EE 462

ELECTRIC DRIVES
Dr Francis Boafo Effah
Senior Lecturer
Department of Electrical & Electronic Eng.
Faculty of Computer & Electrical Eng.
College of Engineering
1
Room BKB 13, Bamfo Kwakye Building
Email: fbeffah.coe@knust.edu.gh
 Unit 2
DC Motor Drive Systems
2a DC Motors

2
 DC Drive Systems
▪Basic operation of a DC machine
•electrical equations
•mechanical equations
•four quadrant operation
•field weakening

3
 DC Drive Systems
▪Operation from a DC to DC converter
•1 Quadrant Drive
•2 and 4 Quadrant Drives
▪Operation from an AC to DC rectifier
▪Control of DC Drives
▪Current Control
▪Speed Control
4
 Basic Operation of a DC Machine
▪Separately Excited DC Motor – separate
voltage supplies for the stator (field/flux) and
rotor (armature) circuits
▪The armature (rotor) circuit is accessed by a
commutator which ensures that the current
flow in the armature winding is at right angles
(orthogonal) to the flux

5
 Basic Operation of a DC Machine
▪ Faraday’s Law states:
𝑻𝒐𝒓𝒒𝒖𝒆 = 𝒌. 𝑭𝒍𝒖𝒙 × 𝒂𝒓𝒎𝒂𝒕𝒖𝒓𝒆 𝒄𝒖𝒓𝒓𝒆𝒏𝒕

(k = a constant determined by physical layout, winding

distribution, etc.)
▪ The torque direction can be changed by either
1) changing the direction of the armature current
(common)
2) reversing the polarity of the field (less often)

6
 DC Motor Equivalent Circuit - Field
▪The magnetic field (flux) is usually set
up by the winding on the stator – an
electromagnet
▪Field winding usually has many turns
and therefore draws only a small
current

7
 DC Motor Equivalent Circuit - Field
▪ Field winding has resistance 𝑅𝑓 and self-inductance 𝐿𝑓
▪ 𝑉𝑓 = 𝐼𝑓 𝑅𝑓 + 𝐿𝑓 𝑑𝐼𝑓 /𝑑𝑡
▪ 𝑉𝑓 =applied field voltage, 𝐼𝑓 = resultant field
current
▪ Flux = 𝜙𝑓 = 𝑘𝑓 ∙ 𝐼𝑓
▪ Rated 𝐼𝑓 gives rated flux – usually applied and kept
constant up to “base” speed. Reduce 𝐼𝑓 when field
weakening is required.
8
 DC Motor Equivalent Circuit - Armature
▪𝑉𝑎 - external voltage source used to
cause current to flow in the armature
circuit
▪Armature has winding resistance (𝑅𝑎 )
and self inductance (𝐿𝑎 )

9
 DC Motor Equivalent Circuit - Armature
▪When the armature rotates in an
electromagnetic field (flux), an EMF is
induced in the armature coils (Lenz’s
Law)
▪Back EMF (𝑬𝒃 )
▪𝐸𝑏 opposes the armature current when
the machine is acting as a motor
10
 Motor Operation
▪Consider motor at standstill (𝐸𝑏 = 0)
▪Apply DC voltage to field winding
▪Field current builds up and sets up
flux within motor
▪Apply DC voltage to armature
winding
11
▪Armature current builds up
▪𝑉𝑎 = 𝐼𝑎 𝑅𝑎 + 𝐿𝑎 𝑑𝐼𝑎 /𝑑𝑡 (at a standstill)
▪Current flowing in a conductor
orthogonal to a magnetic flux creates
a force
▪ The sum of the forces on the
conductors causes a torque which
causes the armature to turn
12
▪The torque developed by the motor (𝑇𝑒 )
𝑇𝑒 = 𝑘 ∙ 𝜙𝑓 ∙ 𝐼𝑎
▪ 𝑘 ∙ 𝜙𝑓 is commonly called the machine or
torque constant (𝑘𝑚 )
▪ 𝑘𝑚 varies with flux level!
▪BUT we usually operate the machine at
constant flux
▪𝑇𝑒 = 𝑘𝑚 ∙ 𝐼𝑎

13
▪Because the motor turns in a magnetic
flux, an EMF is induced in the armature
conductors which opposes the applied
voltage (and tries to reduce 𝐼𝑎 )
𝑘∙𝑑𝜙𝑓
𝐸𝑏 = = 𝑘 ∙ 𝜙𝑓 ∙ 𝜔𝑟 = 𝑘𝑚 ∙ 𝜔𝑟
𝑑𝑡
▪Note that 𝑘𝑚 is the same machine
constant in the torque equation

14
 Torque and Back EMF
▪ Torque
▪ Due to the interaction of the flux and the
armature current
𝑇𝑒 = 𝑘𝑚 ∙ 𝐼𝑎
▪ Back EMF
▪ Caused by the rotation of armature conductors
in the flux
𝐸𝑏 = 𝑘𝑚 ∙ 𝜔𝑟
15
 Torque and Back EMF
▪ For these equations to hold, you MUST use the correct
measurement units
▪ Torque - 𝑇𝑒 in Nm
▪ Current - 𝐼𝑎 in A
▪ Flux - 𝜙 in Wb
▪ Back EMF - 𝐸𝑏 in V
▪ Speed - 𝜔𝑟 in radians/sec (NOT rev/min or Hz)
▪ Also, output (mechanical) power = 𝑇𝑒 ∙ 𝜔𝑟 = 𝐸𝑏 ∙ 𝐼𝑎
16
 Four Quadrant Operation
▪ Steady state operation
𝑹𝒂 (𝑑𝐼𝑎 /𝑑𝑡 = 0), 𝜔𝑟 , 𝐸𝑏 constants
▪ 𝐸𝑏 > 0 ⇒ Forward
𝑽𝒂 𝑬𝒃 ▪ 𝐸𝑏 < 0 ⇒ Reverse

𝑰𝒂
▪ 𝑉𝑎 > 𝐸𝑏 ⇒ Motoring
▪ 𝑉𝑎 < 𝐸𝑏 ⇒ Generating
𝐸𝑏 ∝ 𝜔𝑟 : Therefore, the polarity of 𝐸𝑏 defines the polarity of the speed.
𝐼𝑎 ∝ 𝑇𝑒 : Therefore, the polarity of 𝐼𝑎 defines the polarity of the torque.
17
 Operation within the Torque - Speed
𝑻𝑳
Speed drop Envelope
Due to 𝑰𝒂 𝑹𝒂
▪ Consider operation at 50%
base speed.
50%
Base speed
𝝎𝒓 ▪ At no-load and steady state
𝑹𝒂 ▪ 𝑇𝑒 = 0, 𝐼𝑎 = 0, 𝑑𝐼𝑎 /𝑑𝑡 = 0
▪ applied voltage = 50% rated
𝑽𝒂 𝑬𝒃 voltage = 𝐸𝑏
▪ motor runs at half base-speed
𝑰𝒂

