Transcript DC Motor Drive Updated

EET 421
POWER ELECTRONIC DRIVES
Indra Nisja
• General Concept
• Speed Control
• SCR Drives
• Switched-mode DC Drives
Advantages of DC motor :
 Ease of control
 Deliver high starting torque
 Near-linear performance
Disadvantages:
 High maintenance
 Large and expensive (compared to induction
motor)
 Not suitable for high-speed operation due to
commutator and brushes
 Not suitable in explosive or very clean environment
• The DC drive is relatively simple and cheap (compared to
induction motor drives). But DC motor itself is more expensive.
• Due to the numerous disadvantages of DC motor (especially
maintenance), it is getting less popular, particularly in high
power applications.
• For low power applications the cost of DC motor plus drives
is still economical.
• For servo application, DC drives is still popular because of
good dynamic response and ease of control.
• Future Trend? Not so bright prospect for DC, especially in
high power drives.
• The field windings is used to excite the field flux.
• Armature current is supplied to the rotor via brush
and commutator for the mechanical work.
• Interaction of field flux and armature current in the
rotor produces torque.
• When a separately excited motor is excited by a
field current of if and an armature current of ia flows in
the circuit, the motor develops a back emf and a
torque to balance the load torque at a particular speed.
• The if is independent of the ia .Each windings are
supplied separately. Any change in the armature
current has no effect on the field current.
• The if is normally much less than the ia.
Field and armature equations
Instantaneous field current :
where Rf and Lf are the field resistor and inductor, respectively
Instantaneous armature current :
where Ra and La are the armature resistor and inductor,
respectively
The motor back emf, which is also known as speed voltage,
is expressed :
eg = Kv ω if
Kv is the motor voltage constant (in V/A-rad/s)
and ω is the motor speed (in rad/sec)
Basic Torque Equation
The torque developed by the motor is :
Td = Kt if ia
Where (Kt = Kv) is the torque constant in V/A – rad/s
Sometimes it is written as :
Td = Kt Φ ia
For normal operation the developed torque must be equal to the load
torque plus the friction and inertia, i.e, :
where
B : viscous friction constant (N-m/rad/s)
TL : load torque (N-m)
J : inertia of the motor (kg.m2)
Under steady-state operation, time
derivatives is zero. Assuming the motor
is saturated
For field circuit,
Vf = If Rf
The back emf is given by:
Eg = Kv ω if
The armature circuit
Va = Ia Ra + Eg
Va = Ia Ra + Kv ω If
Steady-state Torque and Speed
The motor speed can be easily derived :
If Ra is a small value (which is usual), or when the motor is slightly
loaded, i.e, Ia is small
That is if the field current is kept constant, the motor speed depends
only on the supply voltage.
The developed torque is :
Td = Kt If Ia = B ω + TL
The required power is :
Pd = T d ω
 From the derivation, several important facts can be
deduced for steady-state operation of DC motor.
 For a fixed field current, or flux (If) , the torque demand
can be satisfied by varying the armature current (Ia).
 The motor speed can be varied by:
– controlling Va (voltage control)
– controlling Vf (field control)
 These observations leads to the application of
variable DC voltage to control the speed and torque
of DC motor.
Consider a 500V, 10kW , 20A rated- DC motor with armature resistance
of 1 ohm. When supplied at 500V, the unloaded motor runs at 1040
rev/min, drawing a current of 0.8A
– Estimate the full load speed at rated values
– Estimate the no-load speed at 250V.
Va = Ia Ra + Kv ω If
At full load and rated value,
Kv I f 
Va  I a Ra

Kv I f 
At no load and voltage at 250 V
(Note : in reality, this equation strictly rad/sec)
500  0.8(1)
 0.48
1040

Family of steady-state torque speed curves for a range
of armature voltage can be drawn as above.


