DC Generator

A self-excited DC shunt generator builds up its own field current from the small voltage due to residual magnetism, with no external DC source.

Voltage build-up process

     +---------+-----------o +
     |         |
   ( G )     [Rsh]  field         V
     |         |  winding
     +---------+-----------o -

1. The poles keep a little residual magnetism. When the armature is driven at rated speed, a small EMF $E_r$ (a few volts) is induced.
2. This EMF drives a small current through the shunt field winding (connected across the armature).
3. If the field is connected so that its MMF aids the residual flux, the flux increases, so the induced EMF increases.
4. The higher EMF drives more field current, which increases flux further, and so on (cumulative process).
5. Build-up stops where the OCC (magnetisation curve, $E$ vs $I_f$) meets the field resistance line ($V = I_fR_f$), because at that point the generated EMF just equals the drop across the field circuit. Saturation of the iron makes the OCC bend over, so the two curves intersect at a definite point.

  E ^           field resistance line
    |         /     (slope = Rf)
    |        /   ___....--- OCC
    |       / .-'
    |      /.'   <- operating point
    |     .'      (OCC meets Rf line)
    |   .'/
    |  / /  steps of build-up
    | / /
    |//
 Er +------------------------> If

Critical field resistance

The critical field resistance $R_c$ is the maximum shunt-field circuit resistance at which the generator will just build up voltage at a given speed. It equals the slope of the tangent to the linear part of the OCC drawn from the origin. If $R_f > R_c$, the field line lies above the OCC, they meet only near the residual voltage, and the machine fails to build up.

Critical speed

The critical speed $N_c$ is the minimum speed at which a shunt generator with a given field resistance $R_f$ will just build up. At that speed the tangent of the OCC equals $R_f$. Since EMF ∝ speed:

$$\frac{N_c}{N} = \frac{R_f}{R_c} \quad\Rightarrow\quad N_c = N\,\frac{R_f}{R_c}$$

where $R_c$ is the critical resistance at speed $N$.

Example: if $R_c = 200\ \Omega$ at 1000 rpm and the field resistance is $150\ \Omega$, then $N_c = 1000 \times \frac{150}{200} = 750$ rpm.

Conditions for build-up

• Residual magnetism must be present.
• Field MMF must aid the residual flux (correct field connection / direction of rotation).
• Field circuit resistance must be less than critical resistance.
• Speed must be above critical speed.
• At start, load resistance should be high (no load), so the armature voltage is not pulled down.

In a short-shunt compound generator the shunt field is connected directly across the armature, and the series field is in series with the load.

       Ia            Rse = 0.05     IL = 50 A
   +---->----+-------/\/\/\-----+--->---o +
   |         |                  |
 ( E )     [Rsh]                       V = 200 V
 Ra=0.08   100 ohm                load
   |         |                  |
   +---------+------------------+-------o -

Step 1: Voltage across the shunt field

$$V_{sh} = V + I_LR_{se} = 200 + 50 \times 0.05 = 202.5\ \text{V}$$

Step 2: Shunt field and armature currents

$$\begin{aligned} I_{sh} &= \frac{V_{sh}}{R_{sh}} = \frac{202.5}{100} = 2.025\ \text{A} \\ I_a &= I_L + I_{sh} = 50 + 2.025 = 52.025\ \text{A} \end{aligned}$$

Step 3: Generated EMF

$$\begin{aligned} E_g &= V_{sh} + I_aR_a = 202.5 + 52.025 \times 0.08 \\ &= 202.5 + 4.162 = 206.66\ \text{V} \end{aligned}$$

Step 4: Losses

The no-load (rotational: iron + friction) loss of 300 W is taken as the stray loss; the field and armature copper losses are added separately.

$$\begin{aligned} \text{Armature Cu loss} &= I_a^2R_a = 52.025^2 \times 0.08 = 216.53\ \text{W} \\ \text{Series field loss} &= I_L^2R_{se} = 50^2 \times 0.05 = 125\ \text{W} \\ \text{Shunt field loss} &= V_{sh}I_{sh} = 202.5 \times 2.025 = 410.06\ \text{W} \\ \text{Stray (no-load) loss} &= 300\ \text{W} \\ \text{Total loss} &= 216.53 + 125 + 410.06 + 300 = 1051.59\ \text{W} \end{aligned}$$

Check: $E_gI_a = 206.66 \times 52.025 = 10751.6$ W = output + copper losses ($10000 + 751.6$).

Step 5: Efficiency

$$\begin{aligned} \text{Output} &= VI_L = 200 \times 50 = 10000\ \text{W} \\ \eta &= \frac{10000}{10000 + 1051.59}\times 100 = 90.48\% \end{aligned}$$

Answer: Generated EMF $E_g =$ 206.66 V; efficiency $\approx$ 90.48%.

Given: $I_L = 120$ A, $V = 120$ V, $R_{sh} = 30\ \Omega$, $R_{se} = 0.05\ \Omega$, $R_a = 0.1\ \Omega$. Brush drop is neglected.

(i) Long-shunt connection

The series field is in series with the armature; the shunt field is across the load terminals.

       Ia    Rse              IL = 120 A
   +--->---/\/\/--+------+--->---o +
   |              |      |
 ( E )          [Rsh]  load     V = 120 V
   Ra             |      |
   +--------------+------+-------o -

$$\begin{aligned} I_{sh} &= \frac{V}{R_{sh}} = \frac{120}{30} = 4\ \text{A} \\ I_a &= I_L + I_{sh} = 120 + 4 = 124\ \text{A} \\ E_g &= V + I_a(R_a + R_{se}) = 120 + 124 \times (0.1 + 0.05) \\ &= 120 + 18.6 = 138.6\ \text{V} \end{aligned}$$

(ii) Short-shunt connection

The shunt field is across the armature; the series field carries the load current.

