Power electronics Solution to examination 2009-12 … Tenta 20091214.pdf · Industrial Electrical Engineering and Automation. Lund University, Sweden. Power electronics Solution to

Industrial Electrical Engineering and AutomationLund University, Sweden

Power electronics Solution to examination 2009-12-14

© Namn Namn Föredragstitel

Examination 20091214 1a


Examination 20091214 1b

VVUdc 5662400max Maximum dc voltage

Average dc voltage VVU avedc 54024003_

Margin to maximum voltage 1500V = 934 V


Examination 20091214 1c

Phase current

-500-400-300-200-100

0100200300400500

0 0,002 0,004 0,006 0,008 0,01 0,012 0,014 0,016 0,018 0,02

time [s]

[A]


Examination 20091214 1c cont’d

A

dxxdxxI rmsdiode

1152

)0sin(02

)2sin(112

400

2)2cos(1

122400)(sin400

02.000167.02

00

22

AdxxdxxIavediode

4.42cos0cos3200)sin(

3200)sin(400

02.000167.02

00

Rectifier diode rms-current

Rectifier diode average current


Examination 20091214 1c cont’d

Rectifier diode Threshold voltage 0.95 V Differential resistance 2.5 mohm Irms 115 A Average current 42,4 A

WIRIVP rmsdiffavethresholddioderectifier 7.731150025.04.4295.0 22 Rectifier diode power loss


Examination 20091214 1d

0

0,5

1

1,5

2

2,5

3

0 0,002 0,004 0,006 0,008 0,01 0,012 0,014 0,016 0,018 0,02

AdxxIdc 127cos0cos200)sin(02.0

40000167.06

0

Dc current to the dc link and to the 4QC

Duty cycle 80% IGBT and diode on state current 127/0.8= 159A Conduction percentage of IGBT transistor (incl freewheeling) 80+20/2= 90% Conduction percentage of IGBT diode (when freewheeling) 20/2= 10%

Switching frequency 2,5 kHz


Examination 20091214 1d cont’dIGBT transistor Threshold voltage 1.3 V Differential resistance 15 mohm On state voltage at 159 A 3.68 V Turn on energy at 900 V and 180 A 63 mJ Turn off energy at 900 V and 180 A 80 mJ IGBT diode Threshold voltage 1.05 V

Differential resistance 9.7 mohm On state voltage at 149 A 2.59 V Turn on energy at 900 V and 180 A 0 mJ Turn off energy at 900 V and 180 A 20 mJ Power loss

WP

WP

lossdiode

losstrans

68180900

15954002.025001.015959.2

716180900

159540143.025009.015968.3

_

_


Examination 20091214 1e

Rectifier diode (6) Loss each 73,7W Rth diode 0.13 K/W Temp diff 9.6 deg

IGBT transistor (2) Loss each 716 W Rth trans 0.069 K/W Temp diff 49 deg

Heatsink Contribution fron 6 rectifier diodes and from two IGBT. Ambient temperature 42 deg Total loss to heatsink 6*73,7+2*715+2*68=2008 W Temp diff over heatsink 2008*0.03=60 .2 deg

Junction temperature Rectifier diode 42+60.2+9.6=112 deg IGBT diode 42+60.2+17=119 deg IGBT transistor 42+60.2+49=151 deg

IGBT diode (2) Loss each 68 W Rth diode 0.25 K/W Temp diff 17 deg

Examination 20091214 2a The buck converter, ”step down chopper”

Vload

VDc

Examination 20091214 2b The buck converter with RCD snubber

D R

C

T

FD

i

At turn off of transistor T, the current i commtutates over to the capacitor C via diode D. The capacitor C charges until the potential of the transistor emitter reduces till the diode FD becomes forward biased and thereafter the load current iload flows through diode FD and the current i=0. A turn on of the transistor T, the capacitor C is discharged via the the transistor T and resistor R. The diode FD becomes reverse biased and the current i commutates to the transistor T.

iload

Examination 20091214 2c Fly-back converter with snubbers and

common mode (CM) and differential mode (DM) filter

vD

DM-filter CM-filter

double voltage

Examination 20091214 2d Fly-back converter withSnubber operation

L1

C1

D2

R1

L2

C2

R2

T1

D3

L3

C3

R3

D0

ip is

For the description of the snubber operation the stray inductance L1 between the switch transistor and the supply/dclink, and the transformer leakage inductance, L2 on primary side and L3 on secondary side are added as discrete component in the circuit drawing above.

Examination 20091214 2d cont’d Fly-back converter withSnubber operation

When transistor T1 is conducting, the current ip >0 and is =0 in ideally steady state mode.

