Transcript
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CHAPTER 5
DC-DC CONVERTERS
Problem 5-1
Vs = 220 V, k = 0.8, R = 20 Q and vch = 1.5 V.
a) From Eq. (5-1), Va = 0.8 x 220 - 1.5) = 174.8 V
b) From Eq. (5-2), V0 = V0.8 x 220 = V
c) From Eq. (5-5), P0 = 0.8 x 220 - 1.5) 2 /20 = 3819.4 W
From Eq. (5-6), P I = 0.8 x 220 x 220 -1.5)/20 = 3845.6 W
The chopper efficiency is P 0/Pi = 3819.4/3845.6 = 99.32 d) From Eq. (5-4), R = 20/0.8 = 12.5 Q
e) From Eq. (5-8), the output voltage is
v 0 = [sin 2;rx 0.8) cos 2 ; r x l O O O O / ) +0.69 l x sin 2 ; rx lOOOO/) ]= 8 3 x sin (62832*+ )
where |> = tan' 1 [sin(0.8x27i)/0.691] = 54
The rms value is Vi = 82.32/V2 = 58.2 V
Note: The efficiency calculation, wh ich includes the conduction loss of the
chopper, does not take into account the switching loss du e to turn-on and
turn-off of the converter.
Problem 5-2
Vs = 220 V, R = 10 Q, L = 15.5 mH, E = 20 V, k = 0.5 and f = 5000 Hz
From Eq. (5-15), I 2 = 0.9375 i 1.2496
From Eq. (5-16), i = 0.9375 I 2 - 1.2496
a) Solving these two equations , i = 8.6453 A
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b) I 2 = 9.3547 A
c) Al = I 2 - i = 9.35475 - 8.6453 = 0.7094 A
From Eq. 5-21), Al max = 0.7094 A
and Eq. 5-22) gives the approximately value, Almax = 0.7097 A d) The average load current is approximately,
Ia = h + I i /2 = 9.35475 + 8.6453)/2 = 9 A
e) From Eq. 5-24),-il/2
= 9.002 A
f) I s = k I a = 0.8 x 9 = 7.2 A
and the input resistance is \ V S /IS = 220/7.2 = 30.56 Q
g) From Eq. 5-25), I R = V k I = V0 .8 x 22.1 = 15.63 A
Problem 5-3
Vs = 220 V, R = 0.2 Q, E = 10 V, f = 200 Hz, T = 1/f = 0.005 SAi = 200 x 0.5 = 10 A.
Va = k Vs = R a
The voltage across the inductor is given by
For a linear rise of current, dt = ti = kT and di = Ai
A / = -
For worst case ripple condition: = 0
and this gives, k = 0.5
Ai L = 10 x L = 220 1 - 0.5) 0.5 x 0.005 or L = 27.5 mH43
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Problem 5-4
Vs = 110 V, E = 220 V, Po = 30 kW = 30000 W c) Since the input power must be the same as the output power,
Vs I s = Po or 110 x I s = 30000 or I s = A
a) The battery current, I b = P 0/E = 30000/220 = 136.36 A
I b = 1 - k) I s or k = 136.36/272.73 - 1 = 0.5
b) Rch = 1 - k) E/Is =d 0.5) x 220/272.73 = 0.4033 Q
Problem 5-5
Vs = 110 V, L = 7.5 mH, E = 220 V
From Eq. 5-28), U t) = 110 x 10 3 /7.5) t + i
From Eq. 5-29),
i2 t = [ 110 - 220) x 10 3 /7.5) t + I 2 = - 110 x 10 3 /7.5) t + I 2
where I = ii t=kT) = 110 x 10 3 /7.5) kT + i
L = | 2 [t= l-k) kT] = -110 x 10 3 /7.5) l-k)kT + I
Solving for i and I yields i = 0, I = 110 x 10 3 /7.5) kT
ii t) = 110 x 10 3 /7.5) t, for 0 < t < kT
I2 t) = - 110 x 10 3 /7.5)t + 110 x 10 3 /7.5) l-k)T, for 0 < t < 1 - k) T.
