Self Excited Generator

When the field winding is supplied from the armature of the generator itself then it is said to be self excited generator. Now without generated e.m.f., field can not be excited in such generator and without excitation there can not be generated e.m.f. So one may obviously wonder, how this type of generator works. The answer to this is residual magnetism possessed by the field poles, under normal condition.
Practically through the generator is not working, without any current through field winding, the field poles possess some magnetic flux. This is called residual flux and the property is called residual magnetism. Thus when the generator is started, due to such residual flux, it develops a small e.m.f. which now drives a small current through the field winding. This tends to increase the flux produced. This in turn increases the induced e.m.f. This further increases the field current and the flux. The process is cumulative and continues till the generator develops rated voltage across its armature. This is voltage building process in self excited generators.
Based on how field winding is connected to the armature to drive its excitation, this type is further divided into following three types.
i) Shunt generator
ii) Series generator
iii) Compound generator
Shunt Generator
When the field winding is connected in parallel with the armature and the combination across the load then the generator is called shunt generator.
The field winding has large number of turns of thin wire so it has high resistance. Let Rsh be the resistance of the field winding.
Fig. 1 Shunt generator


1.1 Voltage and Current Relations
From the Fig. 1, we can write
Ia = IL + Ish
Now voltage across load is Vt which is same across field winding as both are in parallel with each other.
... Ish = Vt /Rsh
While induced e.m.f. E, still requires to supply voltage drop Ia Ra and brush contact drop.
... E = Vt + Ia Ra + Vbrush
Where E = (ΦPNZ)/(60A)
In practical, brush contact drop can be neglected.
Series Generators
When the field winding is connected in series with the armature winding while supplying the load then the generator is called series generator. It is shown in the Fig. 1.
Field winding, in this case is denoted as S1 and S2. The resistance of series field winding is very small and hence naturally it has less number of turns of thick cross-section wire as shown in the Fig. 1.
Fig. 1 Series generators


Let Rse be the resistance of the series field winding.
1.1 Voltage and current Relations
As all armature, field and load are in series they carry the same current.
... Ia = Ise = IL
Where Ise = Current through series field winding.
Now in addition to drop Ia Ra, induced e.m.f. has to supply voltage drop across series field winding too. This is Ise Rse i.e. Ia Rse as Ia = Ise. So voltage equations can be written as,
E = Vt + Ia Ra + Ia Rse + Vbrush
... E = Vt + Ia (Ra + Rse) + Vbrush
where E = (ΦPNZ)/(60A)
Compound Generator
In this type, the part of the field winding is connected in parallel with armature and part in series with the armature. Both series and shunt field windings are mounted on the same poles. Depending upon the connection of shunt and series field winding, compound generator is further classified as : i) Long shunt compound generator, ii) Short shunt compound generator.
1.1 Long Shunt Compound Generator
In this type, shunt field winding is connected across the series combination of armature and series field winding as shown in the Fig. 1.
Fig. 1 Long shunt compound generator


Voltage and current relations are as follows.
From the Fig. 1.
Ia = Ise
and Ia = Ish + IL
Voltage across shunt field winding is Vt.
Ish = Vt /Rsh
where Rsh = Resistance of shunt field winding
And voltage equation is,
E = Vt + Ia Ra + Ia Rse + Vbrush
Where Rse = Resistance of series field winding
1.2 Short Shunt Compound Generator
In this type, shunt field winding is connected, only across the armature, excluding series field winding as shown in the Fig. 2.
Fig. 2 Short shunt compound generator

Voltage and current relations are as follows.
For the Fig. 2, Ia = Ise + Ish
and Ise = IL
... Ia = IL + Ish
The drop across shunt field winding is drop across the armature only and not the total Vt, in this case. So drop across shunt field winding is E -Ia Ra .
Ish = (E - Ia Ra ) / ( Rsh)
Now the voltage equation is E = Vt + Ia Ra + Ise Rse + Vbrush
... Ise = IL
... E = Vt + Ia Ra + IL Rse + Vbrush
Neglecting Vbrush , we can write,
E = Vt + Ia Ra + IL Rse
E - Ia Ra = Vt + IL Rse
Ish = (Vt + IL Rse) / ( Rsh)
Any of the two above expression of Ish can be used, depending on the quantities known while solving the problems.
1.3 Cumulative and Differential Compound Generator
It is mentioned earlier that the two windings, shunt and series field are wound on the same pole. Depending on the direction of winding on the pole, two fluxes produced by shunt and series field may help or may oppose each other. This facts decides whether generator is cumulative or differential compound. If the two fluxes help each other as shown in Fig. 3 the generator is called cumulative compound generator.
Fig. 3 Cumulative compound generator