18
 Operation within the Torque Speed
𝑻𝑳
Speed drop Envelope
Due to 𝑰𝒂 𝑹𝒂 ▪ Increase load torque
▪ 𝑇𝑒 increases as does 𝐼𝑎
50% 𝝎𝒓 ▪ 𝑑𝐼𝑎 /𝑑𝑡 = 0 (considering steady state
Base speed operation)
𝑹𝒂
▪ applied voltage is still 50% rated
voltage.
𝑽𝒂 𝑬𝒃
▪ But 𝐼𝑎 𝑅𝑎 is now significant, 𝐸𝑏 is lower
than the no-load case and the
𝑰𝒂 speed is slower than the no-load
case.
19
 Field Weakening in DC Machines
Aim
▪ To extend operating range [increase speed range]
▪ Assume for this discussion
1) Steady state
2) Ra negligible
∴ 𝑉𝑎 = 𝐸𝑏 = 𝑘 ∙ 𝜙𝑓 ∙ 𝜔𝑟
𝑇𝑒 = 𝑘 ∙ 𝜙𝑓 ∙ 𝐼𝑎

20
 Field Weakening in DC Machines
At base speed
𝑉𝑎−𝑟𝑎𝑡𝑒𝑑 = 𝐸𝑏 = 𝑘 ∙ 𝜙𝑓−𝑟𝑎𝑡𝑒𝑑 ∙ 𝜔𝑟−𝑏𝑎𝑠𝑒
At rated torque
𝑇𝑒−𝑟𝑎𝑡𝑒𝑑 = 𝑘 ∙ 𝜙𝑓−𝑟𝑎𝑡𝑒𝑑 ∙ 𝐼𝑎−𝑟𝑎𝑡𝑒𝑑
Field Control

21
 Field Weakening in DC Machines
If we have applied 𝑉𝑎−𝑟𝑎𝑡𝑒𝑑 to the armature of the
machine, and we want the machine to go faster –
reduce the field (field current)
e.g.
▪ Reduce field current to 50% rated value.
▪ 𝜙𝑓 = 0.5𝜙𝑓−𝑟𝑎𝑡𝑒𝑑
▪ 𝑉𝑎−𝑟𝑎𝑡𝑒𝑑 = 𝐸𝑏 = 𝑘 ∙ (0.5𝜙𝑓−𝑟𝑎𝑡𝑒𝑑 ) ∙ 𝜔𝑟
▪ ∴ 𝜔𝑟 = 2 × 𝜔𝑟−𝑏𝑎𝑠𝑒 i.e. you have double the speed
22
 Field Weakening in DC Machines
But there is always a compromise!
1
𝑇𝑒−𝑚𝑎𝑥 = 𝑘 ∙ (0.5𝜙𝑓−𝑟𝑎𝑡𝑒𝑑 ) ∙ 𝐼𝑎−𝑟𝑎𝑡𝑒𝑑 = 𝑇𝑒−𝑟𝑎𝑡𝑒𝑑
2

23
 Torque Speed Envelope

Max torque
Torque defined by max 𝑰𝒂 𝑽𝒂 at rated value.
Control 𝑰𝒇 (field current) to
Operate in here
Rated Field
torque weakening
Operate in here
by controlling
𝑽𝒂 and 𝑰𝒂
Speed
Base Max
speed speed

24
 Example for a DC Machine
▪A DC machine is rated at 15 kW and is
92% efficient. Its rated armature voltage
is 400 V and its rated speed is 1500 rpm.
Calculate 𝑅𝑎 and 𝑘𝑚 .
▪Answer
▪ 𝑅𝑎 = 0.78 Ω
▪ 𝑘𝑚 = 2.34 V/rad/s
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 𝐿𝑎 𝑑𝐼𝑎
𝑉𝑎 = 𝐼𝑎 𝑅𝑎 + + 𝐸𝑏
𝑑𝑡
Consider steady-state operation at rated conditions
𝑑𝐼𝑎
∴ 𝑉𝑎 = 𝐼𝑎 𝑅𝑎 + 𝐸𝑏 ( = 0)
𝑑𝑡

Consider power
𝑉𝑎 × 𝐼𝑎 = 𝐼𝑎2 𝑅𝑎 + 𝐸𝑏 × 𝐼𝑎
𝑉𝑎 𝐼𝑎 = input power =100% of the power
𝐼𝑎2 𝑅𝑎 = losses in 𝑅𝑎 = 8% of the power
𝐸𝑏 𝐼𝑎 = mechanical output power (15 kW as specified)
= 92% of the power
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 0.92 ∙ 𝑃𝑖𝑛 = 15000
15000
𝑉𝑎 𝐼𝑎 =
0.92
𝑉𝑎 = 400 V ⇒ 𝐼𝑎 = 40.8 A
15000
𝑃𝑖𝑛 = = 16304 W
0.92
𝑃𝑙𝑜𝑠𝑠 = 0.08 × 16304 = 𝐼𝑎2 𝑅𝑎
⇒ 𝑅𝑎 = 0.78Ω

27
𝐸𝑏 = 𝑘𝑚 𝜔𝑟
𝐸𝑏 = 𝑉𝑎 − 𝐼𝑎 𝑅𝑎 = 400 − 40.8 × 0.78 = 368 V

368
𝑘𝑚 = 2𝜋 = 2.34 V/rad/s = 2.34 Nm/A
1500×
60

Rated torque = 𝑘𝑚 × 𝐼𝑎−𝑟𝑎𝑡𝑒𝑑

= 2.34 × 40.8 = 95.47 Nm

28
 Requirements for a Variable Speed Drive
▪ Ideal controllable voltage source
▪ Output voltage = reference voltage under ALL conditions
▪ Change 𝑉𝑎 to change the speed
▪ 𝑉𝑎 ≈ 𝐸𝑏
▪ 𝐼𝑎 = 𝑇𝑒 /𝑘𝑚 ≈ 𝑇𝑙 /𝑘𝑚
▪ Power Converter
▪ Fixed DC to variable DC (chopper)
▪ Fixed AC to variable DC (controlled rectifier)
29
Early Variable Speed Drive:

30
▪ This method is known as Ward Leonard system
▪ It was designed to provide highly accurate speed
and position control of separately excited DC
motor
▪ The motor field current is fed from a constant-
voltage source
▪ The armature is fed from a variable supply
consisting of a motor-generator set
▪ These are two machines of the same power rating
as the motor being controlled
▪ 31
▪ The motor driving the generator is either AC or DC
constant speed drive.
▪ The AC driving motor is either a 3-phase induction
motor or a 3-phase synchronous motor.
▪ The field winding of the separately excited DC
motor is energized even when the motor is at a
standstill unless the stop is of long duration.
▪ Speed control of the separately excited motor is
done by varying the excitation and hence the
voltage of the DC generator

32
▪ The potentiometer regulator indicated permits the
variation of the exciting current from maximum to
zero and in the reverse direction when the direction
of rotation of the motor is to be changed.
▪ The motor-generator set runs continuously, but the
generator is excited only during the actual moments
when the variable voltage motor is working.
▪ They were much used in the past for large reversing
rolling mills and mine hoist motors where
regenerative braking can be applied.

