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DC machines

Electrical and Electronics Engineering (B.E./B.Tech) · Electrical machines · free study notes for exams, placements and interviews

In short

A DC machine converts mechanical energy into direct-current electrical energy (generator) or DC electrical energy into rotation (motor), using a field system, an armature winding and a commutator that mechanically rectifies the alternating EMF induced in the coils. The same machine works either way — only the direction of power flow changes — which is why the EMF equation, armature reaction and commutation are common to both. DC machines carry heavy weight in electrical machines papers, GATE EE (EMF/torque equations, characteristics, speed control, efficiency tests) and in PSU and campus interviews at NTPC, Power Grid, Indian Railways, BHEL, CIL, Siemens, ABB, L&T, Crompton and Kirloskar. Although induction and BLDC drives have replaced DC motors in most new installations, DC machines remain the teaching foundation for all rotating machines and still run cranes, rolling mills, traction legacy stock, electroplating plants and small servo and automotive drives.

Explained

Construction and basic action

Armature windings — lap versus wave

EMF, torque and armature reaction

Commutation

Types of excitation and characteristics

Starting, braking and speed control

Losses, efficiency and testing

Key points

Formulas and examples

Eg = φZNP/(60A) — Eg: generated EMF (V); φ: flux per pole (Wb); Z: total armature conductors; N: speed (rpm); P: poles; A: parallel paths (lap A = P, wave A = 2)
Generator: V = Eg − IaRa − Vbrush; Ia = IL + Ish. Motor: Eb = V − IaRa − Vbrush; Ia = IL − Ish (shunt)
Ta = (P·φ·Z·Ia)/(2πA) N·m; also Ta = 9.55 Eb Ia / N (N in rpm); Pmech = Eb·Ia (W)
N = (V − IaRa)/(kφ) ⇒ N1/N2 = (Eb1/Eb2)·(φ2/φ1)
Series motor (unsaturated): T ∝ Ia², N ∝ 1/Ia; Shunt motor: T ∝ Ia, N ≈ constant
Shunt generator: Ish = V/Rsh; Ia = IL + V/Rsh. Long-shunt compound: Ia = IL + Ish, series field carries Ia
Efficiency η = output/input; ηmax when Ia²Ra = Wconstant ⇒ Ia = √(Wc/Ra)
Speed regulation % = (N_NL − N_FL)/N_FL × 100; Voltage regulation % = (E0 − V)/V × 100
Reactance voltage = L × (2Ic/Tc), Ic = coil current = Ia/A; Tc = brush contact width/peripheral commutator speed
Hysteresis loss Wh = Kh·f·Bmax1.6·V; Eddy loss We = Ke·f²·t²·Bmax²·V

Example 1 — EMF, lap and wave. A 4-pole DC generator has 500 armature conductors, flux 0.03 Wb per pole, running at 1,200 rpm.

Example 2 — Shunt motor: current, torque, speed. A 230 V DC shunt motor has Ra = 0.4 Ω, Rsh = 115 Ω, and draws 32 A from the line at 1,000 rpm. Total brush drop 2 V.

  1. Ish = 230/115 = 2 A → Ia = 32 − 2 = 30 A.
  2. Eb = 230 − (30 × 0.4) − 2 = 230 − 12 − 2 = 216 V.
  3. Mechanical power developed = Eb·Ia = 216 × 30 = 6,480 W.
  4. Armature torque Ta = 9.55 × 6,480/1,000 = 61.9 N·m.
  5. If the load is removed so that Ia falls to 3 A (flux unchanged): Eb2 = 230 − 1.2 − 2 = 226.8 V → N2 = 1,000 × 226.8/216 = 1,050 rpm. Speed regulation = (1,050 − 1,000)/1,000 = 5%.

Example 3 — Series motor torque and speed. A 4-pole 250 V series motor takes 25 A and runs at 600 rpm; Ra + Rse = 0.6 Ω. Flux is proportional to current up to 30 A.

  1. Eb1 = 250 − 25 × 0.6 = 235 V; Ta1 = 9.55 × 235 × 25/600 = 93.5 N·m.
  2. Load increases so the current becomes 30 A: Eb2 = 250 − 30 × 0.6 = 232 V.
  3. N2 = N1 × (Eb2/Eb1) × (φ1/φ2) = 600 × (232/235) × (25/30) = 600 × 0.98723 × 0.8333 = 493.6 rpm.
  4. Torque ratio = (30/25)² = 1.44 → Ta2 ≈ 93.5 × ... check directly: Ta2 = 9.55 × 232 × 30/493.6 = 134.6 N·m; ratio 134.6/93.5 = 1.44, confirming T ∝ Ia².

Example 4 — Starting resistance. The same 230 V shunt motor (Ra = 0.4 Ω, full-load Ia = 30 A) is to be started with the armature current limited to 1.5 × full load.

Example 5 — Swinburne's test and maximum efficiency. A 230 V shunt machine takes 5 A at no load; Ra = 0.4 Ω, Rsh = 115 Ω.

  1. No-load input = 230 × 5 = 1,150 W. Ish = 2 A, Ia0 = 3 A.
  2. No-load armature copper loss = 3² × 0.4 = 3.6 W → constant losses Wc = 1,150 − 3.6 = 1,146.4 W (includes the 460 W field loss, iron, friction and windage).
  3. As a motor at 50 A line: Ia = 48 A; armature Cu loss = 48² × 0.4 = 921.6 W; total loss = 921.6 + 1,146.4 = 2,068 W; input = 230 × 50 = 11,500 W; output = 9,432 W → η = 82.0%.
  4. As a generator delivering 50 A: Ia = 52 A; armature Cu loss = 52² × 0.4 = 1,081.6 W; output = 11,500 W; input = 11,500 + 1,081.6 + 1,146.4 = 13,728 W → η = 83.8%.
  5. Maximum efficiency as a motor: excluding the field loss from Wc for the armature-circuit condition is a common simplification; using the full Wc, Ia = √(1,146.4/0.4) = 53.5 A, so maximum efficiency occurs near 55 A line current.
FeatureShunt motorSeries motorCumulative compound
Field connectionAcross supplyIn series with armatureBoth windings
Starting torqueMedium (T ∝ Ia)Very high (T ∝ Ia²)High
No-load speedDefinite, slightly above FLDangerously highDefinite
Speed-load curveNearly flat (5–10% droop)Steeply drooping hyperbolaIntermediate droop
Safe on no loadYesNoYes
Typical useLathes, fans, pumps, blowersTraction, cranes, hoists, startersRolling mills, presses, lifts
Generator typeExternal characteristicTypical application
Separately excitedSlight droop, easy voltage controlTesting, Ward-Leonard supply
ShuntModerate droop; breakdown at high loadGeneral DC supply, battery charging
SeriesRising then fallingDC series boosters
Level compoundFlat (V_FL = V_NL)Constant-voltage DC bus
Over compoundRising with loadLong feeders (compensates line drop)
Differential compoundSteeply droopingOlder DC arc-welding sets

Where it is used

Interview questions

Common mistakes

Related topics

Transformers · Induction motors · Induction motor testing, starting and speed control · Synchronous machines · Special machines: stepper and BLDC motors · Electrical machine design basics

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