Transcript Speed Control of DC motors (DC Drives) Dynamics of Motor Load Systems dm TJ  TL dt dm TJ  Tm  Bm dt J moment of inertia kg-m2 m instantaneous angular.

Speed Control of DC motors
(DC Drives)
Dynamics of Motor Load Systems
dm
TJ
 TL
dt
dm
TJ
 Tm  Bm
dt
J moment of inertia kg-m2
m
instantaneous angular velocity rad/sec
T developed torque of the motor N.m.
TL
The load torque.
Bm
Viscous Friction torque.
B Viscous Friction coefficient.
Steady State Stability
d m
0
dt
m
dTL
dT

dm dm
T
A
Unstable
TL
TL
B
Stable
T
Torque
Four-quadrant operation.
DC Motors
Steady State Speed Torque Relations
E  Kem
V  E  I a Ra
V  Kem  I a Ra
V
Ra
m 

Ia
Ke Ke
T  K e I a
T
Ia 
Ke
V
Ra
m 

T
2
K e  K e 
Separately Excited DC Motor
E  Km
K  Ke
V  E  I a Ra
T  K Ia
V  Km  I a Ra
V Ra
m   I a
K K
V Ra
m   2 T
K K
T
Ia 
K
Series DC Motor
E  Ke K f I am
V  E  I a Ra
T
  K f Ia
T  Ke I a
2
Ke K f I a
V  K e K f I am  I a Ra
T
V
Ra
I

a
m 

K
K
e
f
Ke K f I a Ke K f
V
1
Ra
m 

Ke K f T Ke K f
Methods of Speed Control
Armature Voltage Control
Field Flux Control
V
Ra
m 

T
2
K e  K e 
Armature Resistance Control
Armature Voltage Control
1.Controlled Rectfier
V
Ra
m 

T
2
K e  K e 
2. Chopper (DC-DC Converter)
Field Flux Control
m
V
Ra
m 

T
2
K e  K e  
m
Decreasing flux
at constant V
Deacreasing 
at constant V
Decreasing V
at Full flux
Deacreasing V
at constant 
T
Separately
Series
Torque
Torque & Power Limitations in Combined
armature Voltage and Field Control
Torque
Pm
Armature Voltage
Control
rated
Speed
Pm
Torq
u
e
Field Control
m
A 230 V 500 rpm, 100 A separetly excited DC motor has Ra  0.1 . The motor is
driving at a rated conditions, a load whose torque is constant and independent of
speed. The speeds below the rated speed are obtained with armature voltage control
(with full field) and the speeds above the rated speed are obtained by field control
(with rated armature voltage).
(a) Calculate the motor terminal voltage when the speed is 400 rpm.
(b) by what amount should flux be reduced to get a motor speed of 800 rpm?
E1  Vt  Ra I a1  230  0.1*100  220
E2 N 2
E2 400
 E2  176V



220 500
E1 N1
V2  E2  I a 2 Ra
 176  100 * 0.1  186V
b.
E3 I f 3 * N 3

E1 I f 1 * N1
E3
800

x
220
500
x
 E3  352 x
T3 I f 3 I a 3
I a3

*
 x*
T1 I f 1 I a1
I a1
I a 3  I a1 / x
100
*
0
.
1
V3  E3  I a 3 Ra
230  352 x 
x
2
312 x  230 x  10  0
 x1  0.61
x2  0.05
If3
I f1
A220 V 600 rpm, 100 A separetly excited DC motor has Ra  0.12 . The motor is driving
at a rated conditions, a load whose torque is constant and independent of speed at speeds
below the rated speed and the mechanical power is constant and independent of speed at
speeds higher than the rated speed. The speeds below the rated speed are obtained with
armature voltage control (with full field) and the speeds above the rated speed are obtained
by field control (with rated armature voltage).
(a) Calculate the motor terminal voltage when the speed is 500 rpm.
(b) by what amount should flux be reduced to get a motor speed of 900 rpm?
E1  Vt  I a1 Ra  220  0.12 *100  208V
E
500
E2 N 2
Then, 2 
Then, E2  173.3V

E1 N1
208 600
Then, V2  E 2  I a 2 Ra  173.3  100 * 0.12  185.33V
E3 I f 3 * N 3
E
900
(b)

Then, 3  x
,
E1 I f 1 * N 1
208
600
Where x 
If3
I f1
Then E3  312 x
(1)
P  T but, as we know from the data P =constant above rated speed.
Then T
1

but  N Then,
T3 N 1

T1 N 3
T3 I f 3 I a 3
I
But

*
 x * a3
T1 I f 1 I a1
I a1
(2)
(3)
Substitute from (2) into (3) we get:
N1
I
 x * a3
N3
I a1
But V3  E3  I a 3 Ra Then, I a 3 
(4)
V3  E3 220  E3

Ra
0.12
(5)
Substitute from (5) into (4) we get the following equation:
600
220  E3
 x*
900
100 * 0.12
Substitute from (1) into (6) then,
600
220  312 x
 x*
900
100 * 0.12
Then, 312 x 2  220 x  8  0
Then, x1  0.6667 and x2  0.0385
(6)
Armature Resistance Control
Separately or shunt field
m
V
Ra  Rext
m 

