DC Generator

A DC generator converts mechanical energy into electrical energy using Faraday's law of electromagnetic induction: when a conductor cuts magnetic flux, an emf is induced in it, $e = Blv$ volts (direction by Fleming's right-hand rule).

Working:

1. The field poles produce a flux $\Phi$ across the air gap.
2. The prime mover rotates the armature, so the armature conductors cut this flux and an emf is induced in each conductor.
3. In a single coil the induced emf is alternating, because each coil side passes under a north pole and then a south pole.
4. The commutator and brushes act as a mechanical rectifier. Each time the coil emf reverses, the segments reverse their connection to the brushes, so the brush terminals give a unidirectional (DC) voltage.
5. With a load connected, current flows out through the positive brush.

        N pole            S pole
      +---------+       +---------+
      |    ^    |       |         |
      |  B -->  |       |         |
      |  ______ |_______|______   |
      | / loop (armature coil) \  |
      +-+-----------------------+-+
        | slip rings/commutator |
     brush A ----[ load ]---- brush B

The machine is a simple loop generator in the figure: one coil rotating between N and S poles with a split-ring commutator (two segments) and two brushes.

Derivation of the emf equation

Let

• $\Phi$ = flux per pole (Wb), $P$ = number of poles
• $Z$ = total armature conductors, $A$ = number of parallel paths
• $N$ = speed in rpm

In one revolution a conductor cuts the flux of all poles, $P\Phi$ weber, in time $60/N$ second.

$$\text{emf per conductor} = \frac{d\phi}{dt} = \frac{P\Phi}{60/N} = \frac{P\Phi N}{60}\ \text{V}$$

The $Z$ conductors form $A$ parallel paths, so each path has $Z/A$ conductors in series. The generated emf equals the emf of one path:

$$E_g = \frac{P\Phi N}{60}\times\frac{Z}{A} = \frac{\Phi Z N P}{60A}\ \text{V}$$

• Lap winding: $A = P$, so $E_g = \dfrac{\Phi Z N}{60}$
• Wave winding: $A = 2$, so $E_g = \dfrac{\Phi Z N P}{120}$

In terms of angular speed, $E_g = \dfrac{ZP}{2\pi A}\Phi\omega = K_a\Phi\omega$.

Voltage build-up in a DC shunt generator

A shunt generator is self-excited, so its voltage builds up from the residual magnetism in the poles.

1. The armature is driven at rated speed. Residual flux induces a small emf $E_r$ (about 1–3 V) in the armature.
2. This emf sends a small current $I_f = E_r/(R_f + R_a)$ through the shunt field winding.
3. If the field connections are correct (the field current aids the residual flux), the flux increases, which raises the emf, which raises the field current, and so on.
4. The build-up stops at the point where the field-resistance line ($V = I_fR_f$) cuts the open-circuit characteristic (OCC). Here the emf generated equals the voltage drop in the field circuit. This is the no-load terminal voltage.

 V (volts)          OCC
  |              _______---
  |           _-'    /  Rf line
  |  b      ,'      /
  |      ,-'      /
  |   ,-'       /   <- Rc (tangent)
  | ,'        /
  |/-- Er  /
  +-------------------------- If

Conditions for build-up: (i) residual magnetism, (ii) field winding connected so it aids the residual flux, (iii) total field circuit resistance less than the critical resistance.

Critical field resistance ($R_c$)

It is the maximum value of field circuit resistance with which the shunt generator will still build up at a given speed. Graphically, it is the slope of the field-resistance line that is tangent to the initial straight part of the OCC. If $R_f > R_c$, the line cuts the OCC only near the residual voltage, so the generator does not build up.

Critical speed ($N_c$)

It is the minimum speed below which a shunt generator will not build up voltage for a given field resistance. The OCC is for one speed; at lower speed the OCC drops (since $E \propto N$). The speed at which the OCC becomes tangent to the given field-resistance line is the critical speed.

Relation: $R_c \propto N$, i.e. $\dfrac{R_{c2}}{R_{c1}} = \dfrac{N_2}{N_1}$.

In a shunt generator the armature current supplies both the load and the field: $I_a = I_L + I_{sh}$. The generated emf must also cover the armature drop and the brush drop.

(a) Armature current

Load current:

$$I_L = \frac{P}{V} = \frac{3000}{120} = 25\ \text{A}$$

Shunt field current:

$$I_{sh} = \frac{V}{R_{sh}} = \frac{120}{60} = 2\ \text{A}$$ $$I_a = I_L + I_{sh} = 25 + 2 = 27\ \text{A}$$

(b) Current in each armature conductor

A wave winding has $A = 2$ parallel paths regardless of the number of poles.

$$I_{\text{conductor}} = \frac{I_a}{A} = \frac{27}{2} = 13.5\ \text{A}$$

(c) Generated emf

Brush drop $= 1\ \text{V} \times 2\ \text{brushes (one positive, one negative)} = 2\ \text{V}$.

$$\begin{aligned} E &= V + I_aR_a + V_{brush} \\ &= 120 + (27)(0.2) + 2 \\ &= 120 + 5.40 + 2 = 127.40\ \text{V} \end{aligned}$$

Answer: (a) $I_a = 27\ \text{A}$; (b) current per conductor $= 13.5\ \text{A}$; (c) $E = 127.40\ \text{V}$.

