DC Machines: Generators and Motors

EMF and torque equations, lap and wave windings, types of excitation, armature reaction and commutation, generator characteristics, motor speed and torque, speed control, starting and losses.

A DC machine converts between mechanical power and DC electrical power through a commutator. The same machine works as a generator or a motor.

EMF and torque

E = P Φ N Z/(60 A) (V)

  • P = poles, Φ = flux per pole (Wb), N = speed (rpm), Z = armature conductors, A = parallel paths: A = P for lap winding, A = 2 for wave winding.
  • Torque T = P Φ Z I_a/(2π A) = E I_a/ω. So T ∝ Φ I_a and E ∝ Φ N.

Types

Separately excited, and self-excited: shunt, series and compound (cumulative or differential).

  • Generator: V = E − I_a R_a. Motor: V = E_b + I_a R_a (E_b = back EMF).
  • Armature reaction distorts and weakens the main flux; interpoles and compensating windings counter it and help commutation.

Generator characteristics

  • Shunt generator builds up only if there is residual flux, the field resistance is below the critical resistance, and the field is connected the right way.
  • Series generators have a rising characteristic; over-compounded generators make up line drop.

Motor characteristics

MotorSpeed–loadTorqueUse
ShuntNearly constant speedT ∝ I_aLathes, fans
SeriesSpeed falls sharply with load; dangerous at no loadT ∝ I_a² (unsaturated)Traction, cranes
Cumulative compoundBetween the two—Presses, shears

Speed: N ∝ E_b/Φ = (V − I_a R_a)/Φ.

Speed control

  • Field control (shunt motor): weakening Φ raises speed above base speed (constant power).
  • Armature voltage / resistance control: reduces speed below base speed (constant torque). Ward–Leonard gives smooth control.

Starting and losses

At standstill E_b = 0, so the starting current V/R_a would be huge: a starter adds series resistance (3-point and 4-point starters).

Losses: copper (I²R), iron, mechanical, brush contact. Maximum efficiency when variable (copper) loss = constant loss.

Worked example

A 4-pole lap-wound generator has 500 conductors and 0.02 Wb per pole at 1200 rpm.

  • E = 4 × 0.02 × 1200 × 500/(60 × 4) = 200 V.

Practice

GATE-styleNAT · numerical answer

A 220 V shunt motor has R_a = 0.5 Ω and draws an armature current of 40 A. What is the back EMF in V?

Show answer and explanationAnswer: 200

E_b = V − I_aR_a = 220 − 20 = 200 V.

GATE-styleNAT · numerical answer

The same motor runs at 1000 rpm. If the flux is reduced by 20% with the same back EMF, what is the new speed in rpm?

Show answer and explanationAnswer: 1250

N ∝ E_b/Φ, so N = 1000/0.8 = 1250 rpm.

GATE-styleNAT · numerical answer

A 6-pole wave-wound armature has 300 conductors, 0.03 Wb per pole and runs at 1000 rpm. What EMF (V) is generated?

Show answer and explanationAnswer: 450

E = PΦNZ/(60A) = 6 × 0.03 × 1000 × 300/120 = 450 V.

GATE-styleMCQ · one correct answer

Why should a DC series motor never be started without load?

Show answer and explanationAnswer: A

At no load the current and flux are tiny, and N ∝ 1/Φ, so the speed runs away.