ELEC4240/9240
Power Electronics
University of New South Wales School of Electrical Engineering & Telecommunications Solution of Tutorial 4 1. id
L V d d
+ voi
+ v L
− C
D
−
io
i L
vo
V o
R (Load)
V d d = 5 - 40 V; f s = 20 kHz; V o = 5 V; I o = 1 A For just continuous conduction; I OB = I LB
=
DT s 2 L
( V d − V o )
The likelihood of discontinuous conduction is higher for lower D lower D..
(ii)
Dmin
=
Lmin
=
5 40
D 2 f s I oB
Given D = I OB
= 0.125
Vo 5 12.6
=
when V d d = 40 V
(Vd − Vo ) =
0.125 2 × 20000 × 1
( 40 − 5 ) = 109.3
µ F
= 5V , L = 109.3 µ H , Vd = 12.5V , I o = 1 A
= 0.397
D 2 f s L
( Vd
− VO ) = 0.68 A < I O
∴ Conduction is continuous.
Solutions to Tutorial 4
ST4-1
M. F. Rahman
ELEC4240/9240
Power Electronics
For continuous conduction, ∆VO
VO
=
∴C =
1 TS 2 ( 1 − D ) 8
LC
= 0.01
(1 − D)
1 8 f
2 S
× L × 0.01
= 172 µ F
Note: In this solution, we have assumed that ∆V O is caused by C only. In practice, the voltage drop across the equivalent series resistance (ESR) of the capacitor may contribute significantly to the ∆V O . This drop has been neglected. di L
(iii) During T on :
dt
di L
During T off :
dt
= =
Vd
− V O L
−V O L
= 0.0695A / µ sec .
= −0.0457 A / µ sec .
iripple
0.69A
69500A/sec
– 45700A/sec
- 0.69A
t
DT s D
= 5 / 12.5 = 0.397
∆i L
= DTS
Vd
− V O L
= 0.397 ×
1 f s
× 0.0695 × 10 6
= 1.38A i L i F0
- 45700 A/sec
0
∴
1 2
∆i L
t on
T s
t
= 1.38 / 2 = 0.69 A
Solutions of Tutorial 4
ST4-2
M. F. Rahman
ELEC4240/9240
Power Electronics
During tON , i F0
iL
= −0.69 + 69500t A
= 45700 × DTS + 0.69 = 45700 × 0.397 ×
1 20000
+ 0.69 = 1.597 A
∴ i L during off time = 1.597 − 45700 t 1
∆ I Lrms =
∫
TS
TS
0
=
iL2 dt
1
[
T S
∫
TON
0
( −0.69 + 69500t ) 2 dt
T S
+ ∫T
( 1.597
− 45700t ) 2 dt
ON
= 0.63A (iv)
i Lpeak i L
I o
0 t 1
t 2
di
During turn on,
∴
Vd
− VO L
i Lpeak
=
= IO
t1
DT S L
dt
(Vd
=
Vd
− V O L
and t1 =
− VO );
t
∆1T s
t on = DT s
(i)
T s
LI O
Vd
− V O
t2
= DTS +
L( i Lpeak
− IO )
V O
DT S (V − V ) − I Vo DTS (Vd − VO ) − LI OVO + L(Vd − VO ) × d O O L ∴ t2 − t 1 = (Vd − VO ) × VO ∴ ∆VO =
∴ ∆V O =
∆Q
C
=
1 C
1
× × ( iLpeak − I O ) × ( t 2 − t1 ) = 2
1 DT S 2C
L
(Vd
− VO ) − I O ( t 2 − t 1 )
[ DTS (Vd − VO ) − LI O ] × DTS (Vd − VO )VO − LI OVO + (Vd − VO ) {DTS (Vd − VO ) − LI O }
Solutions to Tutorial 4
2LCVO ( Vd ST4-3
− VO ) M. F. Rahman
ELEC4240/9240
