UNIVERSITY OF EDINBURGH

School of Engineering

Induction Machines

Dr Quan Li
Faraday Building 3.092
Quan.Li@ed.ac.uk

1
Power Systems, Power Electronics and Machines 3
Induction Motor

Induction motors are used in a wide range of applications

• Industrial
• Medical
• Transportation
• Integrated applications
• Communications
• Household applications

2
Induction Motors In Daily Life

3
 Induction Machine
• 90% of all installed industrial motors
 Structure

1. Shaft; 2. Bearing; 3. Rotor; 4. Stator

5
 Michael Faraday - 1831
• Electromagnetic Induction
– Faraday’s Law:
dφtotal
E ∝
dt
 Magnetic Flux

Current
flowing out

Magnetic field due to a single coil

Current
flowing in
 Electromagnetic Force

• A current carrying conductor I, of length L, in a

magnetic field of strength B, experiences a force
F.

φ 2
𝑭𝑭 = 𝑩𝑩 � 𝑰𝑰 �L B= (Wb/m )(T )
A
 Fleming’s Left Hand Rule
• Field: First Finger
• Current: Second Finger
• Motion: Thumb
• All at right angles

Magnet Field, B Magnet Field, B

How Electric Machines Work?

10
Stator Rotating Magnetic Field
Current into a
x

•
Current out of a

11
Current out of b Current out of c

• •

x x

Current into b Current into c

12
13
 Rotor
“squirrel cage rotor”

14
 Stator field N rev/min
From Faraday’s Law, the induced
emf across that single conductor
will be:
dφtotal
E∝
dt
𝑭𝑭 = 𝑩𝑩 � 𝒊𝒊 � 𝒍𝒍

Stator field N rev/min 3 phase supply If the rotor is disconnected from any
mechanical load, it will revolve at
nearly synchronous speed
N rev/min

Rotor Nr rev/min When mechanical load is applied,

rotor slows down to
Nr
Flux linkage (N - Nr) rev/min induces rotor 15
emf, current and torque 3 phase supply
Synchronous Speed, Shaft Speed and Slip
2πf 2πf 60 60f
ω= rad / s ⇒ N= × = rev / min
p p 2π p

Note: p is pole-pairs

To produce torque, the rotor must revolve more slowly than

the synchronous speed of the magnetic field.

N − NR
s= and N R = N(1 − s) rev/min
N
At any speed NR, with slip s, the induced frequency on the
rotor may be expressed
fR = s.fsupply at standstill fR = s.fsupply at speed N fR = 0
16
Torque and Power

17
Torque-Speed Variation
The rotor accelerates towards N and reaches no-load
speed at around 1% slip, where E, I and T are just
sufficient to overcome the no-load losses incurred by
friction and windage on the rotor.

18
 Further application of mechanical load causes the rotor to
slow down to NR. Slip s increases, and T rises to balance
load torque. Full Load Torque (FLT):
FLT at ≈ 90% N Pull-out T at ≈ 80%N ≈ 200-300%FLT

stable operation

19
Operation near the pull-out torque (POT) is dangerously unstable.
Added load causes rotor to slow down and T to fall.
Torque continues to fall further as speed reduces.
Speed falls further and torque decreases again.

unstable operation

20
 Power Flow and Losses

Active power delivered by supply Pin = 3.Vline.Ilinecosφ

21
Stator Rotor
COPPER LOSS, Psl COPPER LOSSES, Pc = s.Pr

IRON LOSS, Pf FRICTION AND WINDAGE LOSSES, Pw

The AIRGAP POWER is established Pr = Pin - (Psl + Pf) 22

ROTOR COPPER LOSSES

Pc ∝ E - which is directly proportional to slip.

At exactly synchronous speed (s=0),
E and I = 0
Pc = 0
At standstill (s = 1)
no power is transmitted to the load, and all of Pr is dissipated
as rotor copper losses
Pc = Pr

Rotor losses vary linearly with slip as: Pc = s.Pr

23
GROSS MECHANICAL POWER
(Pl + Pw) = Pr - Pc = Pr - sPr = Pr (1-s )
SHAFT OUTPUT POWER
Pl = Pr (1-s ) - Pw
AIRGAP POWER
( Pl + Pw )
Pr =
(1 − s )
INPUT POWER
( Pl + Pw )
Pin = Pr + Psl + Pf = + Psl + Pf
(1 − s ) 24
Efficiency

Shaft output power

η=
Electrical input power
Pload (mechanical)
=
Pin (electrical )
Pin - losses
=
Pin
Pin - ( Pf + Psl + Pc + Pw )
= 25
Pin
Example:

Taking the input power as a nominal 100% and the efficiency of

a typical well designed motor as 94%, the losses in the machine
might comprise, in percentage terms.

Input power 100%

less Stator iron losses -1.0%
Stator copper losses -1.5%
Rotor copper losses -2.5%
Friction and windage -1.0%
Leaving
Output power 94%
26
How to calculate and analyze power flow:
Equivalent Circuit

27
Motor Rotor
Consider the machine as a motor initially. All circuit
diagrams represent one phase of the machine, and all
parameters are phase values unless otherwise stated.

R2 represents the rotor winding resistance

X2 is the standstill reactance of the rotor winding 28
 2 2
Z 2 = R 2 + j(sX 2 ) = R + (sX 2 ) ∠ + φ 2
2

The rotor current I2 that flows at any value of slip may be

calculated from:

sE oc
I2 = ∠ − φ2
R 22 + (sX 2 ) 2 29
 Change E2 = Eoc

The rotor current I2 that now flows may be calculated

E oc
I2 = ∠ − φ2
2
 R2  2
  + X2
 s  30
Remember from previous lecture:

Pc = s.Pr (PL + Pw) = (1-s ) Pr PC s

=
(PL + Pw) : Pc : Pr = (1-s) : s : 1 PL + PW (1 − s )

To represent copper loss and gross mechanical power

R2/s = R2 + R2(1-s)/s
31
Loss in R2 ⇔ rotor copper losses Pc

Loss in R2(1-s)/s ⇔ GMP Pl+Pw

Total loss ⇔ 2 R2
Pr = 3I
2
s
Pc = 3I 22 R 2

2 (1 − s)
Pl + Pw = 3I R 2
2
PC s s
=
PL + PW (1 − s )

32
Referred Rotor Values
Consider constant speed rotation:

The same flux links the stator and the rotor

 the same rate of change of flux dφ/dt in each winding
dφ
 E=N induces emf either side of airgap in
dt
stator/rotor

dφ dφ
E2 = N2 E 1 = N1
dt dt
33
 dφ E 1 E 2 E 1 N1
= = ⇒ =
dt N1 N 2 E2 N2
E2 in rotor ⇔ secondary emf
E1 in stator ⇔ primary emf in airgap