       Ia            Rse       IL = 120 A
   +--->---+-------/\/\/---+--->---o +
   |       |               |
 ( E )   [Rsh]           load    V = 120 V
   Ra      |               |
   +-------+---------------+-------o -

$$\begin{aligned} V_{sh} &= V + I_LR_{se} = 120 + 120 \times 0.05 = 126\ \text{V} \\ I_{sh} &= \frac{126}{30} = 4.2\ \text{A} \\ I_a &= I_L + I_{sh} = 120 + 4.2 = 124.2\ \text{A} \\ E_g &= V_{sh} + I_aR_a = 126 + 124.2 \times 0.1 \\ &= 126 + 12.42 = 138.42\ \text{V} \end{aligned}$$

Answer: (i) Long shunt: $E_g =$ 138.6 V; (ii) Short shunt: $E_g =$ 138.42 V.

Why a shunt generator is not started on load

A shunt generator builds up from residual magnetism, and its field current is taken from its own terminal voltage ($I_f = V/R_f$).

• If a load (low resistance) is connected at starting, the small residual EMF drives most of its current through the load, and the armature drop $I_aR_a$ becomes large compared with the tiny EMF.
• The terminal voltage applied to the field therefore stays very low, so the field current, and hence the flux, cannot grow.
• In terms of the characteristic, the total resistance seen by the field (load in parallel with field) is less than the critical value and the external resistance is below the critical load resistance, so the generator fails to build up.

Therefore the load switch is kept open until the generator reaches rated voltage, and only then is the load connected.

Voltage build-up process

   +---------+------o/ o----+
   |         |    switch    |
 ( G )     [Rsh]   (open)  load
   |         |              |
   +---------+--------------+

1. Residual flux: the field poles keep a small residual magnetism. Driving the armature at rated speed induces a small EMF $E_r$ (2-5% of rated).
2. Field current starts: $E_r$ drives a small current $I_{f1} = E_r/R_f$ through the shunt field.
3. Flux increases: if the field winding is connected so that its MMF aids residual flux, flux rises and the EMF rises to $E_1$ (read from the OCC at $I_{f1}$).
4. Cumulative rise: $E_1$ drives a larger $I_{f2}$, giving a higher $E_2$, and so on, step by step.
5. Stable point: the process stops where the OCC meets the field resistance line $V = I_fR_f$. Iron saturation bends the OCC, so the intersection is definite.

  E ^              Rf line
    |             /
    |            /  _____-- OCC
    |           /.-'
    |          /'  <- final voltage
    |       .'/
    |     .' /
    |   .'  /
    | .'   /
 Er +-----/----------------> If

Conditions for build-up: residual magnetism present, field MMF aiding residual flux, field resistance below critical resistance, speed above critical speed, and no (or light) load at starting.

Armature reaction

Armature reaction is the effect of the magnetic field set up by the armature current on the distribution of the main field flux in a DC machine. On no load only the main field acts; on load the armature conductors carry current and produce a cross field that distorts and weakens the main field.

   main flux (N to S) --->
       N |  ____  | S
         | /    \ |        GNA: geometric neutral axis
  --------(  arm. )------  MNA shifts in the direction of
         | \____/ |        rotation (generator) or against
                           rotation (motor)
   armature flux acts along the brush axis (90 deg elec.)

Effects

1. Cross-magnetising effect: armature MMF is at 90° to the main field. It strengthens flux at one pole tip and weakens it at the other, distorting the flux.
2. Shift of magnetic neutral axis (MNA): the MNA shifts forward in the direction of rotation for a generator (backward for a motor). If brushes stay on the geometric neutral axis, coils undergoing commutation have EMF induced in them, causing sparking.
3. Demagnetising effect: if brushes are shifted to the new MNA, part of the armature MMF directly opposes the main field, reducing net flux. This reduces the generated EMF (generator) or increases speed (motor).
4. Because of saturation at the strengthened pole tip, the increase there is less than the decrease at the other tip, so the net flux per pole falls even without brush shift.
5. Uneven flux raises voltage between adjacent commutator segments, risking flashover, and increases iron loss in teeth.

Measures to reduce armature reaction

• Compensating winding: placed in slots in the pole faces, connected in series with the armature, with current opposite to the adjacent armature conductors. It cancels the cross field under the poles (used in large machines).
• Interpoles (commutating poles): small poles between main poles, in series with the armature. They neutralise the cross field in the commutating zone and induce the reversing EMF for sparkless commutation.
• High reluctance in the cross-flux path: pole tips chamfered or made with laminations having slots, or a larger air gap at pole tips.
• Strong main field: design the main field MMF to be large compared with the armature MMF.
• Brush shift: moving brushes to the new MNA reduces sparking (only works for a fixed load; not used in modern machines).

Critical field resistance

The critical field resistance $R_c$ is the maximum resistance of the shunt field circuit at which a self-excited shunt generator will just build up voltage at a given speed. Graphically it is the slope of the tangent drawn from the origin to the initial (linear) part of the OCC.

  E ^      Rc line (tangent)
    |     /      Rf < Rc line
    |    /      /
    |   /     /   ____---- OCC (speed N)
    |  /    / .--'
    | /   /.-'   builds up here
    |/  ./
    | ./     Rf > Rc: no build-up
    |/
    +----------------------> If

• $R_f < R_c$: field line cuts the OCC at a high voltage, so the machine builds up.
• $R_f > R_c$: field line lies above the OCC; voltage stays near residual value.

