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Controlled-Rectifier Fed Drive
Armature
Field
1
Chopper-Fed Drive
Armature
Field
2
Equivalent Circuit of a Separately-Excited DC
Motor
3
v f  Rf i f  Lf
di f
dt
dia
va  Ra ia  La
 eg
dt
eg  K vi f
Td  K t i f ia
d
Td  J
 B  TL
dt
4
Steady  state
d
0
dt
Vf  Rf I f
E g  K v I f
Va  Ra I a  Eg
Va  Ra I a  K v I f
Td  K t I f I a
Td  B  TL
Pd  Td 
5
Solve for the motor speed
Va  Ra I a  K v I f
Va  I a Ra Va  I a Ra


Kv I f
 Vf 
Kv 

R 
 f 
6
Control of the motor speed
• Control armature
voltage, Va
– Voltage control
• Control the field
current, If
– Current control
• Control the armature
current, Ia
Va  I a Ra

 Vf 
Kv 

R 
 f 
7
8
9
Equivalent circuit of a DC Series Motor
10
Eg  K v I a
Va  ( Ra  R f ) I a  Eg
Va  ( Ra  R f ) I a  K v I f
Td  Kt I a I f
Td  B  TL

Va  ( Ra  R f ) I a
Kv I f
11
a
a
f
a
g
V

(
R

R
)
I

K

I
a
a
f
a
v
Controlling the Motor Speed
• Control the armature
voltage, Va
Td  Kt I a I f
Td  B  TL

Va  ( Ra  R f ) I a
Kv I f
• Control the armature
current, Ia
12
13
Motoring Mode
“Back emf”, Eg is < supply voltage Va
Both Ia and If are positive
14
Regenerative Braking Mode
Field Terminals Reversed
Motor acts as a Generator
Eg becomes > supply Voltage Va
Ia becomes negative
15
Dynamic Braking
Replace the supply voltage with a resistor
Power dissipated in the resistor rather than given back to
the source
16
Plugging
Reverse the armature terminals while running
Va and Eg act in the same direction
Ia is reversed, producing “braking” torque
17