AC and DC servo motors

Why it matters

Every CNC axis, robot joint, pick-and-place head and packaging machine positions itself with a servo: a motor, a feedback sensor and a drive that closes the loop. Choosing between a DC and an AC servo, reading a datasheet (torque constant, back-EMF constant, time constants) and predicting how fast an axis responds are everyday tasks for a mechatronics engineer and a standard source of numerical questions.

Key ideas

What makes a motor a "servo". A servo motor is not a special kind of motor so much as a motor built and used for closed-loop control of position, speed or torque. It has low rotor inertia (fast acceleration), a smooth and nearly linear torque–current relation, a high peak-to-continuous torque ratio, and a feedback device (encoder, resolver, tachogenerator or, in small hobby servos, a potentiometer). The drive compares the command with the feedback and adjusts the voltage or current to drive the error towards zero.

Cascaded control loops. Industrial servo drives use three nested loops: an inner current (torque) loop, a speed loop around it, and an outer position loop. The inner loops are the fastest. Speed and current loops are usually PI; the position loop is often P or PID with velocity and acceleration feed-forward.

DC servo motor. A permanent-magnet (or separately excited) DC motor with a low-inertia armature: slotted iron, slotless, disc (printed) or coreless (shell) armature. Two control methods:

• Armature control — field constant, armature voltage varied. The back EMF provides inherent speed feedback (damping), the response is fast and the torque–speed lines are linear and parallel. This is the standard servo arrangement.
• Field control — armature current held constant, field voltage varied. Used only for small motors; the field time constant is large and there is no back-EMF damping, so the system is slower and harder to stabilise.

Drawbacks of the DC servo: brushes and commutator wear, sparking, a commutation limit on peak current and speed, and heat generated in the rotor (harder to remove).

AC servo motors. Two families:

• Two-phase induction servo (classical): a reference winding on a fixed AC voltage and a control winding fed with a variable voltage in quadrature. The rotor is a thin squirrel cage or a drag cup with high resistance, so the X/R ratio is low. This gives a torque–speed curve that slopes downward almost linearly from standstill (positive damping) and ensures the motor stops when the control voltage is removed (no single-phasing). Used for low-power instrument servos.
• Permanent-magnet synchronous servo (PMSM, "brushless AC servo"): magnets on the rotor, three-phase winding on the stator, a high-resolution encoder or resolver, and a drive using field-oriented control (vector control). This is what "AC servo" means in modern machine tools and robots. No brushes means little maintenance, high speed, and heat generated in the stator where it is easily conducted away; the cost is a more complex drive.

Feedback devices. An incremental encoder with N lines per revolution gives 4N counts per revolution with quadrature (×4) decoding. Absolute encoders keep position through power loss. Resolvers are rugged analogue sensors for hot or high-vibration places. A tachogenerator gives a voltage proportional to speed.

Link to neighbouring topics. The motor equations are those of the DC motor; the BLDC motor is the trapezoidal-commutation cousin of the PMSM; motor selection uses the stall torque, no-load speed and time constants derived here.

Formulas

T = K_t · I_a

• T: motor torque (N·m); K_t: torque constant (N·m/A); I_a: armature current (A). Valid while the magnetic circuit is not saturated.

E_b = K_b · ω

• E_b: back EMF (V); K_b: back-EMF constant (V·s/rad); ω: speed (rad/s). In SI units K_t = K_b numerically for an ideal DC motor.

V = I_a · R_a + L_a · dI_a/dt + K_b · ω

• V: applied armature voltage (V); R_a: armature resistance (Ω); L_a: armature inductance (H).

ω = (V − I_a · R_a) / K_b and ω = V/K_b − (R_a / (K_t · K_b)) · T (steady state)

• The torque–speed line: no-load speed ω_0 = V/K_b, stall torque T_s = K_t · V / R_a.

G(s) = θ(s)/V(s) = K_t / ( s · [ (R_a + s·L_a)(J·s + B) + K_t · K_b ] )

• J: total inertia referred to the motor shaft (kg·m²); B: viscous friction (N·m·s/rad). Armature-controlled DC servo.

τ_m = J · R_a / (R_a · B + K_t · K_b) ≈ J · R_a / (K_t · K_b) when B is small

• τ_m: mechanical time constant (s). With L_a neglected, ω(s)/V(s) = K_m / (τ_m · s + 1) with K_m = K_t / (R_a · B + K_t · K_b) (rad/s per V).

τ_e = L_a / R_a

• τ_e: electrical time constant (s); usually much smaller than τ_m.