Critical speed

The critical speed $N_c$ is the speed at which a given field resistance $R_f$ becomes the critical resistance, i.e. the minimum speed at which the generator just builds up. Since the OCC ordinates are proportional to speed:

$$\frac{N_c}{N} = \frac{R_f}{R_c} \quad\Rightarrow\quad N_c = N\,\frac{R_f}{R_c}$$

  E ^        Rf line
    |       /     OCC at N
    |      /   ___----
    |     /.--'
    |    //    OCC at Nc: tangent = Rf line
    |   //  __---
    |  //.-'
    | /.'     OCC below Nc: no build-up
    |/
    +----------------------> If

Example: $R_c = 220\ \Omega$ at 1500 rpm, $R_f = 176\ \Omega$: $N_c = 1500 \times 176/220 = 1200$ rpm.

Performance characteristics of DC shunt generator

1. Open-circuit characteristic (OCC, $E_0$ vs $I_f$): starts at the residual voltage, rises linearly, then bends due to saturation.

2. External characteristic ($V$ vs $I_L$):

  V ^
 E0 +----.___
    |        ''--.__        (a) drop: IaRa
    |               '--.    (b) drop: armature reaction
    |                   '.  (c) drop: lower If
    |                     \
    |              Ic   ,-'  turn-back
    |           <------'
    +------------------------> IL
    0     (short circuit current small)

• Terminal voltage falls as load increases, because of (a) armature resistance drop $I_aR_a$, (b) armature reaction weakening flux, and (c) reduced field current since $I_f = V/R_f$ falls with $V$.
• Beyond a critical load current $I_c$ (load resistance below critical), the curve turns back: further decrease in load resistance lowers both $V$ and $I_L$. On a dead short circuit, the current is small (only residual EMF acts), so a shunt generator protects itself.

3. Internal characteristic ($E$ vs $I_a$): lies above the external curve by $I_aR_a$, showing the effect of armature reaction and reduced field current only.

Comment: the shunt generator gives an almost constant voltage for normal loads (voltage drop about 5-15% from no load to full load), so it is used for battery charging, excitation of alternators and lighting where moderate regulation is acceptable.

Given: open-circuit (generated) EMF $E = 125$ V (constant, since speed and field current are constant), $R_a = 0.02\ \Omega$, speed $N = 3000$ rpm.

$$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 3000}{60} = 314.16\ \text{rad/s}$$

Take armature current positive when it flows out of the machine (generator action): $I_a = \frac{E - V_t}{R_a}$.

(i) Terminal voltage 128 V

$$I_a = \frac{125 - 128}{0.02} = -150\ \text{A}$$

The negative sign means current flows into the armature: the machine runs as a motor, taking 150 A from the 128 V supply.

$$\begin{aligned} \text{Terminal power} &= V_tI_a = 128 \times 150 = 19200\ \text{W} = 19.2\ \text{kW (input)} \\ \text{Electromagnetic power} &= EI_a = 125 \times 150 = 18750\ \text{W} = 18.75\ \text{kW} \\ T &= \frac{EI_a}{\omega} = \frac{18750}{314.16} = 59.68\ \text{N·m} \end{aligned}$$

The torque is a motoring torque (in the direction of rotation). The difference $19200 - 18750 = 450$ W $= 150^2 \times 0.02$ is the armature copper loss.

(ii) Terminal voltage 124 V

$$I_a = \frac{125 - 124}{0.02} = 50\ \text{A}$$

Positive, so the machine runs as a generator, delivering 50 A.

$$\begin{aligned} \text{Terminal power} &= 124 \times 50 = 6200\ \text{W} = 6.2\ \text{kW (output)} \\ \text{Electromagnetic power} &= 125 \times 50 = 6250\ \text{W} = 6.25\ \text{kW} \\ T &= \frac{6250}{314.16} = 19.89\ \text{N·m} \end{aligned}$$

This torque opposes rotation and must be supplied by the prime mover. Loss $= 6250 - 6200 = 50$ W $= 50^2 \times 0.02$.

Answer: (i) $I_a = 150$ A (motoring), $P_t = 19.2$ kW, $P_{em} = 18.75$ kW, $T = 59.68$ N·m; (ii) $I_a = 50$ A (generating), $P_t = 6.2$ kW, $P_{em} = 6.25$ kW, $T = 19.89$ N·m.

This shows that a DC machine changes from generator to motor simply when the terminal voltage exceeds the generated EMF.

Process of voltage build-up

A self-excited shunt generator uses its own output to supply its field.

1. The field poles retain a small residual flux $\phi_r$. When the armature is driven, a small EMF $E_r = \frac{P\phi_rZN}{60A}$ (a few volts) appears.
2. $E_r$ drives a small current through the shunt field, which is connected across the armature.
3. If this current's MMF aids the residual flux, flux rises, so EMF rises.
4. Higher EMF gives more field current, which gives more flux and more EMF: the voltage builds up cumulatively.
5. Build-up stops at the point where the OCC cuts the field resistance line ($V = I_fR_f$). Beyond that point, the EMF generated is less than that needed to drive more field current, so the voltage settles there.

  E ^           Rf line
    |          /
    |         /   ___---- OCC
    |        / .-'
    |       /.'  <-- operating voltage
    |     .'/
    |   .' /
    | .'  /
 Er +'---/------------------> If

Conditions for voltage build-up

1. Residual magnetism must be present in the poles. (If lost, "flash" the field from a battery.)
2. Correct field connection: field MMF must aid the residual flux; otherwise the residual flux is wiped out. If wrong, reverse field terminals or the direction of rotation.
3. Field resistance less than critical resistance $R_c$.
4. Speed above critical speed $N_c$ for the given field resistance.
5. No load or high-resistance load at starting (load resistance above critical load resistance), so that the terminal voltage is not pulled down.
6. Brushes on the correct neutral axis with good contact.

