Outline
Electrical Drive Systems 324
1
DC Motors & Generators
Dr. P.J. Randewijk
Stellenbosch University
Dep. of Electrical & Electronic Engineering
Stephan J. Chapman
Chapter 8 (5th Edition)
Chapter 8 – DC Motors and Generators
8.1 – Introduction to DC Motors
8.2 – The Equivalent Circuit of a DC Motor
8.3 – The Magnetization Curve of a DC Machine
8.4 – Separately Excited and Shunt DC Motors
8.5 – The Permanent-Magnet DC Motor
8.8 – DC Motor Starters
8.10 – DC Motor Efficiency Calculations
8.11 – Introduction to DC Generators
8.12 – The Separately Excited DC Generator
8.13 – The Shunt DC Generator
1 / 48
8.1 Introduction to DC Motors
8.2 DC Motor Equivalent Circuit
The speed regulation (SR) of a DC motor is defined as
SR =
nm,nl − nm,fl
× 100%
nm,fl
(8–2)
with nl ≡ no-load & fl ≡ full-load
The equivalent electrical circuit of the DC machine look
as follows,
F1
A1
LA
RF
The different types of DC motor that exist are:
Ê
Ë
Ì
Í
Î
2 / 48
RA
separately excited
shunt
permanent-magnet
series
compound
+
LF
−
+
EA =K φωm
φ =⇒
−
ωm
F2
A2
LA , RA & EA is the Thévenin equivalent circuit of the
armature, and
RF & LF is the equivalent impedance of the field
3 / 48
4 / 48
8.2 DC Motor Equivalent Circuit (cont.)
8.2 DC Motor Equivalent Circuit (cont.)
The dynamic model, including that for the mechanical
shaft, will look as follows:
A1
For the mechanical side, we can thus write:
τind = τb + τJ + τload
dωm
+ τload
= bωm + J
dt
iA
+
iA RA
RA
−
+
LA
diA
dt
LA
+
−
−
+
m
For steady state operation, i.e. dω
dt =0 (no acceleration),
the above equation simplifies to
VT
EA =K φωm
τind = bωm + τload
−
τind =K φiA
ωm
A2
b
For no-load operation, the only load is the viscous
friction component, so that at no-load
τb =bωm
dωm
τJ =J
dt
J
τind = bωm
τload
5 / 48
8.2 DC Motor Equivalent Circuit (cont.)
6 / 48
8.2 DC Motor Equivalent Circuit (cont.)
VT = EA + Vbrush + IA RA
All the DC machine’s mechanical (i.e. windage & friction)
losses and core/iron (i.e. eddy & hysteresis) losses can
usually be modelled a viscous frictional losses. . .
If we were to ignore the brush losses, the above
equation simplifies to
Pwindage & friction + core = PNL
VT = EA + IA RA
= Pviscous
The steady state equivalent circuit of the DC machine
can thus be simplified to:
2
= bωm
For the electrical side, we can thus write (KVL):
VT = EA + Vbrush + iA RA + LA
F1
diA
dt
A1
RF
For steady state operation, i.e. didtA =0 (with iA =IA
constant), the above equation simplifies to
+
LF
RA
−
+
EA =K φωm
φ =⇒
−
ωm
F2
7 / 48
A2
8 / 48
8.3 DC Machine Magnetisation Curve
8.3 DC Machine Magnetisation Curve (cont.)
With an increase in field current, the flux in the DC
machine would increase linearly if saturation were to be
ignored
The flux produced by the field of the DC machine will
depend on the MMF of the field circuit,
F = NF IF
The relationship would ultimately be determined by the
magnetic core’s B–H relationship
And the total reluctance, R of the DC machine, so that
we can write
F
R
N I
= F F
R
NF VF
=
RRF
φ=
with VF the voltage applied to the field circuit and RF the
resistance of the field circuit
9 / 48
8.3 DC Machine Magnetisation Curve (cont.)
10 / 48
8.3 DC Machine Magnetisation Curve (cont.)
If we were to drive the DC machine at a constant speed
(e.g. ωm =ω0 ) whilst varying the field current
The internal generated voltage will exhibit the same
B–H relationship as shown on the previous slide. . .
