: the two inductances

A three-phase motor has winding inductance, but on a salient-rotor machine that inductance is not a single number. It depends on where the rotor sits. To make sense of this, we rotate into a frame that spins with the rotor and split the inductance along two axes:

• $L_d$, the direct-axis inductance, measured along the line through the rotor’s magnetic poles (where the permanent-magnet flux points).
• $L_q$, the quadrature-axis inductance, measured ninety electrical degrees away, in between the poles.

On a surface-mounted-magnet rotor the magnets sit on the outside and behave almost like air to the stator field, so $L_d \approx L_q$ and the motor is non-salient. On an interior permanent-magnet (IPM) rotor the magnets are buried in steel, and the steel between them offers a low-reluctance path for flux in the quadrature direction. The result is that

$$L_q \gt L_d \quad \text{(salient IPM rotor)}$$

This is not a rounding error. The ratio $L_q / L_d$ (the saliency ratio) can be 1.5 to 3 or more, and field-oriented control mines that difference for reluctance torque on top of the usual magnet torque. To do that, the controller has to know both numbers.

How do you get them? Two practical ways, both done with the rotor held still:

1. LCR meter. Lock the rotor so the d-axis lines up with one phase pair, measure the inductance across the terminals, then rotate the rotor ninety electrical degrees and measure again. The two readings give you $L_d$ and $L_q$ directly. Quick, but the small AC excitation of an LCR meter does not represent the deep saturation a motor sees at rated current.
2. Current-step test. Hold the rotor, apply a voltage step to a phase pair, and record how the current rises. Current in an inductor climbs as $i(t) = (V/R)\,(1 - e^{-tR/L})$, so the time constant $\tau = L/R$ of that rise hands you the inductance once you know the resistance from Lesson 38. Repeat aligned to the d-axis and to the q-axis. This one can be run at realistic current levels, so it captures saturation that the LCR meter misses.

Ke and Kt: two faces of one coupling

Now the star of the lesson. A spinning motor generates a voltage that opposes the supply, the back-EMF, and its size is proportional to speed. The constant of proportionality is the back-EMF constant $K_e$:

$$V_{\text{bemf}} = K_e\,\omega$$

where $\omega$ is the mechanical angular speed in radians per second and $V_{\text{bemf}}$ is in volts. You measure $K_e$ the most direct way imaginable: leave the terminals open, spin the shaft (a drill or a second motor works), and plot the line voltage you read against the speed. The slope is $K_e$, in volts per (radian per second), which is the same as $\text{V}\cdot\text{s/rad}$.

The torque constant $K_t$ lives on the other side of the translator. It says how much torque you get per amp of current:

$$\tau = K_t\,I \qquad \Rightarrow \qquad I = \frac{\tau}{K_t}$$

with $\tau$ in newton-metres and $I$ in amperes, so $K_t$ is in $\text{N}\cdot\text{m/A}$.

Here is the punchline. In a consistent SI unit system, those two constants are numerically equal:

$$K_t = K_e$$

That looks like a coincidence until you check the units. One $\text{N}\cdot\text{m/A}$ is one joule per ampere-second, which is one volt-second. So $\text{N}\cdot\text{m/A}$ and $\text{V}\cdot\text{s/rad}$ are the same physical dimension. The deeper reason is energy conservation: the electrical power the back-EMF removes, $V_{\text{bemf}}\,I = K_e\,\omega\,I$, must equal the mechanical power the torque delivers, $\tau\,\omega = K_t\,I\,\omega$. Cancel $\omega\,I$ from both sides and you are left with $K_e = K_t$. They were never two constants. They are one electromechanical coupling, read once as voltage-per-speed and once as torque-per-current.

Pole-pairs and cogging

The last two facts are geometric. A motor’s rotor has some number of magnetic poles, always even, and the pole-pair count $P$ is half of that. Pole-pairs set the gearing between the electrical angle the controller commands and the physical angle the shaft actually turns:

$$\theta_{\text{elec}} = P\,\theta_{\text{mech}} \qquad \omega_{\text{elec}} = P\,\omega_{\text{mech}}$$

A 14-pole hub motor has 7 pole-pairs, so the field rotates electrically seven times for every one mechanical revolution. Miss the pole-pair count and your controller commutates at the wrong rate, the motor stutters or runs backward, and the speed it reports is off by a factor of $P$. You count poles by slowly rotating the rotor through one mechanical turn while watching the back-EMF: the number of electrical cycles you see is the pole-pair count.