The speed of DC motor can simply be set by applying
the correct voltage.
Note that speed variation from no-load to full load (rated) can
be quite small. It depends on the armature resistance.
R
V
m 
 a Ia
K e K e
Or
Ra
V
m 

T
2
K e ( K e )
Shunt and Separately Excited Motor
With a constant field current, the flux can be assumed to be constant. Let
K e  K (Constant)
R
V
m 
 a Ia
K
K
Series Motor
m 
R
V
 a2 T
K K
Base Speed and Field-weakening
• Base speed: ωbase
the speed which correspond to the rated Va, rated Ia and
rated If.
• Constant Torque region ( w > wbase)
Ia and If are maintained constant to met torque demand.
Va is varied to control the speed. Power increases with speed.
• Constant Power region ( w > wbase)
Va is maintained at the rated value and if is reduced to increase speed.
However, the power developed by the motor (= torque x speed) remains
constant. Known as field weakening.
DC Shunt Motor
Motor efficiency
Pmech
R 
Pelec  Pcontrol
Pelec  Vin xIaR
Pcontrol  Vin xI f
Example
A 500-V, 60-hp, 600-rev/min d.c. shunt motor has a full-load efficiency of 90%. The
resistance of the field itself is 200 ohm and rated field current is 2 A. Ra = 0.2 ohm.
Calculate die full-load (rated) current IaR and in subsequent calculations, maintain this
value. Determine the loss torque.
The speed is to be increased up to 1000 rev/min by field weakening. Calculate the
Extra resistance, over and above the field winding itself to cover the range 600-1000
rev/min. Determine the output torque and power at the top speed, assuming that the
loss torque varies in proportion to speed. For the magnetisation curve use the empirical
expression below, which is an approximation to the curve shape.
where the flux ratio is that between a particular operating flux and rated flux .
The field-current ratio is that of the corresponding field currents.
• Say the motor running at position A. Suddenly va is reduced (below eg).
The current ia will reverse direction. Operating point is shifted to B.
• Since ia is negative, torque Te is negative.
• Power is also negative, which implies power is “generated” back to
the supply.
• In other words, during the deceleration phase, kinetic energy from the
motor and load inertia is returned to the supply.
• This is known as regenerative braking-an efficient way to brake a motor.
Widely employ in electric vehicle and electric trains. If we wish the motor
to operate continuously at position B, the machine have to be driven by
mechanical source.
• The mechanical source is a “prime mover”.
• We must force the prime mover it to run faster so that the generated eg
will be greater than va.
Braking circuits
Example 3.20
A 250V, 500rev/min d.c. separately excited motor has an armature
resistance of 0.13 ohm and takes an armature current
of 60A when delivering rated torque at rated flux. If flux is
maintained constant throughout, calculate the speed at which a
braking torque equal in magnitude to the full-load torque is
developed when:
(a) regeneratively braking at normal terminal voltage;
(b) plugging, with extra resistance to limit the peak torque on
changeover to 3 per unit;
(c) dynamically braking, with resistance to limit the current to
2 per unit;
(d) regeneratively braking at half rated terminal voltage.
(e) What terminal voltage would be required to run the motor in
reverse rotation at rated torque and half rated speed?
• SCR “phase-angle controlled” drive
- By changing the firing angle, variable DC output voltage can be
obtained.
– Single phase (low power) and three phase (high and very high power)
supply can be used
– The line current is unidirectional, but the output voltage can reverse
polarity. Hence 2- quadrant operation is inherently possible.
– 4-quadrant is also possible using “two sets” of controlled rectifiers.
• Switched-mode drive
– Using switched mode DC-DC converter. Dc voltage is varied by
duty cycle.
– Mainly used for low to medium power range.
– Single-quadrant converter (buck): 1- quadrant
– Half bridge: 2-quadrant
– Full bridge: 4-quadrant operation
• Mains operated.
• Variable DC voltages are obtained from SCR firing angle control.
• Slow response.
• Normally field rectifier have much lower ratings than the armature
rectifier. It is only used to establish the flux.
Continuous/Discontinuous current
• The key reason for successful DC drive operation is due to the large
armature inductance La.
• Large La allows for almost constant armature current (with small ripple)
due to “current filtering effect of L”. (Refer to notes on Rectifier).
• Average value of the ripple current is zero. No significant effect on the
torque.
• If La is not large enough, or when the motor is lightly loaded, or if
supply is single phase (halfwave), discontinuous current may occur.
• Effect of discontinuous current: Output voltage of rectifier rises; motor
speed goes higher. In open loop operation the speed is poorly regulated.
• Worthwhile to add extra inductance in series with the armature
inductance.
Armature
For continuous current, armature voltage is :
Armature (DC) current is :
Field voltage
Field
1. Single-Phase Half-Wave Converter Drives
Vm
V 
(1  cos  )
2
0  
for
2. Single-Phase Semiconverter Drives
Vm
for
V
(1  cos  )
0  

