Post on 10-Apr-2015
description
Power Electronics
Professor Mohamed A. El-Sharkawi
El-Sharkawi@University of Washington
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Power Control
Period (τ)
On-time(ton)
Off-time(toff)
Time
Power
P
Ps
PtP ons τ
=
ρ
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Load Switching
Period (τ)
On-time(ton)
Off-time(toff) Time (t)
Power
P
toffton
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Duty Ratio (K)
Energy Consumption (E)Energy Consumption (E)
tPE ≡ tPttPE onss τ
==
Period (τ)
On-time(ton)
Off-time(toff) Time (t)
Power
P
Ps
Ideal Switch
Vsw
iR
vs
vsw
ivt
+
-vs
Rvs
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N
NP
(C)
(B)
(E)
Collector
Emitter
Base
(C)
(B)
(E)
(C)
(E)
(B)
IB
IC
IE
VCE
VCB
VBE
BC II β≈
CBE III +=
BECBCE VVV +=
Bi-polar Transistor
IB
VBE 0.6
IB1
IB2< IB1
I = 0B
Linear Region
Saturation Region
Base CharacteristicsBase Characteristics Collector CharacteristicsCollector Characteristics
Cut Off Region
IC
VCE
Characteristics of BiCharacteristics of Bi--polar polar TransistorTransistor
(C)
(E)
(B)IB
IC
IE
VCE
VCB
VBE
VCE
IC
VCC
RL
IB max
IB = 0
(1)
(2)
IC
R L
V CC
V CE
IB
CLCECC IRVV +=VCC
At point (1)VCE is very small
L
CCC R
VI ≈
At point (2)IC is very small
CCCE VV ≈
ClosedClosedswitchswitch
OpenOpenswitchswitch
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VCE
IC
VCC
RL
IB max
IB = 0
(1)
(2)
IC
10Ω
100VV CE
IB max=2A
VCC
Example• Estimate the losses of the transistor at point 1 and 2. Also calculate the
losses at a mid point in the linear region where IB=0.1A. The current gain in the saturation region is 4.9 and in the linear region is 50.
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VCE
ICVcc
RL
IB max
IB = 0
1
2
IC
10Ω
100VVCE
IB max=2A
VCC
Solution
At point 1Total losses = base loses + collector losses
3
[ ] ( ) W**.**..*
V*IV*I CECBEmaxB
2110294100294702
losses Total 11
=−+
+=
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VCE
ICVcc
RL
IB max
IB = 0
1
2
IC
10Ω
100VVCE
IB max=2A
VCC
Solution
At point 2Total losses = collector lossesAssume VCE=0.99 VCC
3
( ) W*.**.
V*I CEC
1010099010
100990100
losses Total 22
=⎥⎦⎤
⎢⎣⎡ −
=
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VCE
ICVcc
RL
IB max
IB = 0
1
2
IC
10Ω
100VVCE
IB max=0.1A
VCC
Solution
At point 3Total losses = base loses + collector losses
3
[ ] ( ) W.*.**.*.*.
V*IV*I CECBEB
0725010105010010507010
losses Total 333
=−+
+=
Power transistors cannot operate in the linear region
Thyristors [Silicon Controlled Rectifier (SCR)]
AK
VBO
IA
V
VRB
Anode (A)
Cathode (K)
Gate (G)
VTO
Ig > 0 Ig = 0
Ig = max
Ih
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Closing Conditions of SCRClosing Conditions of SCR
1. Positive anode to cathode voltage (VAK)
2. Maximum triggering pulse is applied (Ig)
Anode (A)
Cathode (K)
Gate (G)
Closing angle is α
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Opening Conditions of SCROpening Conditions of SCR
1. Anode current is below the holding value (Ih)
AK
IA
V
VRB
Ig = 0
Ih
Opening angle is β
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Power Converters
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Power Converters
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AC/DC Converters
Single-Phase, Half-Wave
i Rvtvs
+
-
)t sin(V v maxs ω=
)closed is SCRonly when(Rvi s=
)closed is SCRonly when(vRiv st ==
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i
αωt
vs
vt
β
i Rvtvs
+
-Rvi t=
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i
αωt
vs
vt
∫∫ ==β
α
π
ωπ
ωπ
tdvtdvV stave 21
21 2
0
β
Average Voltage Across the Load
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i
αωt
vs
vt∫=π
α
ωωπ
tdtVVave )(sin21
max
∫∫ ==π
α
β
α
ωπ
ωπ
tdvtdvV ssave 21
21
)cos1(2
VV maxave α
π+=
β
RVI ave
ave =
Load voltage
)cos1(2
VV maxave α
π+=Vave
π α
πmaxV
π2Vmax
2π
Root-Mean-Squares (RMS)
∫π
ωπ
2
0
.21 td
2(.)
