Electrical Power

AC Power Generation
Voltage Regulation

1
 AC main generation in aircraft
PMG : Permanent magnet generator

Figure : Electrical power system in

Figure : integrated drive generator (IDG)
aircraft
2
AC power generation and voltage regulation
- Introduction
- Synchronous generator construction
- Speed of rotation of a synchronous generator
- Internal generated voltage of a synchronous generator
- Equivalent circuit of a synchronous generator
- Phasor diagram of a synchronous generator
- Power and torque in synchronous generators
- Measuring synchronous generator model parameters
- Synchronous generator operating alone
- Voltage regulation
- AC Main generation and voltage regulation in aircraft

3
 Introduction

Figure : power rating of the main generators of some common aircraft (in red
medium to long range aircraft ,in black short to medium range aircraft)
4
 Introduction

• AC generator are lighter and smaller in comparison to equal rated power DC

generators.
• The absence of commutators on AC generators improved the maintenance and
lifetime performance
• A relevant of using AC generator is the potentially much higher operating voltage.
In particular, a considerable cabling weight reduction was accomplished by
increasing the operating voltage.
• Using AC generator is also introduced new challenges : the need to manage the
reactive power, the choice of an appropriate frequency.

5
 Synchronous generator construction

• Synchronous generators are synchronous machines used to

convert mechanical power to ac electric power.
• In a synchronous generator, a rotor magnetic field is
produced either by designing the rotor as a permanent
magnet or by applying a dc current to a rotor winding to
create an electromagnet.
• The rotor of the generator is then turned by a prime mover,
producing a rotating magnetic field within the machine.
This rotating magnetic field induces a three-phase set of
voltages within the stator windings of the generator.
• Two terms commonly used to describe the windings on a
machine are field windings and armature windings :
 The term field windings applies to the windings that
produce the main magnetic field in a machine
 The term armature windings applies to the windings where Figure : synchronous generator
the main voltage is induced.

6
 Synchronous generator construction

• The rotor of a synchronous generator is essentially a large electromagnet. The

magnetic poles on the rotor can be of either salient or non salient construction.

Figure : a nonsalient two pole rotor for a

Figure : a salient six pole rotor for a
synchronous machine
synchronous machine
7
 Synchronous generator construction

• A dc current must be supplied to the field circuit on the rotor if it is an

electromagnet. Since the rotor is rotating, a special arrangement is required to get
the dc power to its field windings. There are two common approaches to supplying
this dc power :
 Supply the dc power from the external dc source to the rotor by means of slip
rings and brushes.
 Supply the dc power from a special dc power source mounted directly on the
shaft of the synchronous generator.
• Slip rings and brushes create a few problems when they are used to supply dc
power to the field windings of a synchronous machine. They increase the amount
of maintenance required on the machine, since the brushes must be check for wear
regularly. In addition, brush voltage drop can be the cause of significant power
losses on machines with larger field currents. Despite these problems, slip rings
and brushes are used on all smaller synchronous machines, because no other
method of supplying the dc field current is cost effective.

8
 Synchronous generator construction

• On larger generators and motors, brushless

exciters are used to supply dc field current
to the machine. A brushless exciter is a
small ac generator with its field circuit
mounted on the stator and its armature
circuit mounted on the rotor shaft.
• Three phase output of the exciter generator
is rectified to direct current by a three
phase rectifier circuit also mounted on the
shaft of the generator, and then fed into the
main dc field circuit.
• Since no mechanical contacts ever occur
between the rotor and stator, a brushless
requires less maintenance than slip rings
and brushes.
Figure : a brushless exciter circuit
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 Synchronous generator construction

• To make the excitation of a generator

completely independent of any external
power sources, a small pilot exciter is
often included in the system. a pilot
exciter is a small ac generator with
permanent magnets mounted on the rotor
shaft and a three-phase winding on the
stator. It produces the power for the field
circuit of the exciter, which in turn
controls the field circuit of the main
machine.
• If a pilot exciter is included on the
generator shaft, then no external electric
power is required to run the generator.