Cogging is the other geometric quirk. Even with no current flowing, you can feel a motor want to settle into preferred positions as you turn the shaft by hand, a faint notchiness. That is the permanent-magnet flux preferring to line up with the stator teeth, and it produces torque ripple that has nothing to do with the current you command. You characterize it by measuring torque versus angle at zero current. Good FOC can pre-compensate a known cogging profile, which matters enormously for a robot finger that must close smoothly onto something fragile.

See it / Try it

The constants only become real when you can convert between them. The widget below packs three tools, but for this lesson focus on the Ke ↔ Kt panel at the bottom. Pick which constant you know, type its value, and read the other one out. Because the coupling is one number wearing two hats, the box reports $K_e$ and $K_t$ as equal in SI, and also gives you $K_e$ in the datasheet-friendly $\text{V/krpm}$.

Start by typing a $K_e$ of 0.1, which means 0.1 volts of back-EMF per radian per second. The tool reports $K_t = 0.1\ \text{N}\cdot\text{m/A}$ (identical, as promised) and about $10.5\ \text{V/krpm}$. Now flip the known constant to $K_t$ and watch the equality hold in reverse. The middle panel, copper temperature correction, is your reminder from Lesson 38 that the winding resistance under all of this drifts with heat. The top panel relates wye and delta winding resistance, which is how you got the per-phase resistance you need for the current-step inductance test in the first place. One small calculator, the whole characterization chain.

?An LCR meter reads 1.8 mH along the rotor's pole axis and 3.6 mH ninety electrical degrees away. What kind of rotor is this, and which reading is Lq?

• A Surface-magnet rotor; the 1.8 mH reading is Lq
• B Salient interior-PM rotor; the 3.6 mH reading is Lq
• C Salient interior-PM rotor; the 1.8 mH reading is Lq
• D The readings are equal within tolerance; saliency cannot be determined

?A datasheet gives a motor's back-EMF constant as Ke = 0.05 V·s/rad. Without any further measurement, what is its torque constant Kt in SI units?

• A It cannot be known without a separate torque test
• B 0.05 N·m/A, because Kt equals Ke in consistent SI units
• C 0.05 divided by the pole-pair count
• D 0.05 multiplied by the winding resistance

Lab: characterize a motor end to end

On the bench, run the chain in order. First, resistance: measure line-to-line with a four-wire DMM and use the wye/delta panel to get per-phase $R$. Second, inductance: lock the rotor, apply a small voltage step on a phase pair, scope the current rise, and read $\tau = L/R$ at the d-axis alignment, then rotate ninety electrical degrees and repeat for $L_q$ (an LCR meter is the faster first pass, the current-step is the honest one). Third, Ke: spin the open-circuited motor with a drill at a few known speeds, read line voltage on a scope, and take the slope, then convert to clean SI and you have $K_t$ for free. Fourth, pole-pairs: hand-turn one mechanical revolution and count the back-EMF cycles. Fifth, cogging: with leads open, feel and, if you can, log the torque-versus-angle detents. Write every number down with its units. Those five rows are the entire motor model the next lesson’s controller will trust.

↘ Why back-EMF exists at all, and why it is the same coupling as torque

Back-EMF is not the inductor’s voltage. An inductor opposes a changing current by Faraday’s law, $v = L\,di/dt$, and it does so even when the shaft is locked. The motor’s back-EMF is different: it appears whenever the rotor moves, even at perfectly constant current, and it comes from the armature conductors sweeping through the magnetic field. By Lenz’s law the induced voltage opposes the applied voltage, and it grows in proportion to speed. As the motor spins up, the back-EMF rises until the net voltage across the winding resistance shrinks, the current falls, and the motor settles at the speed where back-EMF nearly balances the supply. That is why a stalled (locked-rotor) motor draws a frightening current: with no rotation there is no back-EMF, and only the tiny winding resistance limits the draw.

The equality $K_t = K_e$ is energy conservation made into algebra. The electrical power the back-EMF takes out of the circuit is $P_{\text{elec}} = V_{\text{bemf}}\,I = K_e\,\omega\,I$. The mechanical power the shaft delivers is $P_{\text{mech}} = \tau\,\omega = K_t\,I\,\omega$. A lossless coupling demands $P_{\text{elec}} = P_{\text{mech}}$:

$$K_e\,\omega\,I = K_t\,I\,\omega \quad \Longrightarrow \quad K_e = K_t$$

The $\omega$ and the $I$ cancel cleanly, leaving the constants equal regardless of speed or load. The real losses (resistance, iron, friction) live outside this ideal coupling and are bookkept separately, which is exactly why the equality survives them. The same algebra, run on Park’s d-q axes, gives the IPM torque equation $\tau = P\,[\,\lambda_m\,i_q + (L_d - L_q)\,i_d\,i_q\,]$, where the first term is the magnet torque set by $K_t$ and the second is the reluctance torque the saliency $L_d \ne L_q$ unlocks. That second term is the practical payoff of measuring both inductances rather than one.