Finding critical field resistance and critical speed

No numerical OCC data is given, so the standard graphical method is described.

Critical field resistance: plot the OCC at rated speed $N$. Draw a tangent from the origin to the initial straight part of the OCC. Its slope is the critical resistance:

$$R_c = \frac{\Delta E}{\Delta I_f}\Big|_{\text{tangent}}$$

Critical speed: for the actual field resistance $R_f$, draw the field line of slope $R_f$. Take any current $I_f$ on the linear part; read the tangent voltage $E_c$ (= $R_cI_f$) and the field-line voltage $E_f$ (= $R_fI_f$). Since EMF ∝ speed:

$$N_c = N\,\frac{E_f}{E_c} = N\,\frac{R_f}{R_c}$$

Example: OCC at 1000 rpm gives 120 V at $I_f = 0.6$ A on the linear part, so $R_c = 120/0.6 = 200\ \Omega$. With $R_f = 160\ \Omega$, $N_c = 1000 \times 160/200 = 800$ rpm. The generator builds up only above 800 rpm, or at 1000 rpm only if $R_f < 200\ \Omega$.

In a shunt generator the armature current supplies both the load (lamps) and the shunt field. The lamps are treated as fixed resistances found from their rating, and they now work at 220 V.

Given: $V = 220$ V, $R_a = 0.1\ \Omega$, $R_{sh} = 50\ \Omega$, six lamps of 200 V, 600 W, brush drop 1 V per brush.

Lamp current

Resistance of one lamp (from its rating):

$$R_{lamp} = \frac{V_{rated}^2}{P_{rated}} = \frac{200^2}{600} = 66.67\ \Omega$$

At 220 V, current through one lamp $= \dfrac{220}{66.67} = 3.3$ A.

Six lamps are in parallel, so

$$I_L = 6 \times 3.3 = 19.8\ \text{A}$$

Field and armature current

$$\begin{aligned} I_{sh} &= \frac{V}{R_{sh}} = \frac{220}{50} = 4.4\ \text{A} \\ I_a &= I_L + I_{sh} = 19.8 + 4.4 = 24.2\ \text{A} \end{aligned}$$

Generated emf

In a lap winding the brushes of the same polarity are in parallel, so the current path passes through one positive and one negative brush set. Total brush drop $= 2 \times 1 = 2$ V.

$$\begin{aligned} E_g &= V + I_a R_a + V_{brush} \\ &= 220 + 24.2 \times 0.1 + 2 \\ &= 220 + 2.42 + 2 = 224.42\ \text{V} \end{aligned}$$

   +----Ra---+------+--------+
   |         |      |        |
  (Eg)      Rsh   lamps x6   V=220 V
   |         |      |        |
   +---------+------+--------+

Answer: Armature current $I_a = 24.2$ A, generated emf $E_g = 224.42$ V.

(If the lamps were assumed to draw their rated current of 3 A each, $I_L = 18$ A, $I_a = 22.4$ A and $E_g = 224.24$ V. The resistance method above is more exact because the lamps are not at rated voltage.)

In a short-shunt compound generator the shunt field is connected directly across the armature, and the series field carries the load current.

Given: $P = 6$ kW, $V = 250$ V, $R_a = 0.1\ \Omega$, $R_{se} = 0.2\ \Omega$, $R_{sh} = 250\ \Omega$.

     +--Ra--+--------Rse--------+
     |      |                   |  +
    (Eg)   Rsh                 Load  V=250 V
     |      |                   |  -
     +------+-------------------+

Step 1: Load current

$$I_L = \frac{P}{V} = \frac{6000}{250} = 24\ \text{A}$$

Step 2: Voltage across the shunt field

The series field carries $I_L$, so

$$\begin{aligned} V_{sh} &= V + I_L R_{se} \\ &= 250 + 24 \times 0.2 = 254.8\ \text{V} \end{aligned}$$

Step 3: Shunt field current and armature current

$$\begin{aligned} I_{sh} &= \frac{V_{sh}}{R_{sh}} = \frac{254.8}{250} = 1.0192\ \text{A} \\ I_a &= I_L + I_{sh} = 24 + 1.0192 = 25.0192\ \text{A} \end{aligned}$$

Step 4: Generated emf

$$\begin{aligned} E_g &= V_{sh} + I_a R_a \\ &= 254.8 + 25.0192 \times 0.1 \\ &= 254.8 + 2.502 = 257.30\ \text{V} \end{aligned}$$

Brush drop is not given, so it is neglected.

Answer: EMF generated by the armature $E_g \approx 257.3$ V.

Given: $I_L = 150$ A, $V = 230$ V, $R_a = 0.15\ \Omega$, $R_{se} = 0.1\ \Omega$, $R_{sh} = 100\ \Omega$, brush drop 2 V per brush, so total brush drop $= 2 \times 2 = 4$ V (one positive and one negative brush in the current path).