In a long shunt compound generator the shunt field is across the armature plus series field combination (i.e. across the load terminals), so the series field carries the armature current.

       Ia        Rse
  +--->---[~~~~]---+---------+----o +
  |                |         |
 (E,Ra)          [Rsh]     Load
  |                |         |
  +----------------+---------+----o -

Currents

$$I_L = \frac{10000}{250} = 40\ \text{A}, \qquad I_{sh} = \frac{250}{250} = 1\ \text{A}$$ $$I_a = I_{se} = I_L + I_{sh} = 40 + 1 = 41\ \text{A}$$

Generated emf

$$E_g = V + I_a(R_a + R_{se}) = 250 + 41(0.1 + 0.15) = 260.25\ \text{V}$$

Speed with lap winding

Slots $\times$ conductors per slot: $Z = 50\times 6 = 300$. For lap winding $A = P = 4$.

$$E_g = \frac{\Phi Z N P}{60A} \;\Rightarrow\; N = \frac{60AE_g}{\Phi Z P} = \frac{60\times 4\times 260.25}{0.05\times 300\times 4} = 1041.0\ \text{rpm}$$

Speed with wave winding

For wave winding $A = 2$; $E_g$, $\Phi$, $Z$ and $P$ are unchanged.

$$N = \frac{60\times 2\times 260.25}{0.05\times 300\times 4} = 520.5\ \text{rpm}$$

The wave-wound machine needs only half the speed because it has half the parallel paths (twice the conductors in series per path).

Answer: lap wound: $N = 1041.0\ \text{rpm}$; wave wound: $N = 520.5\ \text{rpm}$. (Brush drop is not given, so it is neglected.)

In a long shunt machine the shunt field is across the load terminals, so the series field carries the full armature current.

       Ia        Rse
  +--->---[~~~~]---+---------+----o +
  |                |         |
 (E,Ra)          [Rsh]     Load
  |                |         |
  +----------------+---------+----o -

Currents

$$I_L = \frac{10000}{250} = 40\ \text{A}, \qquad I_{sh} = \frac{250}{250} = 1\ \text{A}$$ $$I_a = I_{se} = I_L + I_{sh} = 41\ \text{A}$$

Generated emf

$$E_g = V + I_a(R_a + R_{se}) = 250 + 41(0.1 + 0.15) = 260.25\ \text{V}$$

Speed

For lap winding $A = P = 4$:

$$E_g = \frac{\Phi Z N P}{60A} \;\Rightarrow\; N = \frac{60AE_g}{\Phi Z P} = \frac{60\times 4\times 260.25}{0.05\times 300\times 4} = 1041.0\ \text{rpm}$$

Answer: generator speed $= 1041.0\ \text{rpm}$ (brush drop not given, so neglected).

In a long shunt compound generator the shunt field is connected across the load terminals, so the series field carries the full armature current.

Given: $I_L = 80$ A, $V = 230$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.04\ \Omega$, $R_a = 0.2\ \Omega$.

       Ia        Rse
  +--->---[~~~~]---+---------+----o +
  |                |         |
 (E,Ra)          [Rsh]     Load
  |                |         |
  +----------------+---------+----o -

Step 1: Shunt field current

$$I_{sh} = \frac{V}{R_{sh}} = \frac{230}{100} = 2.30\ \text{A}$$

Step 2: Armature current

$$I_a = I_{se} = I_L + I_{sh} = 80 + 2.30 = 82.30\ \text{A}$$

Step 3: Generated emf

$$E = V + I_a(R_a + R_{se}) = 230 + 82.30(0.2 + 0.04) = 230 + 19.752 = 249.75\ \text{V}$$

Answer: emf generated by the armature $E = 249.75\ \text{V}$.

In a short shunt compound generator the shunt field is connected across the armature only, so the series field carries the load current $I_L$ and the shunt field sees $V + I_LR_{se}$.

Given: $I_L = 80$ A, $V = 220$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.05\ \Omega$, $R_a = 0.1\ \Omega$.