What is the slip dependence of E2 ? 34

 E1 N1 I1 N 2
= =
E2 N 2 I 2 N1

N1
E2 2
E1 N2  N1  2
Z1 = = = Z 2   = Z 2 ⋅ N eff
I1 N2  N 2
I2
N1

R '2 = N eff
2
⋅R2 2
X '2 = N eff ⋅ X2

35
Equivalent Circuit

Rc represents this as a core-loss resistor

Xm represents the magnetising reactance of the machine

R '2 = N eff
2
⋅R2 X '2 = N eff
2
⋅ X2
36
Approximate Equivalent Circuit

37
Approximate Single-phase Equivalent Circuit

Using this circuit, the performance of the machine can be

modelled at any value of slip 38
 R '2 (X + X '
2)
Z t = (R 1 + ) + j(X1 + X 2 ) ϕ t = tan
' −1 1
s R '2
(R 1 + )
s
R '2 2 V1
Z t = (R 1 + ) + (X1 + X '2 ) 2 ∠ϕ t I = ∠ −ϕt
'
2
s Zt

And phase current I1 = I0 + I2’

R c ⋅ jX m
Z0 =
R c + jX m
V1
and I 0 = ∠ − φ0
Z0
39
Can also (more easily…) determine input current from
phase impedances
Z0 Z t Z0∠ + φ0 ⋅ Z t ∠ + φ t
Zin = =
Z0 + Z t (Z0 cos φ0 + Z t cos φ t ) + j(Z0 sin φ0 + Z t sin φ t )
Z 0 ⋅ Z t ∠ + (φ 0 + φ t )
=
(Z0 cos φ0 + Z t cos φ t )2 + (Z0 sin φ0 + Z t sin φ t )2 ∠ + φc
Z0 sin φ0 + Z t sin φ t
φc = tan −1
Z0 cos φ0 + Z t cos φ t

Zc = (Z0 cos φ0 + Z t cos φ t )2 + (Z0 sin φ0 + Z t sin φ t )2

Z0 Z t
Zin = ∠(ϕ0 + φ t − φc ) = Zin ∠φin
Zc
V1
I phase = I1 = ∠ − φin and power factor = cos(−φin )
Zin 40
The voltage-current phasor diagram becomes

41
Calculation of Performance

active phase power Pphase = V1I1cos(−φin )

Total 3-phase power Pin = 3Vline I line cos(−φin )
Reactive input power Q in = 3Vline I linesin( −φin )

Overall efficiency:
Pload ' 2 ' (1 − s)
ζ =η= , where Pload = 3I 2 R 2 − PW
Pin s

42
Performance as a Motor

(1 − s)
'2
Gross mechanical power
'
(PL + PW ) = 3 ⋅ I R 2 2
s
2πN R (TL + TW )
(PL + PW ) = , ( from P = ωT )
60
60 60(PL + PW )
(TL + TW ) = (PL + PW ) =
2πN R 2πN(1 − s)
2
60 ⋅ 3 ⋅ I '2 R '2 (1 − s) 180 R '2 '2
(TL + TW ) = = I2
2πN(1 − s) s 2πN s 43
 '2
60 ⋅ 3 ⋅ I R '2 (1 − s) 180 R '2 '2
(TL + TW ) = 2
= I2
2πN(1 − s) s 2πN s
The referred rotor current was derived earlier as
'
V R
I '2 = 1 ∠ − ϕ t where Z t = (R 1 + 2 ) 2 + (X1 + X '2 ) 2 ∠ϕ t
Zt s

180 R '2 V12

(TL + TW ) =
2πN s R '2 2
(R 1 + ) + (X1 + X '2 ) 2
s
gives an expression for airgap torque in terms of
machine constants and slip.

180V12 sR '2
(TL + TW ) =
2π N (sR 1 + R '2 ) 2 + s 2 (X1 + X '2 ) 2
44
Torque-slip Curve

45
Maximum torques
180V12 1 1
(TL + TW ) = ±
2π N 2 R ± R 2 + (X + X ' ) 2
1 1 1 2

± R '2
when s=
2
R 1 + (X1 + X '2 ) 2

In practice
180V12 1
(TL + TW ) max =±
2πN 2X '2

R '2
when s=± '
X2 46
Performance as a Motor
Using the foregoing information, the performance of the
machine may now be modelled to determine and plot
the variation against slip of:
• Gross mechanical power - PL+PW
• Airgap torque - TL+TW
• Referred rotor current - I2'
• Line current - I1
• Power factor - cosφin
• Active input power - Pin
• Reactive input power - Qin
• Efficiency - η 47
Example:

A single-phase equivalent circuit is shown below. Calculate the line

current and active power drawn from the supply and the power factor,
when the variable resistor is 1.0 Ω. Draw the voltage and current
phasor diagram showing all components of current.

48
UNIVERSITY OF EDINBURGH
School of Engineering

Induction Machines

Dr Quan Li
Faraday Building 3.092
Quan.Li@ed.ac.uk

49
Power Systems, Power Electronics and Machines 3
UNIVERSITY OF EDINBURGH
School of Engineering

Induction Machines

Dr Quan Li
Faraday Building 3.092
Quan.Li@ed.ac.uk

1
Power Systems, Power Electronics and Machines 3
Induction Machines:
Motors or Generators

2
Operation in the Motor/Generator Modes

3
Motor

Slip s = (N-NR)/N and N > NR

Rotor rotates at N(1-s) rev/min

How about rotor magnetic field?

4
 Example
A six-pole machine on a 50Hz supply operating at 5% slip
would have the rotor turning at 950 rev/min.

Stator field cuts rotor conductors at 50 rev/min.

Induced frequency fr on the rotor = s.fsupply = 2.5Hz.

Relative to rotor conductors, rotor flux pattern revolves at

60f r
= 50 rev/min
p
Rotor revolves at 950 rev/min and induced magnetic flux on
rotor revolves at 950+50 = 1000 rev/min relative to the stator –
i.e., at the synchronous speed
5
Electricity supply:
Pf , Pc , Psl, Pw
P = 3.Vline.Ilinecosφ
Which comprises:
stator iron losses Pf ,
rotor copper losses Pc ,
stator copper losses Psl,
friction and windage losses Pw
Leaving mechanical load: PL

Network has to magnetise machine:

Q = 3.Vline.Ilinesinφ
6
No Load Pf , P w

At the synchronous speed:

• engine supplies friction & windage losses
• electricity supply provides no load losses (iron
losses only, zero copper losses) 7
Generator

Slip s = (N-NR)/N and N < NR

Rotor rotates at N(1-s) rev/min

How about rotor magnetic field? 8

 Example
A six-pole machine on a 50Hz supply being driven at 105% of
synchronous speed would operate at -5% slip and would have
the rotor turning at 1050 rev/min.

Stator field cuts rotor conductors at -50 rev/min.

Induced frequency fr on the rotor = s.fsupply = -2.5Hz.

Relative to rotor conductors, rotor flux pattern revolves at

60f r
= −50 rev/min
p
Rotor revolves at 1050 rev/min and induced magnetic flux on
rotor revolves at 1050-50 = 1000 rev/min relative to the stator –
i.e., at the synchronous speed.
9
Engine provides (mech PL):
Pf , Pc , Psl, Pw
P = 3.Vline.Ilinecosφ
Which comprises:
stator iron losses Pf ,
rotor copper losses Pc ,
stator copper losses Psl,
friction and windage losses Pw
Leaving Pout transferred to the supply
network.
Network still has to magnetise machine:
Q = 3.Vline.Ilinesinφ 10
Performance as a Motor

(1 − s)
'2
Gross mechanical power
'
(PL + PW ) = 3 ⋅ I R 2 2
s
2πN R (TL + TW )
(PL + PW ) = , ( from P = ωT )
60
60 60(PL + PW )
(TL + TW ) = (PL + PW ) =
2πN R 2πN(1 − s)
2
60 ⋅ 3 ⋅ I '2 R '2 (1 − s) 180 R '2 '2
(TL + TW ) = = I2
2πN(1 − s) s 2πN s 11
 '2
60 ⋅ 3 ⋅ I R '2 (1 − s) 180 R '2 '2
(TL + TW ) = 2
= I2
2πN(1 − s) s 2πN s
The referred rotor current was derived earlier as
'
V R
I '2 = 1 ∠ − ϕ t where Z t = (R 1 + 2 ) 2 + (X1 + X '2 ) 2 ∠ϕ t
Zt s

180 R '2 V12

(TL + TW ) =
2πN s R '2 2
(R 1 + ) + (X1 + X '2 ) 2
s
gives an expression for airgap torque in terms of
machine constants and slip.