With the relationship between the internal generated
voltage and speed given by
EA = K φωm
(7–38)
The value of K φ for each value of the field current (IF
can be determined from the graph
As long as IF remains constant, K φ will remain constant
as K φ is only a function of field current, i.e. K φ = f (IF )
Thus with IF constant, we can determine EA at a
different speed directly from (7–38)
Alternatively, with IF constant, if we know EA at one
speed,
EA|1 = K φωm|1
11 / 48
12 / 48
8.3 DC Machine Magnetisation Curve (cont.)
8.4 SEPEX and Shunt DC Motors
We can also determine EA at a different speed, by using
the following relationship
For a Shunt DC Machine, the field circuit is connected in
shunt / parallel with the armature circuit
EA|2 = K φωm|2
EA|1
ω
=
ωm|1 m|2
= EA|1
The Terminal Characteristic of a Shunt DC Motor
nm|1 ( 2π
60 )
+ A SEPEX DC Machine’s characteristics will be the
same. . .
We are interested in the output speed versus torque of
the DC machine, i.e. torque as a function of speed. . .
nm|2 ( 2π
60 )
nm|1
= EA|1
nm|2
of
nm|2 = nm|1
EA|1
EA|2
For a Separately Excited (SEPEX) DC Machine, the
field circuit is connected to a separate supply
(8–9)
Ê Write down the Kirchoff’s voltage law (KVL) equation for
the armature circuit
Ë Make use of (7–39) & (7–49) to substitute EA & IA
Ì Rewrite the equation so that we have ωm as a function of
τind
13 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
14 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
Thus:
Speed Control of Shunt DC Motors
VT = EA + IA RA
τ
= K φωm + ind RA
Kφ
(8–3)
(8–6)
resulting in the following equation
ωm =
RA
VT
−
τind
K φ (K φ)2
(8–7)
Speed control of SEPEX and Shunt DC Motors can
accomplished by:
Ê Inserting an external resistor in series with the armature
circuit
Ë Adjusting the terminal voltage applied to the armature
circuit
Ì Inserting an external resistor in series with the field
circuit in order to change the field current and hence the
field flux
which is the equation for a straight line, see Fig. 8–6 (a)
+ We will ignore the effect of armature reaction, see Fig.
8–6 (b)
Nonlinear Analysis of a Shunt DC Motor – ignore
15 / 48
16 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
8.4 SEPEX and Shunt DC Motors (cont.)
Equation (8–7) can thus be rewritten as
INSERTING A RESISTOR IN SERIES WITH THE
ARMATURE CIRCUIT
An external variable resistor / rheostat is inserted in
series with the armature circuit
ωm =
The y–intersect is not affected by RA|ext , only the slope
of the graph as RA|ext is adjusted (see next slide)
RA|ext
F1
A1
The advantage of rheostatic control is
RF
VF
+
−
+
RA + RA|ext
VT
−
τind
Kφ
(K φ)2
4 simple and easy to implement
RA
LF
+
−
+
−
EA =K φωm
φ =⇒
VT
The disadvantage of rheostatic control is
8 it is usually manual control, i.e. difficult to automate
8 not very energy efficient, i.e. additional IA|ext 2 RA losses
−
ωm
F2
A2
17 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
18 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
Sometimes it is more convenient to express torque as a
function of speed
2.0
1.8
Speed [p.u.]
1.6
RA|ext
1.4
τind =
1.2
1.0
Now the x–intersect is constant, but again with a
changing slope as RA|ext is adjusted (see next slide)
0.00 p.u.
0.8
0.05 p.u.
0.10 p.u.
0.6
0.15 p.u.
Specifically were one has load with a known torque vs.
speed characteristic (e.g. a pump or a fan, were we
know τload =k 2 ωm )
0.20 p.u.
0.4
0.25 p.u.
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
(K φ)2
Kφ
VT −
ωm
(RA + RA|ext )
(RA + RA|ext )
This is done in order to determine the operating point of
the system, i.e. were the load’s characteristic curve
intersects the motor’s characteristic curve
2.0
Torque [p.u.]
19 / 48
20 / 48
0.05 p.u.
8.4 SEPEX and Shunt DC Motors (cont.)
CHANGING THE ARMATURE VOLTAGE
0.00 p.u.
0.10 p.u.
0.15 p.u.
0.20 p.u.
0.25 p.u.
8.4 SEPEX and Shunt DC Motors (cont.)
This requires that the voltage applied to the terminals of
the armature circuit, VT , be variable
2.0
1.8
Torque [p.u.]
1.6
F1
1.4
A1
RA|ext
1.2
1.0
RF
0.8
VF
0.6
+
−
+
RA
LF
+
−
+
−
EA =K φωm
φ =⇒
VT
−
0.4
ωm
0.2
0.0
F2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
A2
2.0
Speed [p.u.]