       Ia                 Rse   IL
  +--->---+----------+--[~~~]--->--+--o +
  |       |          |            |
 (E,Ra)   |        [Rsh]        Load
  |       |          |            |
  +-------+----------+------------+--o -

Step 1: Voltage across the shunt field

$$V_{sh} = V + I_LR_{se} = 220 + 80(0.05) = 224.00\ \text{V}$$

Step 2: Currents

$$I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{224.00}{100} = 2.240\ \text{A}, \qquad I_a = I_L + I_{sh} = 80 + 2.240 = 82.240\ \text{A}$$

Step 3: Generated emf

$$E = V_{sh} + I_aR_a = 224.00 + (82.240)(0.1) = 232.22\ \text{V}$$

Answer: generated emf $E = 232.22\ \text{V}$.

For a shunt generator $I_a = I_L + I_{sh}$ and $E = V + I_aR_a$.

(i) Generated emf

$$I_L = \frac{30000}{200} = 150\ \text{A}, \qquad I_{sh} = \frac{200}{50} = 4\ \text{A}$$ $$I_a = I_L + I_{sh} = 154\ \text{A}$$ $$E = V + I_aR_a = 200 + (154)(0.05) = 207.70\ \text{V}$$

(ii) Copper losses

$$\begin{aligned} \text{Armature copper loss} &= I_a^2R_a = (154)^2(0.05) = 1185.8\ \text{W} \\ \text{Shunt field copper loss} &= I_{sh}^2R_{sh} = (4)^2(50) = 800.0\ \text{W} \\ \text{Total copper loss} &= 1185.8 + 800.0 = 1985.8\ \text{W} \end{aligned}$$

(iii) Efficiency

Total losses $=$ copper loss $+$ iron and friction loss $= 1985.8 + 1000 = 2985.8\ \text{W}$.

$$\eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{30000}{30000 + 2985.8}\times 100 = 90.95\%$$

Answer: (i) $E = 207.70\ \text{V}$; (ii) copper loss $= 1985.8\ \text{W}$; (iii) $\eta = 90.95\%$.

For a shunt generator $I_a = I_L + I_{sh}$ and $E = V + I_aR_a$.

(i) Generated emf

$$I_L = \frac{30000}{200} = 150\ \text{A}, \qquad I_{sh} = \frac{200}{50} = 4\ \text{A}$$ $$I_a = I_L + I_{sh} = 154\ \text{A}$$ $$E = V + I_aR_a = 200 + (154)(0.05) = 207.70\ \text{V}$$

(ii) Copper losses

$$\begin{aligned} \text{Armature copper loss} &= I_a^2R_a = (154)^2(0.05) = 1185.8\ \text{W} \\ \text{Shunt field copper loss} &= I_{sh}^2R_{sh} = (4)^2(50) = 800.0\ \text{W} \\ \text{Total copper loss} &= 1185.8 + 800.0 = 1985.8\ \text{W} \end{aligned}$$

(iii) Efficiency

Total losses $=$ copper loss $+$ iron and friction loss $= 1985.8 + 100 = 2085.8\ \text{W}$.

$$\eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{30000}{30000 + 2085.8}\times 100 = 93.50\%$$

Answer: (i) $E = 207.70\ \text{V}$; (ii) copper loss $= 1985.8\ \text{W}$; (iii) $\eta = 93.50\%$.

Principle

A DC generator works on Faraday's law of electromagnetic induction. Whenever a conductor moves in a magnetic field so that it cuts flux, an emf is induced:

$$e = Blv\ \text{volts}$$

The direction of the induced emf is given by Fleming's right-hand rule (thumb = motion, forefinger = field, middle finger = emf).

Working with the figure

        N pole            S pole
      +---------+       +---------+
      |    ^    |       |         |
      |  B -->  |       |         |
      |  ______ |_______|______   |
      | / loop (armature coil) \  |
      +-+-----------------------+-+
        | slip rings/commutator |
     brush A ----[ load ]---- brush B

1. The stator poles (N and S) set up a field. The armature coil is rotated by a prime mover (turbine, engine or motor).
2. Sides of the coil cut the flux, so an emf is induced in the coil. When a coil side is under the N pole the emf is in one direction; under the S pole it is in the opposite direction, so the coil emf is alternating.
3. The coil ends are joined to a split-ring commutator (two half rings insulated from each other). The brushes press on the commutator.
4. When the coil emf reverses, the commutator segments also change brushes. Brush A always touches the segment connected to the coil side under the N pole, so brush A stays positive and brush B stays negative.
5. The voltage across the brushes is therefore unidirectional but pulsating. A practical machine uses many coils and many commutator segments, so the output is nearly steady DC.

Key point: the generator itself produces AC in the armature conductors; the commutator rectifies it.

A DC generator converts mechanical energy into DC electrical energy by electromagnetic induction.

