ECE 320 Energy Conversion and Power Electronics

Dr. Tim Hogan

Chapter 4: Concepts of Electrical Machines: DC Motors (Textbook Sections 3.1-3.4, and 4.1-4.2)

Chapter Objectives

DC machines have faded from use due to their relatively high cost and increased maintenance requirements. Nevertheless, they remain good examples for electromechanical systems used for control. We’ll study DC machines here, at a conceptual level, for two reasons:

1. DC machines although complex in construction, can be useful in establishing the concepts of

emf and torque development, and are described by simple equations.

2. The magnetic fields in them, along with the voltage and torque equations can be used easily to develop the ideas of field orientation.

In doing so we will develop basic steady state equations, again starting from fundamentals of the electromagnetic field. We are going to see the same equations in ‘Brushless DC’ motors, when we discuss synchronous AC machines.

4.1 Geometry, Fields, Voltages, and Currents The geometry shown in Figure 1 describes an outer iron frame (stator), through which (i.e. its

center part) a uniform magnetic flux, , is established. The flux could be established by a current in a coil or by a permanent magnet for example.

φ

In the center part of the frame there is a solid iron cylinder (called rotor), free to rotate around its axis. A coil of one turn is wound diametrically around the cylinder, parallel to its axis, and as the stator and its coil rotate, the flux through the coil changes. Figure 2 shows consecutive locations of the rotor and we can see that the flux through the coil changes both in value and direction. The top graph of Figure 3 shows how the flux linkages of the coil through the coil would change, if the rotor were to rotate at a constant angular velocity, ω.

( )tωφλ cosˆ= (4.1)

Figure 1. Geometry of an elementary DC motor.

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1 2 3 4 5 Figure 2. Flux through the stator coil of the simple dc motor shown in Figure 1.

Since the flux linking the coil changes with time, than a voltage will be induced in this coil, vcoil, as shown below:

( tdtdvcoil ωφ )λ sinˆ−== (V) (4.2)

-1.5

-1

-0.5

0

0.5

1

1.5

0 100 200 300 400 500 600 700

λ coil

Angle (º)

1

2

3

4

5

-1.5

-1

-0.5

0

0.5

1

1.5

0 100 200 300 400 500 600 700

v dcoi

l

Angle (º)

1

2

3

4

5

Figure 3. Flux and voltage in a coil of the motor in Figure 2 with coil positions 1-5.

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The points marked on the cosine and sine waveforms of Figure 3 correspond to the positions of the rotor as shown in Figure 2.

Torque on the coil follows the Lorentz Force Law: which gives the force, F, in newtons on a charge, q, in coloumbs that is exposed to a vectoral electric field, , in volts per meter and magnetic flux density, B, as

E

( )BvF ×+= Eq (4.3)

where v is the velocity of a charged particle q. The first term gives the force on the electron that moves it through the wire or the voltage applied to the wire. The second term gives the force on the charge due to the magnetic field in which the wire is placed. With many electrons contributing to the electrical current, I, (as a vector) in the wire, the force on the wire is

BIF ×= (4.4) The force on the full coil is then twice this value to account for each of the wires shown in Figure 2.

To maintain this torque in a direction that continues the rotary motion of the rotor, the ends of the rotor coil are connected to metal electrode ring segments called a commutator as shown in Figure 4. These ring segments are attached to the rotor so as to rotate with it, and are electrically contacted using two stationary brushes typically made of carbon and copper. The brushes are spring loaded and pushed against the commutator. As the rotor spins, the brushes make contact with the opposite segments of the commutator and they switch between segments of the commutator just as the induced voltage goes through zero and switches sign.

B

ω

commutat

or

brush

brush

Figure 4. A coil of a DC motor and brushes that are pushed up against the rotating

commutator (wire of the coil is highlighted red and soldered to the commutator segments).

Because the current switches direction through the coil each time the brushes rotate to the

opposite segment of the commutator (and that this occurs as the voltage goes through zero), the 1- 3

voltage at the terminals of the brushes is a rectified version of the voltage induced in the coil as shown in Figure 5.