Given: V O Vd
= 0.4 A
I O D
= 5V ,
VO
Vd T S 2L
2 × 20 × 10 3
= 2.889D ∆1
× 109 × 10 −6
0.4
=
2.889D
∴ 1.52D =
= 12.6V
(see Lecture notes or Mohan's equ 7.14)
D ∆1
12.6 × D ∆1
∴ ∆1 =
Vd
∴ ∆1 = 1.52D
D + ∆1
= 0.4 =
I O
=
=
Power Electronics
0.1384 D
0.1384
or D
D
0.1384
=
1.52
= 0.3018
= 21.5mV
∆VO
2. D
i L
id +
−
v L
For Vd
+
ic C
V d
I O
i D
V o
T
=
5 24
= 0.2083A
= 8V
∴
V O Vd
=
VO 24 8
= 24V ,
=3=
= 8 − 16V
Vd
fS
I o R (Load)
= 20kHz
1 1− D
∴ D = 2 / 3 = 0.667 I OB
=
VO T S 2L
D( 1 − D )2
From this, using I OB
(see Lecture notes or Mohan's equ 7.29)
= 0.2083A
L = 213µ H , For Vd
= 16V ,
Solutions of Tutorial 4
∴
V O Vd
=
24 16
=
1 1− D
,
D
= 0.333
ST4-4
M. F. Rahman
ELEC4240/9240
Power Electronics
∴ I OB = 0.2083 =
TSVO D( 1 − D ) 2 2L
L = 427 µ H To ensure that conduction is continuous, the desired L > 427 µ H . For this L, and D = 0.67, I OB
=
24 × 0.67 × (1 − 0.67 ) 2 20000 × 2 × 427 × 10 −6
= 0.102 A, which
is less than 0.2083 A.
Thus, with L = 427 µH, conduction will also be continuous when V d = 8V. Note that if discontinuous conduction is desired for all conditions of input, L < 213 µH must be chosen. For V d = 16V, D
I OB
∴
=
= 0.33
. If we choose L = 213 µH
24 × 0.33 × ( 1 − 0.33 ) 2 20000 × 2 × 213 × 10 −6
= 0.40 A > 0.208 A
Discontinuous conduction will guaranteed
when L is so chosen that conduction is
discontinuous when V d 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) Vd I OB
= 12V
VO
= 24V ,
L
= 427 µ H
f S
= 20kHz
= 5 / 24 = 0.2083A
Assuming continuous conduction, V O Vd
=
I OB =
24
1
12 1 − D VO T S 2L
,
∴ D = 0.5
D(1 - D)
2
=
24 × 0.5 × ( 1 − 0.5 ) 2 20000 × 2 × 427 × 10 −6
= 0.175 A < 0.75 A
∴ The inductor current is continuous . Diode current waveform is as shown below.
i D Solutions to Tutorial 4
i Dmax = 1.8512A
ST4-5
M. F. Rahman
i Dmin = 1.1488 A
ELEC4240/9240
1 2
∆ I L
=
=
I o
Now I L
Power Electronics
1 Vd
− V O
2
L
1− D
=
(1 − DTs ) = 0.3512A
0.75
= 1.5 A
1 − 0.5
and i L max
1
= I L + ∆i L = 1.5 + 0.3512 = 1.8512 A. 2
∴ i D max = iLmax = 1.8512 A = I L − 0.3512 = 1.1488
i D min
∆Q = Note:
∆V o Vo
C
=
A
( i D max − I o + iD min )( 1 − D ) Ts 2
1.5 × 0.5 × T s
=
2
= 18.75 × 10 −6 C
∆Q = I o × DTs = 18.57 × 10 −6 C
= 0.01 =
∆Q C
18.75 × 10−6 0.01 × 24
≈ 78
µ F
(iii) Taking the t = 0 reference at DT S , when the diode current begins, the diode current is given by:
i D
= iD max − 0.0281 × 106 t
∴ I DRMS =
1
T ∫
0
S
Or
I DRMS
=
(1 − D )T S
1 50 × 10
Solutions of Tutorial 4
−6
∫
( iD max
− 0.0281 × 10 6 t ) 2 dt
0.5×50×10 −6
0
( 1.8512 − 0.0281 × 10 6 t ) 2 dt
ST4-6
M. F. Rahman
ELEC4240/9240
1
=
50 × 10
−6
1
=
=
Power Electronics
∫
25×10 −6
0
( 3.4269 − 0.104 × 10 6
× t + 0.00079 ×10 12 × t 2 )dt 25×10 −6
3.4269t − 0.104 × 10 × t / 2 + 0.00079 × 10 × t / 3 0 6
50 × 10 −6
2
12
3
1
3.4269 × ( 25 × 10 −6 ) − 0.104 × 10 6 × ( 25 × 10 −6 ) 2 / 2 + 0.00079 ×10 12 ×( 25 ×10 −6 ) 3 / 3 50 × 10 −6
1
=
50 × 10
−6 ( 85.67
× 10 −6 − 32.5 × 10 −6 + 4.1145 × 10 −6 )
= 1.146 A = 1.07 A 2 I DRMS
=
I D,ripple, RMS
− I o2
=
1.07 2
− 0.752 = 0.763 A
(iv)
I dMAX I d I O
t 1 DT S
i Dmax
V d
=
L
di
During t off ,
Now
DT S
=
dt
− IO
i D max
∆V O
L( i D max
=
Vd ∆Q
C
− V O L
=−
t1
∴ t 1 =
Vd
di dt
=−
Vd
− VO
L
− IO )
− V O
=
Solutions to Tutorial 4
1 ( i D max
− I O )t1
2
C
− LI O )2 = 2LC(VO − Vd ) (Vd DTS
ST4-7
M. F. Rahman
ELEC4240/9240
With D
Power Electronics
= 0.5;
I OB
= 0.175 A . Since I O = 0.1 A < I OB , conduction must be discontinuous. For
continuous conduction. V O Vd
=
24 12
=
1
∴ D = 0.5
1−D
∴ D ≠ 0.5 For discontinuous conduction, D is given by,
D
=
4 VO VO ( 27 Vd Vd
I OBmax occurs for D
∴ I OBmax = = 0.074 ×
− 1)
I O I OB max
= 0.33
2 V O
[from equation 7.31, Mohan]
27 f S L 24
20000 × 427 × 10 −6
= 0.207 A
∴ D = 0.378
∴ ∆VO
− LVO )2 ( 12 × 0.378 × 50 ×10 −6 − 427 ×10 −6 ×0.1 ) 2 = = = 0.042 2LC(VO − Vd ) 2 × 427 × 10 −6 ×78 × 10 −6 ×( 24 − 12 ) (Vd DTS
Solutions of Tutorial 4
ST4-8
V
M. F. Rahman
ELEC4240/9240
3.
Power Electronics
Buck-Boost converter:
T V d
D
i D
i L
−
C
L id
v L
V o
R
+
I o
i L
V d
I L= I d + I o
T s
t
−V 0 t off
t on = DT s
Vd
= 40V
V O
=
VO
= −50V ,
PO
= 75W
fS
= 40kHz
∆VO
= 0.01VO
(i)
Vd (ii)
D 1− D
=
50 40
∴ D = 0.55
We assume that the inductor current is just continuous when the converter supplies 75W or I oB =
75 50
= 1.5A .
Note that discontinuous mode operation is normally preferred.
Solutions to Tutorial 4
ST4-9
M. F. Rahman
ELEC4240/9240
Power Electronics
∴ I OB = 1.5 =
V O f S 2L
VO ( 1 − D )2
L =
2 f S
1.5
( 1 − D )2
=
50 × ( 1 − 0.55 )2 40000 × 2 × 1.5
= 82.3µ H
Note that for large values of L, I OB would be smaller and operation will be in continuous mode when the load current is 1.5A.