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Problem 5-6
Vs = 600 V, R = 0.25 Q L = 20 mH, E = 150 V, k = 0.1 to 0.9 and f = 250
Hz
For k=0.1, the load current is discontinuous
From Eq. 5-15), I = 8.977
ro Eq. 5-16), i = 0, AI = 8.977 A and I a = 4.4885 A
For k=0.2, the load current is discontinuous
I 2 = 17.9103 A Ii = 0 A A I = 17.9103 A and Ia = 8.955 A
For k = 0.3
I 2 = 0.9851 Ii 26.7985, Ii = 0.9656 I 2 - 20.6367
I 2 = 132,64 A Ii = 107.44 A A I = 25.2 A and I a = 120.04 A
For k = 0.4
I 2 = 0.9802 Ii 35.64, I x = 0.97044 I 2 - 17.733
I 2 = 374.42 A Ii = 345.62 A A I = 28.8 A a n d Ia = 360.02 A
Fork = 0.5
I2 = 0.9753 Ii 44.44, Ii = 0.97045 I 2 - 14.814I 2 = 615 A Ii = 585 A A I = 30 A and Ia = 600 A
F o r k = 0.6
I 2 = 0.97044 Ii 53.2, Ii = 0.9802 I 2 - 11.881
I 2 = 854.38 A Ii = 825.58 A A I = 28.8 A and I a = 840 A
For k = 0.7
I 2 = 0.9656 Ii 61.91, Ii = 0.9851 I 2 - 8.933
I 2 = 1092.6 A Ii = 1067.4 A A I = 25.2 A and I a = 1080 A
For k = 0.8
I 2 = 0.9608 Ii 70.58, Ii = 0.99 I2 - 5.97
I 2 = 1329.6 A Ii = 1310.4 A A I = 19.2 A and Ia = 1320 A
For k = 0.9
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I 2 = 0.956 Ii + 79.2, Ii = 0.995 I 2 - 2.99
I 2 = 1565.4 A, Ii = 1554.6 A, A l = 10.8 A and I a = 1560 A
Problem 5-7
Vs = 600 V R = 0 25 Q L = 20 mH E = 150 V k = 0.1 to 0.9 and f = 250
Hz
The maximum ripple occurs at k = 0 5
From Eq. 5-21), Al max = 600/0.25) tanh [0.25/ 4 x 250 x 0.02)] =
29.9985 A .
From Eq. 5-22), Al max = [600/ 4 x 250 x 0.02)] = 30 A
Problem 5 8
Vs = 10 V , f = 1 kHz, R = 10 Q, L = 6.5 mH , E = 5 V and k = 0.5.
V = 10 R : = 1 0 L : = 6 . 5 - 1 0 ~3 f := 1000o
E : = 5 k:=0.5 T := - T-Rf / L z=1.54
From Eq. 5-35), we get
V s-k -z e -0-k)-z V S -E
R i_ e -( 1 - k ) z4 R I] = 1.16 A
From Eq. (5-36), we get
12 : ~ R . -d-k)-z + R 12 = 1-93 A1 C
AI = 0.77
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Problem 5-9
Vs = 15 V, A VC = 10 mV, AI = 0.5 A, f = 20 kHz, Va = 5 V and I a = 0.5 A
(a) From Eq. (5-48), Va = k Vs and k = Va /Vs = 5/15 = 0.3333
(b) From Eq. (5-52), L = 5 (15 - 5)/(0.5 x 20000 x 15) = 333.3
(c) From Eq. (5-53), C = 0.5/(8 x 10 x 10' 3 x 20000) = 312.5 n
d) u aV,
R = 10
From Eq. (5-56) L c k) = IOJMHO 4 = 166.75
From Eq. 5-89) c,(k) := l 16-Lc 0.333).f2 C r 0.333).10 6 = 0.