3. Single-Phase Full-Converter Drives
V 
2Vm
cos 
for
0  
V 
2Vm
cos 
for
0  

4. Single-Phase Dual-Converter Drives

Armature voltage :
Armature (DC) current is :
If single phase is used for field is :
1. Three-Phase Half-Wave Converter Drives
3 3Vm
for
V 
cos 
2
2. Three-Phase Semiconverter Drives
0  
3 3V m
for
V 
(1  cos  )
2
3. Three-Phase Full-Converter Drives
0  
V 
3 3V m

cos 
for
0  
4. Three-Phase Dual-Converter Drives
V 
3 3V m

cos 
for
0  
A separately excited DC motor has a constant torque load of 60 Nm. The
motor is driven by a full-wave converter connected to a 240 V ac supply.
The field constant of the motor KIf = 2.5 and the armature resistance is 2
ohm. Calculate the triggering angle for the motor to operate at 200 rpm.
Assume the current is continuous.
For continuous current,
and
Va  E g  I a Ra
Where Eg is the back emf,i.e
and
T  KI f I a
Va 
2Vm

cos  a
E g  KI f   2.5 I f
 T 

2Vm
 Ra  KI f  
cos  a  

 KI f 

A rectifier-DC motor drive is supplied by a three-phase, full controlled SCR
bridge 240Vrms/50Hz per-phase. The field is supplied by a single-phase 240V
rms/50Hz, with uncontrolled diode bridge rectifier. The field current is set as
maximum as possible.
The separately excited DC motor characteristics is given as follows :
Armature resistance:Ra = 0.3 ohm
Field resistance: Rf =175 ohm
Motor constant: KV =1.5 V/A-rad/s
Assume the inductance of the armature and field circuit is large enough to ensure
continuous and ripple-free currents. If the delay angle of the armature converter (αa) is
45 degrees and the required armature current is 30A,
• a) Calculate the developed torque, Td.
• b) Speed of the motor, ω (rad/s)
• c) If the polarity of the field current is reversed, the motor back emf will reverse.
For the same armature current of 30A, determine the required delay angle of the
armature converter.
Since field current is maximum, α = 0.
(b) Motor speed
The armature is supplied by three-phase with αa = 45o,
Now the polarity of field is reversed, then
and
also
• DC motor in inherently bi-directional. Hence no problem to reverse
the direction. It can be a motor or generator.
• But the rectifier is unidirectional, because the SCR are unidirectional
devices.
• However, if the rectifier is fully controlled, it can be operated to become
negative DC voltage, by making firing angle greater than 90 degrees,
• Reversal can be achieved by:
– armature reversal using contactors (2 quadrant)
– field reversal using contactors (2-quadrant)
– double converter (full 4-quadrants)
Reversal using armature or field contactors
DRIVE REVERSING USING ARMATURE OR FIELD CONTACTORS
CONTACTOR AT THE ARMATURE SIDE
(SINGLE PHASE SYSTEM)
Reversing using double converters
Principle of reversal
Practical circuit
• Supply is DC (maybe from rectified-filtered AC, or some other DC sources).
• DC-DC converters (choppers) are used.
• suitable for applications requiring position control or fast response, for
example in servo applications, robotics, etc.
• Normally operate at high frequency
– the average output voltage response is significantly faster
– the armature current ripple is relatively less than the controlled rectifier
• In terms of quadrant of operations, 3 possible configurations are possible:
– single quadrant,
– two–quadrant
– and four–quadrant
• Unidirectional speed. Braking not required.
For 0 < t < T
The armature voltage at steady state :
Armature (DC) current is :
Ia 
Va  E g
t
t
1 on
Va   V .dt  on V  DV
T 0
T
Ra
And speed can be approximated as :

