DC Machines: Generators and Motors

What to remember

• EMF equation: E = PΦNZ/(60A) volts. Torque: T = PΦZIa/(2πA) N·m. Lap winding has A = P; wave winding has A = 2.
• A DC motor develops back EMF Eb = V − Ia Ra. Speed N ∝ Eb/Φ, torque T ∝ Φ Ia. A series motor must never start without load; a shunt motor must never lose its field.
• Commutation and armature reaction are the main problems. They are cured by interpoles (commutating poles) and compensating windings.

1. Construction

• Yoke (frame): outer body; carries flux and gives mechanical support. Made of cast steel or rolled steel.
• Field poles and pole shoes: carry the field winding; the shoe spreads flux over the air gap.
• Armature core: laminated silicon steel with slots, to reduce eddy-current loss. It rotates in the flux.
• Armature winding: coils placed in slots, joined to the commutator.
• Commutator: hard-drawn copper segments insulated by mica. It converts alternating EMF in the armature to unidirectional output (generator) or reverses the armature current at the right time (motor).
• Brushes: carbon or graphite, held by brush holders; they take current to the outside circuit.

2. Armature windings

EMF equation: E = PΦNZ/(60A), where P = poles, Φ = flux per pole (Wb), N = speed (rpm), Z = total armature conductors, A = parallel paths.

Equivalent forms: E = KaΦω, where ω is angular speed in rad/s.

Torque equation: T = (PZ/(2πA)) Φ Ia = Ka′ΦIa. Armature power Eb·Ia = T·ω.

3. Types of DC generators

By field excitation:

• 1. Separately excited: field from an independent source.
• 2. Self excited: shunt, series, compound (short-shunt and long-shunt; cumulative and differential).

Terminal voltage: V = E − Ia Ra − brush drop. Shunt generator: Ia = IL + Ish. Series generator: Ia = IL = Ise.

Voltage build-up in a self-excited shunt generator requires:

• residual magnetism in the poles;
• field winding connected so that its flux aids the residual flux;
• field resistance less than the critical resistance (the slope of the tangent to the OCC);
• speed above the critical speed.

A series generator cannot build up on open circuit.

Characteristics:

• OCC (magnetisation curve): E against If at constant speed.
• Internal characteristic: E against Ia. External characteristic: V against IL.
• A shunt generator's voltage drops as the load increases, because of the armature drop, armature reaction and the fall in field current.
• A series generator's voltage rises with load until saturation, so it is not used for constant-voltage supply.
• Cumulatively compounded generators: over-compounded (rising voltage), flat-compounded, under-compounded. Differential compounding gives a drooping curve, used for arc welding.

Parallel operation: Shunt generators work well in parallel. Series and compound generators need an equaliser connection to share load stably.

4. Armature reaction and commutation

Armature reaction is the effect of armature flux on the main field flux. It causes (a) a cross-magnetising effect, which distorts flux and shifts the magnetic neutral axis (MNA) in the direction of rotation for a generator and against it for a motor, and (b) a net reduction in flux (weakening) because of saturation. Remedies: compensating windings (in the pole faces, carrying armature current in series), interpoles, and brush shift.

Commutation is the reversal of current in a coil as it passes under a brush. Poor commutation gives sparking. The reactance voltage (L di/dt) opposes the reversal of current. Improving commutation:

• Interpoles (commutating poles): small poles between the main poles, in series with the armature. They produce an EMF that cancels reactance voltage. For a generator, the interpole polarity is the same as the next main pole in the direction of rotation. For a motor, it is the same as the previous main pole.
• Resistance commutation using high-resistance carbon brushes.
• Brush shift (older method).

5. DC motors

Back EMF: Eb = V − Ia Ra. At start Eb = 0, so the starting current is very high (V/Ra). A starter adds resistance to limit it.

Speed relation: N ∝ Eb/Φ. Torque relation: T ∝ Φ Ia.

Cautions: A series motor must never be started without load or coupled by a belt (runaway speed). A shunt motor must not have its field open-circuited (speed rises dangerously).

