H.M Rashid

8/10/2019 H.M Rashid

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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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H.M Rashid