180V12 sR '2
(TL + TW ) =
2π N (sR 1 + R '2 ) 2 + s 2 (X1 + X '2 ) 2
12
13
Performance as a Generator

180V12 - sR '2
(TL + TW ) =
2π N (-sR1 + R '2 ) 2 + s 2 (X1 + X '2 ) 2
' V1
I =
2
R '2 2
(R 1 − ) + (X1 + X '2 ) 2 t
s
(1 + s)
'2 '
(PL + PW ) = - 3 ⋅ I R2 2 14
s
 Induction Machines:
Determine Impedance from Test Results

15
 No-Load Test
Measure:
• friction and windage losses;
provide information about:
• magnetisation current, and
iron losses

16
 Wattmeters
I1(V1 - V3)

I2(V2 - V3)

P = I1(V1 - V3) + I2(V2 - V3)

17
Operating with very small slip:
• R2'.(1-s)/s very high (tends towards open circuit);
• I2' reduces to almost zero;
• Equivalent circuit becomes magnetising impedance.

18
Ptotal = W1 + W2 = 3.Vline.Iline.cosφin

W1 + W2
Power Factor =
3Vline I line

Phase angle φin

Vphase Vphase
Rc = and X m =
I line cosφ0 jI linesinφ0

19
DC Test: R1

DC power supply is connected to

two of the three line terminals

Vdc
R1 =
2 I dc

20
Locked Rotor Test

Slip
s=1
GMP resistor
R2'.(1-s)/s = 0

GMP = 0
21
 Ptotal = W1 + W2 = 3.Vline.Iline.cosφlr

W1 + W2
Power Factor =
3Vline I line

Phase angle φlr

(impedance angle)

Vline V
Zlr = R lr + jX lr = cosφ lr + j line sinφ lr
3I line 3I line

Vline Vline
'
'
R = cosφ lr − R 1 X1 + X = 2 sinφ lr
2
3I line 3I line 22
The equivalent circuit component values determined as above
may be substituted into the expressions for
• torque
• current
• power factor
• efficiency and
• power
and the variation of each of these parameters evaluated as slip
varies with the load or output power

23
 Induction Machines:
Operation on the Electricity Supply
Network

24
Connection & Starting Induction Motors
Infinite grid

• s = 1, R2'(1-s)/s = 0 and GMP resistor is shorted out

• Z2 reduces from
(20.R2' + j.X2') when s = 0.05 (full load)
to (R2' + j.X2') at s = 1 (starting)
25
 Istart =
V1
∠ tan −1 (X 1 + X 2)
'

(R 1 +R 2) + (X
' 2
1 +X )
' 2
2
(R 1 + R2 )
'

Since V1 has not fallen and starting torque α V12

• τstart will be maintained, motor will accelerate to speed
• NR ⇑ slip ⇓ R2'/s ⇑
• Istart ⇓ to Ifull load to complete start.
26
 When grid cannot be considered as infinity, all transmission lines,
transformers, and cables add their series impedance into Zgrid ⇑

For any value of Istart,

• series voltage drop IstartZgrid ⇑ and V1start ⇓
• τstart ⇓ (α V12), and motor may not accelerate up to rated speed
• may stall, drawing Istart and rapidly burn out
27
Vfull load must always ≥ 0.95pu, or Ifull load Zgrid < 0.05pu

Starting an induction motor on a weak grid:

Istart ≈ 6 Ifull load

IstartZgrid ≈ 6 Ifull load Zgrid ≈ 0.3pu

Vstart = Vgrid - IstartZgrid ≈ 0.7pu

τstart = 0.49 τfull load

28
 Voltage Sag due to Motor Starting

240
Voltage [V]

220

5 6 7
Time [Seconds]
29
Reactive Potential Division
At the instant of switch-on, the magnetising reactance Xm has
to be charged up to produce the rotating magnetic field

Xstart may be estimated from: X start =

(X
1 + X '
)
2 ⋅ Xm
∠ + 90°
'
X1 + X 2 + X m

X start X start
Vstart = Vgrid = 1.0 ⋅
X start + X grid X start + X grid

30
When only nameplate details of motor are available, how to
ESTIMATE the voltage drop at start?
SC SC
MVA start = 3Vline I line = MVA motor
FLC FLC

2 2
kVline kVline Z motor
Zstart = = =
MVA start SC SC
MVA motor
FLC FLC
1.0 FLC FLC MVA grid
Zstart (pu) = = Zstart (pu) = ⋅
SC SC SC MVA motor
FLC to the base MVA of the grid

Zstart 1 1
Vstart (pu) = Vgrid = 1.0 =
Zstart + Zgrid 1.0 SC MVA motor
1+ 1+ ⋅
Zstart FLC MVA grid 31
Control of Max Torque

32
Connection & Starting Induction Motors

X X X Main
circuit
breaker Slip Starting
rings resistors

Stator Rotor Short circuiting

X X X switch

180V12 1 R '2
(TL + TW ) max =± s=± '
2πN 2X '2 X2 33
Power Factor Correction
As induction motor must always draw its magnetising current from
the ac supply, there will always be a fixed amount of lagging VArs
taken from the supply to produce the rotating magnetic field.

For loaded machine:

PLOAD
φ1 = cos −1

ξ 3Vline I line

Qin
φ1 = tan −1
Pin
34
Starting & Connection of Induction
Generators
Grid is strong:
The machine may be started as a motor. When the machine reaches
no load speed (just below the synchronous), the main valve on the
turbine is opened. As the valve is opened further, the machine is
pushed through the synchronous speed and starts to export real
power at a leading power factor.

35
Grid is weak:
The main valve of water turbine would be opened to the 'no-load'
position and the shaft system accelerates up to synchronous speed.
Then connect it to the grid. The machine magnetises and develops a
small amount of real power.
Open the main valve further and the increased power developed by
the turbine causes the speed to increase further beyond synchronous,
exporting real power.

36
Induction Generators – Frequency & Voltage Control
Connected to the grid:
Frequency and voltage are determined by the grid, and the
generated power is determined by the engine input power.

Not connected to the grid:

This need some other control system to create a rotating field
pattern in the stator to allow induction of power across airgap.

37
UNIVERSITY OF EDINBURGH
School of Engineering

Induction Machines

Dr Quan Li
Faraday Building 3.092
Quan.Li@ed.ac.uk

38
Power Systems, Power Electronics and Machines 3
1
Recommended text book. Very good for power electronics fundamentals.