Construction

   +----------- Yoke (frame) ------------+
   |  [Pole core + shoe]    [Pole core]  |
   |   field winding         field wdg   |
   |        ( Armature core + winding )  |
   |        Commutator -- Brushes        |
   +---- Shaft, bearings, end covers ----+

1. Yoke (frame): cast iron or cast steel outer frame. It gives mechanical support and carries the return path of the magnetic flux.
2. Pole core and pole shoe: laminated steel bolted to the yoke. The shoe spreads the flux evenly over the air gap and holds the field coils.
3. Field winding: copper coils on the pole cores that produce the main flux when excited.
4. Armature core: cylindrical, made of thin laminated silicon steel (to reduce eddy-current loss) with slots on the outside to hold conductors.
5. Armature winding: insulated copper conductors in the slots, connected as lap or wave winding.
6. Commutator: cylinder of hard-drawn copper segments insulated by mica. It rectifies the armature emf and collects the current.
7. Brushes and brush gear: carbon brushes in holders press on the commutator and carry current to the external circuit.
8. Shaft, bearings and end covers: support and rotate the armature.

Working principle

When the armature is rotated by a prime mover in the field of the poles, its conductors cut flux and an emf $e = Blv$ is induced (Faraday's law). The emf in each coil is alternating; the commutator and brushes convert it to a unidirectional output at the terminals.

EMF equation

Let

• $\Phi$ = flux per pole (Wb), $P$ = number of poles
• $Z$ = total armature conductors, $A$ = number of parallel paths
• $N$ = speed in rpm

In one revolution a conductor cuts the flux of all poles, $P\Phi$ weber, in time $60/N$ second.

$$\text{emf per conductor} = \frac{d\phi}{dt} = \frac{P\Phi}{60/N} = \frac{P\Phi N}{60}\ \text{V}$$

The $Z$ conductors form $A$ parallel paths, so each path has $Z/A$ conductors in series. The generated emf equals the emf of one path:

$$E_g = \frac{P\Phi N}{60}\times\frac{Z}{A} = \frac{\Phi Z N P}{60A}\ \text{V}$$

• Lap winding: $A = P$, so $E_g = \dfrac{\Phi Z N}{60}$
• Wave winding: $A = 2$, so $E_g = \dfrac{\Phi Z N P}{120}$

In terms of angular speed, $E_g = \dfrac{ZP}{2\pi A}\Phi\omega = K_a\Phi\omega$.

A DC generator converts mechanical energy into electrical energy using Faraday's law of electromagnetic induction: when a conductor cuts magnetic flux, an emf is induced in it, $e = Blv$ volts (direction by Fleming's right-hand rule).

Working:

1. The field poles produce a flux $\Phi$ across the air gap.
2. The prime mover rotates the armature, so the armature conductors cut this flux and an emf is induced in each conductor.
3. In a single coil the induced emf is alternating, because each coil side passes under a north pole and then a south pole.
4. The commutator and brushes act as a mechanical rectifier. Each time the coil emf reverses, the segments reverse their connection to the brushes, so the brush terminals give a unidirectional (DC) voltage.
5. With a load connected, current flows out through the positive brush.

Functions of carbon brushes

1. They provide a sliding contact with the rotating commutator and collect the current from the armature to the external circuit (or supply it, in a motor).
2. They connect the armature coils to the load at the right moment, so that coils undergoing commutation are short-circuited by the brush and the current in them reverses.
3. Carbon is used because it is self-lubricating, has a high contact resistance (which helps commutation and reduces sparking), and is softer than copper, so it wears rather than the commutator.

The brushes are set on the magnetic neutral axis, where the emf in the short-circuited coil is nearly zero, so sparking is minimum.

The emf equation gives the average emf generated in the armature of a DC machine.

Let

• $\Phi$ = flux per pole (Wb), $P$ = number of poles
• $Z$ = total armature conductors, $A$ = number of parallel paths
• $N$ = speed in rpm

In one revolution a conductor cuts the flux of all poles, $P\Phi$ weber, in time $60/N$ second.

$$\text{emf per conductor} = \frac{d\phi}{dt} = \frac{P\Phi}{60/N} = \frac{P\Phi N}{60}\ \text{V}$$

The $Z$ conductors form $A$ parallel paths, so each path has $Z/A$ conductors in series. The generated emf equals the emf of one path:

$$E_g = \frac{P\Phi N}{60}\times\frac{Z}{A} = \frac{\Phi Z N P}{60A}\ \text{V}$$

• Lap winding: $A = P$, so $E_g = \dfrac{\Phi Z N}{60}$
• Wave winding: $A = 2$, so $E_g = \dfrac{\Phi Z N P}{120}$

In terms of angular speed, $E_g = \dfrac{ZP}{2\pi A}\Phi\omega = K_a\Phi\omega$.

Function of the commutator

• It acts as a mechanical rectifier: it converts the alternating emf induced in the armature conductors into a unidirectional (DC) voltage at the brushes.
• It collects the current from the armature conductors and passes it to the brushes.
• It connects the correct coil to the brush at the correct time. The copper segments are separated by mica insulation.