∴ L = 82.3µ H will guarantee discontinuous conduction mode operation up to the 75W of load. ∆VO
= 0.01 =
VO
∴ C =
=
Now PO
DT S RC
D f S × R × 0.01 V 2O
∴R =
R
V 2O P O
= 33.33Ω
∴ C = 41.67 µ F
(iii) i L max
i L
I L
=
Id
+
I O
T S
DT S
−V O i L max
=
Vd DT S L
VDT S
I LB
=
∴
i Lmax
2L
=
,
iL min
= 0A
40 × 0.55 2 × 82.3 × 10 −6 × 40000
= 3.374 A
= 6.748 A
Solutions of Tutorial 4
ST4-10
M. F. Rahman
ELEC4240/9240
Power Electronics
(iv)
6.748A i D
i L
T S
DT S
1
6.748( 1 − D )T S
2
T S
I D
= Io = ×
I O
=
= 1.5 A
4.
(i)
2 15
= 0.133A
This is the current above which the converter enters continuous conduction mode.
∴ I OB = 0.133A =
∴ Lmin =
V O f S 2L
( 1 − D )2
V O ( 1 − D )2 2 fS
I OB
At the continuous /discontinuous boundary, D
=
Also: D
=
V O Vd
+ VO V O
Vd
+ VO
= =
15 15 + 8
= 0.652
15 15 + 40
= 0.273
for
Vd
for
= 8V
Vd
= 40V
Note that the smallest D gives the largest Lmin and hence the smallest I OB . ∴ L has to be selected for the smallest D. Lmin
=
15 × ( 1 − 0.273 ) 2 2 × 20 × 10 3
× 0.133
= 1.49mH
(ii) L
= 150 µ H ,
Vd
= 12V ,
IO
= 0.25 A
With continuous conduction, Solutions to Tutorial 4
ST4-11
M. F. Rahman
ELEC4240/9240
D
=
Power Electronics
V O
+ VO
Vd
V O
I OB =
2 Lf S
15
=
= 0.55
15 + 12
(1 - D)
=
2
15 × ( 1 − 0.55 ) 2 20000 × 2 × 150 × 10 −6
= 0.493
Since I o < I oB, conduction is discontinuous. V O
= I OBmax = 2.5 A
2Lf S
∴ I OB = I OB max ( 1 − D )2 With discontinuous conduction, i L V d
DT s
∆ 1 T s
t
-V o T s
− VO ∆1TS = 0 ,
Vd DTS
From
VO
=
Vd
D ∆1
I O
From
Pd
= PO ,
Also,
I O
= IL − Id =
∴
I d
( D + ∆1 ) I O ∆1
I O
∴
∆1
=
∴ D =
∴ D =
V d V O
= ∆1
so that D
=
=
∆1
D
V O V d
,
Vd D 2 fS L
(D + ∆1 ) -
Vd D V O VO 2 f S L
DI O ∆1
(D + ∆1 )
I OBmax D
I O I OB max
×
Vd VO
Vo
I O
V D
I OB max
Solutions of Tutorial 4
×
1 ∆1
=
IO I OB max
×(
V d VO
)2
×
ST4-12
1 D
M. F. Rahman
ELEC4240/9240
Power Electronics
I OBmax
= 2.5 A
∴ D =
15
0.25
12
2.5
= 0.395
(iii)
I tO i D
i D
,m a x
I O
∆1T S
∴ i D max = ∴ ∆1 =
Vd DT S L
2I D
=
i D max
I D
= Io =
1
∆1
=
2I o
=
i D max
2
=
12 150 × 10
2 × 0.25 1.58
−6
0.395
×
=
I D
T S
= 1.58 A
20000
= 0.316
iD max ∆1Ts / T s 2 × 0.25 1.58
= 0.316
The slope of i D
=−
V O L
∴ I to =
=− V O L
15 150 × 10
( D + ∆1 )TS
∴ I D = 3.55 − 10 5 t ∴ I D ,rms =
1 TS
∫
= 3.55 A
( for DTS
( D + ∆1 )TS
DTS
∴ I D ,ripple,rms = I CRMS =
Solutions to Tutorial 4
= −1 × 10 5 A / sec .