2
This picture shows the mini power electronic devices and circuits in a small IC
chip.

3
An example of large power electronics system of power conversion.

4
5
6
7
Many applications. Over 60% of all power generated in the UK is processed by
Power Electronic circuits (figure steadily rising).

8
9
10
Ensures unbroken supply for critical loads in the event of mains failure:

Mains healthy: Rectifier supplies inverter and recharges battery

Mains failure: Battery supplies power to inverter

11
12
Rectifiers and inverter are used to convert between ac and dc at each end of the
link.

13
Whether the power source is the mains 230 V 50Hz supply, or a battery, or a
solar photovoltaic cell, power electronic circuits are needed to convert and
stabilise the voltage suitable for the load.

Loads are ALL electronic equipment, eg. computers, MP3 players, mobile
phones, televisions, printers, low energy lighting, stereo equipment.

14
15
An old device (1960’s), this has been the workhorse for power electronic
equipment. Now only really used in very high power applications, as newer
devices are better at lower powers.

16
Reverse bias:

Like a diode, it will not conduct if reverse biassed (apart from the reverse leakage
current – microamps).

If the reverse breakdown voltage is exceeded, it will die!

17
Before the gate pulse is applied, it will not conduct in the forward direction
(except forward leakage current – microamps).

If the forward breakover voltage is exceeded, the device will “switch on” (this
does not necessarily destroy the device).

18
A short gate pulse (typically 3 volt, 20 μs, 50 mA to 1A, depending on the device
rating) will reduce the forward breakover voltage to 2-3 volts, thus switching the
thyristor on.

It will only “latch-on” – ie. remain on when the gate pulse is removed - if the
current through it is greater than the “latching current”.

Thereafter, it will stay on as long as the current is greater than the “holding
current” (less than the latching current).

19
A short gate pulse (typically 3 volt, 20 μs, 50 mA to 1A, depending on the device
rating) will reduce the forward breakover voltage to 2-3 volts, thus switching the
thyristor on.

It will only “latch-on” – ie. remain on when the gate pulse is removed - if the
current through it is greater than the “latching current”.

Thereafter, it will stay on as long as the current is greater than the “holding
current” (less than the latching current).

20
The difficulty in switching it off is the main reason why the thyristor is not used so
much nowadays.

21
The Power MOSFET can be easily turned off by removing the gate voltage.
Major attractions are that the gate current is extremely low, and that it can be
switched at high speed. However, it is only available at relatively low power
ratings.

Metal-oxide-semiconductor field-effect transistor

22
The GTO is like an ordinary thyristor, except that it can be turned of by applying a
short negative pulse to the gate. However, this negative pulse needs to be large
– typically ¼ of the main current going through the device.

Not suitable for high frequency applications.

23
A newer device, rapidly taking over from the GTO.

24
25
26
Converts input dc voltage (eg battery) to output dc voltage.
Usually, but not always, Vin > Vout

27
28
The average voltage is such that the yellow area = the grey area.

29
D is the proportion of the cycle that the switch is on.

30
D controls the output voltage

31
32
If the load is inductive, a “freewheeling diode” is required to provide a path for the
current when the switch is opened, otherwise very high voltage spikes will occur
which may destroy the switch, due to:

di
v  L.
dt

33
Here, the current falls to zero during the switch “off” time.

34
Here, the inductance is large enough to maintain the current > 0 during the switch
“off” time.

35
Most real loads have:
• Series resistance and/or inductance
• A constant voltage element (battery/capacitor/EMF source)
• A “useful load” element that absorbs active (real) power

36
Vc = E = dc motor back EMF
RL = open circuit
Rs = armature resistance

37
Vc = E = battery voltage
RL = load on battery (if connected)
Rs = smoothing inductor resistance ≈ 0

38
Vc = voltage on smoothing capacitor
RL = load on power supply
Rs = smoothing inductor resistance ≈ 0

39
40
Note that when the current has fallen to zero, there is no volt drop across Rs and
Ls, therefore v2 = VC.

Thus the average voltage is slightly higher than the previous case, due to the
extra (shaded) area.

The equation Vo = D.Vi does not apply in this case.

41
If Rs is negligible (which is often the case), then the current varies linearly with
time, making the analysis much easier (no exponentials!).

The slope can be calculated for each segment from:

di
vL  L.
dt
Therefore:

di
v2  VC  L.
dt
Where (v2-VC) = voltage across the inductor.
If Rs is ignored:

VC
iaverage 
RL
v2 ( average )  VC

42
For any inductor operating in steady state conditions:

di
vL  L.
dt
Therefore, integrating over one cycle:

 vL .dt   L.di  L.i  L.I 2  I1 

I2

I1
Where I1 = current at start of the cycle
I2 = current at end of the cycle

But under steady state conditions, I1 = I2

Therefore, over one complete cycle, under steady state conditions,

 v .dt  0
L

Therefore:
Average voltage across an inductor over a complete cycle, under steady state
conditions, is zero.

43
Converts input dc voltage (eg battery) to output dc voltage.
Usually, but not always, Vin > Vout

44
In this case the average voltage across the inductor = 0, and the current is in
steady state (current at the end of the cycle = current at the start of the cycle).

45
In the figure above the average voltage is clearly not zero, resulting in the current
steadily rising. Thus the circuit is not in steady state.

46
In the step-up chopper, when switch S is “on” current iL builds up in the inductor,
round loop E, L, S and back to E. No current flows through the diode during this
period. As S is “on”, v2 = 0 for this period.

When S is switched “off”, the only path for the inductor current iL is through the
diode to Vdc. (It is assumed that capacitor C will hold Vdc approximately
constant). As the diode is conducting, v2 = Vdc for this period, giving the
waveform for v2 shown on the next slide.

47
48
For the Step-Up Chopper, if VL(average) = 0, then E is the average of v2.

toff
E  Vdc .
ton  toff

 Vdc .
1  D .T
T
Where T is the switching period.
Hence:
Vdc 1

E 1 D

As D (switch duty ratio) must be between 0 and 1, Vdc > E

ie. This is a step-up chopper.

49
The output voltage is always greater than the input voltage.
However, the output current is less than the input current, so we aren’t getting
energy from nowhere – the laws of physics are not broke.

This is similar to a step-up transformer.

50
The circuit above combines the step-up and step-down choppers, such that iL can
be either positive or negative. If iL is positive, power (=Vdc.iL) is positive, ie.
power is from the motor to Vdc. This is regeneration and, if used with a dc motor,
can be used for regenerative braking, where the motor is slowed down by feeding
the kinetic energy of rotation back to the supply, recovering this energy.

An application of this is in battery powered electric vehicles, where braking is

achieved by feeding the vehicle kinetic energy back to the battery, thus re-
charging the battery and extending the period between charging.

51
During normal motor operation S1 is switched on and off (step-down chopper
mode) with D1 providing a path for the inductor current when S1 is switched off.
During this mode S2 is permanently off.

52
For regenerative braking, S1 is permanently off, and S2 is switched on and off
(step-up chopper mode). When S2 is on, current iL builds up from source E,
through R, L, S2 and back to E. When S2 is switched off, the only path for the
current through the inductor to flow is through diode D2 back to the supply, ie.
feeding energy back to the supply.

For this circuit to work, the supply has to be capable of receiving energy (eg there
must not be a diode blocking reverse current).