Root Mean Squares of f
2)( fStep 1:
∫π
ωπ
2
0
2)(21 tdfStep 2:
∫π
ωπ
2
0
2)(21 tdfStep 3:
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ωtv
v2 Average of v2
Square root of the average of v2
Concept of RMS
Averageof v=0
Root-Mean-Squares (RMS)of a sinusoidal voltage
∫==π
ωπ
2
0avemean td)t(v
21VV
[ ] [ ]∫∫ ==ππ
ωωπ
ωπ
2
0
2max
2
0
2rms td)tsin(V
21td)t(v
21V
RMS of SupplyRMS of SupplyVoltageVoltage
∫∫ −=)]([ 2
= 2π
α
π
αωω
πωω
πtd]t2cos(1[
4VtdtsinVV
2max
2max
rms
])(
+ − [=π
απα
22sin1
2Vmax
rmsV
ωtα
vs
vtiRMS of load RMS of load
voltagevoltage
rmsmax s 2 VV =
Vrms
π α
2Vmax
RVI rms
rms =
Example.2:
An ac source of 110V (rms) is connected to a resistive element of 2 Ω through a single SCR.For α = 45o and 90o, calculate the followings:
a) rms voltage across the load resistanceb) rms current of the resistancec) Average voltage drop across the SCR
Solution:
For α = 45o
This looks likethe negativeof the averagevoltage acrossthe load. Why?Why?
i Rvtvs
+
-
a)( )
V13.742
)90sin(180451
2110sin
2VsV rms
rms =⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+−=⎥⎦
⎤⎢⎣⎡
2)(2
+ − 1=ππ
π
πα
πα
b) A07.37213.74
RVI rms
rms ===
c)
V27.42)]45cos(1[2
1102V
)cos1(2
Vtdvtdv21V
SCR
max
0
2
ssSCR
−=+−=
+−=⎥⎥⎦
⎤
⎢⎢⎣
⎡+= ∫ ∫
π
απ
ωωπ
α π
π
ωtα
vs
vti
Electric Power
R I R
V P 2rms
2rms ==
)](+−([= ααππ
2sin)2R8
V P2
max
Single-Phase, Full-Wave, AC-to-DC Conversion for Resistive Loads
S1 S3
i2
R
S4
vs
A
D
C
B
i1
vt
S2
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Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodes
S1 S2
i2R
D2
vs
A
D
C
B
i1
vt
D1
vti2
ωtvs
α
vti1
S1 S3
i2
R
S4
vs
A
D
C
B
i1
vt
S2
)cos(Vtd)tsin(VtdvV maxmaxtave α
πωω
πω
π
π
α
π
α
+=== ∫∫ 111
vti2
ωtvs
α
vti1
∫∫ ==π
α
π
ωωπ
ωπ
tdtVtdtvVrms2
max
2
0
2 )]sin([1)(21
⎥⎦⎤
⎢⎣⎡ +−=
−== ∫∫
πα
πα
ωωπ
ωωπ
π
α
π
α
2)2sin(1
2VV
td)]t2cos(1[2
Vtd)tsin(VV
maxrms
2max2
2max
rms
)](+−([== ααππ
2sin)2R4
VR
V P2
max2
rms
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Half Wave Versus Full Wave
Half Wave Full Wave
Average Voltage
RMS Voltage
Power)](+−([= ααπ
π2sin)2
4 P
2max
RV)](+−([= ααπ
π2sin)2
8 P
2max
RV
⎥⎦⎤
⎢⎣⎡ +−=
πα
πα
2)2sin(1
2maxVVrms⎥⎦
⎤⎢⎣⎡ +−=
πα
πα
2)2sin(1
2maxVVrms
)cos1(max απ
+=VVave
)cos1(2max απ
+=VVave
Example
A full-wave, ac/dc converter is connected to a resistive load of 5 Ω. The voltage of the ac source is 110 V(rms). It is required that the rms voltage across the load to be 55 V. Calculate the triggering angle, and the load power.