Snubber 1. When transistor T1 is turned off the current ip would directly be equal to zero with a high derivative, and thus a high voltage spike will be found over the transistor T1 due to the stray inductance L1 . However, the RC-snubber R1 /C1 offers an alternative current path, which reduces the current derivative and thus the voltage spike, which reduces the risk of destroyed transistor. When transistor T1 is turned on, the capacitor C1 is discharged via the resistor R1 and the transistor T1 .

Snubber 2 When transistor T1 is turned off the current ip would directly be equal to zero with a high derivative, and thus a high voltage spike will be found over the transformer due to the transformer primary leakage inductance L2 . However, the RCD-snubber R2 /C2 /D2 offers an alternative current path, which reduces the current derivative and thus the voltage spike. The capacitor C2 is charged via the diode D2 . When transistor T1 is turned on, the capacitor C2 is discharged via the resistor R2 , the dclink and the switch transistor T1 .

Snubber 3. When transistor T1 is turned on, , the secondary voltage >0 and diode D0 is reverse biased. Due to reverse recovery a current will flow from cathode to anode in the diode D0 during a short time. When the charged carriers in the diode have disappeared the current would become zero, and the transformer secondary leakage inductance L3 would create a high voltage spike over the diode D0 . However, as the snubber offers an alternative current path R3 /C3 the current derivative is reduced, and the voltage spike is reduced. During this process the capacitor C3 is charged. When transistor T1 is turned off the current ip =0 and is >0, and the latter shall flow through the diode D0 , but as the capacitor C3 is charged the diode D0 is reverse biased. To start the current flows through the snubber path D3 /C3 . During this time the capacitor C3 discharges until the diode D0 is forward biased and thn the current commutates over to diode D0 .

Examination 20091214 2e The MOSFET diffusion layer

Source

Drain

Gate

p n+

body

n+ p

body

n- drift region


Examination 20091214 2f

The drift region, doped n-

Examination 20091214 3a, 3b The buck converter as battery charger

400 V, 11 kW -> I=27.5 A ripple current 4.1 A

540 VL I

Switching frequency 2 kHz Period time 0.5 msDuty cycle withj 27.5 A =0.79 On pulse 0.5 *0.79=0.395 ms

mHLLdtdiLUb 11

1.4000395.05.112

000395.01.45.4275403

R

Examination 20091214 3c The buck converter as battery charger

Assume resistance=1 ohm Phase voltage with 0 A=400 Vdc Duty cycle with 0 A=0.74On pulse 0.5 *0.74=0.37 msRise di/dt with 0 A (540-400)/0.011=12.7Fall di/dt with 0 A -400/0.011=-36.4

Current step to 13.75 APhase voltage with 13.75 A=413.75 V Duty cycle with 13.75 A=0.77 On pulse 0.5 *0.77=0.385 msRise di/dt with 13.75 A (540-4 13.75)/0.011=11.5Fall di/dt with 0 A -4 13.75 /0.011=-37.6


Examination 20091214 4a

)3

4(ˆ

)3

2(ˆ

)(ˆ

tCOSee

tCOSee

tCOSee

c

b

a

tj

tj

j

c

j

ba

eE

eetjte

jttejjtjjte

jttjttte

jttjttte

jtjtteeeeeee

23ˆsincos

23ˆ

23sin

23cos

32ˆ

43

43

43

43sin

43

41

43

411cos

32ˆ

23

21

23sin

21cos

23

21

23sin

21coscos

32ˆ

23

21

34sinsin

34coscos

23

21

32sinsin

32coscoscos

32ˆ

23

21

34cos

23

21

32coscos

32ˆ

32 3

43

2


Examination 20091214 4b

< a/b/c vect

d/q >

d-comp vect q-comp angle

>a/b/cid PI-controller ref act emf

3-phase modulator

3-phase inverter

Ts/2-advance

> d/qVect angle

> d/qVect angle

Flux-angleFlux emf

angle

-1

iq PI-controller ref act emf

3-phase current measurement

e e e

3-phase load

udclink

i*

iact

psi


Examination 20091214 5

NmPTorque

NmPTPTorquee

HzfreqElecfreqswitchingleastatHzfreqElecfreqswitchingleastatd

HzfreqsampingperiodfreqpertwiceSamplecHzpolesfreqElec

HzfreqMechc

AIIassumeb

VUVamplitudevoltagePhasea

rpm

rpm

linetolinerms

8.39

60120002

50000

119

6040002

500005

5400.31800.5

36005180018.

20060

120005

1059.03063

500009.030639.0cos500005

30632

2502502

5005

12000

4000

___

Documents

Power electronics Solution to examination 2009-12 … Tenta 20091214.pdf · Industrial Electrical Engineering and Automation. Lund University, Sweden. Power electronics Solution to