Va
Kv I f

FORWARD MOTORING (T1 and D2 operate)
– T1 on: The supply is connected to motor terminal.
– T1 off: The armature current freewheels through D2.
– Va (hence speed) is determined by the duty ratio.

REGENERATION (T2 and D1 operate)
– T2 on: motor acts as a generator
– T2 off: the motor acting as a generator returns
energy to the supply through D2.
A full-bridge DC-DC converter is used.


T1 and T2 operate; T3 and T4 off.
T1 and T2 turn on together: the supply voltage appear across the motor
terminal. Armature current rises.

T1 and T2 turn off: the armature current decay through D3 and D4



T1, T2 and T3 turned off.
When T4 is turned on, the armature current rises through T4 and D2.
When T4 is turned off, the motor, acting as a generator, returns energy
to the supply through D1 and D2.




T3 and T3 operate; T1 and T2 off.
When T3 and T4 are on together, the armature current rises and flows
in reverse direction.
Hence the motor rotates in reverse direction.
When T3 and T4 turn off, the armature current decays through D1 and D2.



T1, T3 and T4 are off.
When T1 is on, the armature current rises through T2 and D4.
When Q2 is turned off, the armature current falls and the motor returns
energy to the supply through D3 and D4.
When IGBT is turning ON:
V  Ri2  L
di2
 Eg
dt
which initial current
i1 (t  0)  I1
t

V  Eg
i2 (t )  I 1e 
R
gives the load current as:
t

(1  e )
This mode valid for 0
 t  t1
at the end of this mode load current becomes:
i1 (t  t1 )  I 2
The load current for mode 2 can be found from:
0  Ri 2  L
di2
 Eg
dt
with initial current i2 (t
t

 0)  I 2
i1 (t )  I 2 e 
This mod is valid for
Eg
R
and redefining the time origin (i.e t=0) at the beginning of mode 2, we have:
t