Solution
])(
+ − [=π22sin
π1 VsV rmsrms
αα
])(
+ − [=π22sin
π1 11055 αα
22sin
18025.2 )(
− =απα o5.112≈α
W6055
)55(R
VP22
rms ===
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DC/DC Converters
DC-to-DC Conversion
1. Step-down (Buck) converter: where the output voltage of the converter is lower than the input voltage.
2. Step-up (Boost) converter: where the output voltage is higher than the input voltage.
3. Step-down/step-up (Buck-Boost) converter.
Step Down (Buck converter)VS
Vl
ton
τTime
I
ton
τTime
VCEVS
I
+
-
Vl
s
t
0s
onsave VKVtdtV1V
on
=== ∫ ττ
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Example
ms0834.02.0417.0t
417.o125K
VKVtV
on
sson
ave
=×=
==
==τ
?t;V5V;V12V)frequencyswitching(kHz5f
onaves ====
ms2.051
f1
===τSolution
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Step up (Boost converter)
Is
L
vs vt R C
it
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ion
L
vs
Is
L
vs vt R C
it
Keep in mind•Inductor current is unidirectional•Voltage across inductor reverses•Inductor cannot permanently store energy
ioff
L
vs vt R C
50
Timeton toff
Δi ionioff
ion
L
vs
ioff
L
vtRCvs
51
Timeton toff
Δi
Time
von
Inductor current
voff
Inductor voltage
Energy is acquired by inductor
Energy is released by inductor
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on
ons t
iLV Δ=
off
offts t
iLvV
Δ−=
ion
L
VS
ioff
L
vtRCVS
⎟⎟⎠
⎞⎜⎜⎝
⎛+=+=
off
ons
off
offst t
tVti
LVv 1Δoffon ii Δ=Δstatesteady At
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Example• A Boost converter is used to step up 20V
into 50V. The switching frequency of the transistor is 5kHz, and the load resistance is 10Ω. Compute the following:
1. The value of the inductance that would limit the current ripple at the source side to 100mA
2. The average current of the load3. The power delivered by the source4. The average current of the source
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Solution
offt*.t
tt
ttVV
on
off
on
off
onst
51
12050
1
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
ms.f
tt offon 20511
===+
( )ms.
t.*.t*.t onon off
120
205151
=
−==
Part 1
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12010020.
L
tiLVon
ons
=
=Δ
mHL 24=
Part 2
ARVI t
t 51050
=== W*I*VP tt 250550 ===
Part 3
Part 4 A.VPI
ss 512
20250
===
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Buck-Boost converter
vtL RC
itis
vs
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ionLvs vtvtL RRC
ioff
+
-
on
ons t
iLV Δ=
toff
offL v
ti
Lv =−=Δ
off
onst t
tVV −=offon iiif ΔΔ =
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DC/AC Converters
DC/AC Conversion
Q 1
Q2Q3
Q4I1
I2
A B
Q1 and Q 2 are on
Q3 and Q 4 are on
Time
Load voltage
VAB
Q 1
Q 2
Q 3
Q 4
Q 5
Q 6
Q1
Q2
Q3
Q4
Q5
Q6
a b c
a b c
no
Vdc
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Q 1
Q 2
Q 3
Q 4
Q 5
Q 6
0vvvVvvv
Vvvv
acca
dccbbc
dcbaab
=−=−=−=
=−=
Vs c a b
n
Q1 Q5
Q6
I I/2
I/2
I/2 I/2 I
I I
I/2
Q1
Q2
Q3
Q4
Q5
Q6
a b c
a b c
no
Vdc
First Time Interval
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Q 1
Q 2
Q 3
Q 4
Q 5
Q 6
Q1
Q2
Q3
Q4
Q5
Q6
a b c
a b c
no
Vdc
dcacca
cbbc
dcbaab
Vvvvvvv
Vvvv
−=−==−=
=−=0
Vs c a b
n
Q1
Q2 Q6
I
I/2
I
I/2 I I/2
I I
Second Time Interval
vab
vbc
vca
- Vdc
Voltage Waveforms Across Load
• Waveforms are symmetrical and equal in magnitude
• Waveforms are shifted by 120 degrees
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AC/AC Converters
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1. Single-Phase, Bidirectional
i R vt vs
+
-
i1
i2
vt i2
ωt vs α
vt i1
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021 2
=ωπ
= ∫π
α
tdvV tave
∫∫ ==π
α
π
ωωπ
ωπ
tdtVtdtvVrms2
max
2
0
2 )]sin([1)(21
vt i2
ωt vs α
vt i1
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⎥⎦⎤
⎢⎣⎡ +−=
−== ∫∫
πα
πα
ωωπ
ωωπ
π
α
π
α
2)2sin(1
2VV
td)]t2cos(1[2
Vtd)tsin(VV
maxrms
2max2
2max
rms
)](+−([== ααππ
2sin)2R4
VR
V P2
max2
rms
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AC/DC dc Link
iin iout Idc
DC/AC
2. DC Link
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3. Uninterruptible Power Supply (UPS)
AC/DC
dc Link
iin iout Idc
DC/ACIb
AC/DC
dc Link
iin = 0 iout Idc
DC/ACIb