Figure : a brushless excitation scheme

that includes a pilot exciter
10
 Speed of Rotation of a Synchronous Generator

• The rate of rotation of the magnetic fields in the machine is related to the stator
electrical frequency by equation :
𝑛𝑚 𝑃
𝑓𝑠𝑒 =
120
𝑓𝑠𝑒 : electrical frequency, in Hz
𝑛𝑚 : mechanical speed of magnetic field, in r/min ( equals speed of rotor for synchronous
machines )
P : number of poles
• Since the rotor turn at the same speed as the magnetic field, this equation relates
the speed of rotor rotation to the resulting electrical frequency.
• Example :
 To generate 60 Hz power in a two-pole machine, the rotor must turn at 3600 r/min
 to generate 50 Hz power in a four-pole machine, the rotor must turn at 1500 r/min

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 Internal generated voltage of a synchronous generator

• Internal generated voltage of a synchronous generator EA : 𝐸𝐴 = 𝐾ϕω

where K is a constant representing the construction of the machine
• EA depends on the flux ϕ in the machine, the speed of rotation ω, and the machine’s
construction. The field current IF is related to the flux ϕ in the manner shown in figure a.
Since EA is directly proportional to the flux, the internal generated voltage EA is related to
the field current as shown in figure b.

Figure (a): Plot of flux versus field Figure (b) : The magnetization curve for the
current for a synchronous generator synchronous generator
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 Equivalent circuit of a synchronous generator

The equivalent circuit of a synchronous generator :

• This figure shows a dc power source supplying
the rotor field circuit is modeled by a coil’s
inductance and resistance in series. In series with
RF is an adjustable resistor Radj which controls
the flow of field current.
• the rest of the equivalent circuit consists of the
models of each phase. Each phase have an
internal generated voltage with a series
inductance (consisting the sum of the armature
reactance and the coil’s self-inductance).
• the voltages and currents of the three phase are
1200 apart in angle, but other-wise the three
phase are identical
Figure : full equivalent circuit of a
three-phase synchronous generator
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 Equivalent circuit of a synchronous generator
If the three phase are Y-connected, then the
If the three phase are Δ-connected, :
terminal voltage VT (which is the same as line-to-
𝑉𝑇 = 𝑉ϕ
line voltage VL) is related to the phase voltage by :
𝑉𝑇 = 𝑉𝐿 = 3𝑉ϕ

Figure : generator equivalent circuit connected in Y Figure : generator equivalent circuit connected in Δ
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 Equivalent circuit of a synchronous generator

• The fact that the three phase of a synchronous generator are identical in all aspects except of
phase angle normally leads to the use of a per-phase equivalent circuit.
Vϕ = EA − jIA X S − R A IA
Where EA : is the internal generated voltage produced in one phase of a synchronous generator
Vϕ : output voltage of a phase
XS : synchronous reactance
RA : stator resistance

Figure : per-phase equivalent circuit of a

three-phase synchronous generator

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 Phasor diagram of a synchronous generator

• The voltages in a synchronous generator are ac

voltages. They are usually expressed as phasors.
Since the phasors have both a magnitude and an
angle, the relationship between them must be
expressed by a two-dimensional plot.
• When the voltages within a phase (EA , Vϕ , j XS IA
and RA IA ) and the current IA in the phase are
plotted in such a fashion to show the relationships
among them, the resulting plot is called a phase
diagram.
• When the generator is supplying a purely Figure : phasor diagram of a synchronous
resistance load. The total voltage EA differs from generator a purely resistive load
the terminal voltage of the phase Vϕ by the
resistive and inductive voltage drops. All voltages
and currents are referenced to Vϕ , which is
arbitrarily assumed to be at an angle of 00

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 Phasor diagram of a synchronous generator

Vϕ = EA − jIA XS − R A IA Vϕ = EA − jIA XS − R A IA

Figure : phasor diagram of a synchronous Figure : phasor diagram of a synchronous

generator at lagging power factor generator at leading power factor
17

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 Power and torque in synchronous generators