(a) Short-shunt connection

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 230 V
     |      |                 |
     +------+-----------------+

$$\begin{aligned} V_{sh} &= V + I_L R_{se} = 230 + 150 \times 0.1 = 245\ \text{V} \\ I_{sh} &= \frac{245}{100} = 2.45\ \text{A} \\ I_a &= I_L + I_{sh} = 150 + 2.45 = 152.45\ \text{A} \\ E_{g,s} &= V_{sh} + I_a R_a + V_{b} \\ &= 245 + 152.45 \times 0.15 + 4 \\ &= 245 + 22.87 + 4 = 271.87\ \text{V} \end{aligned}$$

(b) Same machine as long shunt (same load: 150 A at 230 V)

     +--Ra--Rse--+------------+
     |           |            |
    (Eg)        Rsh         Load 230 V
     |           |            |
     +-----------+------------+

Now the shunt field is across the terminals and the series field carries $I_a$:

$$\begin{aligned} I_{sh} &= \frac{230}{100} = 2.3\ \text{A} \\ I_a &= 150 + 2.3 = 152.3\ \text{A} \\ E_{g,l} &= V + I_a (R_a + R_{se}) + V_b \\ &= 230 + 152.3 \times 0.25 + 4 \\ &= 230 + 38.075 + 4 = 272.075\ \text{V} \end{aligned}$$

Ratio of generated voltages:

$$\frac{E_{g,l}}{E_{g,s}} = \frac{272.075}{271.8675} = 1.00076$$

Answer: (a) $E_g = 271.87$ V (short shunt). (b) $E_g = 272.08$ V as long shunt; ratio long : short $= 1.00076$ (about 1.0008 : 1). The long-shunt machine needs a slightly higher emf because the series field now carries the larger armature current.

A DC generator needs a magnetic field from its field winding. If the field current comes from an external DC source, the generator is separately excited. If the field current comes from the generator's own armature, it is self-excited; it builds up voltage from the residual magnetism of the poles.

Self-excited vs separately excited

Classification of self-excited generators

Depending on how the field winding is connected to the armature:

1. Shunt generator
2. Series generator
3. Compound generator – (a) long shunt, (b) short shunt; each can be cumulative or differential.

1. Shunt generator

Field winding (many turns of thin wire, high resistance) is connected in parallel with the armature.

   +---Ra---+------+
   |        |      |
  (Eg)     Rsh   Load  V
   |        |      |
   +--------+------+

$$I_a = I_L + I_{sh}, \quad I_{sh} = \frac{V}{R_{sh}}, \quad E_g = V + I_a R_a + V_b$$

2. Series generator

Field winding (few turns of thick wire, low resistance) is in series with the armature and load.

   +---Ra---Rse---+
   |              |
  (Eg)          Load  V
   |              |
   +--------------+

$$I_a = I_{se} = I_L, \quad E_g = V + I_a (R_a + R_{se}) + V_b$$

3. Compound generator

Has both shunt and series fields.

Long shunt: shunt field across the series combination of armature and series field.

   +--Ra--Rse--+------+
   |           |      |
  (Eg)        Rsh   Load
   |           |      |
   +-----------+------+

$$I_{sh} = \frac{V}{R_{sh}}, \quad I_a = I_{se} = I_L + I_{sh}, \quad E_g = V + I_a (R_a + R_{se}) + V_b$$

Short shunt: shunt field directly across the armature only.

   +--Ra--+----Rse----+
   |      |           |
  (Eg)   Rsh        Load
   |      |           |
   +------+-----------+

$$I_{se} = I_L, \quad I_{sh} = \frac{V + I_L R_{se}}{R_{sh}}, \quad I_a = I_L + I_{sh}, \quad E_g = V + I_L R_{se} + I_a R_a + V_b$$

In a cumulative compound generator the series flux aids the shunt flux; in a differential one it opposes it. Here $V_b$ is the total brush drop.

Efficiency is output divided by input, where input = output + all losses. The driving engine must supply this input, so its B.H.P. is the input power in horsepower.

Given: $P_{out} = 48.75$ kW, $V = 250$ V, $R_a = 0.02\ \Omega$, $R_{sh} = 50\ \Omega$, no-load (iron + friction) loss = 950 W.

Assumption: the 950 W no-load loss is the iron and mechanical (stray) loss; shunt field copper loss is added separately.

Step 1: Currents

$$\begin{aligned} I_L &= \frac{48750}{250} = 195\ \text{A} \\ I_{sh} &= \frac{250}{50} = 5\ \text{A} \\ I_a &= I_L + I_{sh} = 200\ \text{A} \end{aligned}$$

Step 2: Losses

$$\begin{aligned} P_{cu,a} &= I_a^2 R_a = 200^2 \times 0.02 = 800\ \text{W} \\ P_{cu,sh} &= V I_{sh} = 250 \times 5 = 1250\ \text{W} \\ P_{stray} &= 950\ \text{W} \\ P_{loss} &= 800 + 1250 + 950 = 3000\ \text{W} \end{aligned}$$

Step 3: Efficiency

$$\begin{aligned} P_{in} &= 48750 + 3000 = 51750\ \text{W} \\ \eta &= \frac{48750}{51750} \times 100 = 94.20\% \end{aligned}$$

Step 4: B.H.P. of the engine

Taking 1 hp = 746 W:

$$\text{B.H.P.} = \frac{51750}{746} = 69.37\ \text{hp}$$

(With metric hp, 735.5 W, this is 70.36 hp.)

Answer: Efficiency $\eta = 94.2\%$; B.H.P. of driving engine $\approx 69.37$ hp.

Excitation is the process of producing the main magnetic flux in a DC generator by passing direct current through the field winding on the poles. The way this field current is supplied decides the type of generator.

Methods of excitation

1. Separate excitation: field current from an independent external DC source (battery, another generator, rectifier).
2. Self-excitation: field current taken from the generator's own armature. Voltage builds up from residual magnetism: the small residual flux induces a small emf, which sends a small field current, which increases the flux, and so on until the voltage settles where the field resistance line cuts the OCC.

Conditions for self-excitation: residual magnetism must exist, field connection must aid residual flux, field resistance must be less than critical resistance, and speed must be above critical speed.