53
1
Converts ac to dc.

2
3
The single phase diode bridge is the most common circuit for converting ac to dc.
Diodes D1 and D4 conduct in the positive ½ cycle of the input voltage, and diodes
D2 and D3 in the negative ½ cycle. The output is a set of ½ sine waves.

4
An alternative to the Bridge Rectifier is the Centre-Tapped Transformer Rectifier

5
During the positive ½ cycle, current flows into the dot on the transformer primary
(remember the transformer dot notation), therefore it must flow out of the dot on
the secondary.

Thus current flows out of the dot of the top half, through D1, down through the
load and back to the centre tap on the transformer. Current cannot flow out of
the dot on the lower half, as current would then be going backwards through
diode D2, therefore no current flows in the lower half of the transformer.

6
During the negative ½ cycle, current flows out of the dot on the transformer
primary therefore it must flow into the dot on the secondary.

Current cannot flow into the dot on the upper half, as current would then be going
backwards through diode D1. However, current can flow into the dot on the lower
half of the transformer secondary, forwards through D2, up to the positive rail and
down through the load and back to the centre tap.

Note that current is flowing down through the load in both half cycles, ie.
rectification has taken place.

7
The output waveform is exactly the same as that of the diode bridge rectifier.

8
A transformer is required for this circuit, which can add to the size and weight
considerably. However, a transformer may be required for other reasons (eg. for
isolation, or to step-down the voltage), and the circuit uses only 2 diodes
compared to 4 in the bridge rectifier.

9
The inclusion of the switch and the centre-tapped capacitor network allows the
power supply to operate either form a 230 volt input (eg in Europe) or 110 volt
input (eg in the USA).

10
11
If a 110 (or 115) volt input is used, then the switch should be moved to position B.
In this case the upper capacitor will be charged during the positive half cycle, and
the lower capacitor during the negative half cycle.

12
The total output voltage is approximately the same as for a 230 volt supply with
the switch in position A, therefore any loading on the power supply can be
identical.

13
14
15
16
The output voltage vo, the charging current idc and the ac current iac are shown
above. It can be seen that idc is a very narrow current spike. If a smoother dc
supply is required, the capacitor C must be made larger. However, this also has
the effect of making the current spike narrow, and hence taller (to pass the same
energy.

17
The size of the capacitor depends on the level of smoothing required. The
capacitor discharges from Vmax to Vmin through the load in approximately half an
ac cycle, ie in time 1/(2f) where f is the supply frequency.

Power 
Vmax  Vmin   I
load
2
In a capacitor, Q = CV, therefore as the capacitor discharges:

Q  C.V
 C.Vmax Vmin 
Iload

2f
(charge = current x time)
Combining the above:

power 1
C 
Vmax Vmin  Vmax  Vmin  f
power 1
 
V  Vmin f
2 2
max 

18
New EC regulations have been introduced which severely limit the harmonic
current equipment sold in the EU is allowed to draw from the mains supply.

This necessitates either large, expensive passive filters (L-C), or active filtering,
which is far more complex.

19
Electromagnetic interference

20
Converts ac to dc.

21
22
The single diode ½-wave rectifier blocks the negative ½ cycle.

23
24
The average dc voltage is such that the shaded area = the hatched area (ie. the
area under the curve over one cycle = area under Vdc(av) over one cycle).

Vdc  2  area under curve

1   2
Vdc   0 VM .sint.d t    0.d t 
2  
V
 M   cost t 0  0


2
V
 M  1 1
2
V
Vdc  M


25
The diode can be replaced by a thyristor, which will block (even in the forward
direction) until it is switched on at the appropriate time by the control circuit
sending a pulse to the thyristor gate.

The delay angle α is the angle between when a diode would start to conduct and
when the thyristor is fired.

Thus a diode is similar to a thyristor with α = 0.

26
If the load is resistive, V = IR, therefore the current waveshape is the same as
the voltage waveshape.

27
As with the single diode, the average dc voltage is such that. the area under the
curve over one cycle = area under Vdc(average) over one cycle).

28
A diode is similar to a thyristor with α = 0: same average voltage.

29
Voltage across inductor: di
vdc  vL  L 
dt
Supply voltage: vac = VM.sin (ωt)
When the diode is conducting, vac = vdc
Therefore:

di VM
  sint 
dt L
Integrating:
VM
i  cost   C
L

At t = 0, i = 0, therefore:
VM
C
L
Thus:
VM
it    1 cost 
L
Thus i never falls below zero, so the DIODE IS NEVER BLOCKING!!
Therefore the inductor voltage vdc = VM.sin (ωt)

30
31
 di
vdc  vL  L 
dt
Thus when v is positive I increases, and when v is negative I decreases

32
Real loads always include some resistance.

33
An R-L load is between a pure inductive load (top waveform) and a pure resistive
load (bottom waveform)

34
Inductive load:
i peaks at π
i falls to zero at (2 π-α)
Resistive load:
i peaks at (π/2)
i falls to zero at π
Resistive+Inductive load:
i peaks between (π/2) and π
i falls to zero between π and (2 π-α)

Exactly where i peaks and i falls to zero depends on whether the load is mainly
resistive or mainly inductive (ie. on the L/R time constant).

Don’t try and calculate this – it’s a messy calculation!

35
36
We’ve looked at this single phase diode bridge earlier in this module.

37
This is the same circuit (just drawn with vertical/horizontal lines rather than
diamond-shaped), but with the diodes replaced by thyristors.

Now we can control when the device turns on (gate pulse applied at angle α),
thus controlling the output voltage.

This is drawn for a pure resistive load, so i = v/R.

Thus i = 0 when v = 0
Thus the thyristor turns off at v = 0 (ie. when ωt = π )

38
The average dc voltage across the load is calculated in the usual way,
ie. the average dc voltage is such that. the area under the curve over half a cycle
= area under Vdc(av) over half a cycle.
(The calculation is over just half a cycle, as that is the repeated period: clearly the
average voltage over ½ a mains cycle is the same as over a full cycle in this
case.)

39
If the load is inductive + resistive, then the current will still be positive after the
voltage has fallen to zero.
Therefore the thyristors will remain on beyond ωt = π , resulting in a negative
voltage across the load between ωt = π and ωt = π +α

40
41
Once switched on (“fired” or “triggered”), the thyristor will stay on even if the gate
signal is removed. Therefore short pulses can be used, reducing the gate power
(the power drawn from the control circuit).

A thyristor will turn-on in about 1 μs, but if the load is inductive, a pulse of 50 μs is
typically used to allow the current to reach the “latching current” (otherwise the
thyristor will not latch-on).

42
High frequency pulse transformers are often used to provide isolation between
the high voltage power circuit and low voltage control circuit, to avoid damaging
the latter.

43
Gate drive circuits tend to be specific to each manufacturer, and for the rest of
this course will be omitted from the circuit diagrams.

However, above is an example of a typical control circuit using discrete

analogue/digital components.

The control could also be done using programmable components (eg. PIC,
FPGA).

44
The waveforms at each node in the above circuit are shown on the right hand
side.

Α is controlled by moving the slider on the potentiometer (bottom left) up/down,

hence moving the crossover point on the comparator inputs (D and E) left/right.

Waveform H is applied to thyristors T2 and T3 (via an isolating transformer),

waveform I is applied to thyristors T1 and T4. Multiple pulses are applied, in case
the thyristor doesn’t switch on with the first one.

