DC Generators

LessonAbout 2 h

DC Generators

DC generator is an electrical machine that converts mechanical power into direct-current electrical power by electromagnetic induction.

1. Types of Armature Windings

Armature winding is an arrangement of conductors placed in slots on the armature core to generate induced electromotive force.

1.1 Lap Winding

Lap winding forms a loop as it expands around the armature core with coil ends connected to adjacent commutator segments.

  • Suitable for high-current and low-voltage applications
  • Equalizer rings prevent circulating currents
  • Back pitch \(Y_b\) and front pitch \(Y_f\) must be odd numbers

Lap Winding Pitch Formulas

\[ Y_b = Y_f \pm 2m \]

Where: \(Y_b\) = back pitch (odd integer), \(Y_f\) = front pitch (odd integer), \(m\) = multiplicity factor (\(1\) for simplex, \(2\) for duplex)

EE Board Exam March 1998

Example. A four-pole, DC generator with lap winding has 48 slots and 4 elements per slot. How many coils does it have?

Solution.

\[ n = \dfrac{48 \times 4}{2} = 96\ \text{coils} \]

Answer: \(96\ \text{coils}\)

1.2 Wave Winding

Wave winding forms a wave as it expands around the armature core, spanning approximately two pole pitches.

  • Suitable for high-voltage and low-current applications
  • Requires only two brushes regardless of pole count
  • Average pitch \(Y\) must be an integer

Wave Winding Pitch Formulas

\[ Y = \dfrac{Z \pm 2m}{P} \]
\[ Y = \dfrac{Y_b + Y_f}{2} \]

Where: \(Y\) = average pitch (integer), \(Z\) = total number of winding elements, \(P\) = number of poles, \(m\) = multiplicity factor

1.3 Minimum Number of Elements or Conductors per Slot

Elements per slot specifies the number of coil sides placed inside each armature slot.

  • Simplex winding: \(2\) elements per slot
  • Duplex winding: \(4\) elements per slot
  • Triplex winding: \(6\) elements per slot
  • Quadruplex winding: \(8\) elements per slot

1.4 Total Number of Elements or Conductors

Total armature conductors represents the total count of active electrical conductors on the armature surface.

Total Conductors Formula

\[ Z = (\text{elements / slot})(\text{total number of slots}) \]

Where: \(Z\) = total number of conductors, \(\text{elements / slot}\) = conductors per slot

1.5 Number of Commutator Segments

Commutator segments are insulated copper bars connected to armature coil ends.

Commutator Segments Formula

\[ N_c = \dfrac{Z}{2} \]

Where: \(N_c\) = number of commutator segments, \(Z\) = total number of conductors

1.6 Number of Armature Current Paths

Parallel current paths determines the division of total armature current among internal parallel branches.

Parallel Current Paths Formulas

\[ a_{\text{lap}} = m P \]
\[ a_{\text{wave}} = 2 m \]

Where: \(a\) = number of parallel paths, \(P\) = number of poles, \(m\) = multiplicity factor (\(1\) for simplex, \(2\) for duplex)

EE Board Exam October 1996

Example. If the armature of an eight-pole machine were wound with a simplex wave winding, how many parallel paths would there be?

Solution.

\[ a = 2 m = 2(1) = 2\ \text{paths} \]

Answer: \(2\ \text{paths}\)

1.7 Coil Pitch

Coil pitch is the distance between two sides of a coil relative to the pole pitch.

Coil Pitch Formula

\[ Y_s = \dfrac{\text{coil span in slots}}{\text{slots per pole}} \]

Where: \(Y_s\) = fractional coil pitch, \(\text{slots per pole}\) = total slots divided by \(P\)

2. Generated EMF Equation

Generated electromotive force is the internal voltage induced in armature conductors rotating through a magnetic field.