ΦT = Φsh + Φse
Where Φsh = Flux producd by shunt
Φse = Flux produced by series, field winding
If the two windings are wound in such a direction that the fluxes produced by them oppose each other then the generator is called differential compound generator. This is shown in the Fig. 4.
ΦT = Φsh - Φse
Where Φsh = Flux produced by shunt field winding.
Φse = Flux produced by series field winding.
Fig. 4 Differential compound generator

Separately Excited Generator

When the field winding is supplied from external, separate d.c. supply i.e. excitation of field winding is separate then the generator is called separately excited generator. Schematic representation of this type is shown in the Fig.1.
Fig. 1 Separately excited generator


The field winding of this type of generator has large number of turns of thin wire. So length of such winding is more with less cross-sectional area. So resistance of this field winding is high in order to limit the field current.
1.1 Voltage and Current Relations
The field winding is excited separately, so the field current depends on supply voltage and resistance of the field winding.
For armature side, we can see that it is supplying a load, demanding a load current of IL at a voltage of Vt which is called terminal voltage.
Now Ia = IL
The internally induced e.m.f. E is supplying the voltage of the load hence terminal voltage Vt is a part of E. But E is not equal to Vt while supplying a load. This is because when armature current Ia flows through armature winding, due to armature winding resistance Ra ohms, there is a voltage drop across armature winding equal to Ia Ra volts. The induced e.m.f. has to supply this drop, along with the terminal voltage Vt. To keep Ia Ra drop to minimum, the resistance Ra is designed to be very very small. In addition to this drop, there is some voltage drop at the contacts of the brush called brush contact drop. But this drop is negligible and hence generally neglected. So in all, induced e.m.f. E has three components namely,
i) Terminal voltage Vt
ii) Armature resistance drop Ia Ra
iii) Brush contact drop Vbrush
So voltage equation for separately excited generator can be written as,
E = Vt + Ia Ra + Vbrush
Where E = (ΦPNZ)/(60A)
Generally Vbrush is neglected as is negligible compared to other voltages.

Current Division in Parallel Circuit of Resistors

Consider a parallel circuit of tow resistor and connected across a source of V volts.

       Current through R₁ is I₁ and R₂ is I₂  , while total current drawn from source is I_T.

.^..                                I_T = I₁  + I₂ 

But                             I₁  = V/R₁   , I₂  = V/R₂

i.e.                              V =I₁ R₁  = I₂  R₂

.^..                                 I₁  = I₂   (R₂/R₁)

       Substituting value of I₁ in I_T,

Key point : In general, the current in any branch is equal to the ratio of opposite branch resistance to the total resistance value, multiplied by the total current in the circuit.

Example : Find the magnitude of total current, current through and if , R₁ = 10 Ω , R₂= 20 Ω and V = 50V.

Solution :

      The equivalent resistance of tow is,

                            R_eq =(R₁ R₂ ) / ( R₁ + R₂ ) = (10 x 20)/ (10 + 20) = 6.67 .

                            I_T = V/R_eq = 50/6.67 = 7.5 A

       As per the current distribution in parallel circuit,

It can be verified that  :         I_T = I₁  + I₂ .

Voltage Division in Series Circuit of Resistors

 Consider a series circuit of tow resistors R₁ and R₂ connected to source V volts.

       As tow resistors are connected in series, the current flowing through both the resistors is same, i.e. I. Then applying KVL, we get,

                                                                 V= I R₁ +I R₂ 

       Total voltage applied is equal to the sum of voltage drops V_R1 and V_R2 across R₁and R₂ respectively.

.^..                     V_R1 = I R₁ 

Similarly,             V_R2  = I R₂ 

 

       So this circuit is a voltage divider circuit.

Key point : So in general, voltage drop across any resistor, or combination of resistors, in a series circuit is equal to the ratio of that resistance value to the total resistance, multiplied by the source voltage.

Example : Find the voltage across the three resistances shown in the Fig.

 Solution : 

Key point : It can be seen that voltage across any resistance of series circuit is ratio of that resistance to the total resistance, multiplied by the source voltage.

Source Transformation

Consider a practical voltage source shown in the Fig. 1.(a) having internal resistance R_se, connected to the load having resistance R_L .

Fig. 1(a  

       Now we can replace voltage source by equivalent current source.

Key point : The tow sources are said to be equivalent, if they supply equal load current to the load, with same load connected across its terminals.

       The current delivered in above case by voltage source is,

                           I = V/ (R_se  + R_L ) , R_se  and R_L  are in series             ..............(1)

        If it is to be replaced by a current source then load current must be V/(R_se  +R_L).

       Consider an equivalent current source shown in the Fig. 1(b).

Fig 1 (b) .

       The total current is T.

       Both the resistances will take current proportional to their values.

       From the current division in parallel circuit we can write,

                                                                                                                                ............(2)
       Now this I_L  and V/(R_sh  +R_L  )must be same, so equating (1) and (2),

       Let internal resistance be,  R_se  =  R_sh  = R  say.