Power and torque in synchronous generator :

• A synchronous generator convert mechanical power to three-phase electrical power. Not
all the mechanical power going into a synchronous generator becomes electrical power
out of the machine. The difference between input and output power presents the losses of
the machine.
• The input mechanical power is the shaft
power in the generator :
𝑃𝑖𝑛 = 𝜏𝑎𝑝𝑝 𝜔𝑚
• Power converted from mechanical to electrical
form internally is given by :
𝑃𝑐𝑜𝑛𝑣 = 𝜏𝑖𝑛𝑑 𝜔𝑚 = 3𝐸𝐴 𝐼𝐴 𝑐𝑜𝑠𝛾
Where γ is the angle between EA and IA Figure : flow power diagram of
a synchronous generator

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 Power and torque in synchronous generators

• The real electrical output power of the synchronous can be expressed in line quantities and
in phase quantities as :
𝑃𝑜𝑢𝑡 = 3𝑉𝐿 𝐼𝐿 𝑐𝑜𝑠θ = 3𝑉ϕ 𝐼𝐴 𝑐𝑜𝑠𝜃
• the reactive power output can be expressed in line quantities and in phase quantities as :
𝑄𝑜𝑢𝑡 = 3𝑉𝐿 𝐼𝐿 𝑠𝑖𝑛θ = 3𝑉ϕ 𝐼𝐴 𝑠𝑖𝑛𝜃

Figure : flow power diagram of a synchronous generator

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 Power and torque in synchronous generators

• If the armature resistance RA is ignored (since XS >>

RA ) , then a very useful equation can be derived to
approximate the output power of the generator :
𝑃𝑐𝑜𝑛𝑣 = 𝑃𝑜𝑢𝑡
𝐸𝐴 𝑠𝑖𝑛𝛿
𝐼𝐴 𝑐𝑜𝑠𝜃 =
𝑋𝑆
3𝑉ϕ 𝐸𝐴
𝑃𝑜𝑢𝑡 = 3𝑉ϕ 𝐼𝐴 𝑐𝑜𝑠𝜃 = 𝑠𝑖𝑛𝛿
𝑋𝑆
• The power produced by a synchronous generator
depends on the angle 𝛿 between Vϕ and EA
• The angle δ is known as the internal angle or torque
angle of the machine. The maximum power that the
generator can supply occurs when 𝛿=900 : Figure : simplified phasor diagram with
3𝑉ϕ 𝐸𝐴 armature resistance ignored
𝑃𝑜𝑢𝑡𝑚𝑎𝑥 =
𝑋𝑆

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 Power and torque in synchronous generators

• If the armature resistance RA is ignored (since

XS >> RA ),
𝑃𝑜𝑢𝑡 = 𝑃𝑐𝑜𝑛𝑣 + 𝐼 2 𝑅𝐴 = 𝑃𝑐𝑜𝑛𝑣
𝑃𝑐𝑜𝑛𝑣 = 𝜏𝑖𝑛𝑑 𝜔𝑚
3𝑉ϕ 𝐸𝐴
𝑃𝑜𝑢𝑡 = 3𝑉ϕ 𝐼𝐴 𝑐𝑜𝑠𝜃 = 𝑠𝑖𝑛𝛿
𝑋 𝑆
• The induced torque can be expressed as :

𝑃𝑐𝑜𝑛𝑣 3𝑉ϕ 𝐸𝐴
𝜏𝑖𝑛𝑑 = = 𝑠𝑖𝑛𝛿
𝜔𝑚 𝜔𝑚 𝑋𝑆
Figure : flow power diagram of a synchronous generator

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 Measuring synchronous generator model parameters

• Measuring synchronous generator model

parameters :
 The relationship between field current IF
and flux ( and therefore between the field
current IF and EA )
 The synchronous reactance
 The armature resistance

Figure : per-phase equivalent circuit of a

three-phase synchronous generator

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 Measuring synchronous generator model parameters

Open circuit test on the generator :

• The generator is turned at the rated speed, the
terminals are disconnected from all loads, and the
field current is set to zero.
• Then the field current is gradually increased in
steps, and the terminal voltage is measured at
each step along the way. With the terminals open
IA =0 , so :
Vϕ = EA − jIA XS − R A IA = 𝐸𝐴
• We can construct a lot of EA versus IF from this
information. This plot is called open-circuit
characteristic (OCC) of a generator. With this
characteristic, it is possible to find the internal
generated voltage of the generator for any given
field current. Figure : Open circuit characteristic (OCC)
of a synchronous generator.