Generated EMF Formula

\[ E = \dfrac{P N Z \Phi}{60 a} \]
\[ E = k N \Phi \]

Where: \(E\) = generated emf (V), \(P\) = number of poles, \(N\) = speed of rotation (rpm), \(Z\) = total conductors, \(\Phi\) = flux per pole (Wb), \(a\) = parallel current paths, \(k\) = machine constant

EE Board Exam October 1998

Example. A 4-pole DC generator with duplex lap winding has 48 slots and four elements per slot. The flux per pole is \(2.5 \times 10^6\ \text{maxwells}\) and it runs at \(1500\ \text{rpm}\). What is the output voltage?

Solution.

\[ Z = 48 \times 4 = 192 \qquad\qquad a = m P = 2(4) = 8 \qquad\qquad \Phi = 2.5 \times 10^6 \times 10^{-8} = 0.025\ \text{Wb} \]
\[ E = \dfrac{P N Z \Phi}{60 a} = \dfrac{4(1500)(192)(0.025)}{60(8)} = 60\ \text{V} \]

Answer: \(60\ \text{V}\)

3. Types of DC Generator According to Excitation

Excitation method classifies DC generators based on the source supplying direct current to field windings.

3.1 Separately-Excited Generator

Separately-excited generator has its field winding energized from an independent external DC voltage source.

  • Field current is independent of armature terminal voltage
  • Used in precise control systems and laboratory testing
  • Stable operation across full terminal voltage range

Separately-Excited Generator Equations

\[ I_L = \dfrac{P_L}{V_L} \qquad\qquad I_a = I_L \]
\[ E = V_L + I_a R_a' \]
\[ P_g = E I_a \]

Where: \(I_L\) = load current (A), \(I_a\) = armature current (A), \(V_L\) = terminal voltage (V), \(E\) = generated emf (V), \(R_a'\) = total armature circuit resistance (\(\Omega\)), \(P_g\) = generated power (W)

Example. A separately excited generator when running at \(1200\ \text{rpm}\) supplies \(200\ \text{A}\) at \(125\ \text{V}\) to a circuit of constant resistance. Armature circuit resistance is \(0.10\ \Omega\). Determine the terminal voltage when the speed is dropped to \(1000\ \text{rpm}\) assuming field current is unaltered.

Solution.

\[ E_1 = V_{L1} + I_a R_a = 125 + 200(0.1) = 145\ \text{V} \]
\[ E_2 = E_1\left(\dfrac{N_2}{N_1}\right) = 145\left(\dfrac{1000}{1200}\right) = 120.833\ \text{V} \]
\[ R_L = \dfrac{125}{200} = 0.625\ \Omega \implies I_{a2} = \dfrac{120.833}{0.1 + 0.625} = 166.66\ \text{A} \]
\[ V_{L2} = I_{a2} R_L = 166.66(0.625) = 104.16\ \text{V} \]

Answer: \(104.16\ \text{V}\)

3.2 Self-Excited Generator

Self-excited generator energizes its field winding using the electromotive force generated in its own armature.

  • Requires residual magnetism in pole cores to build voltage
  • Voltage builds up progressively as speed increases
  • Types include shunt, series, and compound generators

4. Shunt Generator

Shunt generator has its field winding connected in parallel with the armature winding and load.

  • Field winding consists of many turns of fine wire
  • Terminal voltage drops slightly as load current increases
  • Shunt field current remains nearly constant

Shunt Generator Formulas

\[ I_{\text{sh}} = \dfrac{V_L}{R_{\text{sh}}} \qquad\qquad I_L = \dfrac{P_L}{V_L} \]
\[ I_a = I_L + I_{\text{sh}} \]
\[ E = V_L + I_a R_a' \]
\[ P_g = E I_a \]

Where: \(I_{\text{sh}}\) = shunt field current (A), \(R_{\text{sh}}\) = shunt field resistance (\(\Omega\)), \(I_L\) = load current (A), \(I_a\) = armature current (A), \(V_L\) = terminal voltage (V), \(E\) = generated emf (V)

Example. A \(50\ \text{kW}, 220\ \text{V}\) shunt generator has an armature resistance of \(0.026\ \Omega\) and a field resistance of \(52\ \Omega\). Calculate the generated emf at rated load.

