Download Tutorial final Questions

Q1
A four-quadrant chopper as shown in Figure below is fed from a 800 Vdc source and
operates at switching frequency, fsw of 1kHz to supply a series DC motor. This chopper
is designed to produce output voltage 90% of the input voltage and the chopper operates
at fourth quadrant. The motor resistance (Ra) 0.50  and motor inductance (La) 3mH. The
machine emf back is 80V and emf back constant, Kv=0.81 V/A rad/s. Total line
inductance (Lb) 8 mH and resistance (Rb) 0.1  series with the supply. This series DC
motor is used to power up a locomotive electric train. The mass of the fully loaded train
is 90 tons and its resistance to motion on level track is 100N. Motor is geared to the
wheel of the motor coach by a 5:1 ratio and the coach wheel tread diameter is 1.25m.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Which switches will be operating?
Calculate the duty cycle of the switch.
Draw the equivalent circuit during ON TIME of the switch (ton) and OFF TIME
of the switch (toff).
Calculate the maximum and minimum current in ampere, semiconductor loss is
neglected.
Draw the output voltage and output current waveform in one graph.
Determine the motor torque when motor is running at 800rpm.
Determine the train rate acceleration (in m2 ).
s
Rb
Lb
T1
D3
D1
Ia = If
Ra
T3
La
Eg
Vdc=800V
+
T4
Q2
D4
Va
_
D2
T2
A separately excited DC motor has rating 220 hp, 230Vdc and 2000 rpm. Armature
voltage supplied by full bridge control rectifier with input voltage:
vs  346.5 sin 314tVolts
Field voltage supplied by full bridge control rectifier with input :
vs  346.5 sin 314tVolts
constant voltage Kv = 0.8 V/A
- armature resistance Ra = 5 Ohm
- field resistanced Rf = 150 Ohm
- Back EMF ,Eg=0.1 Nm
a. Determine the rated load torque of the motor (TL)
b. Determine the rated armature current (Ia,rated)
c. Calculate the load requiring torque if output voltage armature winding rectifier and
field winding rectifier is
 a  00
d. If armature voltage reduced such that the motor run at a speed of 1200 rpm,
calculate the value of
and developed torque. Armature current is 10 A
if
ia
Ra
245Vrms
50Hz
La
Eg
Rf
L
f
245Vrms
50Hz
Principle of Regenerative Brake Control
In regenerative braking the motor acts as a generator and the kinetic energy of the motor
and load are returned back to the supply.
The application of DC to DC converter in
regenerative braking can be explain with Figure below. Let us assume that the armature
of a separately excited DC motor is rotating due to the inertia of motor (and load), and
incase of transportation system, the kinetic energy of the vehicle or train would rotate the
armature shaft. Then if the transistor switched ON, the armature current rises due to short
circuiting of the motor terminal. If the DC to DC converter is turned OFF, diode Dm
would be turn ON and the energy stored in the armature circuit inductance would be
transferred to the supply, provided that the supply is receptive.
is
+
ia
+
Dm
ic
Vs
if
+
Ra
Q1 V
ch
Rf
La
+ Lf
Vf
Eg
_
_
_
_
(a) Circuit
Regenerative braking of DC separately excited DC motor
Va
-Ia
(b) Quadrant
ia
Ia
t
0
is
Ia
0
ic
kT
T
kT
T
kT
T
t
Ia
0
t
Vch
Vs
t
(c) Waveform
The average voltage across the DC to DC converter is:
Vch  (1  k )Vs
(1)
If Ia is average armature current, the regenerated power can be found from:
Pg  I aVs (1  k )
(2)
The voltage generated by the motor acting as a generator is:
Eg  K v I f 
 Vch  Rm I a  (1  k )Vs  Rm I a
(3)
where Kv is machine constant and  is the machine speed in rads per second. The
equivalent load resistance of the motor acting as a generator is:
Req 
Eg
Ia

Vs
(1  k )  Rm
Ia
(4)
By varying the duty cycle k, the equivalent load resistance seen by motor can be varied
V
from Rm to ( s  Rm ) and the regenerative power can be controlled.
Ia
The condition for permissible potentials and polarity of the two voltages:
0  ( Eg  I m I a )  Vs
(5)
Which gives the minimum braking speed of the motor as:
Eg  K vmin I f  Rm I a
or
min 
Rm I a
Kv I f
(6)
and   min . The maximum braking speed of a series motor can be found from
Vs  K vmax I f  Rm I a
or
max 
Vs
R I
 m a
Kv I f Kv I f
(7)
and
  max
The regenerative braking would be effective only if motor speed in between these two
speed limit(e.g., min    max )
Example (Q3)
A dc-dc converter is used in regenerative braking of a dc series motor similar to the
arrangement shown in figure above. The dc supply voltage is 600V. The armature
resistance Ra = 0.02 ohm and field resistance is Rf = 0.03 ohm. The back emf constant is
Kv=15.27 mV/A rad/s. The average armature current is maintained constant at Ia=250A.
The armature current is continuous and has negligible ripple. If the duty cycle of the dcdc converter is 60%, determine (a) the average voltage across the dc-dc converter Vch; (b)
the power generated to the dc supply, Pg; (c) the equivalent load resistance of the motor
acting as a generator, Req; (d) the minimum permissible braking speed min ; (e) the
maximum permissible braking speed max ; and (f) the motor speed.
Principle of Rheostatic Brake Control
Rheostatic braking is also known as dynamic braking. In dynamic braking the energy is
dissipated in a rheostat and it may not be a desirable feature.
ib
ia
+
Dm
ic
Rb
if
Q1
+
Ra
vch = vb
Rf
La
+ Lf
Vf
Eg
_
_
(a) Circuit
Va
-Ia
(b) Quadrant
_
ia
Ia
t
0
Vb
0
ic
RbIa
kT
T
kT
T
t
Ia
0
t
(c) Waveform
Rheostatic braking of DC separately excited DC motor
The average current of braking resistor is,
I b  (1  k ) I a
and the average voltage across the braking resistor is,
Vb  (1  k ) I a Rb
The equivalent load resistance of the generator,
V
Req  b  Rb (1  k )  Rm
Ia
The power dissipated in the resistor Rb is,
Pb  (1  k ) I 2 a Rb
Example: (Q4)
A dc-dc converter is used in rheostatic braking of a dc separately excited dc motor as
shown in Figure above. The armature resistance Ra = 0.05 ohm. The braking resistor is Rb
= 5 ohm. The back emf constant is Kv=1.527 mV/A rad/s. The average armature current
is maintained constant at Ia=150A. The armature current is continuous and has negligible
ripple. The field current is If =1.5A. If the duty cycle of the dc-dc converter is 40%,
determine (a) the average voltage across the dc-dc converter Vch; (b) the power dissipated
in the braking resistor, Pb; (c) the equivalent load resistance of the motor acting as a
generator, Req; (d) the motor speed; and (e) the peak dc-dc converter voltage Vp.