Presentation 2 : Power Electronics

Power Electronics Professor Mohamed A. El-Sharkawi

Power Control

ton Ps = P τ

Power On-time

P

(ton)

ρ

Ps Off-time

(toff)

Time

τ)

Period (

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Load Switching Power P

On-time (ton)

Off-time

(toff)

Time (t)

τ)

Period (

toff

ton El-Sharkawi@University of Washington

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Energy Consumption (E) Power P

On-time (ton)

Ps Off-time

Time (t)

(toff)

τ)

Duty Ratio (K)

Period (

E ≡ Pt

ton Es = Ps t = Pt τ El-Sharkawi@University of Washington

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vsw

Ideal Switch +

i vs

vt -

R

i

vs R

vs Vsw

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Bi-polar Transistor (C)

(C)

IC

Collector Base (B)

(C)

VCB

N

IB (B)

P N

VCE

(B) VBE

Emitter

IE

(E) (E)

IC ≈ β I B I E = I B + IC

(E)

VCE = VCB + VBE

(C)

Characteristics of Bi-polar Transistor

IC VCB IB VCE

(B) VBE IE

IB

Saturation Region IC

(E)

IB1 IB2< IB1 Linear Region IB= 0

V 0.6 BE Base Characteristics

Cut Off Region

VCE

Collector Characteristics

IC

IB max

IC RL

VCC RL

IB

(1)

V CE V CC (2) IB = 0

VCC = VCE + RL I C At point (1) VCE is very small

VCC IC ≈ RL

Closed switch Open switch

VCC

At point (2) IC is very small

VCE ≈ VCC

VCE

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.

IC

IB max

IC 10Ω

VCC RL

IB max=2A

(1)

V CE 100V (2) IB = 0 VCC El-Sharkawi@University of Washington

VCE 13

Solution IC 10Ω IB max=2A

IC Vcc RL

VCE

IB max 1 3

100V 2 IB = 0

At point 1 Total losses = base loses + collector losses

VCC VCE

Total losses = I B max * VBE + I C1 * VCE1

2* 0.7 + [4.9 * 2]* (100 − 4.9 * 2 * 10 ) = 21 W

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Solution IC 10Ω IB max=2A

IC Vcc RL

VCE

IB max 1 3

100V 2 IB = 0

At point 2 Total losses = collector losses Assume VCE=0.99 VCC

VCC VCE

Total losses = I C 2 * VCE 2 ⎡100 − 0.99 * 100 ⎤ * (0.99* 100) = 10 W ⎥ ⎢⎣ 10 ⎦ El-Sharkawi@University of Washington

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Solution IC 10Ω IB max=0.1A

IC Vcc RL

VCE

IB max 1 3

100V 2 IB = 0

At point 3 Total losses = base loses + collector losses

VCC VCE

Total losses = I B 3 * VBE + I C 3 * VCE 3

0.1* 0.7 + [50 * 0.1]* (100 − 50 * 0.1* 10 ) = 250.07 W

Power transistors cannot operate in the linear region El-Sharkawi@University of Washington

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Thyristors [Silicon Controlled Rectifier (SCR)] Anode (A)

IA

Ig = max Ig > 0 Gate (G)

VRB

Ig = 0

Ih V AK

Cathode (K)

V

TO

VBO

Closing Conditions of SCR Anode (A)

1. Positive anode to cathode voltage (VAK) 2. Maximum triggering pulse is applied (Ig)

Gate (G)

Cathode (K)

Closing angle is α

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Opening Conditions of SCR 1. Anode current is below the holding value (Ih) VRB

IA

Ig = 0 Ih V AK

Opening angle is β El-Sharkawi@University of Washington

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Power Converters

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Power Converters

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AC/DC Converters

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Single-Phase, Half-Wave

vs

i

+ vt -

R

vs = Vmax sin( ω t )

vs i = ( only when SCR is closed ) R vt = i R = vs ( only when SCR is closed )

vt i= R vt

α

i

i

vs

+ vt -

R

ωt

β

vs El-Sharkawi@University of Washington

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Average Voltage Across the Load 1 Vave = 2π

β

2π

1 ∫0 vt dω t = 2 π α∫ vs dω t

i

vt

α

ωt

β

vs El-Sharkawi@University of Washington

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β

Load voltage

π

1 1 Vave = vs dω t = v s dω t ∫ ∫ 2π α 2π α π

i

vt

1 Vave = Vmax sin (ω t ) dωt ∫ 2π α

Vmax Vave = ( 1 + cos α ) 2π

α I ave

Vave = R

ωt

β

vs El-Sharkawi@University of Washington

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Vave

Vmax Vave = ( 1 + cos α ) 2π

Vmax

π

Vmax 2π

π 2

π

α

Root-Mean-Squares (RMS)

(.) 1 2π

2π

∫ 0

. dωt

2

Root Mean Squares of f 1 Step 2: 2π

2π

∫

( f ) dωt 2

0

Step 1: Step 3:

1 2π

2π

( f ) d ω t ∫ 2

0

(f)

2

Concept of RMS Average of v2

v2

Square root of the average of v2 El-Sharkawi@University of Washington

ωt v

Average of v=0

30

Root-Mean-Squares (RMS) of a sinusoidal voltage 1 Vmean = Vave = 2π 1 Vrms = 2π

2π

∫ [v( t )]

0

2

2π

∫

v( t ) dωt

0

1 dωt = 2π

2π

∫ [Vmax sin( ωt )]

0

2

dωt

RMS of load voltage

i

vt

α

ωt vs

2 π Vmax Vrms = [ sin(ω t )]2 dω t =

2π

Vrms

∫

α

Vmax = 2

Vmax =

2 π Vmax [ 1 − cos( 2ω t ] dωt

4π

∫

α

α sin ( 2α ) [1 − + ] π 2π

2 Vs rms

RMS of Supply Voltage

Vrms Vmax 2

Vrms I rms = R

π

α

Example.2: An ac source of 110V (rms) is connected to a resistive element of 2 Ω through a single SCR. o o For α = 45 and 90 , calculate the followings: a) b) c)

rms voltage across the load resistance rms current of the resistance Average voltage drop across the SCR

i

vs

Solution:

+ vt -

R

For α = 45

o

a)

Vrms =

Vsrms 2

⎡ α sin(2α ) ⎤ 110 ⎢⎣1 − π + 2π ⎥⎦ = 2

)

V 74.13 I rms = rms = = 37.07 A R 2

b) ⎡α

c)

(

⎛ 45 π ⎞ ⎜ 1− 180 + sin( 90 ) ⎟ = 74.13 V ⎜⎜ π 2 π ⎟⎟ ⎝ ⎠

2π

1 ⎢ v s dωt + ∫ v s dωt VSCR = 2π ⎢ ∫ π ⎣0 VSCR = −

vt

⎤ V ⎥ = − max ( 1 + cos α ) 2π ⎥⎦

2 110 [ 1 + cos( 45 )] = − 42.27 V 2π

α

This looks like the negative i of the average voltage across the load. Why? ωt

vs

Electric Power 2 Vrms 2 P = = I rms R

R

2 Vmax P = [2(π − α ) + sin( 2α )]

8π R

Single-Phase, Full-Wave, AC-to-DC Conversion for Resistive Loads C

i1 S1

S3 vs

i2 R

A S4

B S2 D

vt

Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodes C

i1 S1

S2 vs

i2 R

A D2

B

vt

D1

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C

i1 S1

S3 vs

i2 R

A

B S2

S4

D

i1

vt

vt

α

i2

ωt vs

vt

i1

vt

i2

vt

α

ωt vs

Vave =

1

π

Vrms =

π

∫α v

t

1 2π

dωt =

1

π

2π

∫ v(t ) 0

2

π

∫α V

max

dωt =

sin( ωt ) dωt =

1

π

[V ∫ πα

max

Vmax

π

( 1 + cos α )

sin(ωt )] dωt 2

Vrms =

2 Vmax

π

π

π

α

α

2 V 2 max [ 1 − cos( 2ωt )] dωt sin( t ) d t ω ω = ∫ 2π ∫

Vmax Vrms = 2

⎡ α sin( 2α ) ⎤ ⎢⎣1 − π + 2π ⎥⎦

2 2 Vrms Vmax P = = [2(π − α ) + sin( 2α )]

R

4π R

Half Wave Versus Full Wave Half Wave Average Voltage RMS Voltage Power

Full Wave

Vmax Vave = (1 + cos α ) 2π Vrms =

Vmax 2

⎡ α sin( 2α ) ⎤ ⎢1 − π + 2π ⎥ ⎣ ⎦

2 Vmax P = [2(π − α ) + sin ( 2α )] 8π R

Vave = Vrms

Vmax

Vmax = 2

π

(1 + cos α )

⎡ α sin( 2α ) ⎤ ⎢1 − π + 2π ⎥ ⎣ ⎦

2 Vmax P = [2(π − α ) + sin ( 2α )] 4π R

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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 Vrms = Vsrms 55 = 110 π

α

sin( 2α ) [1 − + ] π 2π α sin( 2α )

[1 −

π

+

sin( 2α ) 2.25 = α − 180 2

2π

]

α ≈ 112.5 o

2 Vrms ( 55 )2 P= = = 605 W R 5

DC/DC Converters

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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 voltage is higher than the input voltage. 3. Step-down/step-up (Buck-Boost) converter.

output

Step Down (Buck converter) VS

I Vl ton

Time

τ

VCE

VS

I + Vl -

ton

Time

τ

Vave =

1

τ

t on

∫ Vs dt =

0

ton

τ

Vs = K Vs

Example f = 5 kHz ( switching frequency ) Vs = 12 V ; Vave = 5 V ; ton = ? Solution