       Then, V= I x R_sh  = I x R

       or       I= V/R_sh

.^..              I = V/R = V/R_se 

Key point : If voltage source is converted to current source, then current source I = V/ R_se with parallel internal resistance equal to R_se .

Key point : If current source is converted to voltage source, the voltage source V = I R_sh  with sereis internal resistance equal to R_sh .

       The direction of current of equivalent circuit source is always from -ve to +ve, internal to the source. While converting current source to voltage source, polarities of voltage is always as +ve terminal at top of arrow and -ve terminal at bottom of arrow, as direction of current is from -ve to +ve, internal to the source. This ensure that current flows from positive to negative terminal in the external circuit.

       Note the direction of transformed sources, shown in the Fig. 2(a), (b), (c) and (d).

Fig. 2

Example 1 : Transform a voltage source of 20 volts with an internal resistance of 5 Ω to a current source.

Solution : Refer to the Fig. 3(a)

Fig. 3

       The current of current source is, I = V/Rse  = 20/5 = 4 A with internal parallel resistance same as Rse .

       Equivalent current source is as shown in the Fig. 3(b).

Example 2 : Convert the given current source of 50 A with internal resistance of 10 Ω to the equivalent voltage source.

Fig. 4

Solution : The given values are,          I= 50 A and  R_sh = 10 Ω.

       For the equivalent voltage source, 

                                                     V= I x = 50 x 10 = 500 V

                                                    R_se = R_sh = 10 Ω in series.

       The equivalent voltage source is shown in the Fig. 4(a).

Fig 4 (a)

       Note the polarities of voltage source, which are such that +ve at top of arrow and -ve at bottom.

Circle Diagram for a Series R-L Circuit

  Consider a series R-L circuit with a variable R as shown in the Fig. 1. It is excited by an alternating source of V volts. The frequency of the source is f Hz.

Fig. 1

       Let             I = Current flowing through the circuit

                        Z = Impedance of the circuit

                        Z = R + j X_L        where X_L = 2 fL

        Now R is variable while X_L is fixed.

       The phasor diagram is shown in the Fig. 2(a). The current I lags voltage V by angle as the circuit is inductive. The impedance triangle is shown in the Fig. 2(b).

Fig. 2

       From the impedance triangle we can write,

                            sin Φ = X_L/Z

       Substituting in the expression for I,

               I = (V/X_L) sinΦ                                              ..............(1)

       This is the equation of a circle in polar co-ordinates with a diameter equal to (V/X_L).

       When the resistance R =0, then Φ = 90^o hence sin Φ  = 1.

.^..             I = I_m = (V/X_L)

       This is the maximum value of current.

       As R resistance, the phase angle decreases thus decreasing sin. Effectively current I also decrease. When R→ ∞ the Φ→0^o and current becomes zero.

       The locus obtained of extremities of a current phasor plotted for various values of R is a semicircle. The semicircle is shown in the Fig. 3. The voltage axis is taken as vertical axis as a reference, with respect to which the various current phasors are plotted.

Fig.  3  Circle diagram

       The power factors at various conditions are cosΦ₁, cosΦ₂ etc. As Φ varies only from 0^oto 90^o, the diagram is semicircle, infact it is a half part of a circle hence it is known as circle diagram.

       This theory of series R-L circuit can be easily extended to a three phase induction motor.

Circle Diagram of Induction motor

 By using the data obtained from the no load test and the blocked rotor test, the circle diagram can be drawn using the following steps :

Step 1 : Take reference phasor V as vertical (Y-axis).

Step 2 : Select suitable current scale such that diameter of circle is about 20 to 30 cm.

Step3 : From no load test, I_o and are Φ_o obtained. Draw vector I_o, lagging V by angle Φ_o. This is the line OO' as shown in the Fig. 1.

Step 4 : Draw horizontal line through extremity of I_o i.e. O', parallel to horizontal axis.

Step 5 : Draw the current I_SN calculated from I_sc with the same scale, lagging V by angle Φ_sc, from the origin O. This is phasor OA as shown in the Fig. 1.

Step 6 : Join O'A is called output line.

Step 7 : Draw a perpendicular bisector of O'A. Extend it to meet line O'B at point C. This is the centre of the circle.

Step 8 : Draw the circle, with C as a center and radius equal to O'C. This meets the horizontal line drawn from O' at B as shown in the Fig. 1.

Step 9 : Draw the perpendicular from point A on the horizontal axis, to meet O'B line at F and meet horizontal axis at D.

Step 10 : Torque line.

      The torque line separates stator and rotor copper losses.

      Note that as voltage axis is vertical, all the vertical distances are proportional to active components of currents or power inputs, if measured at appropriate scale. 

