EE466/EE966 Part 2: Power Electronics

Topic 9: Control of DC machines

Control of DC machines

1. Introduction
2. Power semiconductor devices
3. Non-isolated DC-DC power converters
4. Isolated-output DC-DC power converters
5. Thermal management
6. Passive components for power electronics
7. DC-AC power conversion
8. AC-DC power conversion
9. Control of DC machines
Control of DC machines

In this topic we will:

• Review DC machine operating principles

• Show how power electronics may be used to control a DC machine

• Finally, we’ll assemble and walk through a model of a drive system

including the
• Machine
• Power electronics
• Control system
In other words, we’ll look at the power electronics in the context of an
overall drive system.
 Control of DC machines

• To recap, the average output voltage in any switching interval T is set by the
modulation index M
• M can be made to be time variant
T

VDC

Ton(t)

t
Ton (t )
m(t ) 
T
Vo (t )  m(t ) *Vdc
DC motor: Torque and back EMF
production (Revision)
Rotor torque
Rotor MMF
Force = BiL

Stator flux

δ
θ
B
θ
r X
NR (Rotor Turns)

Rotor Force = BiL

Airgap
Machine length L
Control of DC machines

Torque production
Force, F, on a number of current-carrying conductors, NR:

F    i  l   N R
Force, FN, acting in the direction of rotor rotation:

FN  F  cos( )
Torque, Tr, acting on rotor

Tr  2  FN  r  2  r   i  l N R  cos( )
Control of DC machines

• The term NR×I represents an MMF (MR) which acts perpendicularly to the
plane of the rotor coil.

• The magnetic field Β is the result of an MMF (MS) produced by current in

the stator windings

• The resultant torque can therefore be expressed as:

Tr  K  M S  M R  sin( )
torque angle    / 2  

where K is a constant for the machine.

Control of DC machines

Back EMF
The voltage induced on the armature coil is given by:

d d ( B  A)
EN N
dt dt

Here Β is constant and the area for flux linkage varies according to the
position of the rotor coil

A  2  r  L  sin( )   t

d ( B  A) d (sint )
EN  2 r l  B  2  r  l  B    cos( )
dt dt
 Control of DC machines

• Torque is produced through the interaction of the rotor and stator MMFs

• Maximum torque is produced for a torque angle (δ) of π/2.

• In a DC machine the optimum torque angle is achieved by the action of the

commutator.

• The rotor will experience a back EMF, resulting from the movement of the
rotor relative to the stator field.

• Both the torque and the EMF are proportional to the stator field.
Control of DC machines

Controlled by power electronics Armature

current
Ra La Back EMF, E

Va E

Te

Electrical Torque
Applied Voltage
 Control of DC machines

Rotor (armature) equations

dia
Note that the series
inductance of the machine Va  ia Ra  La E
here acts as the power dt
converter’s choke.

So, unlike a buck converter

we do not need to include a E  k   
discrete choke.

T  k    ia
 Control of DC machines

Applied voltage

• The armature voltage can be applied by a power converter, for example,

a self-commutated chopper
• Under steady state conditions the average voltage over the machine’s
series inductance must be zero thus:

 Ton 
V a  Vdc    I a Ra  E
 T 

The assumes that the machine mechanical time constant is much longer than T,
i.e., rotor speed does not change within each switching period T.
 Control of DC machines
A note on the machine’s equivalent circuit: In any reasonably efficient machine
the voltage drop across Ra is small. For the purpose of calculating δ we can
neglect it as a first approximation:

La

 Ton 
V a  Vdc  E
Va T 

We can see that the average applied voltage is equal to the back emf E: this is necessary to
ensure zero voltage over the series inductance La.

On the other hand, for the purpose of working out efficiency : the I2R loss in the armature
resistance is normally the biggest contributor to losses and must be considered.
 Control of DC machines

Mechanical system

d
Te  TLoad  Tlosses  J
dt
J: moment of Inertia of motor

Tlosses = Friction, Windage + Drive chain losses (speed dependent)

 Control of DC machines

Operation of DC machines

k a  Va
Torque Ra k a  Va k a 2 2 
T 
Ra Ra
Armature
Current

Va T

Ra
Ia
T = KaΦIa
Va
k a E = kaΦω
rpm

Va = IaRa + E
DC machine operating regions
k a  Va k a 2  2 
T 
Maximum Ra Ra
Torque
Torque Base speed is the maximum
Rated Torque
speed at which rated torque can
be achieved.

Region 1 Region 2

Base Rated
Speed Speed
Speed

• In Region 1 the maximum torque is limited to the rated value. This is fixed
by the mechanical properties of the machine and the electrical (heating)
properties of the armature and field windings.

