Motor Design

Kv, Kt and Ke: one motor constant in three units

Motor datasheets list a speed constant in rpm per volt, a back-EMF constant in volts per krpm, and a torque constant in mN·m or oz·in per amp. They look like three separate specifications. They are one number in different units, and the only real trap is which volts and which amps the number was measured in. This article shows why the constants are linked, how to convert them for brushed DC and BLDC motors, where the factors of √3, √2 and 1.5 come from, and how to use them to choose a winding for a robot joint. Two real datasheets are worked through along the way: the maxon RE 40 brushed DC motor and the ODrive D5065 270KV brushless motor.

Quick reference
  • In SI units Kt (N·m/A) and Ke (V·s/rad) are numerically equal, and Kv is their inverse (rad/s per volt). Reason: E·I = T·ω.
  • Brushed DC: Ke = Kt = 9.549 / Kv with Kv in rpm/V. Speed constant [rpm/V] × torque constant [mN·m/A] = 9,549.
  • BLDC / PMSM, sinusoidal (FOC) drive: line-to-line peak Ke = 9.549 / Kv; Kt per peak phase ampere = 8.27 / Kv; Kt per RMS phase ampere = 11.70 / Kv.
  • Mixing conventions costs 15 % (peak phase vs. DC-link amps) or 41 % (peak vs. RMS). Always record which convention a number uses.
  • Rewinding with N times the turns multiplies Ke and Kt by N, divides Kv by N and multiplies resistance by N². The motor constant Km = Kt/√R (for a three-phase motor, Kt,pk/√(1.5Rph)) does not change: Kv picks a voltage and current level, Km measures how good the motor is.

One constant, three names

A motor converts electrical power into mechanical power. On the electrical side that power is the back-EMF E times the current I. On the mechanical side it is torque T times angular speed ω. Resistive heating, friction and iron loss are handled separately, so the conversion itself loses nothing:

E · I = T · ω E = back-EMF (V)  •  I = current (A)  •  T = electromagnetic torque (N·m)  •  ω = mechanical speed (rad/s)

Two proportionalities are enough. Back-EMF grows linearly with speed, E = Keω, and torque grows linearly with current, T = KtI. Substituting both gives KeωI = KtIω, so the speed and current cancel:

Kt = Ke     Kv = 1 / Ke Kt in N·m/A and Ke in V·s/rad are the same number • Kv in rad/s per volt is the no-load speed per volt

Ke and Kt are one constant seen from the electrical side and from the mechanical side. If a datasheet shows two numbers that disagree after unit conversion, the cause is a unit or a convention, not physics. The hobby unit for Kv is rpm per volt, which is the SI value multiplied by 60/(2π) = 9.549.

ConstantSI unitDatasheet unitsConversion
Back-EMF constant KeV·s/radV per 1000 rpmKe [V/krpm] = 104.72 × Ke [V·s/rad]
Torque constant KtN·m/AmN·m/A, oz·in/A× 1,000 for mN·m/A  •  × 141.61 for oz·in/A
Speed constant Kvrad/s per Vrpm/VKv [rpm/V] = 9.549 / Ke [V·s/rad]
Product—rpm/V × mN·m/A= 30,000/π = 9,549.3
US servo shortcut—oz·in/A and V/krpmKt [oz·in/A] = 1.352 × Ke [V/krpm]

maxon calls these the speed constant kn (rpm/V), the torque constant kM (mN·m/A) and the generator constant kG (V per 1000 rpm), and notes that kn is essentially the inverse of kM and kG = 1/kn [1]. The 9,549 product above is not a vendor number: it is 60/(2π) × 1000, the unit conversion that makes the SI equality hold.