21 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
22 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
Equation (8–7) can thus be used as is
2.0
1.8
V
RA
ωm = T −
τind
K φ (K φ)2
We can now see that only the y –intersect is affected by
adjusting VT , whilst the slope remains constant (see
next slide)
The advantage of armature terminal voltage control is
Spoed [p.e.]
1.6
VT
1.4
1.2
1.0
1.00 p.e.
0.8
0.90 p.e.
0.80 p.e.
0.6
0.70 p.e.
0.60 p.e.
0.4
4 more energy efficient that rheostatic control
4 there is a “nice” linear relationship between terminal
voltage and speed (will be shown later)
0.50 p.e.
0.2
0.0
0.0
The disadvantage of armature terminal voltage control
is
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
Draaimoment [p.e.]
8 a variable voltage source is not that trivial and
8 it can be expensive
23 / 48
24 / 48
1.0 p.e.
0.9 p.e.
0.8 p.e.
0.5 p.e.
Again sometimes it is more convenient to express
torque as a function of speed
0.7 p.e.
8.4 SEPEX and Shunt DC Motors (cont.)
0.6 p.e.
8.4 SEPEX and Shunt DC Motors (cont.)
2.0
τind =
(K φ)2
Kφ
VT −
ωm
RA
RA
We can now see that only the x–intersect is affected by
adjusting VT , whilst again the slope remains constant
(see next slide)
Draaimoment [p.e.]
1.8
1.6
1.4
VT
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
Spoed [p.e.]
25 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
Yet another way of looking at armature terminal voltage
control is by plotting VT as a function of ωm for different
values of τm (see next slide)
Again we start by writing down the KVL equation and
then substituting (7–38) & (7–49) into it
IA
EA
z}|{
z }| {
τm
VT =
RA + K φωm
Kφ
2.0
1.0 p.u.
0.5 p.u.
1.8
Terminal Voltage [p.u.]
8.4 SEPEX and Shunt DC Motors (cont.)
26 / 48
0.0 p.u.
1.6
τm
1.4
1.2
1.0
0.8
0.6
0.4
N.B.
0.2
There is a “nice” linear relationship between the speed of a
DC motor and the applied Armature-Terminal Voltage
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
Speed [p.u.]
+ If RA → 0, the the three graphs will be on top of one
another. . .
27 / 48
28 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
8.4 SEPEX and Shunt DC Motors (cont.)
Equation (8–7) can again be used as is
CHANGING THE FIELD RESISTANCE (OR FIELD
VOLTAGE)
This requires that the voltage applied to the terminals of
the field circuit, VF , be variable, or an external resistor in
series with the field circuit being used (as shown in Fig.
8–7)
F1
VF
+
−
+
LF
RA
+
−
+
−
EA =K φωm
φ =⇒
VT
but now with
We can now see that both the y–intersect and the slope
is affected by the change in flux, i.e. the change in IF
(see next slide)
−
ωm
F2
VT
RA
−
τind
K φ (K φ)2
K φ = f (IF )
V
= f( F )
RF
A1
RF
ωm =
A2
29 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
30 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
The advantage of field current control is
2.0
4 can operate above rated speed
4 easier to control due to the lower current value of the
field current compared to the armature current – the
power and current rating of RF |ext or VF is much smaller
than that of RA|ext or VT
IF
1.8
Speed [p.u.]
1.6
1.4
0.60 p.u.
1.2
0.70 p.u.
The disadvantage of field current control is
0.80 p.u.
1.0
0.90 p.u.
1.00 p.u.
8 if the field current and hence the flux goes to zero, i.e.
φ → 0, the speed of the machine can “run away”, i.e.
ωm → ∞
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
Torque [p.u.]
31 / 48
32 / 48
0.6 p.u.
1.0 p.u.
0.9 p.u.
Again sometimes it is more convenient to express
torque as a function of speed
0.8 p.u.
8.4 SEPEX and Shunt DC Motors (cont.)
0.7 p.u.
8.4 SEPEX and Shunt DC Motors (cont.)
2.0
1.8
(K φ)2
Kφ
VT −
ωm
RA
RA
1.6
Torque [p.u.]
τind =
but again now with
K φ = f (IF )
V
= f( F )
RF
IF
1.4
1.2
1.0
0.8
0.6
0.4
0.2
We can again now see that both the y–intersect and the
slope is affected by the change in flux, i.e. the change in
IF (see next slide)
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
Speed [p.u.]