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 Synchronous generator operating alone

• The synchronous generator operating alone : to understand the operating characteristics of a synchronous
generator operating alone, examine a generator supplying a load.
• What happens when we increase the load in this generator ?
• The speed of generator will be assumed constant, and all terminal characteristics are drawn assuming
constant speed. Also, the rotor flux in the generators is assumed constant unless their field current is
explicitly changed.
• An increase in the load is an increase in the real and/or reactive power drawn from the generator. Such a
load increase increases the load current drawn from the generator. Because the field resistor has not been
changed, the field current is constant, and therefore the flux ϕ is constant. Since, the prime mover also
keeps a constant speed ω, the magnitude of the internal generated voltage EA =Kϕω is constant.

Figure : a single generator supplying a load

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 Synchronous generator operating alone

• First, examine a generator operating at a EA , IA , XS : are parameters of generator at the

lagging power factor. The current IA lags first state.
phase to the voltage Vϕ by an angle θ. E’A , I’A ,X’S : are parameters of generator at
• If more load is added at the same power the second state ( when the load is increased)
factor, then 𝐼𝐴 increases but remains at
the same angle θ with respect to Vϕ as
before. Therefore, the armature reaction
voltage jXSIA is larger than before but at
the same angle.
• Now since: 𝐸𝐴 = 𝑉𝜙 + 𝑗𝐼𝐴 𝑋𝑆
𝐸′𝐴 = 𝑉′𝜙 + 𝑗𝐼′𝐴 𝑋′𝑆
The magnitude of the internal generated
voltage EA is constant . So, the magnitude
of voltage Vϕ decreases rather sharply.
Figure : the effect of an increase in generator loads
at constant power factor upon its terminal voltage
(case of lagging power factor)
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 Synchronous generator operating alone

• If the generator is loaded with unity EA , IA , XS : are parameters of generator at the

power factor loads. The current IA is in first state.
E’A , I’A ,X’S : are parameters of generator at
phase with the voltage Vϕ . the second state ( when the load is increased)
• If more load is added at the same
power factor, then 𝐼𝐴 increases.
Therefore, the armature reaction
voltage jXSIA increases.
• Now since: 𝐸𝐴 = 𝑉𝜙 + 𝑗𝐼𝐴 𝑋𝑆
𝐸′𝐴 = 𝑉′𝜙 + 𝑗𝐼′𝐴 𝑋′𝑆
the magnitude of the internal
generated voltage EA is constant . So,
the magnitude of voltage Vϕ decreases Figure : the effect of an increase in generator loads
only slightly. at constant power factor upon its terminal voltage
(case of unity power factor)
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 Synchronous generator operating alone

• If the generator is loaded with leading EA , IA , XS : are parameters of generator at the

power factor loads. The current IA leads first state.
phase to the voltage Vϕ . E’A , I’A ,X’S : are parameters of generator at
the second state ( when the load is increased)
• If more load is added at the same power
factor, then 𝐼𝐴 increases. Therefore, the
armature reaction voltage jXSIA increases.
• Now since: 𝐸𝐴 = 𝑉𝜙 + 𝑗𝐼𝐴 𝑋𝑆
𝐸′𝐴 = 𝑉′𝜙 + 𝑗𝐼′𝐴 𝑋′𝑆
the magnitude of the internal generated
voltage EA is constant . And, the magnitude
of voltage Vϕ increases.
Figure : the effect of an increase in generator loads
• In this case, an increase in the load in the at constant power factor upon its terminal voltage
generator produced an increase in the (case of leading power factor)
terminal voltage.

