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Presentation Summary • Different types of dc motors •Separately excited dc motor steady state equations •Speed control methodology by armature voltage control •Speed control methodology by flux control •Steady-state operating regions of a drive •Numerical example on steady-state operating region •Dynamic model of separately excited dc motor with constant flux •Numerical example on transfer function of the dynamic model •Block diagram for dc motor speed control
DC motor speed control Two types of dc machines are commonly used for speed control: 1) Separately excited dc machine . 2) Series excited dc machine . For this course we will discuss only about the separately excited dc motor.
Separately excited dc machine Under steady state,
V𝑎 = 𝐼𝑎𝑅𝑎 + 𝐸𝑏 : Armature equation (i)
𝐸𝑏 = 𝐾𝑚𝜔: Back EMF equation(𝐾𝑚 = constant =𝐾𝑒Φ) (ii)
𝑇𝑚 = 𝐾𝑚 𝐼𝑎 : Torque equation (iii)
𝑉𝑎=𝑅𝑎𝑇𝑚
𝐾𝑚+ 𝐾𝑚𝜔 (Using (ii) and (iii) in (i))
𝜔 =𝑉𝑎
𝐾𝑚−
𝑅𝑎
𝐾𝑚2 𝑇𝑚
Here 𝑉𝑎 = armature voltage in V, 𝐼𝑎 = armature current in A, , 𝐸𝑏 = back emf in V, 𝑇𝑚
= motor torque in N − m, 𝜔 = speed inrad
s, 𝐾𝑒 = machine constant (unitless),Φ =
flux
polein Wb., 𝑅𝑎 = armature resistance in Ω. These are steady-state values.
Armature voltage control
Speed below base (rated) speed is controlled by controlling Va. . Speed can be varied
by varying Va , keeping flux rated value.
Tm
ω Va =Va1
Va =Va2
Va =Va3
Va1>Va2>Va3F (Km) is constant
Flux control
Speed above base (rated) speed is controlled by changing flux Φ (or
Km) and keeping armature voltage constant at rated value.
Km2
Tm
ω
Km1
Km3
Km1<Km2<Km3
Va is at rated value
Steady-state operating regions of a drive
Base Speed
Torque limit
Power limit
Armature Voltage
Control
Flux
Control
• By controlling Va smoothly one can obtain speed control for any torque. This is
the preferred method of speed control below base speed (i.e.) constant torque,
variable power region.
• Above base speed the speed control is achieved by reducing flux of the machine
(flux weakening) at rated armature voltage. This means reducing Φ (or Km )
(constant power, variable torque region).
•Time constant of the armature circuit is much smaller compared to the armature
in a separately excited dc motor . Hence Va control is faster than F(Km ) control.
Example on operating regions
In a separately excited dc motor, the speed is varied from 0-1500 rpm
(base speed) by varying the terminal voltage from 0-500V, and keeping
the machine flux constant. The rated torque is 300 N-m.
(a) Find the output power at (i) 750 rpm and (ii) 1500 rpm if the torque
is held constant at 300 N-m up to base speed.
(b) Above base speed, the armature voltage is kept constant at 500V and
the flux is weakened to achieve speed control. Find the motor torque at 3000 rpm.
Solution of example
(a) Torque at 750 rpm = torque 1500 rpm = 300 N-m.
750 rpm = 2π.750
60= 78.54 rad/s.
1500 rpm = 2π.1500
60= 157.08
rad
s.
∴ Power at 750 rpm = 78.54 ∗ 300 = 23.56 kW.
∴ Power at 1500 rpm = 157.08 ∗ 300 = 47.124 kW
(b) This constant power region. ∴ output power = rated power= 47.124 kW.
Hence torque = 47124
2𝜋 .3000
60
= 150 N-m.
Separately excited D.C. motor dynamic model
eb
Ra La
+
-
load
ω
LfRf
ef
+
-
va
if
ia
Field equation:
𝑒𝑓 = 𝑖𝑓𝑅𝑓 + 𝐿𝑓𝑑𝑖𝑓
𝑑𝑡 (1)
∅𝑓 = 𝐾𝑓 𝑖𝑓 [Assuming linear magnetic characteristics] (2)
Armature equation:
𝑣𝑎 = 𝑅𝑎 𝑖𝑎 + 𝐿𝑎𝑑𝑖𝑎
𝑑𝑡+ 𝑒𝑏 (3)
𝑒𝑏 = 𝐾𝑒∅𝑓𝜔 (4)
where 𝜔 is the speed in rad/s and 𝐾𝑒 is the machine constant.
Torque equation:
𝑇𝑚 = 𝐾𝑒∅𝑓𝐼𝑎 (5)
Equation of motion:
𝑇𝑚 = 𝑇𝐿 + 𝐵𝜔 + 𝐽𝑑𝜔
𝑑𝑡 (6)
Dynamic model with constant flux
Our objective is to develop a dynamic model for the separately excited motor assuming
constant flux operation.