Solution.

\[ I_L = \dfrac{50000}{220} = 227.27\ \text{A} \qquad\qquad I_{\text{sh}} = \dfrac{220}{52} = 4.23\ \text{A} \]
\[ I_a = 227.27 + 4.23 = 231.5\ \text{A} \]
\[ E = 220 + 231.5(0.026) = 226.02\ \text{V} \]

Answer: \(226.02\ \text{V}\)

5. Series Generator

Series generator has its field winding connected in series with the armature winding and the load.

  • Field winding consists of few turns of heavy wire
  • Terminal voltage rises with load current
  • Used as voltage boosters in transmission feeders

Series Generator Formulas

\[ I_a = I_{\text{se}} = I_L \]
\[ E = V_L + I_a(R_a' + R_{\text{se}}) \]
\[ P_g = E I_a \]

Where: \(I_{\text{se}}\) = series field current (A), \(R_{\text{se}}\) = series field resistance (\(\Omega\)), \(I_a\) = armature current (A), \(V_L\) = terminal voltage (V), \(E\) = generated emf (V)

Example. A DC series generator is supplying a current of \(5\ \text{A}\) to a series lighting system through a feeder of total resistance \(1.0\ \Omega\). The terminal voltage is \(2500\ \text{V}\). The armature and series field resistances are respectively \(18\ \Omega\) and \(15\ \Omega\). A \(30\ \Omega\) diverter resistance is shunted across the series field. Determine the power developed in the armature of the generator.

Solution.

\[ R_{\text{equiv}} = \dfrac{15(30)}{15 + 30} = 10\ \Omega \]
\[ E = V_L + I_a(R_a + R_{\text{equiv}} + R_{\text{feeder}}) = 2500 + 5(18 + 10 + 1) = 2645\ \text{V} \]
\[ P_g = E I_a = 2645(5) = 132250\ \text{W} = 132.25\ \text{kW} \]

Answer: \(132.25\ \text{kW}\)

6. Long-Shunt Compound Generator

Long-shunt compound generator has its shunt field winding connected across the series combination of armature and series field.

  • Contains both series and shunt field windings
  • Series field current equals armature current (\(I_{\text{se}} = I_a\))
  • Provides stable flat or over-compounded voltage control

Long-Shunt Compound Generator Formulas

\[ I_{\text{se}} = I_a \qquad\qquad I_L = \dfrac{P_L}{V_L} \qquad\qquad I_{\text{sh}} = \dfrac{V_L}{R_{\text{sh}}} \]
\[ I_a = I_{\text{sh}} + I_L \]
\[ E = V_L + I_a(R_a' + R_{\text{se}}) \]
\[ P_g = E I_a \]

Where: \(I_{\text{se}}\) = series field current (A), \(I_{\text{sh}}\) = shunt field current (A), \(I_a\) = armature current (A), \(V_L\) = terminal voltage (V), \(R_{\text{se}}\) = series field resistance (\(\Omega\)), \(R_{\text{sh}}\) = shunt field resistance (\(\Omega\))

Example. A long-shunt compound generator delivers \(100\ \text{A}, 250\ \text{V}\) at rated load. The resistances are \(R_a = 0.1\ \Omega\), \(R_{\text{ip}} = 0.02\ \Omega\), \(R_{\text{sh}} = 100\ \Omega\), and \(R_{\text{se}} = 0.025\ \Omega\). If stray power losses are \(1.8\ \text{kW}\), solve for the overall efficiency.