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 Synchronous generator operating alone

• Conclusions :
 If lagging loads (+Q or inductive reactive power loads) are added to a generator,
Vϕ and the terminal voltage VT decrease significantly.
 If unity-power-factor loads (no reactive power) are added to a generator, there is a
slight decrease in Vϕ and the terminal voltage.
 If leading loads (-Q or capacitive reactive power loads) are added to a generator,
Vϕ and the terminal voltage will rise.

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 Voltage regulation

• A convenient way to compare the voltage behavior of two generators is by their voltage
regulation. The voltage regulation (VR) of a generator is defined by the equation :
𝑉𝑛𝑙 −𝑉𝑓𝑙
𝑉𝑅 = × 100%
𝑉𝑓𝑙
Vnl : is the no-load terminal voltage of generator
Vfl : is the full-load terminal voltage of generator
• A synchronous generator operating at a lagging power factor has a fairly positive
voltage regulation
• A synchronous generator operating at a unity power factor has a small positive
voltage regulation
• A synchronous generator operating at a leading power factor has a negative
voltage regulation

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 Voltage regulation

• It is desirable to keep the voltage supplied to a load constant, even through the
load itself varies.
• The obvious approach is to vary the magnitude of EA to compensate for changes
in the load. Recall that 𝐸𝐴 = 𝐾ϕω. Since the frequency should not be changed in
a normal system, EA must be controlled by varying the flux in the machine.
• For example, suppose that a lagging load is adding to a generator. then Vϕ
decrease. To keep Vϕ in constant RF should be decreased. IF will increase
the flux ϕ will increase EA will increase Vϕ will be kept in constant.

Keep Vϕ in constant
under the variation of
load

Figure : per-phase equivalent circuit of a three-phase synchronous generator

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 Voltage regulation

• The idea can be summarized as follows :

 Decreasing the field resistance in the generator increases its field current
 An increase in the field current increases the flux in the machine
 an increases in the flux increases the internal generated voltage 𝐸𝐴 = 𝐾ϕω
 An increase in EA increases Vϕ and the terminal voltage of the generator.
• The process can be reversed to decrease the terminal voltage. It is possible to regulate the
terminal voltage of a generator throughout a series of load changes simply by adjusting the
field current

Figure : per-phase equivalent circuit of a three-phase synchronous generator

31
 AC Generator

• Example 1: A 480-V 60-Hz, Δ-connected, four-pole synchronous generator has the Open circuit
characteristic shown in Figure a. This generator has a synchronous reactance of 0.1 Ω and an
armature resistance of 0.015 Ω. At full load, the machine supplies 1200 A at 0.8 power factor
lagging. Under full-load conditions, the friction and windage losses are 40 kW, and the core losses
are 30 kW. Ignore any field circuit losses
a. What is the speed of rotation of this generator?
b. How much field current must be supplied to the generator to make the terminal voltage 480 V at
no load?
c. If the generator is now connected to a load and the load draws 1200 A at 0.8 power factor
lagging, how much field current will be required to keep the terminal voltage equal to 480 V?
d. Suppose that the generator is connected to a load drawing 1200 A at 0.8 power factor leading.
How much field current would be required to keep VT at 480 V?

32
 AC Generator

Figure : phasor diagram of

synchronous generator

Figure : Open circuit characteristic of

synchronous generator
33
 AC Generator

Solution :
This synchronous generator is Δ connected, so its phase voltage is equal to its line voltage Vϕ = VT ,
while its phase current is related to its line current by the equation IL = 3 Iϕ
a. The relationship between the electrical frequency produced by a synchronous generator and the
mechanical rate of shaft rotation is given by equation :
𝑛𝑚 𝑃 120𝑓𝑠𝑒 120(60 𝐻𝑧)
𝑓𝑠𝑒 = therefore 𝑛𝑚 = = = 1800 𝑟/𝑚𝑖𝑛
120 𝑃 4 𝑝𝑜𝑙𝑒𝑠
b. In this machine, Vϕ = VT . Since the generator is at no load , IA = 0 and Vϕ = EA . Therefore, Vϕ =
VT = EA = 480 V. and from the open circuit characteristic, IF = 4.5 A

