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.