Solution.

\[ I_{\text{sh}} = \dfrac{250}{100} = 2.5\ \text{A} \qquad\qquad I_a = 100 + 2.5 = 102.5\ \text{A} \]
\[ P_L = 250(100) = 25000\ \text{W} = 25\ \text{kW} \]
\[ P_a = (102.5)^2(0.1) = 1.05\ \text{kW} \qquad\qquad P_{\text{ip}} = (102.5)^2(0.02) = 0.21\ \text{kW} \]
\[ P_{\text{sh}} = 250(2.5) = 0.625\ \text{kW} \qquad\qquad P_{\text{se}} = (102.5)^2(0.025) = 0.263\ \text{kW} \]
\[ \eta = \dfrac{25}{25 + 1.05 + 0.21 + 0.625 + 0.263 + 1.8} \times 100\% = 86.36\% \]

Answer: \(86.36\%\)

7. Short-Shunt Compound Generator

Short-shunt compound generator has its shunt field winding connected directly across the armature terminals, in parallel with the series field and load.

  • Series field current equals load current (\(I_{\text{se}} = I_L\))
  • Shunt field voltage equals armature voltage drop (\(V_{\text{sh}} = V_L + I_L R_{\text{se}}\))
  • Provides flexible compounding adjustments

Short-Shunt Compound Generator Formulas

\[ I_{\text{se}} = I_L \qquad\qquad I_L = \dfrac{P_L}{V_L} \qquad\qquad I_{\text{sh}} = \dfrac{V_L + I_{\text{se}} R_{\text{se}}}{R_{\text{sh}}} \]
\[ I_a = I_{\text{sh}} + I_L \]
\[ E = V_L + I_a R_a' + I_{\text{se}} R_{\text{se}} \]
\[ P_g = E I_a \]

Where: \(I_{\text{se}}\) = series field current (A), \(I_{\text{sh}}\) = shunt field current (A), \(I_a\) = armature current (A), \(V_L\) = terminal voltage (V), \(E\) = generated emf (V)

EE Board Exam April 1992

Example. A compound dynamo supplies a load of \(20\ \text{kW}\) at \(400\ \text{V}\) through a feeder having a resistance equal to \(0.1\ \Omega\). The shunt and series field resistances are \(85\ \Omega\) and \(0.06\ \Omega\) respectively. Determine the total electromotive force when the generator is connected short shunt. Assume a brush contact drop of \(1\ \text{V}\) per brush and an armature resistance of \(0.1\ \Omega\).

Solution.

\[ I_L = \dfrac{20000}{400} = 50\ \text{A} \qquad\qquad V_{\text{sh}} = 400 + 50(0.06 + 0.1) = 408\ \text{V} \]
\[ I_{\text{sh}} = \dfrac{408}{85} = 4.8\ \text{A} \qquad\qquad I_a = 50 + 4.8 = 54.8\ \text{A} \]
\[ E = V_{\text{sh}} + I_a R_a + V_b = 408 + 54.8(0.1) + 2 = 415.48\ \text{V} \]

Answer: \(415.48\ \text{V}\)

8. Voltage Regulation

Voltage regulation is the percentage change in terminal voltage when full load is removed while speed and field excitation remain constant.

Voltage Regulation Formula

\[ \%\text{VR} = \dfrac{V_{\text{NL}} - V_{\text{FL}}}{V_{\text{FL}}} \times 100\% \]

Where: \(\%\text{VR}\) = percentage voltage regulation, \(V_{\text{NL}}\) = no-load terminal voltage (V), \(V_{\text{FL}}\) = full-load terminal voltage (V)

9. Power Losses and Efficiency

Power losses in a DC generator convert input mechanical energy into heat and stray losses during operation.

9.1 General Equations

Total generator loss is the sum of copper losses, iron core losses, mechanical friction losses, and brush losses.