33
Mine shaft elevator
34
Steel mills
35
▪ The advantages of the Ward-Leonard
system include:
▪ A wide range of speed from zero to
high speed in either direction
▪ The main motor characteristics are
similar
▪ It can provide regenerative braking

36
▪ Its disadvantages include:
▪ Only speeds below base speed are
obtainable
▪ High initial cost of three machines
▪ Low overall system efficiency because
three machines are used in chain for
the energy conversion.

37
▪ The Ward Leonard drives have been
superseded by thyristor converters which
offer
▪ lower initial cost,
▪ higher efficiency (typically over 95%),
▪ smaller size,
▪ reduced maintenance, and
▪ faster response to changes in set speed.
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 Topic 2
DC Motor Drive Systems
2b Chopper Fed Drives

39
 Chopper Fed DC Drive
▪Chopper = DC to DC converter
▪Used where a fixed DC supply is available
▪ battery, overhead catenary
▪ tram, power steering, robot
▪Or where very fast torque control is required
▪ compare with controlled rectifiers later in the course
▪Configurations for 1, 2, and 4 Quadrant operation
▪ tend to employ fast switches (low power) i.e. MOSFET or
IGBTs

40
 Chopper Basics
▪DC to DC converter
▪supply is a fixed DC voltage 𝑉𝑑𝑐
▪usually use fixed switching
frequency (1/T) and vary switching
time (duty cycle 𝛿)
▪𝛿 = 𝑇𝑜𝑛 /𝑇

41
 Chopper Basics
▪ Output voltage has a DC component and a large ripple
component
▪ BUT this usually feeds an inductive load
▪ L limits di/dt, therefore current is much smoother than
the voltage
▪ 𝐼𝑑𝑐 large compared to ripple component so it is
reasonably efficient
▪ Vary 𝛿 to vary the mean output voltage
▪ NOTE if the output current is CONTINUOUS
▪ operating voltage 𝑉𝑜𝑢𝑡 = 𝛿 ∙ 𝑉𝑑𝑐
▪ i.e., ideal controlled voltage source
42
 Assumptions
▪ Usual assumptions (steady state operation)
▪ constant DC supply voltage (𝑉𝑑𝑐 or 𝑉𝑠𝑢𝑝 )
▪ fixed switching period (T), variable duty cycle 𝛿
▪ ideal diode, switch
▪ Initially ignore motor resistance for the switching analysis
over one switching period – the inductance dominates
over this small timescale
▪ assume motor inertia is large so the speed and motor
back emf (𝐸𝑏 ) stay constant during a switching period
▪ the net change in armature current is zero (over a
switching period)
43
One Quadrant Chopper Drive
▪Speed - only forward
▪Torque - only one
direction (𝐼𝑎 cannot
reverse)
▪1Q - 1 quadrant forward
motoring only

44
 One Quadrant Chopper Drive
▪ Operation with continuous current
Q1
Q1 𝑻𝑶𝑵 𝑻𝑶𝑭𝑭 𝑻𝑶𝑵 𝑻𝑶𝑭𝑭 control

𝑰𝒂 ∆𝑰𝒂 ത𝑰𝒂 (mean)

Q1 D1 Q1 D1

𝑽𝒂
ഥ 𝒂 (mean)
𝑽

Energy transferred from Energy transferred from

supply to load AND stored stored magnetic energy
𝑰𝒔𝒖𝒑 magnetic energy in La to mechanical load

45
 One Quadrant Chopper Drive
▪ Operation with continuous current
▪ Important variables:
▪ 𝐸𝑏 - a measure of speed
▪ 𝑰ത𝒂 - mean armature current, a measure of mean
electrically developed torque
▪𝑽 ഥ 𝒂 - mean armature voltage ≈ 𝐸𝑏 ⇒ speed
▪ ∆𝐼𝑎 = change in armature current over a
switching period
𝑻𝑶𝑵
ഥ𝒂 =
𝑽 𝑽𝒅𝒄 = 𝜹𝑽𝒅𝒄
𝑻
46
 One Quadrant Chopper Drive
▪ Operation with discontinuous current
Q1
Q1 𝑻𝑶𝑵 𝑻𝑶𝑭𝑭 𝑻𝑶𝑵 𝑻𝑶𝑭𝑭 control

𝑰𝒂 ത𝑰𝒂 (mean)
∆𝑰𝒂
Q1 D1 Q1 D1
𝑽𝒅𝒄 𝑽𝒅𝒄 Coasting
𝒕𝟐
𝑽𝒂
ഥ 𝒂 (mean)
𝑽
𝑬𝒃

Energy transferred from Energy transferred from

supply to load AND stored stored magnetic energy
𝑰𝒔𝒖𝒑 magnetic energy in La to mechanical load

47
 One Quadrant Chopper Drive
▪Operation with discontinuous current
𝐸𝑏
𝜔𝑟 = ഥ 𝒂 (ignoring 𝑅𝑎 ),
𝐸𝑏 = 𝑽
𝑘𝑚

ഥ𝒆
𝑻 ഥ𝑳
𝑻
𝒊𝒂 = =
𝑘𝑚 𝑘𝑚

BUT ഥ 𝒂 ≠ 𝜹𝑽𝒅𝒄 ⇒ no longer have an ideal

𝑽
controlled voltage source
48
 Operation with Discontinuous Current
▪Mean armature current in discontinuous mode

1 𝑇 1
𝑖𝑎ҧ = ‫׬‬ 𝑖 𝑡 𝑑𝑡 = [1/2 ∙ Δ𝑖𝑎 ∙ 𝛿𝑇 + 1/2 ∙ ∆𝑖𝑎 ∙ 𝑡2 + 0] (1)
𝑇 0 𝑎 𝑇

Q1 ON D1 ON

49
Operation with Discontinuous Current
When Q1 is ON,
∆𝐼𝑎 (𝑉𝑑𝑐 −𝐸𝑏 )
𝑉𝑑𝑐 − 𝐸𝑏 = 𝐿𝑎 ⇒ ∆𝐼𝑎 = ∙ 𝛿𝑇 (2)
𝛿𝑇 𝐿𝑎
When D1 is ON,
∆𝐼𝑎 𝑉𝑑𝑐 −𝐸𝑏
0 = 𝐸𝑏 + 𝐿𝑎 − ⇒ 𝑡2 = ∙ 𝛿𝑇 (3)
𝑡2 𝐸𝑏
Substituting (2) and (3) into (1), we have
𝜹𝟐 ∙𝑻∙𝑽𝒅𝒄 𝑽𝒅𝒄
𝒊𝒂ҧ = −𝟏 PROVE!
𝟐𝑳𝒂 𝑬𝒃

50
Operation with Discontinuous Current
This equation is only valid for discontinuous
current (which occurs at low load torque).
We would usually set 𝛿 to a constant value and
expect speed to remain (roughly) constant
BUT when discontinuous current occurs, then as
𝑖𝑎ҧ goes down, 𝐸𝑏 increases for a fixed 𝛿.