Starters: Three-point starter has NVC (no-volt coil) and overload release; the NVC is in series with the shunt field, so field weakening is a problem. The four-point starter connects the NVC separately, which removes that problem. A two-point starter is used for series motors.

Speed control of shunt motors:

• 1. Flux (field) control: a rheostat in the field circuit; speeds above base speed.
• 2. Armature (rheostat) control: resistance in series with the armature; speeds below base speed, wasteful.
• 3. Voltage control: Ward-Leonard system or controlled rectifier (thyristor drive); smooth control in both directions, below and above base speed.

Series motor speed control: field diverter, tapped field, series-parallel (used in traction).

Braking: Plugging (reverse armature connection), rheostatic (dynamic), and regenerative braking.

6. Losses, efficiency and tests

Losses: copper losses (armature, field), iron losses (hysteresis, eddy), mechanical (friction and windage), brush contact loss, stray load loss. Efficiency is maximum when the variable loss equals the constant loss. For a generator, the maximum efficiency is when armature copper loss = constant loss.

Worked example: A 4-pole lap-wound generator has 400 conductors and flux 0.02 Wb per pole at 600 rpm. A = P = 4. E = (4 × 0.02 × 600 × 400)/(60 × 4) = 80 V.

Example 2: A shunt motor on 220 V has Ra = 0.5 Ω and Ia = 20 A. Eb = 220 − 10 = 210 V.

Efficiency example: A shunt motor takes 20 A from a 220 V supply with field current 1 A and Ra = 0.5 Ω. Input = 220 × 20 = 4400 W. Armature current Ia = 19 A. Armature copper loss = 19² × 0.5 = 180.5 W. Field loss = 220 × 1 = 220 W. Eb = 220 − 9.5 = 210.5 V and electrical power developed = 210.5 × 19 = 4000 W. This is the power converted from electrical to mechanical form before friction and iron losses.

Applications of DC machines: DC generators are now rare because AC grids dominate; they survive in excitation systems and small special supplies. DC motors are valued for easy and smooth speed control and high starting torque, which suits electric traction, steel rolling mills, paper mills, cranes and battery-driven vehicles. Brushless DC (BLDC) motors replace the mechanical commutator with electronic switching and are widely used in fans, drives and small appliances.

Exam traps

• 1. Lap winding: A = P. Wave winding: A = 2. Do not exchange them.
• 2. A series generator does not build up without a load; a shunt generator does not build up above its critical resistance.
• 3. Speed rises when flux falls: a shunt motor with an open field races.
• 4. Series motor torque is proportional to Ia² (before saturation); shunt motor torque to Ia.
• 5. Interpoles are in series with the armature; shunt field poles are connected across the supply.
• 6. Swinburne's test gives constant losses at no load, not efficiency at full load directly.
• 7. Four-point starter solves the field-weakening problem of the three-point starter.
• 8. Ward-Leonard control is voltage control, not field control.

One-liners

• 1. Commutator: copper segments separated by mica.
• 2. Brushes are carbon; brush contact drop is small.
• 3. Armature core is laminated to cut eddy loss.
• 4. Starting current of a DC motor is V/Ra.
• 5. Back EMF is zero at standstill.
• 6. Compensating windings are placed in pole-face slots in series with the armature.
• 7. Interpoles neutralise reactance voltage.
• 8. Series motor is used where very high starting torque is required, such as traction.
• 9. Shunt motor has nearly constant speed.
• 10. Hopkinson's test is also called the regenerative test.
• 11. Plugging means reversing armature connections to stop quickly.
• 12. Equaliser connection is needed for parallel operation of series generators.

Practice questions

1. In a lap-wound DC armature with P poles, the number of parallel paths A is

   1. equal to P/2
   2. always 2
   3. equal to 2P
   4. equal to P
   Answer

   D. equal to P

   Lap winding: A = P; wave winding: A = 2.