  • Armature copper loss: \(P_a = I_a^2 R_a\)
  • Shunt field copper loss: \(P_{\text{sh}} = V_L I_{\text{sh}} = I_{\text{sh}}^2 R_{\text{sh}}\)
  • Series field copper loss: \(P_{\text{se}} = I_{\text{se}}^2 R_{\text{se}}\)
  • Stray power loss: sum of friction, windage, and core losses

General Loss Equations

\[ P_{\text{loss}} = P_a + P_{\text{sh}} + P_{\text{se}} + P_{\text{stray}} + P_{\text{brush}} \]
\[ P_{\text{input}} = P_{\text{output}} + P_{\text{loss}} \]

Where: \(P_{\text{loss}}\) = total power loss (W), \(P_{\text{stray}}\) = rotational stray power loss (W), \(P_{\text{input}}\) = brake horsepower input (W)

9.2 Efficiency

Overall commercial efficiency is the ratio of output electrical power delivered to the load to total input mechanical power.

Generator Efficiency Formulas

\[ \eta = \dfrac{P_{\text{output}}}{P_{\text{input}}} \times 100\% = \dfrac{P_L}{P_L + P_{\text{loss}}} \times 100\% \]
\[ \eta_{\text{electrical}} = \dfrac{P_L}{P_g} \times 100\% = \dfrac{V_L I_L}{E I_a} \times 100\% \]

Where: \(\eta\) = overall efficiency, \(\eta_{\text{electrical}}\) = electrical efficiency, \(P_L\) = load output power (W), \(P_g\) = generated power (W)

EE Board Exam October 1980

Example. The field current of a \(100\ \text{kW}, 250\ \text{volt}\) shunt generator is \(5\ \text{amperes}\) when delivering power at rated terminal voltage and rated load current. The combined armature and brush resistance is \(0.01\ \Omega\). Determine the efficiency of the generator.

Solution.

\[ I_L = \dfrac{100000}{250} = 400\ \text{A} \qquad\qquad I_a = 400 + 5 = 405\ \text{A} \]
\[ E = 250 + 405(0.01) = 254.05\ \text{V} \qquad\qquad P_g = 254.05(405) = 102.89\ \text{kW} \]
\[ \eta = \dfrac{100}{102.89} \times 100\% = 97.2\% \]

Answer: \(97.2\%\)

9.3 Maximum Efficiency

Maximum efficiency occurs at the load current where variable copper losses equal constant rotational losses.

Maximum Efficiency Formulas

\[ I_a^2 R_a = P_k \]
\[ \eta_{\max} = \dfrac{P_L}{P_L + 2 P_k} \times 100\% \]

Where: \(P_k\) = constant or rotational losses (W), \(P_L\) = load power at maximum efficiency point (W)

Example. A \(30\ \text{kW}\) series generator has an efficiency of \(88\%\) when operating at rated load. If the stray power loss is \(15\%\) of the full load losses, calculate the maximum efficiency of the generator. Assume that the stray power loss is constant and the other losses vary as the square of the load.

Solution.

\[ P_{\text{lossFL}} = \dfrac{30000}{0.88} - 30000 = 4090.91\ \text{W} \]
\[ P_k = 0.15(4090.91) = 613.64\ \text{W} \qquad\qquad P_{\text{varFL}} = 0.85(4090.91) = 3477.27\ \text{W} \]
\[ \text{Load fraction } x = \sqrt{\dfrac{613.64}{3477.27}} = 0.4201 \qquad\qquad P_{\text{out}} = 0.4201(30000) = 12603\ \text{W} \]
\[ \eta_{\max} = \dfrac{12603}{12603 + 2(613.64)} \times 100\% = 91.12\% \]

Answer: \(91.12\%\)

10. Operating Characteristic Triangle of a DC Generator

Operating characteristic triangle defines the linear slope relationship between change in terminal voltage and change in load current.