Brushed DC motor: the simple case

A permanent-magnet brushed motor has one winding and one current, so there is no convention to choose. The armature voltage equation and the torque equation are

V = Keω + R·I     T = Kt(I − I0) R = armature resistance • I0 = no-load current, which pays for friction and iron loss

Eliminate the current and the motor becomes a straight line from no-load speed to stall torque:

ω = (V − R·I0) / Ke − [R / (KeKt)] · T no-load speed ≈ V / Ke = Kv·V • stall torque = Kt(V/R − I0) • slope = R/Kt² = 1/Km²

This is where Kv gets its meaning: it is the no-load speed per volt, ignoring the small R·I0 drop. The slope depends on R and Kt, not on the supply voltage, so raising the voltage slides the whole line to the right without changing its steepness. The quantity Km = Kt/√R, the motor constant, sets that slope and is the number that says how much torque you get per square root of watt of copper loss.

A real datasheet: maxon RE 40 (part 148877)

The maxon RE 40 is a well-known 150 W graphite-brush motor of the kind used in robotics and laboratory equipment. Its datasheet [2] lists, at 48 V, a torque constant of 60.3 mN·m/A, a speed constant of 158 rpm/V, a terminal resistance of 1.13 Ω and a no-load current of 68.6 mA. Everything else in the table can be rebuilt from just those few numbers plus the supply voltage, which is a good way to check that you have understood them.

Datasheet entryPublishedComputedFrom
Speed constant158 rpm/V158.4 rpm/V9549 / 60.3
No-load speed7,590 rpm7,589 rpm(V − R·I0) / Ke
Stall current42.4 A42.5 AV / R
Stall torque2,560 mN·m2,557 mN·mKt(Istall − I0)
Speed/torque gradient2.97 rpm/mN·m2.97 rpm/mN·mR / (KeKt) = 1/Km²
Max. efficiency92 %92.1 %(1 − √(I0/Istall))²
Nominal current at 187 mN·m3.17 A3.17 AT/Kt + I0
Nominal speed at 187 mN·m7,000 rpm7,034 rpm(V − R·I) / Ke

All eight rows agree to the rounding of the datasheet. Only three motor-specific numbers are doing the work: Kt (= Ke), R and I0. The speed constant is not an independent specification; it is the torque constant read in other units, and the datasheet's 158 rpm/V and 60.3 mN·m/A multiply to 9,527, within 0.3 % of 9,549.

At the nominal point the motor draws 3.17 A and delivers 138 W from 152 W, an efficiency of 90.5 %. The best efficiency of 92 % occurs at only 1.71 A, or 99 mN·m, because the two losses are balanced when R·I² equals the no-load loss. A motor at its best efficiency is lightly loaded and fast, far from the 2.56 N·m stall point, which is why a small fast motor behind a gearbox often beats a large slow one.

025507510003,0006,0009,00000.10.20.30.40.50.60.187 N·m: 7,034 rpm, 90.5 %speed (left axis, rpm)efficiency (right axis, %)shaft torque, N·m
Speed and efficiency of the maxon RE 40 at 48 V, reconstructed from Kt = 60.3 mN·m/A, R = 1.13 Ω and I0 = 68.6 mA. The marker is the nominal 187 mN·m point. Stall (2.56 N·m) lies beyond the right edge.

Two caveats. First, the equality Kt = Ke describes a permanent-magnet motor; in a wound-field motor both constants scale with the field flux, so they change with field current. Second, brush voltage drop shifts the no-load speed slightly, so measure Ke by turning the motor as a generator on open circuit rather than from V/ω0.

BLDC and PMSM: the same balance, with conventions

A three-phase machine obeys the same power balance. The difference is that there are three voltages and three currents, each of which can be quoted as peak or RMS, per phase or line-to-line. Every factor of √3, √2 or 1.5 in this section comes from that bookkeeping, not from new physics.

Let ψ be the peak permanent-magnet flux linkage of one phase, p the number of pole pairs and ωm the mechanical speed. The phase back-EMF is ea = −pωmψ sin θe, the same expression used in the winding sequence article. Its peak is pψωm, and the flux linkage itself comes from the winding: ψ = kw·N·Φ1, the winding factor times the series turns times the fundamental flux per pole.