2. A wave-wound armature always has

   1. two parallel paths
   2. as many paths as poles
   3. paths equal to coil sides
   4. four parallel paths
   Answer

   A. two parallel paths

   Wave winding gives A = 2 irrespective of poles.

3. Wave winding is preferred for

   1. high-voltage, low-current machines
   2. low-voltage, high-current machines
   3. machines needing equaliser rings
   4. very low speed machines only
   Answer

   A. high-voltage, low-current machines

   Few parallel paths give more series conductors, so higher voltage.

4. A 6-pole wave-wound generator has 600 conductors, flux 0.03 Wb per pole, running at 500 rpm. The generated EMF is

   1. 900 V
   2. 450 V
   3. 225 V
   4. 150 V
   Answer

   B. 450 V

   E = PΦNZ/(60A) = 6×0.03×500×600/(60×2) = 450 V.

5. An 8-pole lap-wound generator has 800 conductors and 0.05 Wb per pole at 900 rpm. The EMF is

   1. 75 V
   2. 300 V
   3. 1200 V
   4. 600 V
   Answer

   D. 600 V

   A = 8: E = 8×0.05×900×800/(60×8) = 600 V.

6. A 4-pole lap-wound armature has 400 conductors. The number of conductors in each parallel path is

   1. 400
   2. 200
   3. 50
   4. 100
   Answer

   D. 100

   A = 4, so 400/4 = 100 conductors per path.

7. The commutator of a DC machine is made of

   1. Carbon blocks
   2. hard-drawn copper segments insulated by mica
   3. Aluminium segments insulated by paper
   4. Laminated silicon steel
   Answer

   B. hard-drawn copper segments insulated by mica

   Copper conducts; mica separates the segments.

8. The armature core of a DC machine is laminated to reduce

   1. eddy-current loss
   2. friction loss
   3. copper loss
   4. brush loss
   Answer

   A. eddy-current loss

   Thin insulated laminations break eddy-current paths.

9. A DC shunt generator delivers a load current of 100 A with a field current of 5 A. Its armature current is

   1. 100 A
   2. 20 A
   3. 105 A
   4. 95 A
   Answer

   C. 105 A

   Ia = IL + Ish = 105 A.

10. A DC generator has E = 230 V, armature current 50 A and Ra = 0.2 Ω. Ignoring brush drop, its terminal voltage is

    1. 240 V
    2. 220 V
    3. 230 V
    4. 210 V
    Answer

    B. 220 V

    V = E − IaRa = 230 − 10 = 220 V.

11. A shunt generator will fail to build up voltage if

    1. the speed is above the critical speed
    2. residual magnetism is present
    3. its field resistance exceeds the critical resistance
    4. the field aids the residual flux
    Answer

    C. its field resistance exceeds the critical resistance

    Field resistance above critical makes the field line miss the OCC.

12. Which is NOT needed for a self-excited shunt generator to build up?

    1. Field resistance below critical value
    2. Field connections aiding residual flux
    3. Residual magnetism
    4. A separate DC source for the field
    Answer

    D. A separate DC source for the field

    Self-excitation uses its own output.

13. Statements about generators: 1. A series generator builds up voltage on open circuit. 2. A shunt generator's terminal voltage falls with increasing load.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    B. 2 only

    A series generator needs load current for its field; shunt generator voltage drops with load.

14. Differential compound generators, having a drooping voltage curve, are suited to

    1. arc welding
    2. lighting supply
    3. traction
    4. battery charging at fixed voltage
    Answer

    A. arc welding

    Drooping curve limits short-circuit current in welding.

15. Armature reaction in a DC generator causes

    1. creation of eddy currents in field
    2. reduction in armature resistance
    3. distortion of main flux and shift of the neutral axis
    4. increase in speed
    Answer

    C. distortion of main flux and shift of the neutral axis

    Cross-magnetisation shifts the MNA and, with saturation, weakens flux.

16. The MNA in a DC generator due to armature reaction shifts

    1. by 90° to the poles
    2. against the direction of rotation
    3. not at all
    4. in the direction of rotation
    Answer

    D. in the direction of rotation

    In a motor, the shift is against the direction of rotation.