Characteristic Triangle Equations

\[ \dfrac{\Delta V}{\Delta I} = \dfrac{V_{\text{NL}} - V_{\text{FL}}}{I_{\text{FL}}} \]
\[ \dfrac{\Delta V}{\Delta I} = \dfrac{\%\text{VR}(V_{\text{FL}})(100)}{I_{\text{FL}}} \]

Where: \(\Delta V\) = change in bus terminal voltage (V), \(\Delta I\) = change in generator load current (A), \(V_{\text{NL}}\) = no-load voltage (V), \(V_{\text{FL}}\) = full-load voltage (V), \(I_{\text{FL}}\) = full-load current (A)

11. Parallel Operation of DC Generators

Parallel operation connects two or more DC generators to a common bus bar to share total load current.

  • Terminal voltages must be equal in magnitude
  • Polarities must match strictly before closing bus switch
  • Drooping external characteristics ensure stable load division

11.1 Case 1: External Characteristics Given

External characteristics method evaluates parallel load distribution using speed-regulation slopes and bus voltage droop.

Parallel Load Sharing Equations

\[ I_t = I_a + I_b \]
\[ \Delta I = \Delta I_a + \Delta I_b \]
\[ V_{\text{busNEW}} = V_{\text{bus}} \pm \Delta V \]

Where: \(I_a, I_b\) = currents delivered by generators A and B (A), \(I_t\) = total bus current (A), \(\Delta V\) = change in bus operating voltage (V)

Example. Two \(250\ \text{V}\) DC generators are operating in parallel and taking equal shares of a \(3000\ \text{A}, 250\ \text{V}\) bus load. The ratings of the generators are: Generator 1: \(500\ \text{kW}\), Regulation \(2\%\); Generator 2: \(400\ \text{kW}\), Regulation \(4\%\). If the load drops to \(2100\ \text{A}\), determine the new operating bus voltage.

Solution.

\[ I_{\text{rated1}} = \dfrac{500000}{250} = 2000\ \text{A} \qquad\qquad I_{\text{rated2}} = \dfrac{400000}{250} = 1600\ \text{A} \]
\[ \Delta I_1 = \left[\dfrac{2000}{0.02(250)}\right]\Delta V = 400 \Delta V \qquad\qquad \Delta I_2 = \left[\dfrac{1600}{0.04(250)}\right]\Delta V = 160 \Delta V \]
\[ 400 \Delta V + 160 \Delta V = 3000 - 1500 = 1500 \implies \Delta V = 2.678\ \text{V} \]
\[ V_{\text{busNEW}} = 250 + 2.678 = 252.678\ \text{V} \approx 253\ \text{V} \]

Answer: \(253\ \text{V}\)

11.2 Case 2: Internal Characteristics Given

Internal characteristics method solves parallel load division using Nodal Analysis (KCL) with internal generated emfs and resistances.

Parallel Internal Nodal Equation

\[ I_{a1} + I_{a2} = I_L \]
\[ \dfrac{E_1 - V_L}{R_{a1}} + \dfrac{E_2 - V_L}{R_{a2}} = I_L \]

Where: \(E_1, E_2\) = generated internal emfs (V), \(R_{a1}, R_{a2}\) = internal armature resistances (\(\Omega\)), \(V_L\) = common bus terminal voltage (V), \(I_L\) = total load current (A)

Example. Two shunt generators operating in parallel are giving a total output of \(320\ \text{kW}\). Generator A has an armature resistance of \(0.02\ \Omega\) and induced emf of \(450\ \text{V}\). Generator B has an armature resistance of \(0.025\ \Omega\) and induced emf of \(460\ \text{V}\). Find the bus bar voltage. Neglect shunt field current.

Solution.

\[ \dfrac{450 - V_L}{0.02} + \dfrac{460 - V_L}{0.025} = \dfrac{320000}{V_L} \]
\[ (22500 - 50 V_L) + (18400 - 40 V_L) = \dfrac{320000}{V_L} \implies 40900 V_L - 90 V_L^2 = 320000 \]
\[ V_L^2 - 454.44 V_L + 3555.55 = 0 \implies V_L = 446.5\ \text{V} \]

Answer: \(446.5\ \text{V}\)