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Motor

Servo Motor Torque & Speed

Calculate servo motor torque, speed, efficiency, and back-EMF from electrical and load parameters.

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Formula

Kt=Ke=Vωnl,T=KtI,ω=V−IRaKe=ωnl(1−TTstall),Tstall=KtVRaK_t = K_e = \frac{V}{\omega_{nl}},\quad T = K_t I,\quad \omega = \frac{V - I R_a}{K_e} = \omega_{nl}\left(1 - \frac{T}{T_{stall}}\right),\quad T_{stall} = \frac{K_t V}{R_a}

Reference: Hughes & Drury, Electric Motors and Drives, ch. 3 (d.c. motors); maxon, Key information on maxon DC motors

T— Torque at the operating current (N·m)
K_t— Torque constant, equal to the back-EMF constant K_e in SI units (N·m/A)
V— Supply voltage (V)
I— Operating current (A)
R_a— Winding resistance (Ω)
ω_nl— No-load angular speed (= 2π × RPM / 60) (rad/s)
ω— Angular speed at the operating point (rad/s)
T_stall— Stall torque at this supply voltage (N·m)

How It Works

This calculator finds the operating point of a brushed permanent-magnet DC servo motor on the linear motor model. From the supply voltage VV, the datasheet no-load speed nnln_{nl} and the winding resistance RaR_a it derives the torque constant and the stall torque, then solves two operating points: one at the operating current II you measure (torque, speed, back-EMF, input and output power, efficiency) and one at the load torque TLT_L your mechanism demands (the current, speed and output power the motor settles at). The same model describes the small DC motor inside a hobby RC servo and a brushed industrial servo motor. The servo's position loop, gearbox and PWM drive are outside it, so enter values at the motor shaft.

The supply voltage divides between the winding resistance and the back-EMF, V=IRa+KeωV = I R_a + K_e \omega, and the torque is proportional to the current, T=KtIT = K_t I. In SI units the torque constant in N·m/A equals the back-EMF constant in V·s/rad. With the no-load current neglected, the back-EMF equals the supply voltage at no load, so Kt=Ke=V/ωnlK_t = K_e = V/\omega_{nl}, where ωnl=2πnnl/60\omega_{nl} = 2\pi n_{nl}/60. At current II the speed is ω=(V−IRa)/Ke\omega = (V - I R_a)/K_e and the output power is the back-EMF times the current, Pout=(V−IRa)I=TωP_{out} = (V - I R_a) I = T\omega; the rest of Pin=VIP_{in} = VI is the copper loss I2RaI^2 R_a. The efficiency, 1−IRa/V1 - I R_a/V, is therefore the operating speed as a fraction of the no-load speed. At zero speed there is no back-EMF, the current is V/RaV/R_a and the torque is the stall torque Tstall=KtV/RaT_{stall} = K_t V/R_a. Every operating point lies on the straight torque-speed line ω=ωnl(1−T/Tstall)\omega = \omega_{nl}(1 - T/T_{stall}), so under a load torque TLT_L the motor draws IL=TL/KtI_L = T_L/K_t and runs at nL=nnl(1−TL/Tstall)n_L = n_{nl}(1 - T_L/T_{stall}). A load torque above the stall torque is reported as a stall: zero speed at the stall current.

The model's main simplification is the no-load current I0I_0, the current the motor draws with nothing on its shaft to supply its friction, brush and iron losses. The calculator neglects it, so the torque it reports is the electromagnetic torque and the shaft delivers about Kt(I−I0)K_t (I - I_0). Efficiency is overstated most at light load, where I0I_0 is a large share of II, and Ke=V/ωnlK_e = V/\omega_{nl} comes out slightly high, because some of the supply voltage drops across RaR_a even at no load. For a motor whose no-load current is a few percent of its operating current, these errors are a few percent. The winding resistance rises about 0.39% per °C, so a hot motor has less stall torque and loses more speed under load than a cold one. Brush voltage drop, winding inductance, magnetic saturation at high current and the driver's current limit are not modelled.

Worked Example

Problem

A 24 V brushed DC servo motor drives a robot joint through a gearbox. Its datasheet gives a no-load speed of 4000 rpm at 24 V and a winding resistance of 1.2 Ω. On the bench the motor draws 2 A under its normal working load, and the joint's worst-case load, referred to the motor shaft, is 0.25 N·m. Find the motor constants, the stall torque and both operating points.