Datasheets and ESCs quote the voltage between two terminals. For a star-connected winding the line-to-line peak back-EMF is √3 times the phase value:

Ke,LL = √3 · pψ     Kv [rpm/V] = 9.549 / Ke,LL Ke,LL = line-to-line peak back-EMF per mechanical rad/s • Kv = rpm per volt of DC bus, assuming the inverter can reach full modulation

The torque constant follows from the balanced three-phase power at the terminals. With current in phase with the back-EMF (the q-axis, id = 0), the power is √3 × VLL,rms × Irms. In peak quantities that is (√3/2) × ELL,pk × Ipk. Setting it equal to Tω gives

Kt,pk = (√3 / 2) · Ke,LL = 1.5 · pψ torque per peak phase current, sinusoidal currents on the q-axis • the same result is T = (3/2)pψIq

Because the derivation uses terminal quantities, it holds for a delta-connected winding as well as a star. Measure at the terminals and the star/delta question disappears.

ConstantDefinitionFrom Kv [rpm/V]Example, Kv = 270
Ke, line-to-line peak√3 · pψ9.549 / Kv0.0354 V·s/rad
Ke, phase peakpψ5.513 / Kv0.0204 V·s/rad
Ke, line-to-line RMS√3 · pψ / √26.752 / Kv0.0250 V·s/rad
Ke, phase RMSpψ / √23.898 / Kv0.0144 V·s/rad
Kt per peak phase amp (FOC)1.5 · pψ8.270 / Kv0.0306 N·m/A
Kt per RMS phase amp1.5√2 · pψ11.70 / Kv0.0433 N·m/A
Kt per DC-link amp, six-step≈ Ke,LL (trapezoidal EMF)9.549 / Kv0.0354 N·m/A

The last row applies to ideal six-step drive of a motor with flat-topped (trapezoidal) back-EMF: only two phases conduct, the voltage across them is the flat line-to-line value, and ELL·Idc = Tω gives Kt = Ke,LL just as in the DC motor. A motor with sinusoidal back-EMF driven by ideal 120° current blocks gets (3/π) × Ke,LL = 0.955 × Ke,LL per ampere of block amplitude, and torque ripple at six times the electrical frequency.

The ratio between the two most common quotes is 9.549/8.270 = 2/√3 = 1.155. A torque estimate made with 9.549/Kv for a controller that regulates peak phase current is therefore 15 % too high; one made with peak where RMS was meant is off by √2. FOC firmware that regulates iq in amplitude-invariant form uses peak phase current, and a multimeter in AC mode reads RMS, so check the convention in your controller's documentation before trusting a number.

A real datasheet: ODrive D5065 270KV

ODrive's motor page [3] lists the D5065 270KV, a 14-pole (7 pole-pair) outrunner popular in hobby and research robot actuators. Its entries are: Kv 270 rpm/V, torque constant 0.031 N·m/A, phase resistance 39 mΩ and phase inductance 16 µH (both phase-to-neutral), 45 A continuous in free air and 85 A for 3 seconds. The footnotes matter as much as the numbers: "All voltage units are line-line amplitude" and "All current units are phase amplitude." That is exactly the line-to-line peak Ke and peak-phase-current Kt of the table above, so the published torque constant should equal 8.27/Kv. The page lists two sibling motors in the same format, which gives three checks:

MotorKv (rpm/V)Published Kt (N·m/A)8.27 / KvDifference9.549 / KvDifference
D5312s 330KV3300.0250.0251+0.2 %0.0289+16 %
D5065 270KV2700.0310.0306-1.2 %0.0354+14 %
D6374 150KV1500.0550.0551+0.2 %0.0637+16 %

The 8.27/Kv column matches all three published values to within their two-figure rounding. The 9.549/Kv column, the DC-motor rule applied to a brushless motor, overstates torque per amp by 14–16 %. This is the 2/√3 factor from the previous section appearing in a real datasheet.