−6
2 D
i dt
2 I Drms
=
< t ≤ ( D + ∆1 )TS ) 1 50 × 10
− I O2 =
−6
∫
0.711×50 ×10
−6
−6 0.395×50 ×10
0.498 2
( 3.55 − 10 5 t )2 dt
= 0.498 A
− 0.25 2 = 0.43A
ST4-13
M. F. Rahman
ELEC4240/9240
Power Electronics
(iv) The diode current i D exceeds the load current I O , during t 1. The capacitor charges up during this time with current i D
− I O .
I D ,m ax
t 1 ∆1T S
t' = 0
i D max
Vd
=
R
The slope of current i D during t 1 i D
= iD max −
V O
i D
=
Vd DTS
−
i D
= IO =
∴ t1 = (
∆V O
=
∴ t1 =
VO L
Vd DTS L
Vd DT S
C
=
V O L
t '
−
VO L
− IO )
L ∆Q
=−
' t ' , by taking the origin of t at the time when i D starts to flow.
L
L
T S
V O
DTS ; I O =
L
I O
t 1
L V O
1 ( i D max
− I O )t1
2
C
i D max − iD di
=
1.58 − 0.25 10
5
=
1 / R( I OVd DTS
− L )(Vd DTS − LI O ) 2LC
= 13.3 × 10 −6 sec
dt ∆Q
∆VO
1
= × 13.3 × 10 −6 ×( 1.58 − 0.25 ) = 8.845 × 10 −6 C 2
=
∆Q
C
=
Solutions of Tutorial 4
8.845 × 10 −6 470 × 10 −6
= 18.8mV
ST4-14
M. F. Rahman
ELEC4240/9240
5.
Power Electronics
Cuk converter C 1
i L1
L1
id
+ v L1
-
L2
-
- v L2 +
+ vc1 -
V d
i L2
vo
D
T
C
V o
R (Load)
+ +
Vd
= 12V
V O
=
Vd
D 1− D V O
I O
=
I O
=
I d
R
∆i L1
I L1
= 0.6 ,
=
0.6
=
D
L1
12
=
= 2mH
L2
= 1mH
C
= 25 µ F
fS
= 25kHz
R
= 12 Ω
∴VO = 18V
1 − 0.6
18
1− D
∴ I d = I O VC1
D
I o
= 1.5 A 1 − 0.6
= 0.67
0.6
/ 0.67
= 2.24 A
= VO + Vd = 18 + 12 = 30V
=
VC1
− V d
L1
( 1 − D )TS
=
30 − 12 2 × 10
−3
× ( 1 − 0.6 ) ×
1 25000
= 0.144 A
= I d = 2.24 A
∴ i L1,MAX = I d +
1
∴ i L1,MIN = I d −
1
∆i L2
=
V O L2
2
2
= 2.24 + × 0.144 = 2.312 A
∆iL1
= 2.24 − × 0.144 = 2.168 A
( 1 − D )TS
Solutions to Tutorial 4
1
∆iL1
2
=
1 2
15 1 × 10 −3
× ( 1 − 0.6 ) ×
1 25000
ST4-15
= 0.24 A
M. F. Rahman
ELEC4240/9240
Power Electronics
C1
i L1 L
i L2 L
1
V d
D
T
−
2
C
R
V O
+ i L1
i L2 DT s
t
(1–D)T s T s
V d
DT s V d – V C1 V C1 – V o
DT s
−V O 1
∴ i L2 ,MAX = I O + ∴ i L1,MIN = I d − I L2
2 1 2
∆i L2
∆iL1
T s
1
= 1.5 + × 0.24 = 1.62 A 2
1
= 1.5 − × 0.24 = 1.38 A 2
= I o = 1.5 A
(iv) Note that the output circuit comprising of D,L2 ,C and R is very similar to the output stage of the buck converter. Thus, from the Buck converter analysis, ∆V O
VO
=
1− D 8L2Cf S 2
∴ ∆VO =
( 1 − 0.6 ) × 15 8 × 1 × 10
Solutions of Tutorial 4
−3
× 25 × 10 −6 × 25 2 × 10 6
ST4-16
= 0.048V = 48mV
peak
− peak
M. F. Rahman
