Electrical Machines · Chapter 92

Servo Motors and Linear Induction Motors

Part 6 · Single-Phase and Special Machines — precise motion in a loop, and the induction motor laid out flat.

Prof. Mithun Mondal Engineering Devotion Digital Textbook
i Learning Objectives

By the end of this chapter you should be able to:

  1. State the requirements of a servo motor and explain how they shape its design.

  2. Describe DC and AC (two-phase) servo motors and their torque–speed characteristics.

  3. Explain the principle of the linear induction motor (LIM) and compute its synchronous velocity, slip and thrust.

  4. Describe end effects and the applications of LIMs.

Section 92-1

What Makes a Motor a Servo Motor?

A servo motor is the actuator in a closed-loop position or speed control system. It is judged not by its efficiency at one operating point but by how quickly and accurately it follows a command.

  • Low inertia — long, slender rotor (or ironless/disc armature) for high acceleration \(T/J\).

  • Linear torque–speed and torque–voltage characteristics over the whole range, including reversal.

  • High torque at zero and low speed, with no tendency to run on (single-phasing) when the control signal is removed.

  • Smooth torque, fast electrical time constant, and a built-in feedback device (encoder, resolver or tachogenerator).

Section 92-2

DC Servo Motors

A DC servo is a separately excited or PM DC motor with low armature inertia. In armature control (the usual method) the field is constant and the armature voltage is the control input:

\[T = K_tI_a, \qquad V_a = I_aR_a + K_b\omega \;\Rightarrow\; \omega = \frac{V_a}{K_b} - \frac{R_a}{K_bK_t}T\]

The torque–speed curves are parallel straight lines, one for each control voltage — ideal for linear control design. Field control is possible but slower, because the field time constant is large.

Section 92-3

AC (Two-Phase) Servo Motors

The classic AC servo is a two-phase induction motor. The reference (main) winding is fed with a fixed voltage; the control winding, in space quadrature, receives a variable voltage 90° phase-shifted from it. The magnitude of the control voltage sets the torque, and its phase (leading or lagging) sets the direction.

⤓
The key design choice
A very high rotor resistance: \(R_2 > X_2\)

An ordinary induction motor peaks at small slip and has a positive-slope (unstable) region. Making \(R_2/X_2 > 1\) pushes the peak torque beyond standstill, so the torque–speed curve has a negative slope everywhere — inherent damping and stability. It also ensures that with zero control voltage the single-phase reference field produces a braking, not driving, torque, so the motor stops rather than single-phasing on.

Thin drag-cup (non-magnetic aluminium cup) rotors give very low inertia for small instrument servos. Modern industrial "AC servos", however, are almost always PMSMs (Chapter 90) under field-oriented control with high-resolution encoders.

Section 92-4

The Linear Induction Motor

Imagine cutting a cage induction motor along a radius and unrolling it flat. The stator becomes the primary — a flat three-phase winding producing a travelling field instead of a rotating one. The rotor becomes the secondary — usually an aluminium or copper sheet, often backed by steel. The force produced is linear thrust.

\[v_s = 2\tau f, \qquad s = \frac{v_s - v}{v_s}, \qquad F = \frac{P_\text{ag}}{v_s}\]

\(\tau\) is the pole pitch in metres. Note that \(v_s\) does not depend on the number of poles — only on the pole pitch and frequency. As in the rotary motor, \(P_\text{mech} = (1 - s)P_\text{ag}\) and secondary copper loss \(= sP_\text{ag}\).

  • Single-sided LIMs have one primary and a steel-backed secondary; double-sided LIMs have primaries on both sides of a thin conducting plate.

  • Either the primary or the secondary can be the moving member (short-primary or short-secondary designs).

1 Worked Example 92.1 — LIM Thrust and Power

Problem. A 50 Hz LIM has a pole pitch of 0.25 m. The vehicle moves at 22 m/s while 20 kW crosses the air gap. Find the synchronous velocity, slip, thrust, mechanical power and secondary copper loss.

\[v_s = 2(0.25)(50) = 25~\text{m/s} = 90~\text{km/h}, \qquad s = \frac{25 - 22}{25} = 0.12\]
\[F = \frac{20\,000}{25} = 800~\text{N}, \qquad P_\text{mech} = Fv = 800(22) = 17.6~\text{kW}\]
\[P_{\text{Cu},2} = sP_\text{ag} = 0.12(20) = 2.4~\text{kW}\]

A slip of 12 % is high for a rotary motor but typical of a LIM, whose large effective air gap and end effects give it a higher slip and poorer efficiency.

Section 92-5

End Effects, Merits and Applications

  • End effects: the primary has a beginning and an end. Secondary material entering the field resists the build-up of flux; material leaving it carries trailing currents. Both reduce thrust and efficiency, especially at high speed.

  • Large air gap: mechanical clearance plus the non-magnetic secondary sheet gives high magnetising current and low power factor.

  • Normal force: single-sided LIMs produce a large attractive force between primary and steel-backed secondary, which the structure must resist (or exploit for levitation).

Merits

Direct linear thrust with no gears, belts or wheel–rail adhesion; no contact, so little wear; simple, rugged secondary; easy speed control with an inverter.

Applications

Metro and airport people-movers, maglev propulsion, sliding doors, conveyors and baggage handling, roller-coaster launch systems, electromagnetic pumps for liquid metals.

Section 92-6

Summary and Common Mistakes

Key Points
  • Servo motors need low inertia, linear characteristics and stability over the whole speed range.

  • Two-phase AC servo: control voltage in quadrature; high \(R_2/X_2\) gives a negative-slope characteristic.

  • LIM: \(v_s = 2\tau f\); thrust \(= P_\text{ag}/v_s\); end effects and large gap reduce performance.

Common Mistakes
  • Using \(120f/P\) for a LIM — its synchronous speed is a velocity, \(2\tau f\).

  • Dividing air-gap power by actual velocity (not synchronous velocity) to get thrust.

  • Designing an AC servo with low rotor resistance, as for a general-purpose motor.

Section 92-7

Chapter Review

Practice Problems
  1. P92.1 A LIM with a pole pitch of 0.3 m is supplied at 40 Hz and runs at 20 m/s. Find \(v_s\) and the slip.

    Show answer
    \(v_s = 2(0.3)(40) = 24\) m/s; \(s = 4/24 = 0.167\).
  2. P92.2 The LIM of P92.1 develops 1.5 kN of thrust. Find the air-gap power, mechanical power and secondary loss.

    Show answer
    \(P_\text{ag} = 1500 \times 24 = 36\) kW; \(P_\text{mech} = 1500 \times 20 = 30\) kW; loss \(= 6\) kW.
Multiple-Choice Questions
  1. MCQ 1. The rotor of a two-phase AC servo motor has:
    (a) low resistance   (b) high resistance   (c) permanent magnets   (d) a commutator

    Show answer
    (b)
  2. MCQ 2. The synchronous velocity of a LIM depends on:
    (a) number of poles only   (b) pole pitch and frequency   (c) secondary resistance   (d) air-gap length

    Show answer
    (b)
Conceptual Questions
  1. Why must an AC servo motor not run on when the control voltage is removed?

  2. Explain the entry and exit end effects in a LIM.

  3. Why is the power factor of a LIM lower than that of a rotary induction motor?

Course Complete

This chapter closes the textbook's survey of electrical machines. Every machine studied — from the transformer to the linear motor — rests on the ideas of Part 1: a magnetic field, an air gap, and the energy stored between them. Return to the course contents for the additional machine topics, labs and solved problems.