1 1 τ = = = 0.2 ms f 5 Vave =

ton

τ

Vs = K Vs

5 K = = o.417 12 ton = 0.417 × 0.2 = 0.0834 ms El-Sharkawi@University of Washington

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Step up (Boost converter) L

it

Is vs

C

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R

vt

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L

it

Is vs

C

R

vt

Keep in mind •Inductor current is unidirectional •Voltage across inductor reverses •Inductor cannot permanently store energy

L

L

ion vs

ioff vs

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C

R

vt

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L

vs

Δi

L

ion

i off

vs

ion

ton

C

R

vt

ioff

toff

Time

50

Inductor current

Δi

ton Inductor voltage

toff

Time

von

voff Time

Energy is acquired by inductor Energy is released by inductor 51

L

L

i on

i off

VS

VS

Vs = L

Δ ion

V s = vt − L

ton

R

C

At steady state Δ ion = Δ ioff

vt =Vs + L

Δ ioff t

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vt

Δ ioff t off

⎛ t on ⎜ = Vs 1 + ⎜ t off ⎝ 52

⎞ ⎟ ⎟ ⎠

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 load 3. The power delivered by the source 4. The average current of the source El-Sharkawi@University of Washington

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Solution Part 1

⎞ ⎛ t Vt = Vs ⎜1 + on ⎟ ⎜ t ⎟ off ⎠ ⎝ ⎞ ⎛ t on ⎟ ⎜ 50 = 20 1 + ⎜ t ⎟ off ⎠ ⎝ ton =1.5 * t off

ton + toff

1 1 = = = 0.2 ms f 5

ton =1.5* t off = 1.5* (0.2 − ton ) = 0.12 ms

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Vs = L

Δ ion ton

100 20 = L 0.12

L = 24 mH

Part 2

Part 3

Vt 50 = =5 A It = R 10 Part 4

P = Vt * I t = 50 * 5 = 250 W

P 250 Is = = = 12.5 A Vs 20 El-Sharkawi@University of Washington

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Buck-Boost converter it

is vs

L

C

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R

vt

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-

vs

L

ion

C

ioff

R

L

vt +

Vs = L

Δ ion ton

if Δ ion = Δ ioff

vL = − L

Δ ioff toff

= vt

ton Vt = −Vs toff El-Sharkawi@University of Washington

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DC/AC Converters

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DC/AC Conversion Q1 A Q3

Load voltage

Q4

I1 B I2

Q2

Q 1 and Q 2 are on

VAB Time

Q 3 and Q 4 are on

Q3

Q1 Vdc a Q4

o

Q5 b

Q6

a

b

c Q2 n

Q 1 Q 2 Q 3 Q 4 Q 5 Q 6

c

Q1

First Time Interval

Q2 Q3

Q3

Q1 Vdc a Q4

b Q6

a

Q5

b

c

Q4 Q5

c

Q6

Q2 n

o

I/2

I I/2

vab = va − vb = Vdc vbc = vb − vc = − Vdc vca = vc − va = 0

I/2 Q5

Q1

Vs

a

b

c

Q6 I

I

I/2

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I

n

I/2

61

Q1

Second Time Interval

Q2 Q3

Q3

Q1 Vdc a Q4

b Q6

a

Q5

b

c

Q4 Q5

c

Q6

Q2 n

o

I I

vab = va − vb = Vdc

Q1

Vs

a

b

vbc = vb − vc = 0 vca = vc − va = − Vdc

c

Q6 I

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Q2

I

I/2

I

I/2

n

I/2

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Voltage Waveforms Across Load vab - Vdc vbc

vca

•

Waveforms are symmetrical and equal in magnitude

•

Waveforms are shifted by 120 degrees

AC/AC Converters

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1. Single-Phase, Bidirectional i1 i2

vs

i

i1

vt

vt

α

vs

+ vt -

R

i2

ωt

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i1

vt

vt

α

Vave

Vrms =

1 2π

2π

∫ v(t ) 0

2

i2

ωt

vs

1 = 2π

dωt =

2π

∫v

t

d ωt = 0

α

1

π

[V ∫ πα

El-Sharkawi@University of Washington

max

sin(ωt )] dωt 2

66

Vrms =

2 Vmax

π

π

π

α

α

2 V 2 max [ 1 − cos( 2ωt )] dωt sin( t ) d t ω ω = ∫ 2π ∫

Vmax Vrms = 2

P =

2 Vrms

R

=

⎡ α sin( 2α ) ⎤ ⎢⎣1 − π + 2π ⎥⎦

2 Vmax

4π R

[2(π − α ) + sin( 2α )]

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2. DC Link iin

Idc AC/DC

dc Link

iout DC/AC

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3. Uninterruptible Power Supply (UPS) iin

Idc AC/DC

Ib

iout DC/AC

dc Link

Idc

iin = 0 AC/DC

Ib

iout DC/AC

dc Link El-Sharkawi@University of Washington

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