(1  e )
t1  t  t2
At the end of this mode the load current becomes:
i2 (t  t 2 )  I1
Example:
An electric train is powered by four 750-V d.c. series motors. The motor resistance
and inductance are respectively 0.25  and 6mH. A total line inductance of 10 mH and
resistance of 0.1 is in series with the supply. The fixed voltage from the traction
supply is regulated by the chopper of Figure 1 below, operating at 200 Hz. When the
machine is running at 800 rev/min, the generated e.m.f. per ampere can be taken
as an average value of 0.79 V/A. If the modulation factor (D) is 80%,
a. calculate the maximum and minimum currents, allowing 2V for semiconductor loss,
and find the average torque of each motor.
b. The mass of the fully loaded train is 120 tonnes and its resistance to
motion on level track is 500N. Each motor is geared to the wheel of the motor
coach by a 3:1 ratio and the coach wheel tread diameter is 1.0m. Estimate the
train's rate of acceleration under the above conditions.
Figure 1: Separately excited DC motor
Solution:
Mode 1 (switch ON)
dia
V  Ri a  L
 Eg  2
dt
750  0.79ia  0.35ia  0.016
dia
2
dt
Rearranging:
748  1.14ia  0.016
dia
dt
For steady state condition:
L
dia
0
dt
So, i
a
 656.14 Amp
Time constant (  
i2 (t )  i1e
t
0.0457
L
)=0.0457
R
 656.14(1  e
t
0.0457
)
At 200Hz chopping frequency with D=0.8
So, tON = 4ms
i2 (t )  i1e
0.004
0.0457
 656.14(1  e
0.004
0.0457
)
i2 (t )  0.9162i1  656.14(1  0.9162)
i2 (t )  0.9162i1  55
(1)
Mode 2 (switch OFF)
di 2
 Eg
dt
di
0  0.79ia  0.25ia  0.006 a  2
dt
0  Ri 2  L
0  1.04ia 0.006
dia
2
dt
For steady state condition:
L
So,
dia
0
dt
ia  1.923 Amp
Time constant ( 
i a (t )  i 2 e
L )=0.024
R
t
0.024
 1.923(1  e
t
0.024
At 200Hz chopping frequency with D=0.8
So, tOFF = 1ms
)
i a (t )  i 2 e
0.001
0.024
 1.923(1  e
i1 (t )  0.959i2  0.0788
0.001
0.024
)
(2)
Eliminate i1, substitute eq (2) into eq (1):
i2 (t )  0.9162(0.959i2  0.0788)  55
i2  452.45 Amp
i1  433.82 Amp
For a series motor, torque =
k q xia 
0.79ia xia
m
Kq=Kv x If
In this case, the average torque will be proportional to the average value of (ia)2, i.e. the
mean-square current. As the electrical L/R time constant is considerably larger than tON or
toff, it is reasonable to assume that the graph of current against time roughly follows straight
lines.
The area under the current squared curve for tON:
t ON (i12  (i1 xi2 )  i22 ) / 3
0.004(433.82 2  (433.82 x452.45)  452.45 2 ) / 3  785.59 A
The area under the current squared curve for tOFF:
t OFF (i12  (i1 xi2 )  i22 ) / 3
0.001(433.82 2  (433.82 x452.45)  452.45 2 ) / 3  196.4 A
The mean current squared is then,
758.59  196.4
 196397.51
0.005
0.79 x196397.51
Torque 
 1,853N  m
2
800 x
60
Tractive force at wheel =
torquexgearratio 1,853 x3

 11,118 N
hheelradiu s
0.5
For four motors, total tractive force = 44,472N. With 500 N rolling resistance, the tractive force available for acceleration is 43972N.
The train mass is 120 tonnes.
Acceleration:
43.972
m
 0.36643 2
120
s
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 314tVoltsField voltage supplied by diode 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
-
a. Calculate the load requiring torque if  a  0 0
b. If armature voltage reduced such that the motor run at a speed of 1200 rpm,
calculate the value of  a and developed torque. Armature current is 10 A
Solution
Output of control rectifier for α=0:
Va 
Va 
2Vm

2 x346.5

cos  a
cos 0
Va  220V
Output of diode rectifier:
Va 
2Vm
Va 
2x346.5


Va  220V
Equation of armature separately excitation DC motor:
Va  E g  I a Ra
E g  Va  I a Ra
(1)
E g  K v I f
(2)
Substitute (2) into (1):
K v I f  V a  I a R a
Equation of field winding of DC motor:
Vf  I f Rf
If 
Vf
Rf

220
 1.467 Amp
150
So,
Va  I a Ra

Kv I f

220  5I a
1.1736
Ia 
 1.1736  220
5
I a  0.23472  44
Torque requiring by the load is:
TL  1.1736(0.23472  44) N  m
b.
from equation (2)
E g  1.1736 x1200
E g  147.4
Va  147.4  50
Va  197.4Volt
2
60
Firing angle (a) is:
 a  Cos 1
197.4
220
 a  26.2 0