33 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
34 / 48
8.4 SEPEX and Shunt DC Motors (cont.)
Below rated / base speed, speed control is done by
TORQUE AND POWER LIMITS OF A DC MOTOR
armature terminal voltage control
armature external resistance (rheostatic) control
In all the graphs shown in die previous slides, there
were a “dot” in the middle. . .
Above rated / base speed, speed control is done by
This is the rated operating point of the DC machine
field current control
Which occurs when,
Electrical motors different somewhat from Internal
Combustion Engines (ICEs) in terms of their
4 the rated terminal voltage is applied,
4 the rated field current is drawn and
4 the rated armature current is drawn,
“maximum developed torque” and
“maximum developed power”
So that the DC machine will be
Defined in Chapman as
+ developing its rated torque and
+ developing its rated power and
+ running at rated (or base) speed
τmax = K φIA|max
(8–14)
Pmax = τmax ωm
(8–15)
and
35 / 48
36 / 48
Loads
8.4 SEPEX and Shunt DC Motors (cont.)
and Power
Electronics
Series
1
8.4 SEPEX and Shunt DC Motors (cont.)
Functional block diagram of power conversion components in generator set.
So that I would redefine (8–14) conditions.
& (8–15) toIt is also possible to control the engine to run
τrated
Prated = τrated ωm|rated
(8–15’)
Gen-Set Operation
The interesting thing about electrical
motors
is that
A block
diagram
of theitelectronic power conversion system
can develop more than its ratedfor
torque.
..
the proof-of-concept
gen-set developed at the Oak
Ridge
National
Laboratory (ORNL) is shown in Figure 1
It however implies that more than
the rated
armature
[6],
[7].
The
military
gen-set uses an internal combustion
current needs to be drawn. . .
(IC) diesel engine to drive a radial-gap permanent magnet
Which could result in the motor (PM)
burning
out. . . at variable speed. The speed of the engine
alternator
8.4
50
45
40
35
30
25
20
15
10
5
0
1,800
Torque
Fuel Consumption
Power
2,200 2,600 3,000 3,400
Engine Speed (r/min)
0.40
0.38
0.36
0.34
0.32
0.30
0.28
0.26
0.24
0.22
0.20
3,800
2
Peak torque, power, and fuel consumption for the diesel
engine in the proof-of-concept gen-set.
is determined from a user-selectable interface that allows
the engine to run at its most
efficient operating point for a
37 / 48
38 / 48
given load and ambient thermal conditions. The variable
frequency, variable voltage produced by the PM alternator
is diode-rectified to dc voltage, and an inverter is used to
SEPEX and Shunt DC Motors
(cont.)
produce selectable-frequency,
controllable ac voltage. 8.4 SEPEX and Shunt DC Motors (cont.)
The user is allowed to select single-phase 120-V,
For a DC motor, the maximum (or
rated) torque–speed
dual-phase
120/240-V, or three-phase 120/208-V. Each of+ Below rated / base speed,
voltage configurations can be generated at 50, 60,
and power–speed curves, looksthese
as follows
Ê the applied armature terminal voltage increases linearly
and 400 Hz such that the unit can be compatible with
with speed,
equipment produced from around the world or for aero­
Ë but the armature current can be at its maximum / rated
space applications. The power-conversion system also in­
value, even from standstill
Ì thus the maximum power increases linearly with speed
corporates a bidirectional dc-dc converter that can charge
Í where as the maximum torque the DC motor can
24-V batteries that are used to start the IC engine and to
develop will be constant
power auxiliary loads [8]. The converter can also draw
power from the batteries to help maintain the dc link dur­
ing severe load transients.
+ Above rated / base speed,
Engine and Alternator Description
Each gen-set size was determined by selecting an advanced
diesel engine that had a high power-to-weight ratio. For
the smallest gen-set, an air-cooled Ruggerini MD 191
rated at 13 kW was used as the prime mover. The engine
39 / 48
Ê the applied armature terminal voltage can’t increase any
more,
Ë neither can the armature current,
Ì thus the maximum power stays constant (it can’t 3
increase)
PM alternator mounted to gen-set engine.