Function of the carbon brushes

• They make sliding contact with the commutator and carry the armature current to the external load.
• Carbon is self-lubricating, with a high contact resistance that reduces sparking and a low wear on the commutator.
• They are placed at the magnetic neutral axis for sparkless commutation.

Why a shunt generator is started without load

A shunt generator is self-excited. The emf builds up from the residual magnetism like this: residual emf $\to$ small field current $\to$ more flux $\to$ more emf, until the field-resistance line meets the magnetisation curve.

If the load is connected during starting:

1. The terminal voltage is very low, so a large part of the armature current goes to the load and the field current is reduced ($I_{sh} = V/R_{sh}$), because the load is in parallel with the field.
2. The voltage drop $I_aR_a$ in the armature takes away a larger share of the small emf, so the terminal voltage stays near zero.
3. The machine may fail to build up its voltage, and a heavy overload current may flow.

Hence the load is connected only after the voltage has reached its rated value.

A shunt generator is self-excited: its field winding is connected across its own armature, so the field current comes from the voltage the machine itself produces.

Voltage build-up: residual flux produces a small emf $\to$ this drives a small field current through $R_{sh}$ $\to$ the field current strengthens the flux $\to$ the emf rises. This continues until the field-resistance line meets the open-circuit characteristic.

Reasons for starting on open circuit:

1. With a load connected, the load draws current from the small initial voltage, and the armature drop $I_aR_a$ reduces the terminal voltage further. The field current $I_{sh} = V/R_{sh}$ therefore stays too small to raise the flux.
2. A low-resistance load across the terminals effectively short-circuits the field, so the machine never builds up (the voltage collapses to the residual value).
3. A heavy load current at low voltage would also overload the armature without delivering useful power.

With the output terminals open, the whole armature current goes into the field, so the voltage builds up quickly to its rated value. The load is switched on afterwards.

Voltage build-up in a DC shunt generator

The shunt field is across the armature, so the generator excites itself from residual magnetism.

1. When the armature rotates, the residual flux induces a small emf $E_r$ (1–3 V).
2. This emf drives a small current $I_f = E_r/(R_f + R_a)$ through the shunt field.
3. If the field connections are such that this current strengthens the residual flux, the flux rises, which raises the emf, which raises $I_f$ further.
4. The process continues until the field-resistance line ($V = I_fR_f$) meets the open-circuit characteristic. The voltage at this point is the no-load voltage.

 E, V
  |           ____---- OCC
  |        .-'    /  V = If.Rf
  |     .-'     /
  |   .'      /
  | Er.-----/
  +--------------------- If

Conditions: residual magnetism present, correct field polarity, field resistance below the critical value, speed above the critical speed.

Why a series generator is not started at no load

In a series generator the field winding is in series with the armature and the load, so the field current equals the load current.

• At no load the circuit is open, so $I_a = I_f = 0$.
• With no field current there is only residual flux, so the generated emf is just the small residual emf (a few volts). The machine cannot build up its voltage.
• Voltage appears only after a load is connected, and it then rises with the load current (until saturation).

So the series generator must be started with a load (a closed circuit) connected to get its normal voltage.

Excitation means supplying current to the field winding to produce the main flux. On this basis DC generators are divided into two groups.

1. Separately excited generator

The field winding is fed from an independent DC source (battery or another generator).

  +----[Rheo]---[ Field ]        Armature
  | Ext. DC                        (E, Ra)---->  Load

• Field current is independent of the armature load, so flux is nearly constant.
• Terminal voltage falls only slightly with load (armature drop and armature reaction).
• Used where wide voltage control is needed (e.g. Ward-Leonard system, testing).

2. Self-excited generator

The field current is taken from the generator's own armature. Residual magnetism is required to start.

(a) Shunt generator: the field winding (many turns of thin wire, high resistance) is connected across the armature.

$$I_a = I_L + I_{sh}, \qquad V = E - I_aR_a$$

It gives nearly constant voltage with moderate drooping; used for battery charging and lighting.

(b) Series generator: the field winding (few turns of thick wire) is in series with the armature and load.

$$I_a = I_{se} = I_L, \qquad V = E - I_a(R_a + R_{se})$$

The voltage rises with load current. Used as a booster and in arc lamps.

(c) Compound generator: has both shunt and series fields.

• Short shunt: the shunt field is across the armature only; the series field is in series with the load.
• Long shunt: the shunt field is across the armature and series field combination.
• Cumulative compound: the series field aids the shunt field (flat, over- or under-compounded).
• Differential compound: the series field opposes the shunt field (drooping voltage; used in welding).

 Short shunt                 Long shunt
 +--[Ra]--+--[Rse]---o       +--[Ra]--[Rse]--+---o
 |        |                  |               |
 |      [Rsh]   Load         |             [Rsh]  Load
 +--------+------------o     +---------------+---o

Types of DC generator

Generators are classified by the way the field is excited.