Starting from just Kv = 270, p = 7, Rph = 39 mΩ and the 45 A rating, you can derive the rest:

The page itself cautions that all performance figures are approximate and depend on cooling, so treat these as first estimates.

Measuring Ke on the bench

The back-EMF test from the winding sequence article also gives Kv. You do not need a tachometer, because the electrical frequency on the scope already encodes the speed: fe = p·n/60.

  1. Disconnect the drive and turn the shaft at a steady speed, for example with a drill.
  2. Read the electrical frequency fe from the scope and compute n = 60·fe/p, then ωm = 2πn/60.
  3. Read the line-to-line peak voltage between two terminals. A true-RMS meter gives the RMS value, which is the peak divided by √2 for a sine wave.
  4. Ke,LL = Vpk/ωm, then Kv = 9.549/Ke,LL and Kt,pk = 0.866 × Ke,LL.
Worked example — what a D5065 270KV (p = 7) should show on the scope

If the motor is turned at 1,000 rpm, the scope should show fe = 7 × 1,000 / 60 = 116.7 Hz, with ωm = 104.72 rad/s. The line-to-line peak should then be about 3.70 V (a true-RMS meter would read 2.62 V). Working back from such a reading:

Ke,LL = 3.70 / 104.72 = 0.0354 V·s/rad
Kv = 9.549 / 0.0354 = 270 rpm/V
Kt,pk = 0.866 × 0.0354 = 0.0306 N·m/A
ψ = 0.0354 / (√3 × 7) = 2.92 mWb

This measures Ke cold and at open circuit. The torque constant under load is usually a little lower, for the reasons in the last section.

Kv is a winding choice, Km is the motor

A common belief is that a high-Kv motor is the more powerful one. It is not. Put N turns in series per phase and everything scales with N: the flux linkage ψ, and therefore Ke and Kt, grow in proportion to N; the wire cross-section shrinks as 1/N for the same slot, so resistance grows as N². Kv falls as 1/N and resistance rises as 1/Kv². The combination that does not move is the motor constant,

Km = Kt,pk / √(1.5 Rph)     PCu = (T / Km)² Rph = phase resistance (half the terminal-to-terminal value for a star) • Km in N·m/√W • for a DC motor Km = Kt/√R

Kv therefore says nothing about how good a motor is. The three ODrive motors above show it:

MotorKvKt (N·m/A)Rph (mΩ)Km (N·m/√W)Torque at continuous currentCopper loss there
D5312s 330KV3300.025370.1060.75 N·m at 30 A50 W
D5065 270KV2700.031390.1281.40 N·m at 45 A118 W
D6374 150KV1500.055390.2272.75 N·m at 50 A146 W

The highest-Kv motor has the lowest Km; the lowest-Kv motor has the highest and is the larger machine. (These are three different motors, not rewinds of one, so their Km values differ. Within one motor, Km stays fixed as the turns change.) For the D5065, with Kt = 8.27/270 = 0.0306 N·m/A and Rph = 39 mΩ (terminal-to-terminal 78 mΩ), Km = 0.127 N·m/√W. The copper loss for any torque is the same whichever winding it carries. What changes is how that torque is paid for: with a low Kv you pay in volts, with a high Kv you pay in amps.

Worked example — three hypothetical windings of a D5065-sized motor for 0.8 N·m at 5,000 rpm on a 24 V bus

The bus voltage must cover the back-EMF plus the resistive drop. With id = 0 and space-vector modulation (phase voltage limit Vdc/√3), a first estimate is

Vdc ≥ Ke,LL·ωm + √3·Rph·Ipk,   Ipk = T / Kt,pk neglects the ωL·I term, so it is a lower bound
WindingKe,LL (V·s/rad)Kt,pk (N·m/A)Rph (mΩ)Ipk (A)Vdc neededPCu (W)
200 Kv0.04770.041371.119.3527.4 V (114 %)39.9
270 Kv0.03540.030639.026.1220.3 V (85 %)39.9
360 Kv0.02650.023021.934.8215.2 V (63 %)39.9