-1.5

-1

-0.5

0

0.5

1

1.5

v coil

Time (arb. units) -1.5

-1

-0.5

0

0.5

1

1.5

v term

inal

Time (arb. units)

Figure 5. Induced voltage in a coil and terminal voltage in an elementary DC machine. If a number of coils are placed on the rotor, as shown in Figure 6, with each coil connected to a

different segment of the commutator, then the total induced voltage to the coils, E, will be:

(4.5) ωφkE =

where k is proportional to the number of coils.

Figure 6. Multiple coils on the rotor of a DC machine.

We stated in equation (2.39) that ieTPP mechmechelec ⋅=⋅⇒= ω thus with the multiple coil DC

machine: ω⋅=⋅ TiE (4.6) (4.7) ωωφ ⋅=⋅ Tik ˆ

(4.8) ikT φ=

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If the DC machine is connected to a load or a source as in Figure 7, then the induced voltage and terminal voltage will be related by: wdgg RiEV −=terminals for a generator (4.9)

wdgmRiEV +=terminals for a motor (4.10)

Load

or

Source

Rwdg

E

gi

mi

Figure 7. Circuit with a DC machine with the current reference directions defined as

indicated for DC machine used as a generator, ig, or as a motor, im. Example 4.1.1

A DC motor, when connected to a 100 (V) source and with no load connected to the motor runs at 1200 (rpm). Its stator resistance is 2 (Ω). What should be the torque and current if it is fed from a 220 (V) supply and its speed is 1500 (rpm)? Assume the field is constant.

With no load connected to the motor, we will assume the torque is zero (assuming no friction in bearings). With the torque equal to zero, the current is zero since: KiikT == φ . This means for this operation:

ωφω KkEV ===

and with ( ) (rad/s) 66.125)s( 60

(min) 1revolution

(rad) 2rpm 1200 =⋅⋅=πω , then

s)(V 796.0

66.125(V) 100

⋅=

⋅=

K

K

Then at 1500 (rpm)

( ) (rad/s) 08.157)s( 60

(min) 1revolution

(rad) 2rpm 1500 =⋅⋅=πωo

(V) 125== ωKE

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For a motor:

(N·m) 81.37

(A) 5.47

(V) 125(V) 220

==

=

+=

+=

KIT

I

IR

IREV

wdg

wdg

4.2 Energy Considerations Conservation of energy governs that the total energy supplied to an electromechanical system

equals the energy output. Some of the energy output could be mechanical, some could be lost as heat, and some could be stored. This is summarized in equation (3.10) of your textbook as:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

heat intoconvertedEnergy

field magneticin the stored

energyin Increase

outputenergyMechanical

sourceselectric from

inputEnergy

In a lossless system, the last term is zero. When the losses are negligible, a differential change in the electrical energy input corresponds to the sum of a differential mechanical energy output and a differential change in the energy stored in the magnetic field, or

fldmechelec dWdWdW += (4.11)

Mechanical energy is force times distance, and the time rate of change of electrical energy is power or and (dteidW =elec 4.11) can be written as:

fldfld dWdxFdtei += (4.12)

where Ffld is the force produced by the magnetic field. From equation (3.1) in the handout notes, we

saw dtde λ

= . Using the linear inductor assumption of (2.19) which was ii

NL λφ== , then (4.12)

becomes: dxFdidW fldfld −= λ (4.13)

The energy in the field is given as the energy stored in an inductor

L

LiW2

2fld 2

121 λ

== (4.14)

The current and force produced by the magnetic field can then be found from (4.13) as:

( )x

xWiλλ

∂∂

=,fld (4.15)

( ) ( )dx

xdLix

xWF2

, 2fld

fld =∂

∂−=

λ

λ (4.16)

If the magnetic force results in a torque as in a rotating mechanical terminal, the mechanical energy from the field is then replaced with torque and angular displacement, θdTdxF fldfld ⇒ , as:

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θλ dTdidW fldfld −= (4.17) thus leading to

( )θθ

ddLiT

2

2fld = (4.18)

In a rotary machine with multiple windings, the inductance in equation (4.18) would contain self and mutual inductances of all coils involved. Chapter Notes: • The field of a DC motor can be created either by a DC current or a permanent magnet.

• The two fields, the one coming from the stator and the one coming from the moving rotor, are both stationary (despite rotation) and they are perpendicular to each other.

• If the directions of current in the stator and in the rotor reverse together, torque will remain in the same direction. Hence if the same current flows in both windings, it could be AC and the motor will not reverse.

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