T
K e  K e 2
m
TR
Increasing
Re
Re
Torque
Armature Resistance Control
V
1
Ra
m 

Series field
m
Ke K f
T
Increasing
Re
Torque
Ke K f
(b) A DC series motor drives an elevator load that requires a constant torque of 200
N.m. The DC supply voltage is 400 V and the combined resistance of the armature and
series field winding is 0.75 ohm. Neglect rotational losses and armature reaction effect.
(i) The speed of the elevator is controlled by variating the supply DC voltage. At 220V
input voltage and 40A motor current, determine the speed and the horsepower output of
the motor and the efficiency of the system.
(ii) The elevator is controlled by inserting resistance in series with the armature of the
series motor. For the speed of part (i), determine the values of the series resistance,
horsepower output of the motor, and efficiency of the system.
(i) Ea  VT  I a Ra  Rs   220  40 * 0.75  190 V
T
200
E a  k a  m , T  k a  I a , k a  

5
Ia
40
Ea 190
m 

 38 rad / sec
 363rpm
k a
5
Ea I a 190 * 40
Pout , HP 

 10.188 HP
746
746
Pin  220 * 40  8800 Watt
Pout 190 * 40


* 100  86.364%
Pin
8800
(ii) Ea  VT  I a Ra  Rs  Rexternal 
400  190
Ra  Rs  Rexternal  
 5.25
40
Rexternal  5.25  0.75  4.5
Pout 190 * 40


* 100  47.5%
Pin 400 * 40
(b) A DC series motor drives an elevator load that requires a constant torque of 200 N.m.
The DC supply voltage is 400 V and the combined resistance of the armature and series
field winding is 0.75 ohms.
(i) The speed of the elevator is controlled by buck converter. At 50% duty cycle (i.e., D=0.5)
of the chopper, the motor current is 40 amps. Determine the speed and the horsepower
output of the motor and the efficiency of the system.
(ii)
Compare between the scheme in part(i) and the method of inserting resistance in
series with the armature of the series motor for the same speed.
(i) I LB 
TS Vd
400 * 0.5 * (1  0.5)
D (1  D ) 
 25 A
2L
2 * 0.0001 * 20000
It is clear that the load current (40A) greater than the inductor boundary current.
So, this chopper works in continuous conduction mode.
VT  DV s  0.5 * 400  200V
Ea  k a  m , T  k a  I a , k a  
T
200

5
Ia
40
E a 170

 34 rad / sec
 324.8rpm
k a
5
E I
170 * 40
Pout , HP  a a 
 9.12 HP
746
746
Pin  200 * 40  8000 Watt
P
170 * 40
  out 
* 100  85%
Pin
8000
m 
(ii) Ea  VT  I a Ra  Rs  Rexternal  ,
400  170
Ra  Rs  Rexternal  
 5.75
40
Rexternal  5.75  0.75  5 ,
Pout 170 * 40


* 100  42.5%
Pin
400 * 40
It is clear that in using the method of inserting
resistance in series with the armature of the series
motor the efficiency get much lower than using
buck converter in DC motor control
The speed of a 125 hp, 680 V, 1200 rpm, separately excited DC motor is controlled by a
single-phase full bridge controlled converter.The converter is operated from a 600 V, 60
Hz supply. The rated armature current of the motor is 170 A. The motor parameters are
Ra  0.1 , and K = 0.4 V/rpm. The converter and ac supply are considered to be ideal.
For firing angle  = 30°,
(i) Draw the converter output voltage and supply current at rated load.
(ii) Find the input power factor and THD of supply current at rated load.
(iii) Find no-load speed, Assuming that, at no load, the armature current is 12% of the
rated current and is continuous.
(iv) Find the firing angle to obtain the rated speed of 1200 rpm at rated motor current.
If we move the axis of I s to the point t  
the supply current become odd function and,
then an coefficients of Fourier series equal zero, an  0 , and
bn 

2


 I o * sin nt dt 
0
2 Io
 cos nt 0
n
2 Io
cos 0  cos n   4 I o for n  1, 3, 5, .......... ...
n
n
Then from Fourier series concepts we can say:
i( t ) 
4 *170

1
1
1
1
* (sin t  sin 3t  sin 5t  sin 7t  sin 9t  ..........)
3
5
7
9
4 * 170
I S1 
 153.054 A
2
I
153.054
Power factor= s1 * cos  
* cos 30  0.7797 Lag
Is
170
2
I 
 THD( I s (t ))   S 
 I S1 
2




I
 1   o   1  48.34%
 4 Io 


 2 
(III) At no load condition the motor terminal voltage is:
Vdc 
2Vm

cos  
2 * 600 * 2

cos 30 o   467.818V
12
No-load current = 170 *
 20.4 A
100
E a  Vt  I a Ra  467.818  20.4 * 0.1  465.778V
Ea
465.778

 1164.45 rpm,
The no-load speed is N 0 
K a
0.4
(IV) At full load condition,
E a  K a * N FL  0.4 * 1200  480 V
Then, Vt  E a  I a Ra  480  170 * 0.1  497 V
Then, Vdc 
2Vm

Then,   23.067
cos  
o
2 * 600 * 2

cos    497V