Types of DC generators

            DC generators
           /             \
   Separately          Self-excited
    excited          /     |       \
                 Shunt  Series  Compound
                                /       \
                         Long shunt  Short shunt
                     (cumulative or differential)

1. Separately excited

Field fed from outside; $I_a = I_L$, $E_g = V + I_a R_a + V_b$. Gives good control of voltage over a wide range; used in Ward-Leonard drives and testing.

2. Shunt generator

Field (many turns, thin wire, high resistance) in parallel with the armature. $I_{sh} = V/R_{sh}$, $I_a = I_L + I_{sh}$, $E_g = V + I_a R_a + V_b$. Voltage falls slightly with load. Used for battery charging, lighting, and as exciters.

3. Series generator

Field (few turns, thick wire, low resistance) in series with armature and load. $I_a = I_{se} = I_L$, $E_g = V + I_a (R_a + R_{se}) + V_b$. Voltage rises with load. Used as boosters in DC feeders.

4. Compound generator

Has both shunt and series fields.

• Long shunt: shunt field across (armature + series field). $I_a = I_{se} = I_L + I_{sh}$.
• Short shunt: shunt field across armature only. $I_{se} = I_L$, $I_{sh} = (V + I_L R_{se})/R_{sh}$.
• Cumulative: series flux aids shunt flux; can be over-, flat- or under-compounded. Flat compound gives nearly constant voltage for supply.
• Differential: series flux opposes shunt flux; voltage drops sharply with load. Used for arc welding.

First find the generated emf from the emf equation, then solve the circuit (armature resistance in series with the load and shunt field in parallel) to get the terminal voltage.

Given: $P = 4$, wave winding so $A = 2$, $Z = 378$, $\phi = 0.02$ Wb, $N = 1000$ rpm, $R_a = 1\ \Omega$, $R_{sh} = 100\ \Omega$, $R_L = 10\ \Omega$.

Step 1: Generated emf

$$\begin{aligned} E_g &= \frac{P \phi Z N}{60 A} \\ &= \frac{4 \times 0.02 \times 378 \times 1000}{60 \times 2} = 252\ \text{V} \end{aligned}$$

Step 2: Equivalent resistance across the terminals

Load and shunt field are in parallel:

$$R_p = \frac{R_L R_{sh}}{R_L + R_{sh}} = \frac{10 \times 100}{110} = 9.091\ \Omega$$

   +---Ra=1---+-------+
   |          |       |
  (252 V)   Rsh=100  RL=10
   |          |       |
   +----------+-------+

Step 3: Armature current and terminal voltage

$$\begin{aligned} I_a &= \frac{E_g}{R_a + R_p} = \frac{252}{1 + 9.091} = 24.97\ \text{A} \\ V &= I_a R_p = 24.97 \times 9.091 = 227.03\ \text{V} \end{aligned}$$

Check: $V = E_g - I_a R_a = 252 - 24.97 = 227.03$ V.

Step 4: Power absorbed by load

$$\begin{aligned} I_L &= \frac{227.03}{10} = 22.70\ \text{A} \\ P_L &= \frac{V^2}{R_L} = \frac{227.03^2}{10} = 5154\ \text{W} \end{aligned}$$

Answer: Power absorbed by the load $\approx 5.15$ kW (5154 W), with $E_g = 252$ V and $V = 227.03$ V.

Given: $P_{out} = 30$ kW, $V = 200$ V, $R_a = 0.05\ \Omega$, $R_{sh} = 50\ \Omega$, iron + friction loss = 1000 W.

Currents

$$\begin{aligned} I_L &= \frac{30000}{200} = 150\ \text{A} \\ I_{sh} &= \frac{200}{50} = 4\ \text{A} \\ I_a &= I_L + I_{sh} = 154\ \text{A} \end{aligned}$$

(i) Generated emf

$$\begin{aligned} E_g &= V + I_a R_a \\ &= 200 + 154 \times 0.05 = 207.7\ \text{V} \end{aligned}$$

(ii) Copper losses

$$\begin{aligned} P_{cu,a} &= I_a^2 R_a = 154^2 \times 0.05 = 1185.8\ \text{W} \\ P_{cu,sh} &= V I_{sh} = 200 \times 4 = 800\ \text{W} \\ P_{cu} &= 1185.8 + 800 = 1985.8\ \text{W} \end{aligned}$$

(iii) Efficiency

$$\begin{aligned} P_{loss} &= 1985.8 + 1000 = 2985.8\ \text{W} \\ P_{in} &= 30000 + 2985.8 = 32985.8\ \text{W} \\ \eta &= \frac{30000}{32985.8} \times 100 = 90.95\% \end{aligned}$$

Answer: (i) $E_g = 207.7$ V, (ii) total copper loss $= 1985.8$ W, (iii) $\eta \approx 90.95\%$.

Function of commutator segments

The commutator is a cylinder made of wedge-shaped, hard-drawn copper segments insulated from each other by mica, mounted on the shaft. Each segment is connected to the end of an armature coil, and carbon brushes press on its surface.

The emf induced in each armature conductor is alternating, because a conductor passes under N and S poles in turn. The commutator, together with the brushes, works as a mechanical rectifier:

1. Rectification: as a coil side moves from under one pole to the next, the coil connection to the brush is reversed at the same instant the emf reverses. So the external circuit always gets current in one direction (unidirectional DC).
2. Collecting current: it provides a sliding contact so current can be taken from the rotating armature to the stationary external circuit through the brushes.
3. Connecting coils in series/parallel paths: the segments join the coils so that their emfs add up between brushes (lap or wave paths).
4. Commutation: it reverses the current in a coil while the coil is short-circuited by the brush, keeping the output steady with many segments (low ripple).

