Tutorial 4 Power Electronics

ELEC4240/9240 Power Electronics

Solutions to Tutorial 4 ST4-1 M. F. Rahman

University of New South Wales School of Electrical Engineering & Telecommunications

Solution of Tutorial 4

1.

Vd

R (Load)

io

VoC

iL L

+ vL −

D

id

+ voi

− vo

Vd = 5 - 40 V; fs = 20 kHz; Vo = 5 V; Io = 1 A

For just continuous conduction;

The likelihood of discontinuous conduction is higher for lower D.

min5D 0.125

40= = when Vd = 40 V

( ) ( )min d os oB

D 0.125L V V 40 5 109.3 F2 f I 2 20000 1

µ= − = − =× ×

(ii) Given o d oV 5V , L 109.3 H , V 12.5V , I 1Aµ= = = =

Conduction is continuous.

OB d O Os

5D 0.39712.6

DI (V V ) 0.68 A I2 f L

= =

= − = <

∴

)VV(L2

DTII ods

LBOB −==


Solutions of Tutorial 4 ST4-2 M. F. Rahman

For continuous conduction,

2O S

O

2S

V T ( 1 D )1 0.01V 8 LC

1 ( 1 D )C 172 F8 f L 0.01

∆

µ

−= =

−∴ = =

× ×

Note: In this solution, we have assumed that OV∆ is caused by C only. In practice, the voltage drop across the equivalent series resistance (ESR) of the capacitor may contribute significantly to the OV∆ . This drop has been neglected.

(iii) During Ton : d OL V Vdi 0.0695A / sec .dt L

µ−= =

During Toff : OL Vdi 0.0457 A / sec .dt L

µ−= = −

6d OL S

s

D 5 / 12.5 0.397

V V 1i DT 0.397 0.0695 10L f

1.38 A

∆

= =

−= = × × ×

=

ton Ts t

iF0

0

- 45700 A/sec

iL

L1 i 1.38 / 2 0.69 A2∆∴ = =

t

- 0.69A

0.69A– 45700A/sec

sDT

69500A/sec

iripple



During AON L t , i 0.69 69500t = − +

S A

during off time

F0

L

1i 45700 DT 0.69 45700 0.397 0.69 1.59720000

i 1.597 45700 t

= × + = × × + =

∴ = −

S ON S

ON

T T T2 2 2Lrms L0 0 T

S S

1 1I i dt [ ( 0.69 69500t ) dt ( 1.597 45700t ) dtT T

0.63A

∆ = = − + + −

=

∫ ∫ ∫

(iv)

(i) During turn on, d OV Vdidt L

−=

d O O1 O 1

d O

V V LIt I and tL V V−

∴ = =−

Lpeak OSLpeak d O 2 S

O

L( i I )DTi (V V ); t DTL V

−= − = +

So S d O O O d O d O O

2 1d O O

DTV DT (V V ) LI V L(V V ) (V V ) I

Lt t

(V V ) V

− − + − × − − ∴ − =− ×

SO Lpeak O 2 1 d O O 2 1

DTQ 1 1 1V ( i I ) ( t t ) (V V ) I ( t t )C C 2 2C L∆∆ ∴ = = × × − × − = − − −

[ ] S d O O S d O O O O d O S d O O

OO d O

DT (V V ) LI DT (V V )V LI V (V V ) DT (V V ) LIV

2LCV (V V )∆

− − × − − + − − − ∴ =−

iLpeak

Ts

ton = DTs s1T∆

iL

t1 t2 t

0

Io



Given: O O dI 0.4 A V 5V , V 12.6V= = =

O1

d 1

V D 1.52DV D

∆∆

= ∴ =+

d SO 1

V TI 0.4 D2L

∆= = (see Lecture notes or Mohan's equ 7.14)

113 6

12.6 D 2.889D2 20 10 109 10

∆ ∆−

×=

× × × ×=

10.4 0.1384

2.889D D∆∴ = =

0.1384 0.13841.52D or D 0.3018D 1.52

∴ = = =

OV 21.5mV∆ = 2.