K_b [V·s/rad] = K_E [V per 1000 rpm] × 60 / (2π × 1000)

• Converts the datasheet back-EMF constant into SI.

Δθ = 360° / (4 · N)

• Resolution of a quadrature-decoded incremental encoder with N lines per revolution.

Worked examples

Example 1 (standard): torque–speed line of a DC servo. Given: permanent-magnet DC servo, V = 24 V, R_a = 2 Ω, K_t = K_b = 0.1 (N·m/A, V·s/rad). Find stall torque, no-load speed, and the speed, output power and efficiency at a load of 0.3 N·m (neglect friction).

1. Stall torque: T_s = K_t · V / R_a = 0.1 × 24 / 2 = 1.2 N·m.
2. No-load speed: ω_0 = V/K_b = 24 / 0.1 = 240 rad/s (≈ 2292 rpm).
3. Current at 0.3 N·m: I_a = T/K_t = 0.3 / 0.1 = 3 A.
4. Speed: ω = (V − I_a·R_a)/K_b = (24 − 3 × 2) / 0.1 = 180 rad/s (≈ 1719 rpm).
5. Output power: T·ω = 0.3 × 180 = 54 W. Input power: V·I_a = 24 × 3 = 72 W.
6. Efficiency: 54 / 72 = 75 % (the 18 W difference is I²R copper loss, 3² × 2 = 18 W).

Example 2 (GATE level): step response and encoder resolution. Given: the motor of Example 1 drives a load so that the total inertia on the motor shaft is J = 2 × 10⁻⁴ kg·m²; B and L_a are negligible. A 24 V step is applied from rest. A 2500-line encoder with ×4 decoding is fitted. Find (a) τ_m, (b) speed at t = τ_m, (c) time to reach 98 % of final speed, (d) angular resolution.

1. Mechanical time constant: τ_m = J·R_a/(K_t·K_b) = 2 × 10⁻⁴ × 2 / (0.1 × 0.1) = 0.04 s (40 ms).
2. Final speed = V/K_b = 240 rad/s; first-order response ω(t) = 240·(1 − e^(−t/τ_m)). At t = τ_m: 240 × 0.632 = 151.7 rad/s.
3. 98 % is reached at about 4τ_m: 4 × 0.04 = 0.16 s.
4. Counts per revolution = 4 × 2500 = 10 000, so Δθ = 360°/10 000 = 0.036°. Check: if B = 1 × 10⁻⁴ N·m·s/rad were included, τ_m = 2 × 10⁻⁴ × 2 / (2 × 10⁻⁴ + 0.01) = 0.0392 s — a 2 % change, which is why B is often neglected.

Common mistakes

• Using K_b in V/krpm or V/(rpm) directly in SI equations. Convert first: 10.5 V/krpm = 0.100 V·s/rad.
• Writing ω = V/K_b for a loaded motor. That is only the no-load speed; the I_a·R_a drop must be subtracted.
• Forgetting to refer load inertia through the gear ratio: load inertia appears on the motor shaft divided by n² (n = motor speed / load speed).
• Taking N lines as N counts. With quadrature decoding the resolution is four times finer.
• Calling field control the normal servo method. Armature control is standard because back EMF gives damping and the response is fast.
• Saying the two-phase AC servo uses a low-resistance rotor. It needs a high-resistance rotor for a negatively sloping, stable torque–speed curve.

For GATE ME

Expect the armature-controlled DC motor transfer function and its time constant, stall torque and no-load speed from a linear torque–speed line, power and efficiency from V, I and R_a, unit conversion of K_b, and reflected inertia through a gearbox. Questions also link the motor to a first-order step response (time to a given fraction of final speed) or to a position loop with proportional gain. Practise deriving G(s) from the electrical and mechanical equations and reducing it to first order.

Quick check

1. Why is armature control preferred over field control for a DC servo?
2. A DC servo has K_t = 0.05 N·m/A. What current gives 0.4 N·m?
3. Why does a two-phase AC servo use a high-resistance rotor?
4. A 1000-line encoder is read with ×4 decoding. What is the angular resolution?
5. What is the mechanical time constant of a motor with J = 1 × 10⁻⁴ kg·m², R_a = 1 Ω, K_t = K_b = 0.05 (B negligible)?

Answers: 1. Back EMF gives inherent damping and the armature circuit has a small time constant, so the response is fast and stable. 2. 8 A. 3. To get a torque–speed curve that falls linearly with speed (stable, damped) and to prevent running on one phase. 4. 0.09°. 5. 0.04 s.