17. Interpoles are connected in

    1. series with the armature
    2. across the brushes
    3. parallel with the shunt field
    4. series with the shunt field
    Answer

    A. series with the armature

    Their flux then varies with armature current, so it cancels reactance voltage at any load.

18. Compensating windings are placed

    1. in slots in the main pole faces, in series with the armature
    2. on the yoke
    3. on the commutator
    4. on the shaft
    Answer

    A. in slots in the main pole faces, in series with the armature

    They neutralise armature mmf under the pole faces.

19. The reactance voltage during commutation is due to

    1. residual flux
    2. brush resistance
    3. armature resistance
    4. self-inductance of the coil undergoing commutation
    Answer

    D. self-inductance of the coil undergoing commutation

    L di/dt opposes reversal of current.

20. The back EMF of a DC motor on 200 V supply with Ia = 25 A and Ra = 0.4 Ω is

    1. 200 V
    2. 190 V
    3. 210 V
    4. 10 V
    Answer

    B. 190 V

    Eb = V − IaRa = 200 − 10 = 190 V.

21. A 220 V DC motor has Ra = 0.5 Ω. The starting current without a starter is

    1. 44 A
    2. 22 A
    3. 440 A
    4. 110 A
    Answer

    C. 440 A

    Ist = V/Ra = 220/0.5 = 440 A.

22. A 200 V shunt motor has Ra = 0.5 Ω. To limit starting current to 20 A, the external starter resistance should be

    1. 9.5 Ω
    2. 10 Ω
    3. 4.5 Ω
    4. 0.5 Ω
    Answer

    A. 9.5 Ω

    Total R = 200/20 = 10 Ω; external = 10 − 0.5 = 9.5 Ω.

23. A DC motor has back EMF 200 V at 1000 rpm with constant flux. At 150 V back EMF, its speed is

    1. 1333 rpm
    2. 500 rpm
    3. 750 rpm
    4. 800 rpm
    Answer

    C. 750 rpm

    N ∝ Eb: 1000 × 150/200 = 750 rpm.

24. In a shunt motor with constant back EMF, flux is cut by 20%. The speed becomes

    1. 1.2 times
    2. 1.25 times
    3. 0.8 times
    4. unchanged
    Answer

    B. 1.25 times

    N ∝ 1/Φ, so 1/0.8 = 1.25.

25. A DC motor armature develops 2000 W (Eb·Ia) at 100 rad/s. The torque developed is

    1. 200 N·m
    2. 2 N·m
    3. 0.05 N·m
    4. 20 N·m
    Answer

    D. 20 N·m

    T = P/ω = 2000/100 = 20 N·m.

26. In an unsaturated DC series motor, doubling the armature current changes the torque by a factor of

    1. 2
    2. 1
    3. 8
    4. 4
    Answer

    D. 4

    Φ ∝ Ia, so T ∝ Ia²; 2² = 4.

27. If flux is halved while armature current stays constant, a DC motor's torque becomes

    1. half
    2. double
    3. one quarter
    4. unchanged
    Answer

    A. half

    T ∝ Φ Ia.

28. A DC series motor must not be started on no load because

    1. it would draw no current
    2. its speed would rise to dangerous levels
    3. its torque would be zero at all times
    4. its field would reverse
    Answer

    B. its speed would rise to dangerous levels

    At no load Ia and flux are small, so speed becomes very high.

29. The speed of a DC shunt motor rises dangerously if

    1. its armature is overloaded
    2. brushes are replaced
    3. its field circuit opens
    4. the supply voltage rises slightly
    Answer

    C. its field circuit opens

    Flux falls to residual and N ∝ 1/Φ.

30. Match motor to use: (a) Series (b) Shunt (c) Cumulative compound 1. Lathe, constant speed 2. Electric traction 3. Rolling mill

    1. a-2, b-1, c-3
    2. a-3, b-2, c-1
    3. a-1, b-2, c-3
    4. a-2, b-3, c-1
    Answer

    A. a-2, b-1, c-3

    Series suits traction; shunt, constant speed; compound, sudden loads like rolling mills.

