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