• In Region 2 the field current is reduced and the available torque

decreases with increasing speed. The maximum (rated) speed is fixed by
the mechanical design of the motor and commutator.
 Control of DC machines

DC machine operating regions

• Base Speed : again, the maximum speed at which the machine can provide
rated torque. Fixed by the maximum permitted armature voltage.

• Rated Speed: The maximum speed at which the machine should operate.

• Constant Torque Region: The region in which full torque can be achieved.

• Constant Power Region: Above base speed the field excitation may be
reduced to allow higher speeds.
 Control of DC machines

Power converters
• The applied voltage may be controlled using any of the converter options
previously considered.

• The use of power electronics will give rise to ripple current. This will give
rise to torque variation.

• Ripple current gives increased vibration and losses in the machine.

– Increasing the switching frequency will reduce ripple current but
increase the losses in the power electronics.

• If the mechanical time constant of the system is much longer than that
of the power electronics - as is normally the case - a linear control model
is usually appropriate.
 Control of DC machines
Single switch

Ia

Reverse Braking Forward Motoring

4 switch
H-Bridge

Va

Reverse Motoring Forward Braking

Two switch chopper

 Control of DC machines
DC machine operating range

Shown here is a two-switch

arrangement capable of
operation in the two quadrants TR1
as highlighted in the previous D1
slide.
L1
As mentioned earlier in these VSUPPLY
slides, the machine’s series
TR2
inductance acts as the “choke”. Vout = machine
D2 back emf, E

For simplicity the winding

resistance has been omitted 0V
Machine
here.
 Control of DC machines
DC machine operating range
• When motoring, TR1 and D2 act as the switch and diode in a buck converter
• When braking – essentially generating – TR2 and D1 act as the switch and diode in a
boost converter and energy is returned to the supply: note the change in the current
direction in L1.

TR1 TR1
D1 D1

L1 L1
VSUPPLY VSUPPLY
iL1 iL1
TR2 TR2
E E
D2 D2

0V 0V
Machine in motoring mode Machine in generating mode
 Control of DC machines
DC machine operating range
The boost converter thus formed may not at first appear obvious when compared with
the diagrams in Topic 3 of this module. However, it is essentially the same. When
transferring power from the machine back into the supply, “E” becomes the input
voltage, Vin, and VSUPPLY becomes the output voltage, Vout. The machine’s inductance is
now acting as the input choke of the boost converter.

TR1
D1 L1 D1

L1
Vin Vout
VSUPPLY TR1
iL1
TR2
E
D2 0V

Boost converter
0V
Machine
Control of DC machines

Control of the DC machine: overview of a

drive system

• The equations shown earlier in these slides show that control of the DC
machine may be controlled by varying the applied armature voltage.

• This may be achieved by controlling the modulation index of the power

converter.
Control of DC machines

DC machine dynamics

Load torque
Ifield


 1
ϕ
ka 
Va Σ
R a  sL a
× Σ

Armature
E circuit

ω 1
kaϕ
F  sJ
Mechanical
system
 Control of DC machines

Closed-loop control
• With a fixed applied armature voltage any change in load torque will
result in a change in motor speed.
• Speed feedback may be used to stabilise the rotational speed in
response to changes in torque.
• Although speed feedback can reduce the speed error, it can result in
high currents at start up and stall conditions.
• The use of an internal current (torque) control loop will improve the
machine dynamics and automatically regulate the armature current.
• Transducers and electronic control will be required to implement the
closed loop control
Speed control
Modulation control to Electrical
power converter Torque (Nm)
Speed demand TL
(control units) Speed error eω
Armature
Voltage (Volts)
-
+ Control Power
ωref Motor
Function converter
+
-
Motor
EMF

Transducer Mechanical
gain System
Speed feedback
ωf (control units)
Shaft Speed ω
Rad/s
Speed control loop

• Closed-loop speed control may be used to regulate the motor speed.

• The motor speed can be measured using a device such as a tacho-

generator or shaft encoder which converts shaft speed into an electronic
value.

• An error signal is generated which is the difference between the

reference speed and the measured value.

• The power converter M is now a function of the error between the

reference signal and the measured value. ( Va=m*Vdc)
Control range (linear controllers)
For a full bridge converter the modulation is limited to +/-1
1 M, Modulation index
Saturation

M = kp*(ωref-ωf) Slope=Kp

Error
Signal Eω

-1

Control Saturation Control range

(M cannot exceed this range)
Control variables

• The controller sees scaled versions of the system quantities and internal
control variables such as the error signal and modulation index
command.

