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1
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2
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3
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4
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5
Controlled-Rectifier Fed Drive
Armature Field
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6
Chopper-Fed Drive
Armature Field
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7
Equivalent Circuit of a Separately-Excited
DC Motor
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8
0
f f f
g v f
a a a g
a a a v f
d t f a
d L
d d
Steady state
d
dt
V R I
E K I
V R I E
V R I K I
T K I I
T B T
P T
ω
ω
ω
ω
−
=
=
=
= +
= +
=
= +
=
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9
a a a v f
a a a a a a
v f f
v f
V R I K I
V I R V I R
K I V
K R
ω
ω
= +
− −= =
÷ ÷
Solve for the motor speed
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10
Control of the motor speed
• Control armature
voltae! "a # "oltae control
• Control the fieldcurrent! $f # Current control
• Control the armaturecurrent! $a
a a a
f
v
f
V I RV
K
R
ω −= ÷
÷
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11
Maneti%ation Characteristic
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12
Characteristics of Separately-Excited Motors
Rated
speed
&se field-current control
&se armature
voltae control
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13
Equivalent circuit of a DC Series Motor
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14
( )
( )
( )
g v a
a a f a g
a a f a v f
d t a f
d L
a a f a
v f
E K I
V R R I E
V R R I K I
T K I I
T B T
V R R I K I
ω
ω
ω
ω
=
= + += + +
== +
− +=
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15
Controllin the Motor Speed
• Control the armature
voltae! "a
• Control the armaturecurrent! $a
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16
Characteristics of DC Series motors
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17
'peratin Modes
• $n varia(le-speed applications! a dc motor
may (e operatin in one or more of the
follo)in *Modes+
• Motorin
• Reenerative (ra,in• Dynamic (ra,in
• luin
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18
Motorin Mode
*.ac, emf+! E is / supply voltae "a
.oth $a and $f are positive
Developed torque meets load demand
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19
0
1
2
3
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20
Reenerative .ra,in Mode
Motor acts as a 4enerator
E (ecomes 5 supply "oltae "a
$a (ecomes neative
6inetic enery of the motor is returned to the supply
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21
Dynamic .ra,in
Replace the supply voltae )ith a resistor
o)er dissipated in the resistor rather than iven (ac,
to the source
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22
luin
Reverse the armature terminals )hile runnin"a and E act in the same direction
$a is reversed! producin *(ra,in+ torque
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23
Four-7uadrant 'peration
0
1
2
3
O i M d
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Operating Modes
Four Quadrants:
• Figure 15.8 shows the polarities of the supply voltage Va, bac e!f "g, a#$
ar!ature curre#t %a for a separately e&cite$ !otor.
• %# forwar$ !otori#g ('ua$ra#t %), Va, "g, a#$ %a are all positive. he tor'ue a#$
spee$ are also positive i# this 'ua$ra#t.
• uri#g forwar$ brai#g ('ua$ra#t %%), the !otor ru#s i# the forwar$ $irectio#
a#$ the i#$uce$ e!f "g co#ti#ues to be positive. For the tor'ue to be #egative
a#$ the $irectio# of e#ergy flow to reverse, the ar!ature curre#t !ust be#egative. he supply voltage Va shoul$ be ept less tha# "g.
• %# reverse !otori#g ('ua$ra#t %%%), Va, "g, a#$ %a are all #egative. he tor'ue a#$
spee$ are also #egative i# this 'ua$ra#t. o eep the tor'ue #egative a#$ the
e#ergy flow fro! the source to the !otor, the bac e!f "g !ust satisfy the
co#$itio# * Va * + * "g *. he polarity of "g ca# be reverse$ by cha#gi#g the$irectio# of fiel$ curre#t or by reversi#g the ar!ature ter!i#als.
• uri#g reverse brai#g ('ua$ra#t %V), the !otor ru#s i# the reverse $irectio#.
Va, a#$ "g co#ti#ue to be #egative. For the tor'ue to be positive a#$ the e#ergy
to flow fro! the !otor to the source, the ar!ature curre#t !ust be positive. he
i#$uce$ e!f "g !ust satisfy the co#$itio# * Va * * "g *.