Therefore equation (1) modifies to
𝑒𝑓 = 𝑖𝑓𝑅𝑓
Applying Laplace transformation to equations (3) to (6)
𝑉𝑎(𝑠) = 𝑅𝑎𝐼𝑎(𝑠) + 𝐿𝑎 𝑠𝐼𝑎 𝑠 − 𝐼𝑎 0 + 𝐸𝑏(𝑠) (7)
𝐸𝑏(𝑠) = 𝐾𝑒∅𝑓Ω(𝑠) =𝐾𝑚Ω(𝑠); 𝐾𝑚 = 𝐾𝑒∅𝑓 (8)
𝑇𝑚 𝑠 = 𝐾𝑒∅𝑓𝐼𝑎 𝑠 = 𝐾𝑚 𝐼𝑎 𝑠 (9)
𝑇𝑚(𝑠) = 𝑇𝐿(𝑠) + 𝐵Ω(𝑠) + 𝐽 𝑠Ω 𝑠 − Ω(0) (10)
Dynamic model block diagram
+-
1
Ra + s.La
Keϕf=Km +-
TL(s)
Ia(s)Va(s)
Eb(s)
Keϕf=Km
1
B + s.J
Tm(s)
Ω (s)
Output of the
motor
Equations (7) to (10) can be represented using the following block diagram.
It is assumed that the motor initial current and speed is zero.
Transfer function Following the previous block diagram the speed to input voltage transfer function
Can be derived as follows:
where seconds
seconds
Example on current and speed response computation from block diagram
Let 𝑅𝑎 = 0.5 Ω, 𝐿𝑎 ≈ 0, 𝐵 ≈ 0, 𝐾𝑚 = 1.05 V.s
rad, 𝐽 = 2.5 kg. m2, 𝑇𝐿 = 0, 𝑣𝑎 =
220𝑉, step input applied through a switch. Find 𝜔 𝑡 , 𝑖𝑎 𝑡 and their final values.
Solution:
From the previous slide:
Ω(𝑠)
𝑉𝑎(𝑠)=
𝐺(𝑠)
1 + 𝐺 𝑠 𝐻(𝑠)=
𝐾𝑚
(𝑅𝑎 + 𝑠𝐿𝑎)(𝐵 + 𝑠𝐽)
1 +𝐾𝑚
2
(𝑅𝑎 + 𝑠𝐿𝑎)(𝐵 + 𝑠𝐽)
= 𝐾𝑚
𝑅𝑎 + 𝑠𝐿𝑎 𝐵 + 𝐽𝑠 + 𝐾𝑚2
Ω(𝑠)
𝑉𝑎 (𝑠)=
𝐾𝑚
𝑅𝑎 𝐽𝑠+𝐾𝑚2 =
1.05
0.5∗2.5𝑠+(1.05)2 = 0.84
𝑠+0.882
Example (2) .
∴ Ω 𝑠 =0.84
𝑠+0.882 . 𝑉𝑎 𝑠 =
0.84
𝑠+0.882.220
𝑠=
184.8
s(s+0.882)= 209.52
1
s−
1
s+0.882
∴ 𝜔 𝑡 = 209.52(1 − 𝑒−0.882𝑡) rad/s.
∴ Final value of speed = 𝜔 ∞ = 209.52 rad/s= 2000 rpm.
Using (7), (8) and the initial condition on current 𝐼𝑎 0 = 0A.
𝐼𝑎 𝑠 = 𝑣𝑎 𝑠 − 𝐸𝑏(𝑠) 1
(𝑅𝑎+𝑠𝐿𝑎 )=
220
𝑠− 𝐾𝑚Ω 𝑠
1
𝑅𝑎 =
220
𝑠−
184.8
s(s+0.882) .
1
0.5 =
440
(𝑠+0.882)
∴ 𝑖𝑎 𝑡 = 440𝑒−0.882𝑡 𝐴. (Note the high value of initial current which is undesirable)
∴ Final value of speed = 𝑖𝑎 ∞ = 0A.
Control block diagram
Ωref (s)
+-
1
Ra + s.La
Keϕf=Km +-
TL(s)
Ia(s)
Va(s)
Eb(s)
Keϕf=Km
1
B + s.J
Tm(s)
Ω (s)
Output of the
motor
CONVERTER
CURRENT
CONTROLLER
WITH LIMIT+
-SPEED
CONTROLLER
WITH LIMIT+
-
Iref (s)
Note : If position control is required then another loop has to be added at
the starting. Position reference will be the input. Speed reference will come
from the output of the position (PI) controller.
Conclusions
Thus for controlling motor up to base speed
For changing armature voltage we need a converter.
To minimize or limit the current flowing in the armature
circuit during transients, we need a current sensing loop
and a current controller with limit.
We need a speed controller which will provide the current
reference needed for the current controller.