 Coil emf (AC)      After commutator (DC)
   /\    /\           /\  /\  /\
  /  \  /  \         /  \/  \/  \
 /    \/    \       /            \
       (reverses)    (always positive)

EMF equation of a DC generator

Let

• $P$ = number of poles, $\phi$ = flux per pole (Wb)
• $Z$ = total number of armature conductors
• $A$ = number of parallel paths ($A = P$ for lap, $A = 2$ for wave)
• $N$ = speed (rpm)

By Faraday's law, average emf in one conductor = flux cut per revolution ÷ time for one revolution.

Flux cut by one conductor in one revolution:

$$d\phi = P\phi\ \text{Wb}$$

Time for one revolution:

$$dt = \frac{60}{N}\ \text{s}$$

So emf per conductor:

$$e = \frac{d\phi}{dt} = \frac{P\phi N}{60}\ \text{V}$$

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

$$\begin{aligned} E_g &= \frac{P\phi N}{60} \times \frac{Z}{A} \\ &= \frac{P \phi Z N}{60 A}\ \text{V} \end{aligned}$$

• Lap winding ($A = P$): $E_g = \dfrac{\phi Z N}{60}$
• Wave winding ($A = 2$): $E_g = \dfrac{P \phi Z N}{120}$

For a given machine $P$, $Z$, $A$ are fixed, so $E_g = k \phi N$ (or $E_g = k\phi\omega$), i.e. emf is proportional to flux and speed.

Example: $P = 4$, lap, $Z = 400$, $\phi = 0.02$ Wb, $N = 1500$ rpm gives $E_g = 0.02 \times 400 \times 1500 / 60 = 200$ V.

Given: $I_L = 50$ A, $V = 200$ V, $R_a = 0.5\ \Omega$, $R_{se} = 0.3\ \Omega$, $R_{sh} = 200\ \Omega$; short-shunt connection.

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 200 V
     |      |                 |
     +------+-----------------+

Step 1: Voltage across shunt field

Series field carries load current:

$$V_{sh} = V + I_L R_{se} = 200 + 50 \times 0.3 = 215\ \text{V}$$

Step 2: Shunt field and armature current

$$\begin{aligned} I_{sh} &= \frac{215}{200} = 1.075\ \text{A} \\ I_a &= I_L + I_{sh} = 50 + 1.075 = 51.075\ \text{A} \end{aligned}$$

Step 3: Generated emf

$$\begin{aligned} E_g &= V_{sh} + I_a R_a \\ &= 215 + 51.075 \times 0.5 \\ &= 215 + 25.54 = 240.54\ \text{V} \end{aligned}$$

Brush drop is neglected as it is not given.

Answer: EMF generated by the armature $E_g \approx 240.54$ V.

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

     +--Ra--Rse--+-----------+
     |           |           |
    (Eg)        Rsh        Load 200 V
     |           |           |
     +-----------+-----------+

In long shunt, the shunt field is across the terminals and the series field carries the armature current.

Step 1: Shunt field current

$$I_{sh} = \frac{V}{R_{sh}} = \frac{200}{100} = 2\ \text{A}$$

Step 2: Armature (and series field) current

$$I_a = I_{se} = I_L + I_{sh} = 40 + 2 = 42\ \text{A}$$

Step 3: Generated emf

$$\begin{aligned} E_g &= V + I_a (R_a + R_{se}) \\ &= 200 + 42 \times (0.4 + 0.5) \\ &= 200 + 37.8 = 237.8\ \text{V} \end{aligned}$$

Brush drop is neglected as it is not given.

Answer: EMF generated $E_g = 237.8$ V.

Given: $P = 25$ kW, $V = 250$ V, $R_{sh} = 125\ \Omega$, $R_{se} = 0.5\ \Omega$, $R_a = 0.1\ \Omega$, total brush drop $V_b = 2$ V; short-shunt connection.

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 250 V
     |      |                 |
     +------+-----------------+

Step 1: Load current

$$I_L = \frac{25000}{250} = 100\ \text{A}$$

Step 2: Voltage across shunt field

$$V_{sh} = V + I_L R_{se} = 250 + 100 \times 0.5 = 300\ \text{V}$$

Step 3: Shunt field and armature current

$$\begin{aligned} I_{sh} &= \frac{300}{125} = 2.4\ \text{A} \\ I_a &= I_L + I_{sh} = 100 + 2.4 = 102.4\ \text{A} \end{aligned}$$

Step 4: Generated emf

$$\begin{aligned} E_g &= V_{sh} + I_a R_a + V_b \\ &= 300 + 102.4 \times 0.1 + 2 \\ &= 300 + 10.24 + 2 = 312.24\ \text{V} \end{aligned}$$

Answer: EMF generated in the armature $E_g = 312.24$ V.

A shunt generator has its field winding connected in parallel with the armature, while a series generator has its field winding connected in series with the armature and load. This one difference changes their field design, characteristics and uses.

Circuit diagrams and equations

Shunt generator

   +---Ra---+------+
   |        |      |
  (Eg)     Rsh   Load  V
   |        |      |
   +--------+------+

$$I_{sh} = \frac{V}{R_{sh}}, \quad I_a = I_L + I_{sh}, \quad E_g = V + I_a R_a + V_b$$

Series generator

   +---Ra---Rse---+
   |              |
  (Eg)          Load  V
   |              |
   +--------------+

$$I_a = I_{se} = I_L, \quad E_g = V + I_a (R_a + R_{se}) + V_b$$

Characteristic curves

Open-circuit (magnetisation) characteristic: $E_0$ vs $I_f$. For a shunt generator it is taken by exciting the field separately; its shape is like a B-H curve starting at the residual voltage. For a series generator it is also taken with separate excitation of the series field.