O O d S5I 0.2083A V 24V , V 8 16V f 20kHz24

= = = = − =

For Od

d

V 24 1V 8V 3V 8 1 D

= ∴ = = =−

D 2 / 3 0.667∴ = =

2O SOB

V TI D( 1 D )2L

= − (see Lecture notes or Mohan's equ 7.29)

From this, using OBI 0.2083A= L 213 H ,µ=

For Od

d

V 24 1V 16V , , D 0.333V 16 1 D

= ∴ = = =−

R (Load)

+

iD

T

+ vL −

iL

Vo

C Vd

ic

D

Io

id



2S O

OBT V D( 1 D )I 0.2083

2L

L 427 Hµ

−∴ = =

=

To ensure that conduction is continuous, the desired L 427 Hµ> . For this L, and D = 0.67,

A which is less than A2

OB 6

24 0.67 ( 1 0.67 )I 0.102 , 0.2083 .20000 2 427 10−

× × −= =

× × ×

Thus, with L = 427 µH, conduction will also be continuous when Vd = 8V. Note that if discontinuous conduction is desired for all conditions of input, L < 213 µH must be chosen. For Vd = 16V, D 0.33= . If we choose L = 213 µH

2

OB 6

24 0.33 ( 1 0.33 )I 0.40 A 0.208 A20000 2 213 10−

× × −= = >

× × ×

∴ Discontinuous conduction will guaranteed when L is so chosen that conduction is discontinuous when Vd is 8V. Hence, L ≥ 427 µH must be chosen for operation in continuous conduction. If discontinuous conduction is required, as is normally for the boost converter, we should choose L < 213 µH. (ii)

d O S

OB

V 12V V 24V , L 427 H f 20kHz

I 5 / 24 0.2083A

µ= = = =

= =

Assuming continuous conduction,

O

d

V 24 1 , D 0.5V 12 1 D

= ∴ =−

A A

The inductor current is continuous

22O S

OB 6

V T 24 0.5 ( 1 0.5 )I = D(1 - D) 0.175 0.752 L 20000 2 427 10

.

−

× × −= = <

× × ×

∴

Diode current waveform is as shown below.

iDmax = 1.8512A

Io = 0.75A

iD

iDmin = 1.1488 A



( )d OL s

V V1 1I 1 DT 0.3512A2 2 L∆

−= − =

Now oL

I 0.75I 1.5A1 D 1 0.5

= = =− −

and L max L L1i I i2

= + ∆ = 1.5 + 0.3512 = 1.8512 A.

D max L maxi i 1.8512 A∴ = = D min Li I 0.3512 1.1488= − = A

( ) ( )D max o D min s 6si I i 1 D T 1.5 0.5 T

Q 18.75 102 2

−− + − × ×∆ = = = × C

Note: 6

o sQ I DT 18.57 10−∆ = × = × C

o

o

V Q0.01V C∆ ∆

= =

618.75 10C 78 F

0.01 24µ

−×= ≈

×

(iii) Taking the t = 0 reference at SDT , when the diode current begins, the diode current is

given by: 6

D D maxi i 0.0281 10 t= − ×

S( 1 D )T 6 2DRMS D max0

S

1I ( i 0.0281 10 t ) dtT

−∴ = − ×∫

60.5 50 10 6 2

DRMS 6 0

1Or I ( 1.8512 0.0281 10 t ) dt50 10

−× ×

−= − ×× ∫



625 10 6 12 26 0

1 ( 3.4269 0.104 10 t 0.00079 10 t )dt50 10

−×

−= − × × + × ×× ∫

625 106 2 12 3

6 0

1 3.4269t 0.104 10 t / 2 0.00079 10 t / 350 10

−×

− = − × × + × × ×

6 6 6 2 12 6 36

1 3.4269 ( 25 10 ) 0.104 10 ( 25 10 ) / 2 0.00079 10 ( 25 10 ) / 350 10

− − −− = × × − × × × + × × × ×

6 6 66

1 ( 85.67 10 32.5 10 4.1145 10 )50 10

− − −−= × − × + ×

×

1.146 A 1.07 A= =

2 2D,ripple,RMS DRMS oI I I= − = 2 21.07 0.75 0.763 A− =

(iv)

dD max S

Vi DTL

=

During offt , d OV Vdidt L

−=

Now D max O d O

1

i I V Vdit dt L− −

= − = −

D max O1

d O

L( i I )tV V

−∴ =

−

2

D max O 1 d S OO

O d

( i I )t (V DT LI )Q 1VC 2 C 2LC(V V )∆∆ − −

= = =−

1t

SDT

dI

dMAXI

OI



With OBD 0.5; I 0.175 A= = . Since O OBI 0.1A I= < , conduction must be discontinuous. For continuous conduction.

