Solution Tenta 20120521.ppt [KompatibilitetslÃ¤ge] Tenta 20120521.pdf · 2018. 5. 28. · &5 5 W W ([DPLQDWLRQ F )RUZDUG FRQYHUWHU ZLWKVQXEEHU Y & 5 ORDG 9 'F & GF Y 6 & ' ' ' =

Industrial Electrical Engineering and AutomationLund University, Sweden

Power electronicsSolution to examination 2012-05-21

© Namn Namn Föredragstitel

Examination 2012-05-21 1a


Examination 2012-05-21 1b

VU QCdc 4304 4QC average bridge voltage

Average dc voltage VVU avedc 54024003

_

4QC duty cycle 8.0540

430D


Solution 2012-05-21 1cRectifier diodeContinous rectifier output current 172 AA rectifier diode is conducting 33% of the time

Rectifier diode current 172 A (constant when conducting)Rectifier diode threshold voltage 1.0 VRectifier diode diff resistance 2.2 mohmRectifer diode thermal resistance 0,12 K/W

Rectifer diode on state voltage 1+172*0.0022=1.38 VRectifier diode power loss 1.38*172*0.33=78 W

WP

AI

AI

methodeAlternativ

onstate

avg

rms

78990022.0157

5717233.0

9917233.0

2

2


Examination 2012-05-21 1c4QC transistor The continous load current 172/0.8=215 A (to maintain the power)4QC transistor current 215A4QC load inductance 5.1 mH

Vdt

di

L

Ui

t

iL

dt

diLUripplecurrentLoad 63.8

2000

8.0

0051.0

430540

AI

AI

AI

AI

avg

rms

1942

7.2103.2199.0

2043

7.2107.2103.2193.2199.0

7.21063.85.0215

3.21963.85.0215

22

min

max


Examination 2012-05-21 1c4QC transistor rms-current 205A4QC transistor avg-current 194A4QC transistor threshold voltage 1.4 V4QC transistor diff resistance 12 mohm4QC transistor turn-on loss 65 mJ 4QC transistor turn-off loss 82 mJ4QC transistor thermal resistance 0,043 K/W

WP

WP

WP

total

switch

onstate

987212776

211900

540

180

3.219082.0

180

7.210065.02000

776012.02051944.1 2


Examination 2012-05-21 1c cont’d 4QC diode 4QC diode threshold voltage 1.1 V4QC diode diff resistance 9.5 mohm4QC diode turn-off losses 25 mJ4QC diode thermal resistance 0.078 W/K

WP

WP

WP

AI

AI

AI

AI

total

switch

onstate

avg

rms

1031.356.67

1.35900

540

180

7.210025.02000

6.670095.0685.211.1

5.212

7.2113.2201.0

683

7.2107.2103.2193.2191.0

7.210

3.219

2

22

min

max


Examination 2012-05-21 1c cont’d Upper left IGBT transistor loss 987 WUpper right IGBT transistor loss 0 WLower right IGBT transistor loss 987 WLower left IGBT transistor loss 0 WUpper right IGBT diode loss 103 WUpper left IGBT diode loss 0 WLower left IGBT diode loss 103 WLower right IGBT diode loss 0 W


Examination 2012-05-21 1d

Rectifier diode (6)Loss each 78 WRth diode 0.12 K/WTemp diff 9.4 oC

IGBT transistor (2)Loss each 987WRth trans 0.043 K/WTemp diff 42.4 oC

HeatsinkContribution fron 6 rectifier diodes and from two IGBT.Heatsink thermal resistance 0.025 K/W Ambient temperature 42 oC Total loss to heatsink 6*78+2*103+2*987=2648 WHeatsink sink temperature 42+2648*0.025=108 oC

Junction temperatureRectifier diode 108+9.4=118 oCIGBT diode 108+8=116 oC IGBT transistor 108+42.4=151 oC

IGBT diode (2)Loss each 103 WRth diode 0.078K/WTemp diff 8.0 oC

Examination 2012-05-21 2a The buck converter with RCD snubber

D R

C

T

FD

i

At transistor T turn off, the current i commtutates over to the capacitor C via diode D. The capacitor C voltages increases until the freewheeling diode FD becomes forward biased and thereafter the load current iload flows through diode FD and the current i=0. At transistor T turn on, the capacitor C is discharged via the the transistor T and resistor R. The diode FD becomes reverse biased and the current iload commutates to the transistor T and the current i= iload.

iload

Examination 2012-05-21 2b i

At transistor T turn on, the current i commutates to the transistor T, and the capacitor C is discharged via the the transistor T and resistor R. As the load voltage is 160V the duty cycle is 64%. The switching frequency is 1 kHz and the on state time is 0.64 ms.The time constant =0.64/3 ms =0.213 ms

D R

CT

FD

iload

At transistor T turn off, the capacitor C charges and its voltage increases until the diode FDbecomes forward biased and thereafter the load current commutates to the freewheeling diode.