Solution
  1. Torque constant: ωnl=4000×π/30=418.88\omega_{nl} = 4000 \times \pi/30 = 418.88 rad/s, so Kt=Ke=24/418.88=0.057296K_t = K_e = 24/418.88 = 0.057296 N·m/A, which the calculator shows as 0.0573. Carry it unrounded into the next steps.
  2. Stall torque: the stall current is V/Ra=24/1.2=20V/R_a = 24/1.2 = 20 A, so Tstall=0.057296×20=1.1459T_{stall} = 0.057296 \times 20 = 1.1459 N·m.
  3. Torque at 2 A: T=KtI=0.057296×2=0.1146T = K_t I = 0.057296 \times 2 = 0.1146 N·m.
  4. Speed at 2 A: the back-EMF is 24−2×1.2=21.624 - 2 \times 1.2 = 21.6 V, so n=4000×21.6/24=3600n = 4000 \times 21.6/24 = 3600 rpm.
  5. Power and efficiency at 2 A: Pin=24×2=48P_{in} = 24 \times 2 = 48 W and Pout=21.6×2=43.2P_{out} = 21.6 \times 2 = 43.2 W, the same as TωT\omega = 0.1146 N·m × 376.99 rad/s. The efficiency is 43.2/48=90%43.2/48 = 90\%, equal to 3600/4000; the other 4.8 W is the copper loss 22×1.22^2 \times 1.2.
  6. Worst-case load: IL=0.25/0.057296=4.363I_L = 0.25/0.057296 = 4.363 A and nL=4000×(1−0.25/1.1459)=3127.3n_L = 4000 \times (1 - 0.25/1.1459) = 3127.3 rpm, delivering PLP_L = 81.87 W.
  7. Jam check: with the load torque set to 1.5 N·m the calculator warns that it exceeds the 1.1459 N·m stall torque: the motor stops and draws its 20 A stall current.

Result: The motor carries the worst-case load at 3127.3 rpm and 4.363 A, 22% of its stall torque. Enter 4.363 A as the operating current to read the efficiency at that load, 78.2%. Compare 4.363 A with the datasheet's maximum continuous current, because the model has no thermal limit, and make sure the 0.25 N·m already includes the gearbox's ratio and efficiency.

Practical Tips

  • ✓Enter the no-load speed at the voltage you apply. No-load speed is proportional to supply voltage while KtK_t is fixed, so a motor rated 4000 rpm at 24 V runs at about 2000 rpm with no load at 12 V; entering the 24 V figure with a 12 V supply halves KtK_t and every torque derived from it.
  • ✓Take RaR_a from the datasheet's terminal resistance, or measure it at the terminals with the rotor held still, and correct it for temperature: copper resistance rises about 0.39% per °C, so a winding 50 °C above the measurement temperature has about 20% more resistance, and the stall torque falls accordingly. The winding resistance calculator does the temperature correction.
  • ✓Check your KtK_t against the datasheet: the value derived here is usually slightly higher, by about the ratio of no-load current to stall current, because the model neglects the no-load current. Then use the two operating points together: the measured current tells you the speed and efficiency you have, and the load torque tells you the current you will need, which must stay within both the motor's and the driver's continuous current ratings.

Common Mistakes

  • ✗Dividing output power by the no-load speed to get torque. Torque is output power divided by the operating speed; at 70% of no-load speed, dividing by ωnl\omega_{nl} instead understates it by 30%. T=KtIT = K_t I gives it directly from the current.
  • ✗Entering the torque at the gearbox output, or an RC servo's rated torque, as the load torque. Both are measured after the gear train; the Load Torque input is at the motor shaft. Divide by the gear ratio and the gearbox efficiency first, and convert kgf·cm to N·m (1 kgf·cm = 0.0981 N·m) with the torque unit converter.
  • ✗Treating stall torque as usable torque. At stall the whole supply voltage sits across the winding: at 24 V and 1.2 Ω the motor draws 20 A and dissipates 480 W as heat while delivering nothing. Size continuous loads against the datasheet's continuous current rating, and use the stall torque only as the ceiling of the torque-speed line.

Frequently Asked Questions

It derives the torque constant from the no-load point. With the no-load current neglected, the back-EMF equals the supply voltage at no-load speed, so K_e = V/ω_nl, and in SI units K_t = K_e. At 24 V and 4000 rpm (418.88 rad/s) that is 0.0573 N·m/A, so each ampere produces 0.0573 N·m. A datasheet K_t is usually a little lower, because a real motor drops part of its supply voltage across R_a even at no load.
The stall torque, T_stall = K_t·V/R_a, is where the torque-speed line reaches zero speed: there is no back-EMF, the current is the stall current V/R_a, and all of the input power heats the winding. It is the most torque the motor can produce at this voltage, not a working rating. A load torque above it cannot be turned, so the calculator reports zero speed, the stall current and zero output power at the load torque, with a warning. It assumes the load does not drive the motor backwards.
The model counts only the copper loss I²·R_a, so its efficiency is 1 − I·R_a/V, the operating speed divided by the no-load speed: 90% for a 4000 rpm motor running at 3600 rpm. A real motor also spends its no-load current on friction, brush and iron losses. That loss weighs most at light load, so a real motor's efficiency peaks somewhat below the no-load speed and falls to zero at no load, where this model approaches 100%. Near its rated load, a motor with a small no-load current comes within a few percent of the figure shown here.

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