When does discontinuous current occur?

51
 Operation with Discontinuous Current
Threshold occurs when 𝑡2 = 𝑇 − 𝛿𝑇

𝑉𝑑𝑐 −𝐸𝑏
Or ∙ 𝛿𝑇 = 𝑇 − 𝛿𝑇
𝐸𝑏

𝐸𝑏
i.e. if > 𝛿, we have discontinuous current
𝑉𝑑𝑐
𝐸𝑏
= 𝛿 is the threshold
𝑉𝑑𝑐

52
 Initial Conclusions
▪We would like the chopper to act like an
ideal controlled voltage source, where
▪Mean armature voltage 𝑽 ഥ 𝒂 = 𝜹𝑽𝒅𝒄
▪If armature resistance is ignored
▪𝑣𝑎 = 𝐸𝑏 = 𝑘𝑚 ∙ 𝜔𝑟
▪i.e., speed is directly controlled by the
duty cycle, and is independent of load.

53
 Initial Conclusions
▪These conditions are true if the armature
current is continuous, BUT NOT TRUE if the
armature current is discontinuous (i.e.,
low load torque)
▪We need to examine this condition
further

54
 Discontinuous Current
▪ We will determine the torque-speed characteristics for
a DC motor fed from a 1Q chopper in discontinuous
mode.
▪ Use a fixed 𝜹, e.g. 0.5
▪ Vary 𝐸𝑏 in the equation and see what value of 𝑖𝑎_𝑚𝑒𝑎𝑛
that gives us
▪ Note: 𝑉𝑑𝑐 is constant, so we will vary the term (𝐸𝑏 /𝑉𝑑𝑐 )
▪ (assume if 𝐸𝑏 = 𝑉𝑑𝑐 , this corresponds to base speed)
▪ Let 𝐾 = (𝛿 2 × 𝑇 × 𝑉𝑑𝑐 )/2𝐿𝑎

55
 Discontinuous Current
▪ CASE 1
▪ 𝐸𝑏 /𝑉𝑑𝑐 = 1 i.e. machine is operating at base
speed
▪ 𝑖𝑎_𝑚𝑒𝑎𝑛 = 0 A
▪ 𝑇𝑒 = 0 Nm
▪ CASE 2
▪ 𝐸𝑏 /𝑉𝑑𝑐 = 0.875 (𝜔𝑟 = 0.875 x base speed)
▪ 𝑖𝑎_𝑚𝑒𝑎𝑛 = 0.143𝐾 A
▪ 𝑇𝑒 = 0.143𝐾. 𝑘𝑚 Nm
56
 Discontinuous Current
▪ CASE 3
▪ 𝐸𝑏 /𝑉𝑑𝑐 = 0.75 (𝜔𝑟 = 0.75 x base speed)
▪ 𝑖𝑎_𝑚𝑒𝑎𝑛 = 0.33𝐾 A
▪ 𝑇𝑒 = 0.33𝐾. 𝑘𝑚 Nm
▪ CASE 4
▪ 𝐸𝑏 /𝑉𝑑𝑐 = 0.5 (𝜔𝑟 = 0.5 x base speed)
▪ 𝑖𝑎_𝑚𝑒𝑎𝑛 = 1.0𝐾 A
▪ 𝑇𝑒 = 1.0𝐾. 𝑘𝑚 Nm

57
Torque Speed Characteristics
(Speed)

Threshold of
Discontinuous
current

Discontinuous Continuous
current current

This would be the

“ideal” torque
speed shape
at low load (Load torque)

58
 Droopy Characteristics
▪ The droopy torque-speed characteristic is bad –
you usually want a constant speed for a constant 𝛿,
irrespective of load. You DO NOT want massive
speed variations from small changes in load
▪ The characteristic can be improved by
▪ reducing the switching period, T (but you get
increased switching losses)
▪ add extra inductance, L (increase size and losses)
▪ add speed transducer and closed-loop control
(extra cost)
59
 Important Conclusions
▪ Simple hardware and control
▪ Current can be continuous or discontinuous
▪ If the current is continuous, calculations are easier
as 𝑽ഥ 𝒂 = 𝜹𝑽𝒅𝒄 i.e., for a given 𝛿, speed is
independent of load
▪ If the current is discontinuous, calculations are more
difficult since 𝑽ഥ 𝒂 ≠ 𝜹𝑽𝒅𝒄
▪ For a given 𝛿, speed is determined by load
▪ we can bring in a true value for 𝑅𝑎 for steady state
calculations if necessary
60
Two Quadrant Chopper Drive
ഥ 𝒂 (𝑬𝒃 ) - Positive only
▪𝑽
▪𝒊𝒂ҧ (𝑻𝒆 ) - Positive and
𝑻 𝒆
negative
𝐗 √
𝝎𝒓

𝐗 √

61
 Principle of Operation

Lockout
▪ Again, we usually employ a fixed switching frequency and vary the
duty cycle
▪ Q1 and Q2 are switched alternately, i.e. the drive signals are a
complementary pair
▪ “Lockout” required – a delay between the turn-off of one device
and the turn-on of the other device to prevent shoot-through (short
circuit of the supply through Q1 and Q2)
62
 Principle of Operation – Q1 on, Q2 off,
motoring (positive 𝑰𝒂 )
▪ Current flows from supply to the
armature.
▪ Energy transferred from supply to motor
to load, and also to armature
inductance as stored magnetic energy
(also to armature resistance as waste
heat)
▪ Note: 𝑉𝑎 is the armature voltage, i.e. the
𝑰𝒂 , 𝑰𝒔 sum of the back emf (𝐸𝑏 ), the armature
resistance drop (𝑖𝑎 𝑅𝑎 ) and the armature
inductance voltage drop (𝐿𝑎 𝑑𝑖𝑎 /𝑑𝑡)
63
Principle of Operation – Q1 off, Q2 on,
motoring (positive 𝑰𝒂 )
𝑰𝒂
▪ Armature current freewheels
through D1
▪ Stored energy transferred from
armature inductance to 𝑅𝑎
(waste heat) and load (useful
work)
▪ If the current reduces to zero,
then the motor back emf forces
a negative current through Q2
64
Waveforms in Motoring Mode
𝛿𝑇