45
Above about 5 kW, 3 phase rectifiers are normally used. While the Three-Phase
Half-Wave Rectifier is very rarely used, it is covered in this course to facilitate
understanding of the Three Phase Bridge Rectifier, which is a very common
circuit.

46
Consider a 3-phase supply va, vb,vc, with a single diode in each phase. This is
essentially very similar to the 1-phase ½ wave rectifier discussed previously.

47
At any instant, whichever phase is the most positive, then the associated diode
will be conducting.

Above, during the shaded period va is the most positive, therefore diode D1 will
be conducting. Thus for this period V+ = va. As va > vb and also va > vc , diodes
D2 and D3 are revere biassed.

48
Here, during the shaded period vb is now the most positive, therefore diode D2
will be conducting. Thus for this period V+ = vb. As vb > va and also vb > vc ,
diodes D1 and D3 are now revere biassed.

49
Finally, during the shaded period vc is now the most positive, therefore diode D3
will be conducting. Thus for this period V+ = vc. As vc > va and also vc > vb ,
diodes D1 and D2 are now revere biassed.

The voltage we are interested in this the voltage across the load, vdc.
Vdc = v+ - v-
As v- = 0 (connected to the star point of the supply), Vdc = v+
Thus Vdc is equal to the positive envelope of the 3-phase supply.

50
The average voltage is calculated in the usual way, but with the repeated period
now (2π/3).

51
52
Note that the delay angle α is measured from π/6, the point where a diode would
come on. In this case it is NOT measured from the origin.

At (5π/6) thyristor T2 has not yet been switched on, so T1 continues to conduct
until T2 is switched on at (5π/6+α).

53
 The average voltage is calculated in the usual way:

5
2 
Vdc   6 VM .sint.d t 
3 6


5
 VM   cost 

6

6

    5 
 VM  cos     cos   
 6   6 
      5   5  
 VM  cos  cos  sin  sin  cos  cos  sin  sin 
 6 6  6  6 
 VM  3  cos

3 3
Vdc  .VM . cos
2
 0.827VM . cos

54
If α>π/6 and the load contains some inductance, then T1 will continue to
conduct beyond α= π, resulting in a negative output voltage for a short period
(either until the current falls to zero if the inductance is small, or until T2 turns on,
whichever occurs the earliest).

55
56
Similar to the 1-phase diode bridge, but with an extra limb of two diodes.

57
Diodes D1, D2 and D3 in the 3-phase bridge rectifier are exactly the same as
diodes D1, D2 and D3 in the 3-phase ½ -wave rectifier.

Thus, the voltage V+ in the 3-phase bridge rectifier is the same as V+ in the 3-
phase ½ -wave rectifier. ie. V+ is the positive envelope of va, vb and vc, where
va, vb and vc are the phase voltages.

58
D4, D5 and D6 are similar to D1, D2 and D3, but the other way round, so V- is the
negative envelope of va, vb and vc.

59
The conduction sequence of the 6 diodes is shown above:
D1, D2 and D3 on the top,
D4, D5 and D6 on the bottom.
There is always one diode conducting on the top, and one conducting on the
bottom.
Although each diode conducts for 120º per cycle, not that a diode is switching
every 60º (either on the top or on the bottom).

60
The voltage we are interested in is the voltage across the load, vdc, where:
vdc = v+ - v-

61
During the shaded period,
vdc = v+ - v- = va – vb
which is the line voltage vab

62
Here in this shaded period, with D6 conducting instead of D5,
vdc = v+ - v- = va – vc
which is the line voltage vac

63
Here in this shaded period, with D2 conducting instead of D1,
vdc = v+ - v- = vb – vc
which is the line voltage vbc

64
The top waveform shows the phase voltages va, vb and vc and the voltages v+
and v-.

The lower trace shows the line voltages vab, vbc and vcd. Note that these are √3
times larger, and lead the phase voltages by 30º.

Also shown are the inverses of these line voltages, vba, vcb and vac.

The load voltage, vdc, (which is what we are interested in) is now the positive
envelope of these 6 waveforms.

65
The average voltage is now calculated in the usual way, noting that the repeated
period is now π/3 (60º)

66
The diodes can be replaced with thyristors to allow he output voltage to be
controlled.

67
The upper figure shows v+ and v- on the phase voltage diagram. The voltage of
interest is the load voltage vdc, which is the difference between v+ and v. During
the shaded period this is va-vb, or vab, which is a line voltage. Thus the output
voltage vdc is segments of the line voltages vab, vbc and vca, and their inverses vba,
vcb and vac, as shown in the lower diagram. The lower diagram is the important
waveform (the upper one is shown simply to show how the lower one is derived).

Note that α is not measured from the origin, but from the crossover point when
the diode would start to conduct.

68
The average output voltage is calculated in the usual way.

69
If α > π/3, then if the load contains inductance then the current will continue to
flow through the thyristors after the voltage has dropped to zero, therefore the
thyristors continue to conduct producing a short negative voltage.

70
4 rectifier circuits have been presented:
1. Single phase, ½ wave rectifier
2. Single phase, bridge rectifier
3. Three phase, ½ wave rectifier
4. Three phase, bridge rectifier

71
Their waveforms are shown above. For simplicity they have been shown with
delay angle α = 0º.
Their relative “goodness” as rectifiers are compared below, in terms of:
1. Cost, in terms of the number of thyristors (and associated control circuitry,
2. Average output voltage,
3. Output ripple voltage, expressed as peak-to-peak ripple (what is needed is a
high average:ripple voltage ratio),
4. Output ripple frequency: the higher the ripple frequency, the easier it is to filter
out the ripple (smaller, cheaper filter),
5. Polarity of the ac supply current. In the ½ wave rectifiers the current in the ac
supply is never negative (current cannot go backwards through a
diode/thyristor). This means that there is a dc component in the ac current,
which is a “bad thing”. As a consequence, the ½ wave rectifiers are very
rarely used.

72
The output voltage in the bridge rectifier is twice that of the ½ wave rectifier.

73
Note:

In the 3 phase ½ wave rectifier VM is the peak PHASE voltage, whereas in the 3
phase bridge rectifier VM is the peak LINE voltage. In a star connected
supply Vline = √3 . Vphase. Therefore the average Vdc in the 3 phase bridge
rectifier is twice that in the 3 phase ½ wave rectifier. [Not surprising, given
that a 3 phase bridge rectifier can be considered as two 3 phase ½ wave
rectifiers.]

74
In the ½ wave rectifiers (left hand side) the current in the ac supply is never
negative (current cannot go backwards through a thyristor), therefore the average
current in the ac supply is not zero (ie. there is a dc component in the supply
current).

In the bridge rectifiers (right hand side) the lower thyristors allow a negative
current in the ac supply, therefore the average current in the ac supply is zero (ie
no dc component).

75
76
77
What is the power factor of a 6 thyristor 3 phase bridge rectifier?

To simplify the analysis, let us assume that there is a big inductance in the load
that keeps the load current Idc more or less constant.

Let’s look at phase a: current can flow in phase a if either T1 is conducting or if

T4 is conducting. If both are off, no current can flow in phase a.

On the top, T1 conducts for 1/3 of a cycle (120º).