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25
Sinle-hase DC Drive
Chane "a (y chanin firin anle α $nductor 8m is a *smoothin+ $nductor to prevent
discontinuous current
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26
Sinle-hase DC Drive
9on-reenerative DC drives are the most conventional type in
common usae: ;hey are a(le to control motor speed and torque in
one direction
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27
Armature Reversal
>ith the addition of an electromechanical
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Field Reversal
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7uadrant of 'peration
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31
Sinle-hase Full-Converter Drives
-cos
Vm
Va α
π =
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7uadrant
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33
Sinle-hase Dual-Converter Drives
Dual converter can operate in all four quadrants
'utput voltae and current can (e positive or neative
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2-7uadrant chopper
34
A t)o-quadrant hard-s)itchin DC motor drive is presented: Devices S>0 and D0 form a
first-quadrant chopper and enery is delivered from the DC source "in to the DC motor
2 and D2 form a secondth-quadrant chopper and enery is delivered from
the DC motor
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?-(ride
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A 230 V 500 rpm, 100 A separetly excited D motor !as Ω= 1.0a R " #!e motor is
dri$ing at a rated conditions, a load %!ose tor&ue is constant and independent o'
speed" #!e speeds (elo% t!e rated speed are o(tained %it! armature $oltage control
)%it! 'ull 'ield* and t!e speeds a(o$e t!e rated speed are o(tained (y 'ield control
)%it! rated armature $oltage*"
)a* alculate t!e motor terminal $oltage %!en t!e speed is +00 rpm"
)(* (y %!at amount s!ould 'lux (e reduced to get a motor speed o' 00 rpm-
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--01001.0-/011 =−=−= aat I RV E
1
-
1
-
N
N
E
E =
500
00
--0-
=∴ E
E 12- =∴
V
R I E V aa
1821.010012
---
=+=+=∴
( // NIE f 800E I
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(:
11
//
1
/
N I
N I
E
E
f
f =
500
800
--0
/ x E =∴
1
/
f
f
I
I x =
x E /5-/ =∴
1
/
1
/
1
/
1
/
a
a
a
a
f
f
I
I x
I
I
I
I
T
T == x I I aa 31/ =
aa R I E V /// += x
x1.0100
/5--/0 +=
010-/0/1- -
=+− x x21.01 =∴ x 05.0- = x
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4--0 V 200 rp!, 100 4 separetly e&cite$ !otor has Ω= 1-.0a R . he !otor is $rivi#g
at a rate$ co#$itio#s, a loa$ whose tor'ue is co#sta#t a#$ i#$epe#$e#t of spee$ at spee$s
below the rate$ spee$ a#$ the !echa#ical power is co#sta#t a#$ i#$epe#$e#t of spee$ at
spee$s higher tha# the rate$ spee$. he spee$s below the rate$ spee$ are obtai#e$ with
ar!ature voltage co#trol (with full fiel$) a#$ the spee$s above the rate$ spee$ are obtai#e$
by fiel$ co#trol (with rate$ ar!ature voltage).
(a) alculate the !otor ter!i#al voltage whe# the spee$ is 500 rp!.
(b) by what a!ou#t shoul$ flu& be re$uce$ to get a !otor spee$ of 600 rp!7
4--0 V 200 rp!, 100 4 separetly e&cite$ !otor has Ω= 1-.0a R . he !otor is $rivi#g
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p , p y a g
at a rate$ co#$itio#s, a loa$ whose tor'ue is co#sta#t a#$ i#$epe#$e#t of spee$ at spee$s
below the rate$ spee$ a#$ the !echa#ical power is co#sta#t a#$ i#$epe#$e#t of spee$ at
spee$s higher tha# the rate$ spee$. he spee$s below the rate$ spee$ are obtai#e$ with
ar!ature voltage co#trol (with full fiel$) a#$ the spee$s above the rate$ spee$ are obtai#e$
by fiel$ co#trol (with rate$ ar!ature voltage).(a) alculate the !otor ter!i#al voltage whe# the spee$ is 500 rp!.
(b) by what a!ou#t shoul$ flu& be re$uce$ to get a !otor spee$ of 600 rp!7
V R I V E aat -081001-.0--011 =−=−=
1
-
1
-
N
N
E
E = (he#,
200
500
-08
- = E
(he#, V E /.1/- =
(he#, V R I E V aa //.1851-.0100/.1/--- =+=+=
(b)///
N I E f = he#600/ x
E = here / f
I x =
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(b)111 N I E f
= he#,200-08
x= , here1 f I
x =
he# x E /1-/ = (1)
ω T P = but, as we #ow fro! the $ata P 9co#sta#t above rate$ spee$.
he#ω
α
1T but N ωα he#,
/
1
1
/
N
N
T
T
= (-)
ut1
/
1
/
1
/
1
/ a
a
a
a
f
f
I
I x
I
I
I
I
T
T == (/)
;ubstitute fro! (-) i#to (/) we get:
1
/
/
1 a
a
I I x
N N = ()
ut aa R I E V /// += he#,1-.0
--0///
/
E
R
E V I
a
a
−=
−= (5)
;ubstitute fro! (5) i#to () we get the followi#g e'uatio#:
1-.0100--0
600200 / E x −= (2)
;ubstitute fro! (1) i#to (2) the#,
1-.0100
/1---0
600
200 x x
−=
h 08--0/1-
-