65
Principle of Operation – Q2 on, Q1 off,
generating (𝑰𝒂 is negative)
𝑰𝒂
▪ 𝐸𝑏 forces current through Q2
▪ Energy is transferred from the load
to the motor and then to stored
energy in the armature inductance
(and heat in 𝑅𝑎 ).
▪ The current magnitude will increase

66
 Principle of Operation – Q2 off, Q1 on,
generating
𝑰𝒂
▪ Current flows from the motor to the supply
via D2
▪ Energy transferred from load, to motor, and
then to the supply. Also, the stored
magnetic energy in 𝐿𝑎 is transferred to the
supply, so the current magnitude will
reduce. 𝑅𝑎 losses as well
▪ If current magnitude reduces to zero, the
supply voltage will force a positive current
through the motor via Q1
67
Waveforms in Generating Mode

68
Waveforms at Low Load

ത𝑰𝒂 (mean)

𝑽𝒂 (𝒕) 𝑽𝒅𝒄
𝟎

ഥ 𝒂 = 𝜹𝑽𝒅𝒄
𝑽
for all conditions

69
 Two Quadrant Chopper
▪ During the interval 𝛿𝑡, the supply is connected to the
armature either through Q1 or D2
▪ During the interval (T − 𝛿𝑡), the armature is short-
circuited either through D1 or Q2
▪𝑽ഥ 𝒂 = 𝜹𝑽𝒅𝒄 (always positive)
▪ 𝒊𝒂ҧ = (𝜹𝑽𝒅𝒄 − 𝑬𝒃 )/𝑹𝒂
▪ 𝒊𝒂 positive, motoring
▪ 𝒊𝒂 negative, generating

70
 conclusions
▪ Two quadrant chopper behaves like an ideal controlled voltage source
▪ Controlled (positive) armature voltage
▪ Armature current (and thus torque) can be positive (motoring) or
negative (generating)
▪ Discontinuous current DOES NOT exist
▪ No droopy torque-speed characteristic
▪ Speed is determined by 𝛿 alone and is independent of torque, except for slight
change due to resistance voltage drop
▪ Further reading
▪ Power Semiconductor Drives, S. B. Dewan, G. R. Slemon and A. Straughen, John
Wiley and Sons
▪ Electric Motors and Drives, Austin Hughes, Butterworth Heinemann, ISBN 0 7506
1741 1
71
 Four Quadrant Chopper

▪ Lots of different modulation strategies, but the easiest

is as follows
72
 Four Quadrant Chopper
▪ Quadrants 1 and 2
▪ Keep Q2 permanently on, Q1 permanently off
▪ Q3 and Q4 form a 2Q chopper for speed in the forward
direction
▪ Drive Q3 and Q4 with fixed frequency and duty cycle
modulation as before
▪ Quadrants 3 and 4
▪ Keep Q1 permanently on, Q2 permanently off
▪ Q3 and Q4 again form a 2Q chopper, but the armature
voltage is now negative, so the speed is in the reverse
direction

73
 Example for Chopper Fed DC Drives
▪A DC motor has rated values of 230 V, 500
rpm, 90 A, R a = 0.115 Ω, and La = 0.011 H
▪A 2Q chopper uses a supply voltage of 230 V
and has a switching frequency of 400 Hz
▪Calculate the speed for motoring at half-
rated torque when 𝛿 = 0.5
▪Calculate the speed at 𝛿 = 0.5 when
generating at rated torque.

74
 Solution
DC motor rated at 230 V, 500 rpm, 90 A, R a =
0.115 Ω, and La = 0.011 H
Chopper VDC = 230 V, fs = 400 Hz, 2 Quadrant
Calculate
1) speed for motoring at half-rated torque
when 𝛿 = 0.5
2) speed at 𝛿 = 0.5 when generating at rated
torque.
75
 Solution
1) Va = Ia R a + Eb + La dIa /dt
dIa
Consider at steady state, therefore ഥa =
= 0. V
dt
δVdc (2Q chopper, therefore valid all the time)
Ia = 45 𝐴 ⇒ half-rated torque.
Eb = Va − Ia R a = 0.5 × 230 − 45 × 0.115 = 109.83 V
At rated conditions:
Eb = Va − Ia R a = 230 − 90 × 0.115 = 219.65 V
Eb ⇒ 500 rpm
76
 Solution
Let X be the speed we want.

𝑋 109.83
= ; Eb ∝ 𝑆𝑝𝑒𝑒𝑑
500 𝑟𝑝𝑚 219.65

X = 250 rpm

77
 Solution
2) 𝛿 = 0.5, Ia = −90 𝐴
Eb = 0.5 × 230 − (−90 × 0.115) = 125.4 V

125.4
𝑆𝑝𝑒𝑒𝑑 = × 500 = 285 rpm
219.65

78
 Topic 2
DC Motor Drive Systems
2c Rectifier Fed Drives

79
 Introduction
▪ Was the most common controlled speed drive in the
industry (80% of the worldwide market in 1986)
▪ Now replaced by V/F induction motor
▪ A low-impedance adjustable DC voltage, directly from an
AC supply
▪ Completely replaced the Ward-Leonard system
▪ cheaper first cost
▪ smaller, quieter, lower maintenance
▪ higher efficiency (>95%)
▪ fast response to changes in set speed

80
 Introduction
▪BUT does have disadvantages
▪not pure DC – significant ripple
▪input current has poor power factor
and distortion
▪is not capable of automatic
regeneration
▪poor overload capability
81
 Revision of Controlled Rectifiers

▪ Fully controlled single-phase rectifier

▪ Switches T1 and T2 fired simultaneously with firing angle 𝛼
▪ Switches T3 and T4 fired half a cycle later
▪ Changing 𝛼, changes the output voltage
▪ IF THE LOAD CURRENT IS NOT CONTINUOUS THERE IS NO SIMPLE
EXPRESSION FOR 𝑉𝑎
82
Inductive Load e.g. DC Motor

83
 Single Phase Controlled Rectifier
▪General points for DC motor drive
▪Rectifier produces crude DC with
pronounced ripple in output voltage (100 Hz)
▪This gives rise to pulsating fluxes and currents;
poles and frames must be laminated to
reduce eddy current losses and commutation
problems
▪Use forced ventilation to allow full torque (𝑖𝑎 )
at low speed i.e., a separate fan
84
 Single Phase Controlled Rectifier
▪Armature inductance makes 𝐼𝑎 smoother
than 𝑉𝑎 therefore torque ripple is reduced
▪Machine inertia is large, therefore speed is
smooth despite torque ripple
▪For continuous current