Load current Iload i=15 ASupply voltage 250 VLoad voltage 160 VCommutation time 0.012 ms250 V

Fdu

dtiC

dt

duCi 72.0

250

101215 6

160 V

2961072.0

102136

6

CRRC

Examination 2012-05-21 2c Forward converter with snubber

vC

Rload

VDcCdc

vS

CD2

D1D3

Z

L

S

N1 N2

C1

R1

Examination 20120521 2d

Examination 20120521 2e

Drift region n-

Examination 20120521 3a The buck converter as battery charger

198 VL I

L 4 mHR 0 ohmSwitching frequency 4 kHzPeriod time 0.25 msSample frequency 8 kHzSample time Ts 0.125 msLoad current 0-10 A

R

a3

Peak rectifier dc-voltage 220*1.414=311 VAverage recfier dc-voltage 311*2/3.14=198 VDuty cycle L 100/198=0.505Duty cycle H 100/311=0.322 On pulse 0.25 *0.505=0.13 ms On pulse 0.25 *0.322=0.08 ms

220 V, 50Hz

© Mats Alaküla L3 – Current Control (DC)

Current controller with fast computer - I

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)1()1()1()1(

kkeT

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T

dtedtdt

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equationGeneral

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Examination 20120521 3c The buck converter as battery charger

Constant 0 AVoltage ref with const 0 A =100 VDuty cycle 100/311=0.32On pulse 0.25 *0.32=0.08 msCurrent ripple =(311-100)/0.004*0.00008=4.24A

Load current 0 to 10 A Inductive voltage drop at current step =98 V Time to reach 10 A t=10*.004/98=0.408 msSample time =0.125 ms, so the current can notBe reached in one sample. Therefore the voltage ref at current step =198 V

Constant 10 ADuty cycle with 10 A = 100/198=0.505On pulse 0.25 *0.505=0.126 msVoltage ref at const 10 A=100 VInductive voltage drop at current step =98 VCurrent ripple=(198-100)/.004*.000126=3.09 A

10A

0A

0.408 msDuty cycle 0.32 Duty cycle 0.505

Uref

198 V

311V

0V

100V

Modulation

Phase current

Ud

4.24A

3.09A

0

dV

1V2V

Fig 1

as

d

sa

d

dddavgavg

ddavg

ddavg

as

d

as

d

as

d

sa

dLL

dddddL

d

dddddavgavg

dddavg

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spers

Lf

V

fL

ViripplecurrentMax

VVVVVe

VVV

VVVatvoltagesPhase

xatLf

V

x

iderivativeondsit

xwhenx

ix

Lf

V

x

iderivativesit

Lf

xxV

f

x

L

xVi

L

tVi

dt

diLVequationviarippleCurrent

xVVxVVeVVeVVinductoroverVoltage

V

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switchrisecurrentAt

VxVVxVVxVVemotoroverVoltage

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VxVvoltagesPhase

f

xTxtdurationpulseOn

xphaseratioControl

xphaseratioControl

2

5.05.012

005.0

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5.05.0max

5.0max04

sec'

5.00212

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212

122

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4

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2

3

4

Examination 20120521 4a


Examination 20120521 4b

a/b< a/b/cvect

d/q >a/bd-comp vectq-compangle

a/b >a/b/cid PI-controllerrefactemf

3-phase modulator

3-phase inverter

Ts/2-advance

a/b > d/qVectangle

a/b > d/qVectangle

Flux-angleFlux emf

angle

-1

iq PI-controllerrefactemf

3-phase current measurement

e e e

3-phase load

udclink

i*ab

iact

psi


Examination 20120521 5a

ym is the permanent magnetization along the positive x-axes

isx is the current along the permanent magnetizationisy is the current perpendicular to the magnetization, /2 in positive direction

Lmx is the inductance in the x-directionLmy is the inductance in the y-direction (The more iron and the smaller the air-gap in in the x- or y-direction, the higher is the inductance in that direction.

The permanent magnetization material has no impact to the inductance)

sysxmymxsymss iiLLxiixT yy


Examination 20120521 5b

sysxmymxsymss iiLLxiixT yy

See the torque equation, the first part of the torque is achieved when the permanent flux ym is multiplied with the current isy .The second part of the torque is the so called reluctance torque. E.g. At high speed the drive system is in field weakening, and the permanent magnetisation must be reduced, which is done with a negative isx. If Lmx<Lmy the difference Lmx-Lmy is negative. When this difference is multiplied with the negative isx and the positive isy the result is a positive torque, called the reluctance torque. The first torque and the reluctance torque are added to the total torque, which can be achieved with different combinations of isx and isy . The combination which gives the lowest sum isx + isy is the optimal combination of isx and isy for a certain torque.


Examination 20120521 5c

1. We want to increase the voltage, more than the available voltage.

2. This can be achieved by weaken the field further, by increasing the negative current isx. However, this results in an increased total current, beyond the max current loci.

3. So, we have to reduce the isy, to fullfill the the maximum current loci.

Documents

Solution Tenta 20120521.ppt [KompatibilitetslÃ¤ge] Tenta 20120521.pdf · 2018. 5. 28. · &5 5 W W ([DPLQDWLRQ F )RUZDUG FRQYHUWHU ZLWKVQXEEHU Y & 5 ORDG 9 'F & GF Y 6 & ' ' ' =