O

d

V 24 1 D 0.5V 12 1 D

= = ∴ =−

D 0.5∴ ≠ For discontinuous conduction, D is given by,

O O O

d d OB max

V V I4D ( 1 )27 V V I

= −

OB maxI occurs for D 0.33=

OOBmax

S

V2I = 27 f L

∴ [from equation 7.31, Mohan]

6

240.074 0.207 A20000 427 10

D 0.378

−= × =× ×

∴ =

2 6 6 2

d S OO 6 6

O d

(V DT LV ) ( 12 0.378 50 10 427 10 0.1)V 0.042 V2LC(V V ) 2 427 10 78 10 ( 24 12 )

∆− −

− −

− × × × − × ×∴ = = =

− × × × × × −



3. Buck-Boost converter:

D

C R L

T

Vd iL −

Vo

+ Io id

iD

d O O S O OV 40V V 50V , P 75W f 40kHz V 0.01V∆= = − = = = (i)

O

d

V D 50 D 0.55V 1 D 40

= = ∴ =−

(ii) We assume that the inductor current is just continuous when the converter supplies 75W or

IoB = 75 1.5A50

= .

Note that discontinuous mode operation is normally preferred.

Vd Li

0V−

sT

ton = DTs toff

t

vL

IL= Id + Io



2OOB

S

2 2O

S

VI 1.5 ( 1 D )f 2L

V ( 1 D ) 50 ( 1 0.55 )L 82.3 H2 f 1.5 40000 2 1.5

µ

∴ = = −

− × −= = =

× ×

Note that for large values of OBL, I would be smaller and operation will be in continuous mode when the load current is 1.5A. ∴ L 82.3 Hµ= will guarantee discontinuous conduction mode operation up to the 75W of load.

O S

O

S

V DT0.01

V RCDC

f R 0.01

∆= =

∴ =× ×

Now2 2

O OO

O

V VP R 33.33R P

Ω= ∴ = =

C 41.67 Fµ∴ = (iii)

d SLmax Lmin

V DTi , i 0 AL

= =

SLB 6

Lmax

VDT 40 0.55I 3.374 A2L 2 82.3 10 40000

i 6.748 A

−

×= = =

× × ×

∴ =

Li

SDTST

L d OI I I= +

OV−

L maxi



(iv)

SD o

S

6.748( 1 D )T1I I 1.5A2 T

−= = × =

4.

O2I 0.133A

15= =

(i) This is the current above which the converter enters continuous conduction mode.

2OOB

S

2O

minS OB

VI 0.133A ( 1 D )f 2L

V ( 1 D )L2 f I

∴ = = −

−∴ =

At the continuous /discontinuous boundary,

Od

d O

V 15D 0.652 for V 8VV V 15 8

= = = =+ +

Also: Od

d O

V 15D 0.273 for V 40VV V 15 40

= = = =+ +

Note that the smallest D gives the largest minL and hence the smallest OBI . ∴ L has to be selected for the smallest D.

2

min 3

15 ( 1 0.273 )L 1.49mH2 20 10 0.133

× −= =

× × ×

(ii) d OL 150 H , V 12V , I 0.25Aµ= = = With continuous conduction,

Li

SDTST

6.748 A

Di



O

d O

V 15D 0.55V V 15 12

= = =+ +

2

2OOB 6

S

V 15 ( 1 0.55 )I = (1 - D) 0.4932 Lf 20000 2 150 10−

× −= =

× × ×

Since Io < IoB, conduction is discontinuous.

OOB max

S

2OB OB max

V I 2.5A2Lf

I I ( 1 D )

= =

∴ = −

With discontinuous conduction,

iLVd

DTs ∆1Ts

Ts

t

-Vo

From d S O 1 SV DT V T 0∆− = ,

O O1

d 1 d

V VD so that DV V

∆∆

= =

From O 1d O

d

IP P ,I D

∆= = ,

Also, d OO L d 1

S 1

V D DII I I D2 f L

∆∆

= − = + ( ) -

d O1O 1

1 O S

V D VD I DV 2 f L

∆ ∆∆+

∴ = +( ) ( )