𝑉𝑑𝑐 = 2 ∙ 2 ∙ 𝑉𝑠𝑢𝑝𝑝𝑙𝑦 ∙ cos 𝛼 /π

▪𝑉𝑠𝑢𝑝𝑝𝑙𝑦 is the supply rms voltage

85
 Three Phase Rectifier

▪ The six thyristors are fired at 60° intervals: 1 – 6 – 2 – 4 – 3 – 5 – 1

▪ Each device always conducts for 120°
▪ Load voltage has 6 pulses per cycle and the lowest ripple
frequency is 300 Hz
▪ Output voltage waveform is much smoother than the single-phase
converter
86
Rectifier Waveforms (1)

87
Rectifier Waveforms (2)

88
Rectifier Waveforms (3)

89
 Rectifier Waveforms (4)
▪ We can assume that the motor current is continuous at all operating
conditions
▪ Ripple voltage magnitude is much lower than single-phase case
▪ Ripple frequency is higher (300 Hz), large motors have bigger 𝐿𝑎
▪ Note, that the firing angle is measured from the point of ‘natural’
commutation of a diode bridge.
𝑉𝑑𝑐 = (3/π) 2 ∙ 𝑉𝐿𝐿_𝑟𝑚𝑠 ∙ cos 𝛼

▪ Ripple is small for small 𝛼, and worst between 𝛼 = 60° and 𝛼 = 90°
▪ Note 𝛼 > 90°, 𝑉𝑑𝑐 is negative – this is called inversion
▪ Firing angle usually limited to 150° - 165°
▪ Ensures correct commutation
90
Supply Current Waveform
Assume ideal DC
load current

91
 Supply Current Waveform
▪ Supply (AC) current needs to be considered if you
have a weak power supply e.g., oil rig, ship.
▪ The harmonic currents of the supply current
(indicated by the distortion factor) can cause
problematic harmonic voltages that affect other
loads (users).
▪ The displacement factor (an indication of Vars)
reduces as 𝛼 increases, meaning the drive
consumes a high AC current for a relatively low
power.
92
 Torque Speed Regions
▪One bridge rectifier provides operation
in quadrants 1 and 4
▪How useful is this?

93
 Torque Speed Regions
▪Use an anti-parallel bridge to operate in
quadrants 2 and 3

94
 Rectifier vs Chopper
▪ During transient operation there has to be a delay (typically 10 ms)
between one bridge being disabled and the other being enabled,
to avoid a possible short circuit
▪ Okay for large drives where the mechanical transient is very slow
▪ For small drives the 4 quadrant chopper gives much better
▪ Generally choppers provide better transient performance
▪ Control action can be altered every 3.3 ms in a controlled
rectifier because that is how quickly the firing angle can be
changed
▪ Control action can be altered every switching period in a
chopper (e.g., 250 µs for 4 kHz switching frequency)
▪ Note this is an oversimplification as we will see in the control
section
95
 Conclusions and References
▪ Much less likelihood of discontinuous current
▪ Can assume the rectifier behaves as an ideal DC
voltage source
▪ Performance calculations are straightforward
▪ Invariably employ closed-loop control
▪ Firing angle adjusted to achieve desired operating
conditions
▪ Supply harmonics 6k ± 1 (5th, 7th, 11th, 13th)
▪ Large drives require special input filters
▪ Power Semiconductor Drives, S. B. Dewan, G. R. Slemon and A. Straughen, John Wiley and Sons
▪ Control of Electrical Drives, W. Leonhard, Springer Verilog, ISBN 0-387-13650-9
▪ Electric Motors and Drives, Austin Hughes, Butterworth Heinemann, ISBN 0 7506 1741 1

96
 Example for Rectifier Fed DC Drives
▪A single phase rectifier (fed from a 260 V AC
supply) feeds a 10-hp, 230 V, 1200 rpm, DC
motor with rated current, 𝐼𝑎 = 38 A, 𝑅𝑎 = 0.3 Ω,
and 𝑘𝑚 = 182 V per 1000 rpm.
▪Assume there is sufficient armature inductance
to give continuous armature current.
▪For 𝛼 = 30° and rated armature current,
calculate
▪a)speed, b) torque, c) supply power factor
97
 Solution
a) Speed

2∙ 2∙𝑉𝑠𝑢𝑝𝑝𝑙𝑦 ∙cos 𝛼 2 2×260×cos 30°

𝑉𝑑𝑐 = = = 202.72 V
π 𝜋

Eb = Va − Ia R a = 202 − 38 × 0.3 = 191.32 V

182
Eb = 𝑘𝑚 𝑁𝑟 ⇒ 191.32 = × 𝑁𝑟
1000
⇒ 𝑁𝑟 = 1051 rpm
98
 Solution
b) Torque
182 × 60
T𝑒 = 𝑘𝑚 𝐼𝑎 = × 38 = 66 𝑁𝑚
1000 × 2𝜋

c) supply power factor

pf = 0.9cos 𝛼 = 0.9 cos 30° = 0.779 lagging

99
 Applications of chopper drives
1) DC wind turbine

1 Quadrant chopper for a generator

100
 Applications of chopper drives
2) Electric Power Steering

▪ Variable speed DC drive

▪ Connected to the steering column via a gearbox
▪ Torque control mode
▪ Responds to a torque demand
▪ Complicated controller
▪ Torque sensor on the steering column to detect the driver’s steering effort

101
 Applications of chopper drives
▪What type of drive?
▪ 4Q chopper
▪ Need forward and backward operation
▪ DC supply (not AC)
▪What are the important specifications we
must consider when designing this?
▪ Must have a stable fast torque response
▪ Stability, no overshoot, no torque ripple (vibration in
the steering wheel)

102
 Applications of chopper drives
▪Safe and reliable
▪Must be small and light (transport
application)
▪Efficient
▪Must be acoustically silent (fs > 20 Hz)
▪Low cost.

103
 Topic 2
DC Motor Drive Systems
2d Control of DC Drives

104
 Control of DC Drives
▪Why do we need closed-loop control?
▪DC motor characteristics
▪S Domain Transfer Function
▪Open loop response
▪Nested (cascade) control structure
▪Current (torque) control loop
▪Speed control loop
105
 Simple Open Loop Control
▪Ideally we want
▪ Fast speed and torque response
▪ Accurate speed holding, stable control
▪What happens if we employ open-loop
control?

106
 Simple Open Loop Control
▪Easy to implement BUT
▪speed not automatically corrected as the
machine is loaded
▪no current control – what happens if you put
a step change on the reference?