• In practice the controller variables will exist as voltages (Analogue

Control) or Binary Numbers (Digital Control). These must be scaled to
match the system quantities.
Example system
Fixed-field separately excited DC machine

• Va(Rated) = 500V Ia(Rated) = 400A

• Rotor equivalent circuit: 4mH, 0.1Ω
• Base speed: 1000rpm
• Machine constant k = 5.3Nm/A
• Moment of inertia: 406kgm2

• Speed is controlled with an analogue controller such that +/-10V demand

equates to +/-1000rpm

• The machine is driven by a four quadrant DC chopper with a supply

voltage of 500V DC.
Linear model of DC motor drive
Negative feedback
with proportional control Kp
Load torque
Power
time converter: Machine
Clock
To Workspace4 Kc 10:500V constant
Step1

5 50 1
5.4
Add Gain Saturation Gain1 0.0041s+0.1
Add1 Add2
Step Transfer Fcn Gain2
Gain3

EMF 5.4
Error Ve Transfer Fcn1
1
Gain4
Speed 405s
-K-
demand Vd ω
Scope
Kf

Mechanical
Gain term to scale speed
inertia
feedback:
1000rpm > 10 input units
 Error*
Current overshoot
≈5000A ( 400A rated current)
Speed demand
Current
speed

Applied Voltage

Load Torque increased at 6s

Error
The steady state error
can be reduced by increasing the
gain*
Description:

• At t = 1s a maximum speed demand signal is applied with a motor speed

of 0. This causes the error signal to rise to its maximum value.
• From t = 1s to t = 4.2s the error signal is saturated and maximum
armature voltage is applied to the motor.
– The armature voltage cause the armature current to flow, this
generates torque causing the motor to accelerate.

• At t = 4.2s the error enters the control band and the armature voltage
starts to decrease.
• For t = 4.2s to t = 6s speed increases more slowly causing the error to
decrease.
– By t = 6s the speed has stabilised at a value that provides the steady
state error.
– The steady state error will be that which allows the load torque to
be supported.
Description 2

• At t = 6s the load torque increases. This causes the machine to slow

down.
– The reduction in ω causes the error signal to increase.

• For t = 6s to t = 7s the increase in speed error causes the armature

voltage to increase.
– The increased armature voltage increases the armature current. This
increases the electrical torque to counter the increase in load torque.

• t > 7s. The motor settles to a new state with an increased speed error
necessary to support the required load torque.
Steady State Error
For the linear controller shown a steady state error must exist
to drive the required armature voltage. This must satisfy the
set of equations given below.

Te
V A  I A  R A  E   R A  K
K
VA  M Vdc  K C  Vm
Vm  Ve  K p  (Vd  V )  K p V  K f  
Te
VA  (Vd  K f   )  K p  K C   RA  K
K
Steady state error and gain

Demand
Speed feedback

Response with speed gain = 5 Response with speed gain = 50

Increasing the gain will reduce the steady state error

Steady state error

The steady state speed error is given below. It can be seen that increasing the
gain of the feedback loop will decrease this speed error.

 TRa 
Vd  K p  K C  
staedy state   K 
(K f  K p  KC  K )

Vd: Demand signal

T: Required Torque (Load Torque)
K: Machine Constant
KC: Converter Gain 50
Kp: Error gain5
Kf: Speed feedback constant 10/1000 V/rpm
Stability Problems

Armature Current Demand

4 6
x 10
500 2

3
0 0

0
0 1 2 3 4 5 6 7 8 9 10
-500 -2
0 1 2 3 4 5 6 7 8 9 10

Speed
Armature Voltage

• Closed loop control can result in oscillation.

• This shows the same system but with armature resistance decreased to 0.01Ω
• Note the oscillation
Internal current control

• Without additional control the speed control loop can result in very large
armature current and may excite oscillation.

• Current control introduces a second “internal” control loop which turns

the speed error into an armature current demand.

• Controlling the armature current directly controls the motor torque and
improves system dynamics.

• The current control loop will also automatically regulate the motor
current since the current demand cannot go above the maximum set by
the controller.
Current control model

Current feedback
time va
Clock error Ia
To Workspace4 To Workspace1
To Workspace3 Step1 To Workspace2

-K- 50 1
5
5.4
Add Gain Saturation Gain1 0.004s+0.01
Add3 Gain5Saturation1 Add2
Step Add1 Transfer Fcn Gain2
Gain3
Gain6
5.4 Transfer Fcn1 Scope1
-K-
1
Gain4 405s
Scope2 -K-

Scope
demand speed

To Workspace5 Current scaling, To Workspace

Rated current scaled

output to be equal to
To Workspace6 maximum demand
signal
Internal current feedback loop

Armature Current Demand

500 500
4

3.5

2.5

1.5

0.5

0 0
0 5 10 15 20 25 30 0
0 5 10 15 20 25 30

Armature Voltage Speed

Same system as previously shown but with armature resistance decreased to 0.01Ω.
Note:
• Armature current is limited
• Oscillation is virtually eliminated
• Increased speed feedback gain can be achieved.