Grounded in Wikipedia: “Counter-electromotive force”, “Motor constants”, “Brushless DC electric motor” (CC BY-SA).

Key takeaways

• ▸**Ld and Lq** are the inductances along and across the rotor poles; a salient interior-PM rotor has **Lq greater than Ld**, and FOC mines that gap for reluctance torque.
• ▸Measure inductance with an **LCR meter** (fast) or a **current-step / time-constant test** (honest at real current); both numbers sag as the iron saturates.
• ▸Get **Ke** by spinning the motor open-circuit and taking the slope of line voltage versus speed.
• ▸In **consistent SI units, Kt = Ke** because N·m/A and V·s/rad are the same dimension; it is energy conservation, not coincidence.
• ▸Pole-pairs set **θ_elec = P · θ_mech**; cogging is zero-current torque ripple. Both must be characterized for smooth, correct commutation.
• ▸The finger motor's **Kt converts grip-force torque into the coil current** the controller must command: I = τ / Kt.

Practice 1 warm-up

A robot-hand finger motor has $K_t = 0.020\ \text{N}\cdot\text{m/A}$. The gripper geometry turns motor torque into fingertip force at a known ratio, and a particular gentle grip needs $0.010\ \text{N}\cdot\text{m}$ of motor torque. What coil current must the controller command?

↘ Show worked solution

Use $I = \tau / K_t$. With $\tau = 0.010\ \text{N}\cdot\text{m}$ and $K_t = 0.020\ \text{N}\cdot\text{m/A}$:

$$I = \frac{0.010\ \text{N}\cdot\text{m}}{0.020\ \text{N}\cdot\text{m/A}} = 0.5\ \text{A}$$

The controller commands 0.5 A. Double the grip torque and you double the current, which is exactly why a current sensor is a perfectly good (indirect) grip-force sensor once you know $K_t$.

Practice 2 core

You spin a motor open-circuit and read these peak line voltages: 4.0 V at 100 rad/s and 8.0 V at 200 rad/s. Find $K_e$, then state $K_t$ in SI units and explain why no torque test was needed.

↘ Show worked solution

$K_e$ is the slope of voltage versus speed. From either point (the data is linear, as back-EMF should be):

$$K_e = \frac{V_{\text{bemf}}}{\omega} = \frac{4.0\ \text{V}}{100\ \text{rad/s}} = 0.040\ \text{V}\cdot\text{s/rad}$$

Checking the second point: $8.0 / 200 = 0.040$, consistent. In consistent SI units the torque constant equals the back-EMF constant, so

$$K_t = K_e = 0.040\ \text{N}\cdot\text{m/A}$$

No torque measurement was needed because $K_t$ and $K_e$ are the same electromechanical coupling: $\text{N}\cdot\text{m/A}$ and $\text{V}\cdot\text{s/rad}$ are identical dimensions, and energy conservation forces the two constants to be numerically equal.

Practice 3 stretch

A hub motor’s controller reports a mechanical speed that is exactly 7 times too high and the motor runs rough at startup. You suspect a pole-pair error. You hand-rotate the shaft through one full mechanical turn and count 7 complete back-EMF cycles. What is the true pole-pair count, how many poles is that, what should the controller have been told, and why does a wrong value cause exactly this symptom?

↘ Show worked solution

The number of electrical cycles per mechanical revolution is the pole-pair count, so $P = 7$ pole-pairs, which means $2P = 14$ magnetic poles. The relationship is $\omega_{\text{elec}} = P\,\omega_{\text{mech}}$, so the controller, working in electrical angle, must divide by $P = 7$ to report mechanical speed. If it was told $P = 1$, it never divides, and the reported mechanical speed comes out $7\times$ too high, exactly the observed factor. The same wrong $P$ also makes the controller advance the commanded field angle at the wrong rate relative to the real rotor, so the commutation slips and the motor stutters at startup. Fixing $P$ to 7 corrects both the reported speed and the rough running.