107
 DC Machine Model
▪Assume the field is constant at the rated
value.
𝐿𝑎 𝑑𝐼𝑎
𝑉𝑎 = 𝐼𝑎 𝑅𝑎 + + 𝐸𝑏
𝑑𝑡

𝐸𝑏 = 𝑘𝑚 ∙ 𝜔𝑟

𝑇𝑒 = 𝑘𝑚 ∙ 𝐼𝑎
108
 DC Machine Model
▪Mechanical system equation
𝑑𝜔𝑟
𝑇𝑒 = 𝑇𝑙 + 𝐽 ∙ + 𝐵 ∙ 𝜔𝑟
𝑑𝑡
▪Laplace transforms

(𝑉𝑎 𝑠 − 𝐸𝑏 (𝑠))/(𝑅𝑎 + 𝑠 ∙ 𝐿𝑎 ) = 𝐼𝑎 (𝑠)

(𝑇𝑒 𝑠 − 𝑇𝑙 (𝑠))/(𝐵 + 𝑠 ∙ 𝐽) = 𝜔𝑟 (𝑠)

109
 Open Loop Response of DC Motor

▪ Open loop DC motor forms a second-order system (with

inherent feedback!)
▪ Relative values of 𝐽, 𝑅𝑎 , 𝐿𝑎 and 𝐵 determine the natural
frequency and damping
▪ A step change in 𝑉𝑎 may causse a very oscillatory response
with very high values of 𝐼𝑎 (not desirable!!)
110
 Typical closed loop control structure

▪ The inner loop (current control) is vitally important. The current

controller ensures the armature current equals the reference
current by adjusting 𝛼 (or 𝛿 for a chopper), to increase or decrease
the armature voltage.
▪ The reference current is the output of the speed controller. Can
clamp the output such that this reference value is limited to a safe
value, no matter how high the speed error is.
111
 Cascade (Nested Loop) Control
▪ Inner (fast) current control loop > 100 Hz
▪ Outer (slow) speed control loop < 10 Hz
▪ Design the current loop first, assuming operation at a constant
speed
▪ Okay since current transients are much faster than any subsequent changes
in speed
▪ Once the current loop is designed, look at speed control and
model the current loop as e.g., a first-order lag
▪ NOTES
▪ most drives employ this structure
▪ speed controller provides torque demand (to create acceleration or
deceleration torque)

112
 Current Controller

▪ Controller – normally proportional plus integral (PI)

𝑘 𝑐 𝑠 + 𝑎𝑐
𝐺𝑐 𝑠 =
𝑠
▪ Initially assume that the converter is just a simple gain 𝑘𝑝𝑐 (e.g.,
converting input reference voltage 0 – 15 V, to say 0 – 600 V
armature voltage gives a gain 𝑘𝑆 of 40)
▪ Set speed to 0 (i.e. 𝐸𝑏 = 0)
113
▪ The forward transfer function G(s) can be written as

𝑘 𝑐 𝑠 + 𝑎𝑐 1/𝐿𝑎
𝐺 𝑠 = ∙ 𝐾𝑝𝑐 ∙
𝑠 𝑠 + 𝑅𝑎 Τ𝐿𝑎
▪ The closed-loop transfer function for this system can be
written as:
𝐺 𝑠
T 𝑠 =
1 + 𝐺 𝑠 𝐻(𝑠)

▪ Where H(s) is the feedback transfer function.

H(s) = 𝐾𝑡 = gain of the feedback transducer

114
▪ Also note that the characteristic equation can be
written as

1+𝐺 𝑠 𝐻 𝑠 =0

Which can be rearranged to give

𝑠2 + 𝐴 ∙ 𝑠 + 𝐵 = 0
Where A and B are functions of known parameters
(𝑅𝑎 , 𝐿𝑎 ) and unknown parameters (𝑘𝑐 , 𝑎𝑐 )

115
▪ Pole Zero Cancellation

▪ To design the controller, the easiest approach is

pole zero cancellation i.e., set the value of the
controller zero to match the plant pole (𝑎𝑝 )
𝑎𝑐 = 𝑎𝑝 = 𝑅𝑎 /𝐿𝑎

𝐾𝐶 𝐾𝑝𝑐
𝑠𝐿𝑎 1
T 𝑠 = =
𝐾𝐶 𝐾𝑝𝑐 1 + 𝑠𝜏
1+
𝑠𝐿𝑎
116
▪ Pole Zero Cancellation

▪ Although this is perhaps the easiest design method,

it is very important to match 𝑎𝑐 to 𝑎𝑝 . Any
mismatch will cause the system response to
deteriorate and perhaps become unstable. Note
also that 𝑅𝑎 will change with load current and
temperature so the system performance may
change.

117
Characteristic Equation
▪ A better design principle is to use the characteristic
equation
𝑠 2 + 𝐴 ∙ 𝑠 + 𝐵 = 0 (previously seen)

▪ The design specification usually requires a system

dominated by a second-order pole pair, with
natural frequency 𝜔𝑜 , and damping factor 𝜉.
▪ Therefore the desired characteristic equation is
𝑠 2 + 2𝜉𝜔𝑜 𝑠 + 𝜔𝑜2 = 0
118
Characteristic Equation
▪Therefore, equating the coefficients of 𝑠1 and
𝑠 0 gives 𝑘𝑐 and 𝑎𝑐 .
▪𝜔𝑜 represents the bandwidth

Note that the presence of a closed loop zero

at −𝑎𝑐 will mean that the closed loop
response will not be exactly that defined by
𝜔𝑜 and 𝜉
119
 Root locus of the simple system

𝑎𝑐 = current controller zero

𝑎𝑆 = pole due to armature (𝑅𝑎 /𝐿𝑎 )
120
Power Converters
▪ Power converters cannot
always be considered purely
as a simple gain. They usually
have a delay associated with
them.
▪ Consider a DC-to-DC
converter (chopper) with a
duty cycle 𝛿 (i.e. 𝑡𝑜𝑛 = 𝛿T).
Assume the chopper is working
with a duty cycle d1 when the
controller commands that it
changes to d2.

121
Chopper
▪ If the controller demands a new duty cycle shortly after
the firing of the IGBT (case 1) there is a delay of almost T
before the chopper can react. In case 2, the delay is
much shorter.
▪ We approximate this delay as a first-order lag with time
constant T/2 – an average delay for a chopper with
switching period T.

i.e. 𝑉𝑎∗ 𝑘𝑝𝑐 𝑉𝑎 for a chopper.