34
 AC Generator

c. if the generator is now connected to a load draws 1200 A at 0.8 power factor lagging, then the
armature current in the machine is :
1200𝐴
𝐼𝐴 = = 692.8 𝐴
3
The power factor lagging : cosθ = -0.8 so θ = -36.870
If the terminal voltage Vϕ is adjusted to be 480 V, the internal generated voltage EA is given by :
𝐸𝐴 = 𝑉ϕ + 𝑅𝐴 𝐼𝐴 + 𝑗𝑋𝑆 𝐼𝐴 = 480∠00 + 0.015Ω 692.8∠ −36.870 + 𝑗0.1Ω 692.8∠ −36.870
= 480∠00 + 10.39∠ −36.870 +692.8∠53.130 = 529.9 + j49.2 = 532∠5.30
To keep the terminal voltage at 480 V, EA must be adjusted to 532 V. Open circuit characteristic of
synchronous generator, the required field current is 5.7 A.

Figure : phasor diagram of the generator

35
 AC Generator

d. The power that the generator is supplying can be found from the equation as below :
𝑃𝑜𝑢𝑡 = 3𝑉𝐿 𝐼𝐿 𝑐𝑜𝑠θ = 3 480𝑉 1200𝐴 cos 36.870 = 798 𝑘𝑊
To determine the power input to the generator, use the flow diagram. The mechanical input power is
given by :
Pin = Pout + Pelec loss + Pcore loss + Pmesh loss+ Pstray loss
The stray losses were not specified here, so they will be ignored. In this generator, the electrical
losses are :
𝑃𝑒𝑙𝑒𝑐 𝑙𝑜𝑠𝑠 = 3𝐼2 𝑅𝐴 = 3. 692.8 2
0.015 = 21.6 𝑘𝑊
The core losses are 30 kW, and the friction and windage losses are 40 kW, so the total input power to
the generator is :
Pin = Pout + Pelec loss + Pcore loss + Pmesh loss+ Pstray loss = 798 +21.6+30+40=889.6 kW
𝑃𝑜𝑢𝑡 798 𝑘𝑊
Therefore, the machine’s overall efficiency is : ƞ = . 100% = 100% = 89.75%
𝑃𝑖𝑛 889.6 𝑘𝑊

36
 AC Generator

e. If the generator’s load were suddenly disconnected from the line, the current IA would drop to zero,
making EA =Vϕ . Since the field current has not changed, so EA has not changed, and Vϕ must rise
to equal to EA .
Therefore, if the load is suddenly dropped, the terminal voltage of the generator would rise to 532 V.
d. If the generator were loaded down with 1200A at 0.8 power factor leading while the terminal
voltage was 480 V:
Cosθ =0.8 so θ=arccos (0.8)=36.870
then the internal generated voltage :
𝐸𝐴 = 𝑉ϕ + 𝑅𝐴 𝐼𝐴 + 𝑗𝑋𝑆 𝐼𝐴 = 480∠00 + 0.015Ω 692.8∠36.870 + 𝑗0.1Ω 692.8∠36.870
= 480∠00 + 10.39∠36.870 + 692.8∠126.870 = 446.7 + j61.7 = 451∠7.10
Therefore, the internal generated voltage EA must be adjusted to provide 451V if VT is to remain
480V. Using the open circuit characteristic, the field current would have to adjusted to 4.1 A.

37
 AC Generator

Example 2 : A three-phase Y-connected synchronous generator is rated 120 MVA, 13.2 kV, 0.8
power factor lagging, and 60 Hz. Its synchronous reactance is 0.9 Ω, and its resistance may be
ignored. What is its voltage regulation?

