tutorial 4 power electronics
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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 −==
ELEC4240/9240 Power Electronics
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
ELEC4240/9240 Power Electronics
Solutions to Tutorial 4 ST4-3 M. F. Rahman
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
ELEC4240/9240 Power Electronics
Solutions of Tutorial 4 ST4-4 M. F. Rahman
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
ELEC4240/9240 Power Electronics
Solutions to Tutorial 4 ST4-5 M. F. Rahman
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
ELEC4240/9240 Power Electronics
Solutions of Tutorial 4 ST4-6 M. F. Rahman
( )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
−× ×
−= − ×× ∫
ELEC4240/9240 Power Electronics
Solutions to Tutorial 4 ST4-7 M. F. Rahman
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
ELEC4240/9240 Power Electronics
Solutions of Tutorial 4 ST4-8 M. F. Rahman
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 )
∆− −
− −
− × × × − × ×∴ = = =
− × × × × × −
ELEC4240/9240 Power Electronics
Solutions to Tutorial 4 ST4-9 M. F. Rahman
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
ELEC4240/9240 Power Electronics
Solutions of Tutorial 4 ST4-10 M. F. Rahman
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
ELEC4240/9240 Power Electronics
Solutions to Tutorial 4 ST4-11 M. F. Rahman
(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
ELEC4240/9240 Power Electronics
Solutions of Tutorial 4 ST4-12 M. F. Rahman
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
∴ =
ELEC4240/9240 Power Electronics
Solutions to Tutorial 4 ST4-13 M. F. Rahman
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
ELEC4240/9240 Power Electronics
Solutions of Tutorial 4 ST4-14 M. F. Rahman
(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
ELEC4240/9240 Power Electronics
Solutions to Tutorial 4 ST4-15 M. F. Rahman
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
∆ −= − = × − × =×
ELEC4240/9240 Power Electronics
Solutions of Tutorial 4 ST4-16 M. F. Rahman
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
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