𝑠𝑇
1+
2

122
Single Phase rectifier

Average delay = 5 ms
𝑉𝑎∗ 𝑘𝑝𝑐 𝑉𝑎
1 + 0.005𝑠

123
Three Phase rectifier

20
▪Max delay = ms (6 thyristors)
6
▪Min delay = 0 ms
▪ Average delay = 1.67 ms

𝑉𝑎∗ 𝑘𝑝𝑐 𝑉𝑎
1 + 0.00167𝑠

124
▪Inherently, a power converter has a delay
between control signal calculation and
actuation signal generation.
▪ We usually approximate this delay to a first-order lag
and set the time constant to the average delay.
▪ Note that a gain will still be associated with the
converter. Therefore, converter circuits can be
represented as follows:
𝐾𝑝𝑐
▪ Chopper: ;
1+𝑠𝑇 Τ2
𝐾𝑝𝑐
▪ Three-phase rectifier at 50 Hz:
1+0.00167𝑠

125
 Characteristic Equation of a more
complicated system
▪ This can be equated to a third-order system

𝑠 3 + 𝐴𝑠 2 + 𝐵𝑠 + 𝐶 = 0

▪ dominated by a second-order pole pair defined by

our design parameters (𝜔𝑜 and 𝜉) and a third, real
pole.

(𝑠 + 𝛼)(𝑠 2 + 2𝜉𝜔𝑜 𝑠 + 𝜔𝑜2 ) = 0

126
▪Therefore, we can equate coefficients
of 𝑠 , 𝑠 , 𝑠 to calculate 𝑘𝑐 , 𝑎𝑐 and 𝛼. 𝛼
2 1 0
should be set greater than 𝜔𝑜 .
▪Note again that the presence of a
closed loop zero will affect the closed
loop response (i.e. not ideal 𝜔𝑜 and 𝜉).
We usually have access to a CAD design package,
e.g. CODAS or MATLAB, and it can be easier to
design using root locus.

127
Root locus of system with a power converter

ac = current controller zero

ap = pole due to armature
(Ra/La)
apc = pole due to power
electronic converter

Make sure 𝜶 > 𝝎𝒐 so that the closed-loop

pole pair dominates the response

128
 Adding a feedback filter
▪ Adding a filter in the feedback path will reduce the noise
on the measured current,
▪ and may also filter out pulsations due to discontinuous
current.
▪ Filter bandwidth should be set according to these
criteria.
▪ The overall effect will be to slow down the system and
destabilize it
▪ Careful selection of controller gain and zero required.
▪ Note, that it may be advisable to use a more
complicated controller.
129
 Speed Controller

▪ The speed controller output must be related to the torque

▪ you must change torque to change speed.
▪ Make the speed controller output the armature current reference.
▪ This value must be limited so that the armature current does not
exceed the rated value for the power converter. This is usually
150% full load current for the motor.
▪ The current control loop can be approximated to a second, or first-
order system, or even just a simple gain if its bandwidth is very high.
130
 Speed Controller Design
▪ Start with a PI controller as before
▪ Incorporate the motor inertia and friction into the load
inertia and friction. Ignore load torque.
▪ The speed transducer can be a tacho-generator which
gives an output voltage proportional to speed.
▪ However, its commutator segments can lead to high-
frequency noise e.g. 20 segments can give 20 × 𝝎𝒐
noise.
▪ This is a fourth-order system so it is potentially unstable –
design carefully.

131
 Example on Control of DC Drives
A DC motor has 𝑅𝑎 = 0.25 Ω, 𝐿𝑎 = 5 𝑚𝐻, inertia - 𝐽 = 5 𝑘𝑔𝑚2 , friction
- 𝐵 = 0.05 𝑁𝑚/𝑟𝑎𝑑/𝑠, motor constant - 𝑘𝑚 = 2.5 𝑉/𝑟𝑎𝑑/𝑠

1) Design a PI controller using a Hall effect current transducer

which gives an output of 0.1 V/A. The PI controller output (𝑉𝑐 ) is
input to a rectifier that powers the armature with the voltage
V, where 𝑉/𝑉𝑐 = 50. The closed loop bandwidth should be 200
rad/s and the damping factor should be 0.7.
2) Stating any assumptions made, design a speed controller for
the drive to give a closed loop bandwidth of 10 rad/s with a
damping factor of 0.7. The speed transducer is a
tachogenerator giving 10 V at 1500 rpm.
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Current loop

Note that the switching frequency is much higher than the desired
bandwidth, so we can ignore the power converter delay.
Therefore, we can use the 2nd-order characteristic equation directly.

1
𝑘𝑐 𝑠 + 𝑎𝑐 𝑉𝑎 𝐿𝑎
1+𝐺 𝑠 𝐻 𝑠 =0⇒1+ ∙ ∙ ∙ 𝐾𝑡𝑟𝑎𝑛𝑠 = 0
𝑠 𝑅
𝑉𝑐 𝑠 + 𝑎
𝐿𝑎
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 𝑘𝑐 𝑠 + 𝑎𝑐 200
1+ ∙ 50 ∙ ∙ 0.1 = 0
𝑠 𝑠 + 50

𝑠 2 + 50𝑠 + 1000𝑘𝑐 𝑠 + 1000𝑘𝑐 𝑎𝑐 = 0

Also
𝑠 2 + 2𝜉𝜔𝑜 𝑠 + 𝜔𝑜2 = 0

Comparing coefficients:

𝑠1
50 + 1000𝑘𝑐 = 2𝜉𝜔𝑜 = 2 × 0.7 × 200 = 280

⇒ 𝑘𝑐 = 0.23
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𝑠0
1000𝑘𝑐 𝑎𝑐 = 𝜔𝑜2 = 2002 = 40,000

⇒ 𝑎𝑐 = 174 rad/s
Speed loop

Assume the current loop response is “ideal” as it is 20 times faster than

the desired speed loop bandwidth.

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The current loop is 20 times faster than the desired speed controller
bandwidth, therefore it can be assumed ideal.

2𝜋 𝑉
𝐾𝑡𝑎𝑐ℎ𝑜 = 10 𝑉 @ 1500 𝑟𝑝𝑚 = 1500 × = 0.0637
60 𝑟𝑎𝑑/𝑠

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Characteristic equation

𝑘𝑐 𝑠 + 𝑎𝑐 𝐾𝑚
1+𝐺 𝑠 𝐻 𝑠 =0⇒1+ ∙ ∙ 𝐾𝑡𝑎𝑐ℎ𝑜 = 0
𝑠 𝐽𝑠 + 𝐵

𝑘𝑐 𝑠 + 𝑎𝑐 2.5
⇒1+ ∙ ∙ 0.0637 = 0
𝑠 5𝑠 + 0.05

𝑘𝑐 𝑠 + 𝑎𝑐 0.5
⇒1+ ∙ ∙ 0.0637 = 0
𝑠 𝑠 + 0.001

⇒ 𝑠 2 +(0.001 + 0.03185𝑘𝑐 )𝑠 + 0.03185𝑘𝑐 𝑎𝑐 = 0

Comparing with
𝑠 2 + 2𝜉𝜔𝑜 𝑠 + 𝜔𝑜2 = 0, we have 𝑘𝑐 = 439.26, 𝑎𝑐 = 7.147
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End of Unit 2

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THANK YOU

ANY QUESTIONS?