1. Separately excited: the field is supplied from an independent DC source. Field current does not depend on the load; the terminal voltage is almost constant (it drops slightly due to $I_aR_a$ and armature reaction).
2. Self-excited: the field is fed from the generator's own armature.
   • Shunt: field winding across the armature. $I_a = I_L + I_{sh}$. The voltage drops slightly with load.
   • Series: field winding in series with armature and load. $I_a = I_L = I_{se}$. The voltage rises with load current, so it is used as a booster.
   • Compound: both fields. It is short shunt (shunt field across armature only) or long shunt (shunt field across armature plus series field), and cumulative (series aids shunt) or differential (series opposes shunt).

Loading (external) characteristic of a compound generator

It is the plot of terminal voltage $V$ against load current $I_L$ at constant speed.

 V |   over-compound
   |      ___----
   |  ___/---- flat
   | /  ----______ under-compound
   |/            \___
   |               \____ differential
   +----------------------------- IL

• Cumulative compound: the series field adds flux as load increases, which compensates for the drop due to armature resistance and armature reaction.
  • Over-compounded: voltage at full load is higher than at no load (series ampere-turns are large).
  • Flat (level) compounded: full-load voltage equals no-load voltage.
  • Under-compounded: full-load voltage is less than no-load voltage, but the fall is less than for a shunt generator.
• Differential compound: the series field opposes the shunt field, so flux decreases with load and the voltage falls rapidly (drooping). It has a nearly constant-current characteristic and is used for arc welding.

Cumulative flat-compounded generators are used for supplying constant voltage at a distance from the generator (such as lighting and power feeders).

Types of armature winding

An armature winding consists of coils placed in slots and joined to the commutator segments. It is classified by how the coils are connected.

1. Lap winding

• The end of a coil is connected to the commutator segment next to the one where the coil began, so successive coils overlap (lap back).
• Number of parallel paths $A = P$ (number of poles).
• Number of brushes = number of poles.
• Needs equaliser rings to balance emf. Used for low voltage, high current machines.

2. Wave winding

• A coil end is connected to a segment about one pole-pair pitch ahead, so the winding progresses like a wave around the armature.
• $A = 2$ always; only two brushes are needed (though $P$ may be used).
• Gives higher emf with fewer turns. Used for high voltage, low current machines.

3. Frog-leg winding: a combination of lap and wave in the same armature, used where equalisers are needed.

Both lap and wave may be simplex, duplex or triplex, and progressive or retrogressive. Coils are normally double-layer (one side in the top of a slot, the other in the bottom).

Lap winding: 12 slots, 2 poles, 12 segments

Each slot has 2 coil sides (double layer), so the number of coils $= 12$ and conductors $Z = 24$.

• Pole pitch $= Z/P = 24/2 = 12$ conductors (6 slots). The coil is made slightly short of full pitch because $y_b$ must be odd.
• Back pitch $y_b = 11$ (odd), front pitch $y_f = 9$ (for progressive lap $y_b - y_f = 2$).
• Commutator pitch $y_c = 1$ segment; $A = P = 2$, so 2 brushes.

Coil $k$ goes from conductor $2k-1$ to $2k+10$ and its finish is joined to the start of coil $k+1$ at segment $k+1$.

 Developed view (coil 1 and 2 shown)
      N                      S
   |--------|-------------|--------|
 slot 1 2 3 4 5 6 7 8 9 10 11 12
 coil 1: top of slot 1 --> bottom of slot 6
 coil 2: top of slot 2 --> bottom of slot 7
 ....
 seg:  1  2  3  4  5  6  7  8  9 10 11 12
 brushes (2): on segments under the neutral axes

Each brush covers a segment, so the coils form 2 parallel paths (6 coils in series per path).

Armature reaction is the effect of the magnetic field set up by the armature current on the main field flux of the machine.

When the generator is loaded, the armature carries current and produces its own mmf. This acts on the main field and has two effects.

1. Distortion (cross-magnetising effect): the armature field is at right angles to the main field. It strengthens the flux under one half of each pole and weakens it under the other half. The flux distribution becomes uneven and the geometrical neutral axis (GNA) shifts to the magnetic neutral axis (MNA) (forward in direction of rotation for a generator).
2. Demagnetising effect: if the brushes are shifted to the new MNA, the armature mmf has a component directly opposing the main field. This reduces the net flux per pole.

   Main flux  -->   |  Armature flux (cross)  |
   Resultant: crowded at one pole tip, weak at other
   GNA ----- shifted to MNA by angle theta

Harmful effects:

• Reduced total flux, so the generated emf is lower.
• Shifted neutral axis, so brushes spark if kept on the GNA.
• Flux crowding at pole tips causes saturation and extra iron loss.