Problem 5-10
Vs = 6 V, Va = 15 V, I a = 0.5 A, f = 20 kHz, L = 250 ^H, and C = 440
(a) From Eq. (5-62) 15 = 6/(l - k) or k = 3/5 = 0.6 = 60
(b) From Eq. (5-67), AI = 6 x (15 - 6)/(20000 x 250 x 10' 6 x 15)
= 0.72 A
(c) From Eq. (5-65), I s = 0.5/(1 - 0.6) = 1.25 A
Peak inductor current, I 2 = I s + Al/2 = 1.25 + 0.72/2 = 1.61 A
(d) From Eq. (5-71), A VC = 0.5 x 0.6/(20000 x 440 x 10' 6) = 34.1 mV
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R := a
R = 30
From Eq. 5-72) L c(k) :=
From Eq. 5-73) C r(k) :=
2 -f
2 - f - R
L r 0.6) -1 0 = 180 jiH
C_(0.6)-10 = 0.5
Problem 5-11
Vs = 12 V, k = 0.6, I a = 1.5 A, f = 25 kHz, L = 250 jiH, and C = 220 ^F
(a) From Eq. (5-78), Va = - 12 x 0.6/(1 - 0.6) = - 18 V
(b) From Eq. (5-87), the peak-to-peak output ripple voltage is
A VC = 1.5 x 0.6/(25000 x 220 x 10' 6) = 163.64 mV
(c) From Eq. (5-84), the peak-to-peak inductor ripple voltage is
AI = 12 x 0.6/(25000 x 250 x 10 6) = 1.152 A
(d) From Eq. (5-81), I s = 1.5 x 0.6/(1 - 0.6) = 2.25 A
Since I s is the average of duration kT, the peak to peak current of transistor,
IP = I s /k + AI/2 = 2.25/0.6 + 1.152/2 = 4.326 A
-V.(e) R: = R =
From Eq. 5-2 - f
From Eq. 5-89) c, k) :=2 - f - R
Lc(0 .6)-10 =
C,(0.6)-10 = i (iF
Problem 5-12
Vs = 15 V, k = 0.4, I a = 1.25 A, f = 25 kHz,
350 nH and C2 = 220 ^F48
= 250 = 400 fiF, L 2 =
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(a) From Eq. (5-100), a = - 0.4 x 15/(1 - 0.4) = - 10 V
(b) From Eq. (5-103), I s = 1.25 x 0.4/(1 - 0.4) = 0.83 A
c) From Eq. (5-106), All =15 x 0.4/(25000 x 250 x 10 6) = 0.96 A
(d) From Eq. (5-112), AVci = 0.83 (1 - 0.4)/(25000 x 400 x 10 6) = 50 mV
(e) From Eq. (5-109), AI 2 = 0.4 x 15/(25000 x 350 x 10 6) = 0.69 A
(f) From Eq. (5-113), AVc2 = 0.69/(8 x 25 0 00 x 220 x 10 6) = 15.58 mV
(g) From Eq. (5-120), AI L2 = 1.25/(1.0- 2 X 0.4) = 6.25 A = Is Ii/2 IL 2 Al2/2 = 0.83 0.96/2 6.25 0.69/2 = 7.91 A
Problem 5-13
Vs = 15 V, k = 0.4, I a = 1.25 A, f = 25 kHz, LI = 250 ^H, Ci = 400 ^F, L 2 =
350 i^H and C2 = 220 jaF
,3V = 15 k := 0.4 I _ : = 1.25b
From Eq. 5-115) L cl k) :=
From Eq. (5-116) L c2 (k) := 1 k) R
From Eq. (5-117) C r1 (k):=
f : = 2 5 - 1 0
4 32
2- f -R
L c2 0.4)-1000-0.14 mH
C cl 0.5)-10 = 0.83 jiF
From Eq. (5-118) C c2 (k) := c c2 (0.5).10 6 = 0.42
Problem 5-14
Vs = 110 V, Va = 80 V, I a = 20 A
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AVC = 0.05 x V a = 0.05 x 80 = 4 V
R = Vg/Ia = 80 20 = 4 Q
From Eq. 5-48), k = Va /Vs = 80/110 = 0.7273
From Eq. 5-49) I s = k I a = 0.7273 x 20 = 14.55 A
AI L = 0 .025 x I a = 0.025 x 20 = 0.5 AA I = 0 1 x l a = 0 1 x20 = 2 A
a) From Eq. 5-51), we get the value of U
fV s x\OkHz x l l
From Eq. 5-128), we get the value of Ce
A 2C 6.25
Assuming a linear rise of load current i L during the time from t = 0 to ti = k
T, Eq. 5-129) gives the approximate value of L as0 7 2 7 3 x 4
AI Lf Q 5xlQkHz
Problem 5-15
PSpice simulation
Problem 5-16
k = 0.4 R = 150 Q, r L = 1 Q and r c = 0.2 Q.
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k := 0.5 R := 150 r L := 1 r c := 0.2
a) Buck
G k ) :=
b)Boost
G k ) :=
k -RR + r L
1-k
1-k) -R
- R + rL + k - l - k ) -
r c + R
c ) B u c k - B o o s t
-kG k ) :=
1-k
k r R
R + rL + k - l -k)-
G 0.5) = 0.5
G 0.5) = 1.95
G 0.5) = -0.97
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