External characteristic ($V$ vs $I_L$):

 V
 |                         Shunt
 |----___
 |       ----____
 |               ---__
 |                    \  (breakdown,
 |       ___           \  curls back)
 |     _/   \__ Series  \
 |   _/        \_        |
 | _/            \       |
 |/________________\_____|_______ IL

• Shunt: $V$ drops slowly as load rises because of $I_a R_a$ drop, armature reaction, and the resulting fall in field current. Beyond a critical load the curve turns back and voltage collapses (short circuit current is small).
• Series: at no load $V$ is only the residual voltage. As load current rises, field current (= load current) rises, so $V$ rises steeply; at very high load, saturation and armature reaction make $V$ fall again.

Comparison

Example: a shunt generator charging a battery keeps nearly constant voltage; a series generator in a long DC feeder raises its voltage as load current rises, compensating line drop.

Given: $I_L = 50$ A, $V = 220$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.05\ \Omega$, $R_a = 0.1\ \Omega$; short-shunt connection.

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 220 V
     |      |                 |
     +------+-----------------+

$$\begin{aligned} V_{sh} &= V + I_L R_{se} = 220 + 50 \times 0.05 = 222.5\ \text{V} \\ I_{sh} &= \frac{222.5}{100} = 2.225\ \text{A} \\ I_a &= I_L + I_{sh} = 50 + 2.225 = 52.225\ \text{A} \\ E_g &= V_{sh} + I_a R_a = 222.5 + 52.225 \times 0.1 \\ &= 222.5 + 5.22 = 227.72\ \text{V} \end{aligned}$$

Brush drop is neglected as it is not given.

Answer: EMF generated $E_g \approx 227.72$ V.

Given: $P = 7.5$ kW, $V = 230$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.3\ \Omega$, $R_a = 0.4\ \Omega$; short-shunt connection.

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 230 V
     |      |                 |
     +------+-----------------+

Step 1: Load current

$$I_L = \frac{7500}{230} = 32.609\ \text{A}$$

Step 2: Shunt field voltage and current

$$\begin{aligned} V_{sh} &= V + I_L R_{se} = 230 + 32.609 \times 0.3 = 239.78\ \text{V} \\ I_{sh} &= \frac{239.78}{100} = 2.398\ \text{A} \end{aligned}$$

Step 3: Armature current and induced emf

$$\begin{aligned} I_a &= I_L + I_{sh} = 32.609 + 2.398 = 35.007\ \text{A} \\ E_g &= V_{sh} + I_a R_a = 239.78 + 35.007 \times 0.4 \\ &= 239.78 + 14.00 = 253.79\ \text{V} \end{aligned}$$

Step 4: Load resistance

$$R_L = \frac{V}{I_L} = \frac{230}{32.609} = 7.053\ \Omega$$

(Or $R_L = V^2/P = 230^2/7500 = 7.053\ \Omega$.)

Answer: Induced emf $E_g \approx 253.79$ V; load resistance $R_L \approx 7.05\ \Omega$.

The commutator is a cylinder of hard-drawn copper segments, insulated from one another by mica, fixed on the armature shaft. Armature coil ends are soldered to the segments and stationary carbon brushes rest on it.

The emf induced in an armature conductor is alternating, since each conductor moves alternately under N and S poles. The commutator converts this into DC at the terminals.

Functions

1. Mechanical rectifier: at the instant the emf in a coil reverses, the coil's connection to the brushes is also reversed, so the brushes always see emf of one polarity. The output current is unidirectional.
2. Current collection: together with the brushes, it provides a sliding contact between the rotating armature and the stationary external circuit.
3. Series addition of coil emfs: segments join the coils so that their emfs add between brushes in each parallel path (lap or wave).
4. Smooth output: with many coils and segments, the rectified emf has very little ripple.

 Coil emf (AC)        Brush output (DC)
   /\                   /\    /\
  /  \                 /  \  /  \
 /    \    /          /    \/    \
       \  /
        \/

Given: $P = 10$ kW, $V = 220$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.3\ \Omega$, $R_a = 0.4\ \Omega$. Brush drop neglected.

Case 1: Short-shunt connection

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 220 V
     |      |                 |
     +------+-----------------+

$$\begin{aligned} I_L &= \frac{10000}{220} = 45.455\ \text{A} \\ V_{sh} &= V + I_L R_{se} = 220 + 45.455 \times 0.3 = 233.64\ \text{V} \\ I_{sh} &= \frac{233.64}{100} = 2.336\ \text{A} \\ I_a &= I_L + I_{sh} = 47.791\ \text{A} \\ E_g &= V_{sh} + I_a R_a = 233.64 + 47.791 \times 0.4 \\ &= 233.64 + 19.12 = 252.75\ \text{V} \end{aligned}$$

Case 2: Reconnected as long shunt, same $E_g$ and $I_a$

     +--Ra--Rse--+-----------+
     |           |           |
    (Eg)        Rsh        Load  V
     |           |           |
     +-----------+-----------+

Now the armature current flows through both $R_a$ and $R_{se}$:

$$\begin{aligned} V &= E_g - I_a (R_a + R_{se}) \\ &= 252.75 - 47.791 \times (0.4 + 0.3) \\ &= 252.75 - 33.45 = 219.30\ \text{V} \end{aligned}$$

Check: $I_{sh} = 219.30/100 = 2.193$ A, $I_L = 47.791 - 2.193 = 45.60$ A, which is consistent.

Answer: Case 1: induced emf $E_g \approx 252.75$ V. Case 2: terminal voltage in long shunt $V \approx 219.3$ V (slightly lower, because the series field now carries the larger armature current).