Remedies:

• Shift brushes to the MNA (small machines).
• Use interpoles (commutating poles) to neutralise cross-magnetisation in the commutation zone.
• Use compensating windings in the pole faces on large and fast-varying machines.
• Increase the air gap or use a laminated pole tip with slots.

Brush drop $= 1\ \text{V}\times 2\ \text{brushes} = 2\ \text{V}$. Load current $I_L = 50$ A, terminal voltage $V = 500$ V.

(i) Long shunt

The shunt field is across the load terminals:

$$I_{sh} = \frac{V}{R_{sh}} = \frac{500}{250} = 2\ \text{A}, \qquad I_a = I_{se} = 50 + 2 = 52\ \text{A}$$ $$\begin{aligned} E &= V + I_a(R_a + R_{se}) + V_{brush} \\ &= 500 + 52(0.05 + 0.03) + 2 \\ &= 500 + 4.16 + 2 = 506.16\ \text{V} \end{aligned}$$

(ii) Short shunt

The series field carries only the load current. The shunt field is across the armature, so its voltage is

$$V_{sh} = V + I_LR_{se} = 500 + 50(0.03) = 501.50\ \text{V}$$ $$I_{sh} = \frac{501.50}{250} = 2.006\ \text{A}, \qquad I_a = I_L + I_{sh} = 50 + 2.006 = 52.006\ \text{A}$$ $$\begin{aligned} E &= V_{sh} + I_aR_a + V_{brush} \\ &= 501.50 + (52.006)(0.05) + 2 \\ &= 501.50 + 2.600 + 2 = 506.10\ \text{V} \end{aligned}$$

Answer: (i) long shunt $E = 506.16\ \text{V}$; (ii) short shunt $E = 506.10\ \text{V}$.

The heater load takes $I_L = 175$ A, so the terminal voltage follows from Ohm's law. In a short shunt machine the series field carries the load current and the shunt field is across the armature.

       Ia                 Rse   IL
  +--->---+----------+--[~~~]--->--+--o +
  |       |          |            |
 (E,Ra)   |        [Rsh]       Heaters
  |       |          |            |
  +-------+----------+------------+--o -

Terminal voltage

$$V = I_L R_L = 175 \times 1.4 = 245.0\ \text{V}$$

Shunt and armature currents

$$V_{sh} = V + I_LR_{se} = 245.0 + 175(0.1) = 262.5\ \text{V}$$ $$I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{262.5}{100} = 2.625\ \text{A}, \qquad I_a = I_L + I_{sh} = 177.625\ \text{A}$$

Generated emf

Brush drop $= 2\ \text{V}\times 2 = 4\ \text{V}$.

$$E = V_{sh} + I_aR_a + V_{brush} = 262.5 + (177.625)(15) + 4 = 2930.9\ \text{V}$$

Copper losses

$$\begin{aligned} P_a &= I_a^2R_a = (177.625)^2(15) = 473259.6\ \text{W} \\ P_{se} &= I_L^2R_{se} = 175^2(0.1) = 3062.5\ \text{W} \\ P_{sh} &= I_{sh}^2R_{sh} = (2.625)^2(100) = 689.1\ \text{W} \\ P_{cu} &= 473259.6 + 3062.5 + 689.1 = 477011.2\ \text{W} \end{aligned}$$

Electrical efficiency

$$\text{Output} = VI_L = 245.0 \times 175 = 42875\ \text{W}$$ $$\eta_{elec} = \frac{\text{output}}{\text{output} + P_{cu}} = \frac{42875}{42875 + 477011.2}\times 100 = 8.25\%$$

Answer: $E = 2930.9\ \text{V}$, copper loss $= 477011.2\ \text{W}$, electrical efficiency $= 8.25\%$. (The armature resistance of $15\ \Omega$ is used as given.)

In a long shunt compound generator the shunt field is connected across the load terminals, so the series field carries the full armature current.

Given: $I_L = 40$ A, $V = 200$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.5\ \Omega$, $R_a = 0.4\ \Omega$.

       Ia        Rse
  +--->---[~~~~]---+---------+----o +
  |                |         |
 (E,Ra)          [Rsh]     Load
  |                |         |
  +----------------+---------+----o -

Step 1: Shunt field current

$$I_{sh} = \frac{V}{R_{sh}} = \frac{200}{100} = 2.00\ \text{A}$$

Step 2: Armature current

$$I_a = I_{se} = I_L + I_{sh} = 40 + 2.00 = 42.00\ \text{A}$$

Step 3: Generated emf

$$E = V + I_a(R_a + R_{se}) = 200 + 42.00(0.4 + 0.5) = 200 + 37.800 = 237.80\ \text{V}$$

Answer: emf generated by the armature $E = 237.80\ \text{V}$.