Í and hence the maximum torque the DC motor can
develop, will decrease hyperbolically (k = xy )
40 / 48
I EEE I NDUSTRY A PPLI CA TI ONS MA GA ZI NE • MA R|A PR 2003 • WWW.I EEE.ORG/I A S
and
where it is most audibly quiet, at its least-polluting operat­
ing point (from an emissions
= K φIA|rated
(8–14’) point of view), or at its most
reliable, stiffest point such that it is less sensitive to load
transients. This article describes a proof-of-concept devel­
opment for a 7.5-kW gen-set in a family of military
gen-sets in the 5- to 60-kW range.
Fuel Consumption (kg/kWh)
For an ICE, this is not possible
Torque (N⋅m) or Power (kW)
I prefer (and it is less confusing) to rather refer to it as
rated torque and rated speed. . .
49
8.5 The Permanent-Magnet DC Motor
8.8 DC Motor Starters
A Permanent-Magnet (PM) DC motor uses PMs to
generate the flux in the motor
From the KVL equation for the armature circuit,
IA =
Thus the flux is fixed and hence K φ is constant
For PM DC motors, manufacturers sometimes refer to
K φ as motor or machine constant, Km
VT − EA
RA
We can see that at start-up,
Advantages of PM DC Motors are
we don’t have to worry about the field
Km = K φ = constant
Disadvantages of PM DC Motors are
PM DC motors are expensive
we can’t operate the PM DC motor above rated speed
because we can’t reduce the flux in the machine
Ignore rest of this section. . .
the speed ωm =0, and thus the back-EMF, EA =0
coupled with the fact that armature resistance, RA , is
usually very small
we can see that the armature current, IA
and hence the torque developed at standstill, τind , will be
extremely high
+ which could lead to “something breaking” on the
mechanical side
Hence a starting resistor in series with armature circuit
or a lower starting armature terminal voltage is required
Ignore rest of this section. . .
41 / 48
8.10 DC Motor Efficiency Calculations
42 / 48
8.11 Introduction to DC Generators
+ Have a look again at Chapman Section 7.7
For DC motor, speed regulation is important, but for a
DC generator (similar to a transformer) voltage
regulation (VR) is important
The efficiency of a DC motor at a certain operating
point, is given by
Pout
× 100%
Pin
Pconv − Pno-load
=
× 100%
Parmature + Pfield
(τ − bωm )ωm
= ind
× 100%
VT IA + VF IF
VR =
η=
Vnl − Vfl
× 100%
Vfl
(8–39)
with nl ≡ no-load & fl ≡ full-load
The different types of DC generators that exist are:
Ê
Ë
Ì
Í
Î
Ï
+ Don’t forget about the losses in the field winding. . .
+ Chapman doesn’t calculate the viscous friction
coefficient, b, and assumes the no-load losses
(Pmech + Pcore ) stays constant and is not affected by
speed
43 / 48
separately excited
shunt
permanent-magnet
series
cumulatively compound
differential compound
44 / 48
8.12 The Separately Excited Generator
8.12 The Separately Excited Generator (cont.)
Also for a motor we were interested in the speed vs.
torque, or torque vs. speed characteristics
The equivalent circuit for the SEPEX DC generator
looks exactly the same as for a SEPEX DC motor
The only difference is that a motor is electrically driven
and mechanically loaded
Where as a generator is mechanically driven and
electrically loaded
F1
from the KVL equation for the armature circuit and with
IA = IL
IL
A1
Where as for a generator, we are interested in the
terminal voltage, VT vs. the load current, IL ,
characteristics
VT = EA − IA RA
RF
VF
+
−
+
LF
RA
−
EA =K φωm
φ =⇒
Furthermore, with EA = K φωm , with the flux being kept
constant, i.e. IF being kept constant, and the rotational
speed, ωm , constant
+
Rload
+
VT
−
−
And hence the terminal voltage VT will vary linearly with
the load current, IL – see Fig. 8–45 (a)
ωm
F2
(8–41)
A2
45 / 48
8.12 The Separately Excited Generator (cont.)
46 / 48
8.13 The Shunt DC Generator
+ The only thing that is important to understand here, is
that the voltage buildup during starting of the shunt DC
generator occurs due to the remanent or residual
magnetic flux in the core of the machine. . .
Control of the Terminal Voltage
Also with EA = K φωm , it follow that the terminal voltage
of the SEPEX generator can be controlled by, either
Ê Changing the speed of rotation
Ë Changing the field current
Nonlinear Analysis of a Separately Excited DC Generator
– ignore (i.e. ignore the effect of armature reaction)
47 / 48
48 / 48