When T1 is conducting, iac = Idc.
[When T1 turns off, T2 conducts for 120º, then T3 conducts for the final 120º.]

78
On the bottom, T4 conducts for 1/3 of a cycle (120º). When T4 is conducting, iac =
-Idc. [When T4 turns off, T5 conducts for 120º, then T6 conducts for the final 120º.]

79
Thus the ac current ia is a square wave as shown:

ia = +Idc when T1 is on
ia = -Idc when T4 is on
ia = 0 when both T1 and T4 are off.

80
If the delay angle α = 0º (as diode bridge case), then the fundamental of the
phase current waveform ia is in phase with the phase voltage va.
[The fundamental is obtained by carrying out a Fourier Series on the current
waveshape. It is effectively the “best sinusoidal approximation” to the current
waveshape.]

Normally, we consider power factor as the cosine of the angle between voltage
and current: hence in this case:
power factor = cos 0º = 1.0 (unity power factor).

81
If the thyristor bridge operates with a delay angle α ≠ 0º , then both thyristors T1
and T4 switch on angle α later: ie. the current ia is shifted angle α to the right.

Thus the phase angle between va and ia is now α, therefore:

power factor = cos α

82
83
The true definition of power factor is NOT the cosine of the angle between
voltage and current.

The true definition is:

Power factor = (real power)/(apparent power)

If (and only if) the waveforms are sinusoidal, this is equal to the cosine of the
angle between voltage and current.

84
So, what is the real power in the system, and what is the apparent power?

Real power is the same on the dc side as on the ac side (if losses are neglected).
Calculating the dc real power is easy – it is just Vdc.Idc.

Apparent power in a 3 phase system = √3.Vline(rms).Iline(rms)

85
In a 3 phase bridge rectifier:

3
Vdc  . 2.Vline(rms) . cos


86
What is the rms of the current ia above?

“rms” stands for “root mean square”

ie. It is the “square root of the mean of the current squared”

2
3
1
Iline(rms)  .2.  I dc2 .d t 
2 0

“square root”: that bit is easy!

“mean of the current squared”:
When the current is non-zero, it is either +Idc or –Idc.
The current squared is for (2π/3) radians due to the “+Idc contribution”.
The “-Idc contribution” is also2 , hence let’s just take the “-Idc contribution” and
multiply by 2.  I dc
To get the “mean” we need to 
2
I dc
average it over one complete cycle, hence this is
all divided by (2π).

87
Now, substituting

2
Iline(rms)  I dc .
3
and

3
Vdc  . 2.Vline(rms) . cos
into 

Vdc.I dc
p. f . 
Gives 3.Vline(rms) .Iline(rms)

3
p. f .   . cos

This is the accurate value of the power factor.

88
The “cos α” term is what we earlier approximated the power factor to be, when
just looking at the fundamental of the current waveform. This term is the
“displacement factor”, and only takes into account the phase difference between
the voltage and current.

The “3/π” term is the “distortion factor”, which takes into account the waveshape.

If the waveforms are all sinusoidal, the distortion factor =1, and the power factor
is then equal to the displacement factor: ie power factor = cosine of the angle
between voltage and current – but this is only true if the waveforms are
sinusoidal.

89
90
EMI = Electromagnetic Interference

91
92
Take a 3 phase bridge:
3
Vdc  .VM . cos

If α < 90º, then Vdc is positive.
Power = Vdc.Idc
which is positive, so power is flowing from the grid to the dc system.

93
 3
Vdc  .VM . cos

If α = 0º, then Vdc is zero.
Power = Vdc.Idc
which is zero, so power no power is flowing.

94
 3
Vdc  .VM . cos

If α > 90º, then Vdc is negative.
Power = Vdc.Idc
which is negative, so power is flowing from the dc system to the grid.
Ie power is flowing in the opposite direction.
[Note that the current continues to flow in the same direction, as current cannot
flow backwards though a thyristor.]

95
Negative power means power flowing in the opposite direction, therefore it is
flowing from the dc side to the ac side.

The converter is therefore converting dc power to ac, therefore it is operating as

an inverter.

This is called a Line Commutated Inverter, as the thyristors are switched off by
the natural voltage/current zero’s due to the sinusoidal ac supply.

[Commutation means switching a device – as in the mechanical commutator in a

dc motor.]

96
For power to flow from dc to ac, there must be an energy source on the dc side
(you can’t get energy from nowhere!)

97
Regenerative braking is when a motor is slowed down by returning the rotational
kinetic energy back to the supply, recovering the energy. This is much more
efficient than dissipating this kinetic energy as heat in mechanical disc brakes.

HVDC is covered later in this course.

98
Let’s now look at the waveforms for Vdc for different values of delay angle α.

99
If α = 0 then the average Vdc = 0

100
If α = 135º then the average Vdc is negative.

Note where α measured from: where a thyristor would come on if it was a diode.

101
102
1
2
3
Thyristors are not normally used due to the difficulty in turning them off. As the
supply to an inverter is dc, there are no natural voltage/current zero’s to turn off
the thyristors.

MOSFETs, IGBTs and GTOs are easily turned off by removing the gate drive
(MOSTFETs and IGBTs) or applying a negative gate pulse (GTOs).

4
The bridge inverter works by switching diagonal transistors together.

Positive ½ cycle:
• T1 and T4 on
• current flows as shown in left top diagram, left to right through the load
• the lpad voltage vac = +Vdc

Negative ½ cycle:
• T2 and T3 on
• current flows as shown in right top diagram, right to left through the load
• the lpad voltage vac = -Vdc

The voltage applied to the load is thus a square wave. Although not a sine wave,
it is still ac.

5
A square wave is OK for some loads, but most loads require a sine wave (or at
least something that resembles a sine wave more closely than a square wave).
The output waveform can be filtered to produce a sine wave.

The average voltage is zero (of course): with ac waveforms we use rms not
average.

The switching of the transistors is controlled by the control circuit. We can

therefore set the frequency to whatever we want – not just 50Hz.

6
We can control the transistor switching to produce a variable pulse width, which
allows us to control the rms output voltage.

This can be used to produce a ac variable voltage from a fixed dc voltage.

Conversely, we can produce a fixed ac rms voltage from a variable dc source (eg
a battery where the dc voltage varies depending on its state of charge). In this
case the pulse width γ would be increased as the dc voltage decreases.

7
With sinusoidal waveforms:
Vpeak = √2.Vrms
But this is not a sine wave, therefore to calculate the rms voltage we need to look
at the true definition of rms.
“rms” stands for “root mean square”
ie. It is the “square root of the mean of the current squared”


1
Vline ( rms )  .2. Vdc2 .d t 
2 0
“square root”: that bit is easy!
“mean of the current squared”:
When the current is non-zero, it is either +Vdc or –Vdc.
The current squared is for γ radians due to the “+Vdc contribution”.
The “-Vdc contribution” is also , hence let’s just take the “-Vdc contribution”
and multiply by 2.  Vdc2
 Vdc2 it over one complete cycle, hence this is
To get the “mean” we need to average
all divided by (2π).

8
As with chopper circuits, a path must be included for the current to flow when the
transistors are switched off if the load has any inductance (otherwise very high
voltage will be produced due to suddenly interrupting current flowing through an
inductor: v = L.di/dt ).

In ac systems, inductance in the load means a lagging power factor.