Short shunt: the shunt field is across the armature and the series field is in the load line. Cumulative: the series field aids the shunt field, which does not change the circuit calculation.

Load current and load resistance

$$I_L = \frac{P}{V} = \frac{7500}{230} = 32.609\ \text{A}, \qquad R_L = \frac{V}{I_L} = \frac{230}{32.609} = 7.053\ \Omega$$

Shunt field and armature currents

$$V_{sh} = V + I_LR_{se} = 230 + (32.609)(0.3) = 239.783\ \text{V}$$ $$I_{sh} = \frac{239.783}{100} = 2.3978\ \text{A}, \qquad I_a = I_L + I_{sh} = 35.007\ \text{A}$$

Induced emf

$$E = V_{sh} + I_aR_a = 239.783 + (35.007)(0.4) = 253.79\ \text{V}$$

Answer: induced emf $= 253.79\ \text{V}$; load resistance $= 7.053\ \Omega$.

In a short shunt compound generator the shunt field is connected across the armature only, so the series field carries the load current $I_L$ and the shunt field sees $V + I_LR_{se}$.

Given: $I_L = 100$ A, $V = 250$ V, $R_{sh} = 130\ \Omega$, $R_{se} = 0.1\ \Omega$, $R_a = 0.1\ \Omega$.

Brush drop $= 2$ V (1 V per brush, 2 brushes).

       Ia                 Rse   IL
  +--->---+----------+--[~~~]--->--+--o +
  |       |          |            |
 (E,Ra)   |        [Rsh]        Load
  |       |          |            |
  +-------+----------+------------+--o -

Step 1: Voltage across the shunt field

$$V_{sh} = V + I_LR_{se} = 250 + 100(0.1) = 260.00\ \text{V}$$

Step 2: Currents

$$I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{260.00}{130} = 2.000\ \text{A}, \qquad I_a = I_L + I_{sh} = 100 + 2.000 = 102.000\ \text{A}$$

Step 3: Generated emf

$$E = V_{sh} + I_aR_a + 2 = 260.00 + (102.000)(0.1) + 2 = 272.20\ \text{V}$$

Answer: generated emf $E = 272.20\ \text{V}$ and armature current $I_a = 102.000\ \text{A}$.

The load (10 Ω) and the field winding (100 Ω) are both across the armature terminals, so they are in parallel. The generated emf comes from the emf equation for a wave winding ($A = 2$).

Step 1: Generated emf

$$E = \frac{\Phi Z N P}{60A} = \frac{0.02\times 378\times 1000\times 4}{60\times 2} = 252\ \text{V}$$

Step 2: Terminal resistance and armature current

Load in parallel with field:

$$R_p = \frac{100\times 10}{100 + 10} = 9.091\ \Omega$$ $$I_a = \frac{E}{R_a + R_p} = \frac{252}{1 + 9.091} = 24.973\ \text{A}$$

Step 3: Terminal voltage

$$V = I_aR_p = (24.973)(9.091) = 227.03\ \text{V}$$

Step 4: Load current and power

$$I_L = \frac{V}{10} = 22.703\ \text{A}, \qquad I_f = \frac{V}{100} = 2.270\ \text{A}$$

(Check: $I_L + I_f = 24.973\ \text{A} = I_a$.)

$$P_{load} = \frac{V^2}{R_L} = \frac{(227.03)^2}{10} = 5154\ \text{W}$$

Answer: power absorbed by the load $\approx 5154\ \text{W}$ ($\approx 5.15$ kW) at a terminal voltage of $227.03\ \text{V}$.

For flat compounding, the series field must supply the extra field ampere-turns that a shunt generator needs to keep the voltage constant from no load to full load. A diverter in parallel with the series field passes the surplus armature current.

        Ia      +----[Rd]----+
   ---->--------+            +----+---> to load
                +--[ Rse ]---+    |
                   Ise            Rsh (across load)

Step 1: Extra ampere-turns needed

$$AT_{extra} = \Delta I_{sh}\times N_{sh} = 0.2 \times 1000 = 200\ \text{AT per pole}$$

Step 2: Series field current required

$$I_{se} = \frac{AT_{extra}}{N_{se}} = \frac{200}{4} = 50\ \text{A}$$

Step 3: Diverter current

The series field is in the armature circuit (long shunt). Taking the full-load armature current of 80 A (the small shunt current is neglected):

$$I_d = I_a - I_{se} = 80 - 50 = 30\ \text{A}$$

Step 4: Diverter resistance

The diverter and series field are in parallel, so their voltage drops are equal:

$$I_dR_d = I_{se}R_{se} \;\Rightarrow\; R_d = \frac{I_{se}R_{se}}{I_d} = \frac{50\times 0.05}{30} = 0.0833\ \Omega$$

Answer: diverter resistance $R_d = 0.0833\ \Omega$.