38
 AC Generator

• Solution :
𝑆 120 𝑀𝑉𝐴
the rated armature current is : 𝐼𝐴 = 𝐼𝐿 = = = 5249 𝐴
3𝑉𝑇 3(13.2 𝑘𝑉)

The power factor is 0.8 lagging, so cosθ = -0.8 and θ = -36.870 , the armature current 𝐼𝐴 = 5248∠ −
36.870
13.2𝑘𝑉
The phase voltage is : 𝑉ϕ = = 7621 𝑉
3
Therefore, the internal generated voltage is :
𝐸𝐴 = 𝑉ϕ + 𝑅𝐴 𝐼𝐴 + 𝑗𝑋𝑆 𝐼𝐴 = 7621∠00 + 𝑗0.9Ω 5249∠ −36.870 = 11120∠19.90 V
The resulting voltage regulation is :
11120 − 7621
𝑉𝑅 = 100% = 45.9%
7621

39
AC main generation in aircraft
PMG : Permanent magnet generator

Figure : integrated drive generator (IDG)

40
 AC main generation in aircraft
PMG : Permanent magnet generator
• Integrated drive generator (IDG)
consists of a constant speed drive
(CSD) and an AC generator mounted
side by side in a single housing. The
CSD components convert a variable
input speed to a constant output
speed.
• The CSD portion of the IDG is a
hydro mechanical device that adds or
subtracts from variable input speed of
the engine gearbox. The CSD
performs this operation by controlled
differential action to maintain the
constant output speed required to
drive the generator.

Figure : integrated drive generator

41
 AC main generation in aircraft

• Main generator : the main generator

consists of a 2-pole rotor and a three-
phase winding stator. As the rotor rotates,
the DC field induces an AC voltage in the
stator winding.
• The exciter stage of the generator
consists of a three-phase winding rotor
which is located on the main generator
shaft. The stator windings and a full wave
bridge rectifier (rotating diode rectifier)
are also located on the rotor. As it rotates
within the stator, the exciter stage
converts a DC field voltage supplied by
the Generator control unit (GCU) to the
exciter stator, to an AC voltage in the
windings of the rotor. The induced AC
voltage is rectified by the rotating diode
rectifier and supplied to the winding of
the main generator.
Figure : AC main generator

42
 AC main generation in aircraft

• Permanent magnet generator (PMG) :

consists of a permanent magnet rotor and
a three-winding stator. As the engine is
running, the PMG rotor induces an AC
voltage in the windings of PMG stator.
• The output of PMG is supplied to the
generator control unit (GCU), which uses
it for the following functions :
 The GCU transforms and rectifies the
voltage via the transformer rectifier to 28
V DC for the internal power supply.
 the GCU sends the PMG voltage via the
generator control relay (GCR) to the
voltage and regulation circuit where it is
rectified and applied to the exciter field
for voltage regulation.

43
 Voltage regulation

Keep the generator output

voltage in constant of 115 Vac

Voltage regulation system

Figure : Electrical power system in aircraft

44
 Voltage Regulation

• The voltage regulation is achieved

by regulating the generator
excitation current. It also supplies
the exciter field through the
generator control relay (GCR),
which is normally closed, and
through a rectifier.
• When the generator starts running,
the PMG supplies the GCU. As the
PMG frequency is below the
threshold, the generator excitation is
reduced through the excitation
control module.

Figure : voltage regulation

45
 Voltage Regulation

• The voltage regulation shutdown

signal, generated by the protection
module, triggers the voltage
regulation shutdown control module
which simulates a high generator
load signal via the current limit
module.
• As soon as the PMG frequency is
over the threshold, the generator
excitation is regulated normally.
• The excitation current depends on
the comparison between the voltage
sensed at the point of regulation
(POR) and a reference voltage.

Figure : voltage regulation

46
 Voltage Regulation

• As the parameters are correct, the

generator line contactor closes.
• The generator excitation also
depends on the load.
• If a fault is detected by the GCU
protection module, the voltage
regulation shutdown control module
simulates a high generator load
signal. The exciter field coils are no
longer supplied. The GCR and GLC
are tripped.

Figure : voltage regulation

47
Thank you for your attention

48