9
Feedback diodes are included across each switch. Thus when T1 and T4 are
switched off, the current can continue to flow through the load in the same
direction via diodes D2 and D3. Thus the voltage across the load is reversed
until the current has fallen to zero.

This is the usual arrangement for single phase bridge inverters.

10
When T1 and T4 are switched off, the current can continue to flow through the
load in the same direction via diodes D2 and D3. Thus the voltage across the
load is reversed until the current has fallen to zero.

11
12
A 3-phase inverter has an extra limb (extra pair of transistors).

13
The capacitors shown above provide a centre-point for the dc supply, to give a
reference point. They are not needed in a real inverter. - they are included here
only to help understand how the circuit works.

Look at the “a” phase:

When T1 is on, va = +½Vdc
When T2 is on, va = - ½Vdc

Note that T1 is on for ½ the cycle, and then T2 is on for the other ½ cycle.

14
Now look at the other two phases – the “b” and “c” phases. The switching of the
“b” phase transistors is delayed by 120º compared to the “a” phase transistors.
The switching of the “c” phase transistors is then delayed by 120º compared to
the “b” phase transistors.

15
The line voltage vab is the difference between va and vb, ie.
vab = va - vb

16
The line voltage vab (bottom trace) is the difference between va and vb, ie.
vab = va - vb

17
The other line voltages vbc and vca can be produced in a similar fashion.

18
`

19
The vast majority of power transmission uses 3-phase ac, but in some cases ac
does have limitations.

20
Take 2 systems interconnected by a long ac overhead line (or a power station
connected to the rest of the electrical network by a long overhead line).

The overhead line has inductance L. The power that is transferred across the
line is given by:

3.E1.E2 .sin
power transfer 
XL
Where δ is the angle between E1 and E2.

21
The maximum power that can be transferred is when δ = 90 º.
If δ > 90º, then the system loses synchronism. Zero power is transferred, the two parts of
the system operate at different frequencies, and there will be very large amounts of
pulsating power.

The maximum power that can be transferred before synchronism is lost is limited by XL
= jωL, which is proportional to the length of the line. Thus very long lines are limited as
to how much power they can transmit.

If the line is dc, there is no such limitation.

22
An underground (or submarine) cable looks like a capacitor to earth (look at its
structure: conductor-insulator-conductor, where the sea/earth is the 2nd
conductor).

The formula for the capacitance of a parallel plate capacitor is:

A
C   0 . r .
In this case dd is the thickness of the insulator. A is the plate area, which in this
case is proportional to the length of the cable. Thus a long cable has significant
capacitance: d may be relatively large for a capacitor, but the cable area can be
huge!

23
The capacitance of the cable can be represented as a set of small capacitors
along the length of the cable between the cable conductor and earth. If the
supply is ac, this capacitance will conduct current Ic1, Ic2 etc. If the cable is more
than 40-50km, the total capacitor current (the “charging” current) can exceed the
cable maximum current rating, which means that no current at all can be drawn
by the load without seriously overloading the cable!

This is a very real problem with cables, and is a major reason why overhead lines
are mostly used in preference to cables: the capacitance of overhead lines is far
less, as the distance between the conductor and earth is far greater.

[The other main reason for preferring overhead lines is cost: a cable is typically
10 times more expensive than an overhead line.]

If the cable is dc instead of ac, the capacitor charging current is not a problem. It
will only flow when first switched on: once the capacitance is charged no further
current will flow through the capacitance.

24
Power electronic converters are needed at the sending end (to convert the ac
power to dc (a rectifier), and also at the receiving end to convert the dc power
back to ac (an inverter).

In most cases it is desirable that power can flow in both directions, so each
converter should be able to operate both as a rectifier and as an inverter.

25
Normally the converters are 6-thyristor bridge rectifiers. In the system above, if
power is flowing from System 1 (on the left) to System 2 (on the right), then
Converter 1 will operate as a rectifier, and Converter 2 as a line commutated
inverter.

Thus in Converter 1:

3
Vdc1  .VM 1 cos1


where α < 90º.

In Converter 2:
3
Vdc2  .VM 2 cos2


where α > 90º.

The dc link current I is thus given by:
Vdc1  Vdc2
I
R
where R/2 is the resistance of each cable.

26
Normally, each converter is made up of two converters in series, one fed from a
star connected winding on the main transformer, and the other from a delta
connected winding on the main transformer. There is a 30º phase shift between
the star line voltages and the delta connected line voltages (check 2nd year
Electrical Power Engineering 2).

27
The 30º phase shift means that there is a resulting 30º phase shift in the ripple
voltages on the dc side of the upper and lower converters. Thus when the two
outputs are added, the ripple frequency is doubled and the ripple voltage is
greatly reduced, from 13.4% of average dc voltage for a 6 thyristor bridge to
3.4% for the combined converter output as shown in the bottom trace above.

This greatly reduces the amount of expensive filtering that is needed to produce
a smooth dc.

28
Another advantage of the converters at each end is that if one cable (say the
lower one) develops a fault, power can still be transferred across the link using
the remaining (upper) cable, using the sea/earth as the return path (sea – salt
water – is a very good conductor). Only ½ the power can be transmitted, but this
is much better than no power at all!

Thus the system reliability is equivalent to a double circuit ac system.

29
Compare the power capability of a dc system versus an ac system, assuming
identical cables.

The cable current rating is I1. If this is exceeded, the cable will overheat.
The cable voltage rating is V1. If this is exceeded, the cable insulation will break
down.

dc system:
2 cables are needed
power transmitted = 2V1I1 (factor 2 due to total sending voltage = 2V1)

30
ac system:
peak ac voltage must not exceed V1
rms current must not exceed I1
assume typical power factor of 0.955
therefore max. power is:

power  3Vphase  I phase  cos

V1
 3  I1  0.955
2
 2 V1  I1

31
Thus the dc system can transmit the same power as the ac system, yet it uses
only two cables compared to the three cables required in a 3-phase ac system.

Thus the total cable costs in the dc system are approximately 2/3 the cost of the
ac system. This is a big saving: long cables are very expensive, particularly if
they are being laid in a trench under the sea.

However, the power electronic converters each end are also very expensive: they
are needed however long the cable are.

32
For a short distance, the cost of the cable is insignificant compared to the cost of
the converters, so the ac system wins.

For a long distance, the cost of the extra cable in the ac system outweighs the
cost of the converters at each end.

33
There can be other reasons for an HVDC link:
1. There is no way that the UK grid system could be kept in synchronism with
the European grid system via a relatively small link. An HVDC link allows the
two systems to be asynchronous: both are nominally 50Hz, but they may
differ slightly.
2. Some HVDC links may be just 10m long! They are included simply to allow
the 2 interconnected systems to run asynchronously.
3. Power flow can be controlled much more accurately and more quickly than in
conventional ac systems, simply by varying the delay angle α at each end.
4. If there is a fault in one system, the link can be shut down very quickly simply
by switching off the converters (by removing their thyristor gate signals).

34
Transformers only work with ac, not dc. Thus ac has a big advantage in that it is
easy to change the voltage from high levels (for transmission levels) to
intermediate voltages (distribution) down to 230V for households etc.

HVDC is good for bulk power transmission from one point to another. However, it
is not easy to tap of from an HVDC line to supply a small town along the route.

35