O dOB max

1 O

I V I DV∆

∴ =

2O d O d

OB max O 1 OB max O

I V I V1 1D ( )I V I V D∆

∴ = × × = × ×

o O

D OB max

V IDV I

∴ =



OB maxI 2.5A=

15 0.25D 0.39512 2.5

∴ = =

(iii)

d SD max 6

V DT 12 0.395i 1.58 AL 150 10 20000−∴ = = × =

×

D1

D max

2I 2 0.25 0.316i 1.58

∆ ×∴ = = =

D o D max 1 s s1I I i T / T2

∆= =

o1

D max

2I 2 0.25 0.316i 1.58

∆ ×= = =

The slope of iD

5O6

V 15 1 10 A / sec .L 150 10−= − = − = − ×

×

Oto 1 S

VI ( D )T 3.55AL

∆∴ = + =

5

D S 1 SI 3.55 10 t ( for DT t ( D )T )∆∴ = − < ≤ +

61 S

6S

( D )T 0.711 50 102 5 2D,rms D 6DT 0.395 50 10

S

2 2 2 2D,ripple,rms CRMS Drms O

1 1I i dt ( 3.55 10 t ) dt 0.498 AT 50 10

I I I I 0.498 0.25 0.43A

∆ −

−

+ × ×

− × ×∴ = = − =

×

∴ = = − = − =

∫ ∫

ST

D ,m a xi

1 ST∆

O DI I=

tOI

Di



(iv) The diode current Di exceeds the load current OI , during t1. The capacitor charges up during this time with current D Oi I− .

d OD max S O

V Vi DT ; IL R

= =

The slope of current Di during O1

VtL

= −

'OD D max

Vi i tL

= − , by taking the origin of 't at the time when Di starts to flow.

'd S OD

V DT Vi tL L

= −

d S OD O 1

V DT Vi I tL L

= = −

d S1 O

O

V DT Lt ( I )L V

∴ = −

D max O 1 O d S d S OO

( i I )t 1 / R( I V DT L )(V DT LI )Q 1VC 2 C 2LC∆∆ − − −

= = =

6D max D1 5

i i 1.58 0.25t 13.3 10 secdi 10dt

−− −∴ = = = ×

6 61Q 13.3 10 ( 1.58 0.25 ) 8.845 10 C2

∆ − −= × × × − = ×

6

O 6

Q 8.845 10V 18.8mVC 470 10∆∆

−

−

×= = =

×

ST1 ST∆

OI

D ,m axI

't 0=

1t



5. Cuk converter

L1 iL1 C1 L2 iL2

+ vL1 - + vc1 - - vL2 +

-vo

+

C R (Load)

+

Vd Vo

id

-

DT

Io

d 1 2 SV 12V D 0.6, L 2mH L 1mH C 25 F f 25kHz R 12µ Ω= = = = = = =

OO

d

V D 0.6 V 18VV 1 D 1 0.6

= = ∴ =− −

OO

O

d

d O

C1 O d

V 18I 1.5AR 12

I 1 D 1 0.6 0.67I D 0.6

I I / 0.67 2.24 A

V V V 18 12 30V

= = =

− −= = =

∴ = =

= + = + =

C1 dL1 S 3

1

L1 d

V V 30 12 1i ( 1 D )T ( 1 0.6 ) 0.144 AL 2 10 25000

I I 2.24 A

∆ −

− −= − = × − × =

×

= =

L1,MAX d L11 1i I i 2.24 0.144 2.312A2 2∆∴ = + = + × =

L1,MIN d L11 1i I i 2.24 0.144 2.168 A2 2∆∴ = − = − × =

OL2 S 3

2

V 15 1i ( 1 D )T ( 1 0.6 ) 0.24 AL 1 10 25000

∆ −= − = × − × =×



L2,MAX O L2

L1,MIN d L1

1 1i I i 1.5 0.24 1.62A2 2

1 1i I i 1.5 0.24 1.38 A2 2

∆

∆

∴ = + = + × =

∴ = − = − × =

L2 oI I 1.5 A= = (iv) Note that the output circuit comprising of 2D,L ,C and R is very similar to the output

stage of the buck converter. Thus, from the Buck converter analysis,

O2

O 2 S

V 1 DV 8L Cf∆ −

=

O 3 6 2 6

( 1 0.6 ) 15V 0.048V 48mV peak peak8 1 10 25 10 25 10

∆ − −

− ×∴ = = = −

× × × × × ×

R O V T

+

D

L1i L2i

C

1 L 2L

Vd

−

C1

L1i

L2i

OV−

Vd – VC1

VC1 – Vo

DTs (1–D)Ts

Vd

Ts

t

Ts

DTs

DTs

Tutorial 4 Power Electronics

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