       Thus the vertical distance AD represents power input at short circuit i.e. W_SN, now which consists of core loss and stator, rotor copper losses.

      Now          FD = O'G

                             = Fixed loss

       Where O'G is drawn perpendicular from O' on horizontal axis. This represents power input on no load i.e. fixed loss.

       Hence            AF α Sum of stator and rotor copper losses

       Then point E can be located as,

      AE/EF = Rotor copper loss / Stator copper loss

      The line O'E under this condition is called torque line.

Fig. 1

 Power scale : As AD represents W_SN i.e. power input on short circuit at normal voltage, the power scale can be obtained as,

      Power scale = W_SN/l(AD)   W/cm

      where l(AD) = Distance AD in cm

Location of Point E : In a slip ring induction motor, the stator resistance per phase R₁ and rotor resistance per phase R₂ can be easily measured. Similarly by introducing ammeters in stator and rotor circuit, the currents I₁ and I₂ also can be measured.

.^..                         K = I₁/I₂ = Transformation ratio

Now   AF/EF = Rotor copper loss / Stator copper loss = (I₂²R₂)/(I₁²R₁) = (R₂/R₂)(I₂²/I₁²) = (R₂/R₂).(1/K²)

But      R₂'= R₂/K² = Rotor resistance referred to stator

.^..        AE/EF = R₂'/R₁

       Thus point E can be obtained by dividing line AF in the ratio R₂' to R₁.

In a squirrel cage motor, the stator resistance can be measured by conducting resistance tset.

.^..        Stator copper loss = 3I_SN² R₁  where I_SN is phase value.

       Neglecting core loss, W_SN = Stator Cu loss + Rotor Cu loss

.^..    Rotor copper loss = W_SN - 3I_SN² R₁

.^..        AE/EF = (W_SN - 3I_SN² R₁)/(3I_SN² R₁)

       Dividing line AF in this ratio, the point E can be obtained and hence O'E represents torque line.

1.1 Predicting Performance Form Circle Diagram

       Let motor is running by taking a current OP as shown in the Fig. 1. The various performance parameters can be obtained from the circle diagram at that load condition.

       Draw perpendicular from point P to meet output line at Q, torque line at R, the base line at S and horizontal axis at T.

       We know the power scale as obtained earlier.

       Using the power scale and various distances, the values of the performance parameters can be obtained as,

       Total motor input = PT x Power scale

       Fixed loss = ST x power scale

      Stator copper loss = SR x power scale

       Rotor copper loss = QR x power scale

      Total loss = QT x power scale

      Rotor output = PQ x power scale

      Rotor input = PQ + QR = PR x power scale

      Slip s = Rotor Cu loss = QR/PR

      Power factor cos = PT/OP

      Motor efficiency = Output / Input = PQ/PT

      Rotor efficiency = Rotor output / Rotor input = PQ/PR

      Rotor output / Rotor input = 1 - s = N/N_s = PQ/PR

      The torque is the rotor input in synchronous watts.

1.2 Maximum Quantities

       The maximum values of various parameters can also be obtained by using circle diagram.

1. Maximum Output : Draw a line parallel to O'A and is also tangent to the circle at point M. The point M can also be obtained by extending the perpendicular drawn from C on O'A to meet the circle at M. Then the maximum output is given by l(MN) at the power scale. This is shown in the Fig. 1.

2. Maximum Input : It occurs at the highest point on the circle i.e. at point L. At this point, tangent to the circle is horizontal. The maximum input given l(LL') at the power scale.

3. Maximum Torque : Draw a line parallel to the torque line and is also tangent to the circle at point J. The point J can also be obtained by drawing perpendicular from C on torque line and extending it to meet circle at point J. The l(JK) represents maximum torque in synchronous watts at the power scale. This torque is also called stalling torque or pull out torque.

4. Maximum Power Factor : Draw a line tangent to the circle from the origin O, meeting circle at point H. Draw a perpendicular from H on horizontal axis till it meets it at point I. Then angle OHI gives angle corresponding to maximum power factor angle.

.^..      Maximum p.f. = cos ∟{OHI}

                               = HI/OH

5. Starting Torque : The torque is proportional to the rotor input. At s = 1, rotor input is equal to rotor copper loss i.e. l(AE).

.^..         T_start = l(AE) x Power scale            ...................in synchronous watts

1.3 Full load Condition 

       The full load motor output is given on the name plates in watts or h.p. Calculates the distance corresponding to the full load output using the power scale.

       Then extend AD upwards from A onwards, equal to the distance corresponding to full load output, say A'. Draw parallel to the output line O'A from A' to meet the circle at point P'. This is the point corresponding to the full load condition, as shown in the Fig. 2.

Fig. 2  Locating full load point

       Once point P' is known, the other performance parameters can be obtained easily as discussed above.