31. Statements about starters: 1. A three-point starter can fail to protect when the field is weakened for speed control. 2. A four-point starter removes this problem by connecting the no-volt coil across the supply.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    In a four-point starter the NVC is not in the shunt field circuit.

32. Which method gives speeds above the base speed in a shunt motor?

    1. Reducing the supply voltage
    2. Field flux control
    3. Plugging
    4. Armature resistance control
    Answer

    B. Field flux control

    Weakening the field raises speed beyond base speed.

33. The Ward-Leonard system gives speed control by

    1. varying field flux only
    2. varying the armature voltage with a motor-generator set
    3. inserting series resistance
    4. changing the number of poles
    Answer

    B. varying the armature voltage with a motor-generator set

    It supplies a variable voltage to the motor, in both directions.

34. Plugging as a method of braking a DC motor means

    1. opening the field
    2. short-circuiting the armature through a resistor
    3. feeding the energy back to the supply
    4. reversing the armature connections while running
    Answer

    D. reversing the armature connections while running

    Reverse torque stops the motor quickly; regenerative braking returns energy.

35. Swinburne's test

    1. is a full-load back-to-back test
    2. is suitable for series motors
    3. is a no-load test that finds the constant losses
    4. finds the temperature rise at full load
    Answer

    C. is a no-load test that finds the constant losses

    It cannot be done on a series motor, as it would race with no load.

36. Hopkinson's test is also called the

    1. regenerative (back-to-back) test
    2. brake test
    3. no-load test
    4. retardation test
    Answer

    A. regenerative (back-to-back) test

    Two identical machines are coupled; supply gives only the losses.

37. Statements about the Swinburne test: 1. It is cheap and easy. 2. It shows the effect of armature reaction on load.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    It is done on no load, so load effects are not shown.

38. A DC generator has maximum efficiency when

    1. armature copper loss is zero
    2. iron loss is zero
    3. field loss is zero
    4. armature copper loss equals constant losses
    Answer

    D. armature copper loss equals constant losses

    Variable loss equals constant loss for maximum efficiency.

39. Equaliser connections are used for parallel operation of

    1. only shunt generators
    2. only separately excited generators
    3. only motors
    4. series and compound generators
    Answer

    D. series and compound generators

    They stabilise load sharing by joining the series field ends.

40. Carbon brushes are used because they

    1. are magnetic
    2. have self-lubricating property and high contact resistance
    3. have zero resistance
    4. can carry unlimited current
    Answer

    B. have self-lubricating property and high contact resistance

    High contact resistance helps resistance commutation.

41. Equalising rings (equalisers) are required in

    1. wave-wound armatures
    2. stator windings of induction motors
    3. lap-wound armatures
    4. transformers
    Answer

    C. lap-wound armatures

    They carry circulating currents from unequal flux between parallel paths.

42. For a DC generator, the interpole polarity in the direction of rotation is

    1. always north
    2. the same as that of the next main pole ahead
    3. opposite to the next main pole
    4. the same as the previous main pole
    Answer

    B. the same as that of the next main pole ahead

    For a motor it is the same as the previous main pole.

43. The four-point starter is used for shunt motors with

    1. field speed control
    2. no field
    3. series windings only
    4. very small ratings only
    Answer

    A. field speed control

    NVC is not affected when field current is reduced.

44. Which losses in a DC machine remain nearly constant from no load to full load?

    1. Armature copper loss
    2. Brush contact loss only
    3. Iron, friction and windage losses
    4. Stray load loss only
    Answer

    C. Iron, friction and windage losses

    They depend on flux and speed, not load current.

45. The brushes of a DC machine are placed on the

    1. shaft
    2. yoke
    3. axis of the poles
    4. magnetic neutral axis
    Answer

    D. magnetic neutral axis

    Coils short-circuited by brushes then have no EMF induced.