Speed control of DC motors

Why it matters

Wide, smooth speed control is the main reason DC motors survive in rolling mills, paper machines, hoists, traction and laboratory drives. Every method follows from one equation, N ∝ (V − Ia·Ra)/Φ, and knowing which method gives constant torque and which gives constant power is what lets you match a drive to a load.

Key ideas

The speed equation. N = (V − Ia·Ra)/(k·Φ), with k = P·Z/(60·A) for N in rpm. Three quantities can be changed: the resistance in the armature circuit, the armature voltage V, and the flux Φ.

1. Armature resistance (rheostatic) control

• An external resistance R in series with the armature: N ∝ V − Ia·(Ra + R). Gives speeds below base speed only.
• Simple and cheap, but the I²R loss in the rheostat is large (efficiency roughly falls in proportion to speed), speed regulation becomes poor because speed now changes a lot with load, and no-load speed is not reduced. Used for short-time duty — cranes, hoists, small motors.

2. Armature voltage control

• Vary V with the field kept at rated value. No-load speed is proportional to V; the speed–torque lines are parallel, so regulation stays good. Speeds below base speed.
• Since Φ is constant and Ia is limited to rated value, the available torque stays constant: constant-torque region.
• Supplies: Ward–Leonard system (motor–generator set feeding the armature; smooth, reversible, regenerative, but costly and bulky), and today controlled rectifiers (thyristor converters) from AC or DC choppers from a DC bus.

3. Field (flux) control

• Reduce field current with a rheostat in the shunt field (or a diverter across a series field): speed rises above base speed. The rheostat carries only the small field current, so losses are low and control is cheap and efficient.
• With Ia at rated value, power V·Ia is constant, so torque falls as speed rises: constant-power region. The upper limit (typically 2–3 times base speed) is set by poor commutation, armature reaction and mechanical stress.
• Never open the field: speed would rise dangerously.

Series motor methods. Field diverter (resistance across the series field → less flux → higher speed), armature diverter (resistance across the armature → for a fixed load torque, more field current → lower speed), tapped field, and series–parallel control of two motors (traction).

Combined control. Drives use armature-voltage control from zero up to base speed and field weakening above it. This gives a constant-torque region followed by a constant-power region, matching most industrial loads.

Electronic drives and braking. Converters also allow regenerative braking (returning energy to the supply), dynamic braking (armature switched across a resistor) and plugging (reversing armature polarity; violent, needs current limiting).

Formulas

• N = (V − Ia·Ra) / (k·Φ) — speed (rpm); k = P·Z/(60·A).
• N2/N1 = [(V2 − Ia2·Ra) / (V1 − Ia1·Ra)] · (Φ1/Φ2) — general ratio between two operating points.
• Ia2 = Ia1·(Φ1/Φ2) — armature current after a flux change at constant load torque (T = Ka·Φ·Ia).
• R_ext = (V − Eb2)/Ia − Ra — armature resistance needed for back EMF Eb2 at current Ia (Ω).
• P_loss,ext = Ia²·R_ext — power wasted in the armature rheostat (W).
• Ish = V / (Rsh + R_field-rheostat) — shunt field current with a field rheostat (A).

Worked examples

Example 1 (standard): armature resistance control. Given: 220 V shunt motor, Ra = 0.5 Ω, armature current 20 A at 1000 rpm. Find the external armature resistance that brings the speed to 750 rpm with the same load torque (flux constant).

1. Same torque and flux ⇒ same armature current, Ia = 20 A.
2. Eb1 = V − Ia·Ra = 220 − 20 × 0.5 = 210 V.
3. Eb ∝ N at constant flux: Eb2 = 210 × 750/1000 = 157.5 V.
4. R_ext = (V − Eb2)/Ia − Ra = (220 − 157.5)/20 − 0.5 = 2.625 Ω.
5. Power wasted in the rheostat = 20² × 2.625 = 1050 W — a quarter of the armature input of 4400 W.

Example 2 (GATE level): field weakening at constant load torque. Given: 250 V shunt motor, Ra = 0.5 Ω, takes armature current 20 A at 1000 rpm. The flux is reduced by 20 % with the load torque unchanged. Find the new speed.

1. Constant torque: Ka·Φ1·Ia1 = Ka·Φ2·Ia2 ⇒ Ia2 = 20 × (1/0.8) = 25 A.
2. Eb1 = 250 − 20 × 0.5 = 240 V; Eb2 = 250 − 25 × 0.5 = 237.5 V.
3. N2 = N1 × (Eb2/Eb1) × (Φ1/Φ2) = 1000 × (237.5/240) × (1/0.8) = 1237 rpm.
4. Note the speed rises by about 24 %, slightly less than 1/0.8 = 1.25 because of the larger Ia·Ra drop.

Common mistakes

• Keeping Ia constant after a flux change when the load torque is constant — Ia must rise as flux falls.
• Assuming speed is exactly proportional to supply voltage; it is Eb, not V, that is proportional to N·Φ.
• Using ratios like kΦ = 1 without units, which gives meaningless speeds; if a machine constant is given, keep track of whether it gives rpm or rad/s.
• Saying field control gives speeds below base speed; with the field already at its rated value, it can only weaken.
• Believing armature resistance control is efficient: the rheostat loss is the price of the speed reduction.
• Forgetting that armature-voltage control is constant torque and field control is constant power.

For GATE IN

Numericals ask for a new speed after changing armature voltage, flux or armature resistance, or for the resistance needed to reach a given speed at a given torque. Conceptual MCQs compare methods (range, efficiency, constant torque versus constant power, Ward–Leonard). Chopper- and rectifier-fed drives connect this topic to the power electronics part of the syllabus.

Quick check

1. Which method gives speeds above base speed?
2. A separately excited motor runs at 1500 rpm with Eb = 220 V. If Eb becomes 180 V at the same flux, what is the speed?
3. Why is the armature-voltage method called constant-torque control?
4. Why is armature-resistance control inefficient?

Answers: 1. Field (flux) weakening. 2. 1500 × 180/220 = 1227 rpm. 3. Flux and maximum armature current are fixed, so maximum torque is the same at every speed. 4. The rheostat carries full armature current and dissipates Ia²·R.