All three windings dissipate 39.9 W, because Km is the same. The 200 Kv winding cannot do the job: it would need 27.4 V, 114 % of the bus. The 360 Kv winding has plenty of voltage (63 % used) but needs 34.8 A peak, which stresses the inverter and wiring. The stock 270 Kv winding uses 85 % of the bus and 26.1 A, a workable fit with about 15 % in hand. As a rule of thumb I would keep 15–25 % of the bus in reserve for battery sag, the inductive drop and inverter dead time, so a higher bus voltage would make this design more comfortable.

01,5003,0004,5006,0007,5009,00000.20.40.60.811.21.4200 Kv270 Kv360 Kvrequired: 0.8 N·m at 5,000 rpmtorque, N·mspeed, rpm
Torque–speed lines for the three windings at a 24 V bus. Changing the turns only shifts the line up or down, because the slope, 595 rpm per N·m, is the same for every winding. The model ignores the ωL·I term and the controller's current limit.

The same logic applies to a gearbox. A ratio N multiplies the torque at the joint and divides the speed, so a high-Kv motor behind a larger ratio can deliver the same joint torque and speed as a low-Kv motor behind a smaller one. Choosing the gear ratio and Kv together is how you move the current/voltage balance to what your battery and controller handle best. Gearbox efficiency is separate and is not included in Kt.

Why a measured Kt comes out lower than the datasheet

Checklist before you trust a number

Frequently asked questions

Is the torque constant Kt equal to the back-EMF constant Ke?

In SI units, yes for a brushed DC motor: Kt in N·m/A equals Ke in V·s/rad because electrical power E·I equals mechanical power T·ω. For a three-phase brushless motor the equality holds only when volts and amps are counted consistently. With line-to-line peak Ke and peak phase current, Kt = (√3/2) × Ke, about 0.866 × Ke.

How do I convert Kv to Kt?

With Kv in rpm/V, the line-to-line peak back-EMF constant is 9.549/Kv in V·s/rad. The torque constant is 9.549/Kv N·m/A for a brushed DC motor, 8.27/Kv N·m per peak phase ampere for a sinusoidally driven BLDC or PMSM, and 11.70/Kv N·m per RMS phase ampere.

What does Kv mean for a BLDC motor?

Kv is the no-load speed in rpm per volt of DC bus voltage. It is the inverse of the line-to-line peak back-EMF constant, so a 270 Kv motor turned at 1,000 rpm shows a line-to-line peak of about 3.70 V.

Does a higher Kv motor have more power?

No. Kv is set by the number of turns in the winding. Rewinding with fewer turns raises Kv and lowers Kt and resistance, but leaves the motor constant Km unchanged, so the torque available for a given copper loss does not change. A higher Kv motor reaches a higher speed per volt and needs more current for the same torque.

Why does my measured Kt not match the datasheet?

Datasheet Kt is the cold, low-current, ideal electromagnetic value. Friction and drag losses (which a shaft torque sensor includes), magnet temperature, core saturation at high current, back-EMF harmonics and the current definition (peak, RMS or DC link) can each move it by several percent.

Model a DC motor from its constants

Enter Kt (= Ke), resistance and supply voltage to see the torque–speed line, current and efficiency. The PMSM tool does the same for a three-phase machine. Both are work in progress.
Open DC Motor Tool →

References

  1. maxon group, Motor constants (support article defining speed, torque and generator constants).
  2. maxon group, RE 40 Ø40 mm, graphite brushes, 150 watt, part 148877. Values quoted at 48 V nominal voltage, read 10 October 2026.
  3. ODrive Robotics, ODrive Motors (D5065 270KV, D6374 150KV and D5312s 330KV tables and footnotes), read 10 October 2026.
  4. D. C. Hanselman, Brushless Permanent-Magnet Motor Design, 2nd ed., The Writing Business, 2003.
← Back to all articles