Buck-Boost Converter

Nama
: Annas Budi Prastyawan
Kelas
: 3 D4 Elektro Industri B
NRP
: 1310171056
6.17 The boost converter of Fig. 6-8 has parameter Vs = 20 V, D = 0,6, R = 12,5 Ω, L = 10uH,
C = 40uF, and the switching frequency is 200 kHz.
(a) Determine the output voltage
(b) Determine the average, maximum, and minimum inductor currents
(c) Determine the output voltage ripple
(d) Determine the average current in the diode. Assume ideal components
Diket : Vs = 20 V, D = 0,6, R = 12,5 Ω, L = 10uH, C = 40uF, f = 200 kHz
(a) Vo = Vs / (1-D)
Vo = 20 / (1-0,6)
= 50 V
(b) IL = Is
IL = Vs / [R(1-D)^2]
IL = 20 / [12,5(1-0,6)^2]
= 10 A
ΔiL = Vs x D x T / L
ΔiL = Vs x D / [L x f]
ΔiL = 20 x 0,6 / [10u x 200k]
=6A
IL max = IL + ΔiL / 2
IL max = 10 + 6 / 2
= 13 A
IL min = IL - ΔiL / 2
IL min = 10 - 6 / 2
=7A
(c) r = D / [R x C x f]
r = 0,6 / [12,5 x 40u x 200k]
= 0,6%
(d) ID = (IL max + IL min) x (1-D)T / 2
ID = (IL max + IL min) x (1-D) / [2 x f]
ID = (13 + 7) x (1-0,6) / [2 x 200k]
= 200 mA
PSIM simulation view
- Circuit diagram
- Output gelombang
=>Vo = 50 V
=> IL (avg) = 10 A
=> ILmax = 13 A
=> ILmin = 7 A
6.18 The boost converter in prob. 6-17, sketch the inductor and capacitor currents. Determine
the rms
value of this currents.
jawab :
Io = Vo/R = 50 / 12,5 = 4 A
IC (avg) = IL (avg) – Io
= 10 – 4
=6A
IC (max) = IL (max) – Io
= 13 – 4
=9A
IC (min) = IL (min) – Io
=7–4
=3A
PSIM simulation view
- Circuit diagram
-
Output
gelombang
=> IC (avg) = 6A
=> IC (max) = 9A
=> IC (min) = 3A
6.19 A boost converter has an input of 5 V and an output of 25 W at 15 V. The minimum inductor
current muss be no less then 50% of the average. The output voltage ripple must be less then
1
percent. The switching frequency is 300 kHz. Determine the duty ratio, minimum inductor
value
and minimum capacitor value.
Diket : Vs = 5 V, Vo = 15 V, Po = 25 W, rVo < 1 %, f = 300 kHz, riL > 50%
jawab :
Vo = Vs / (1-D)
D = 1 - Vs/Vo
D = 1 - 5/15
= 0,667
R = Vo^2 / Po
R = 15^2 / 25
=9Ω
IL = Is
IL = Vs / [R(1-D)^2]
IL = 5 / [9(1-0,667)^2]
= 5,01 A
ΔiL = riL x IL
ΔiL = 50% x 5,01
= 2,5 A
ΔiL > 2,5 A
Metode 2
L = Vs(Vo-Vs) / [f x Vo x ΔiL]
L = 5(15-5) / [300k x 15 x 2,5]
= 4,4 uH
L < 4,4 uH
r = D /[R x C x f]
C = D / [R x r x f]
C = 0,667 / [9 x 0,1% x 300k]
= 24,7 uF
C > 24,7 uF
PSIM simulation view
- Circuit diagram
- Output gelombang
=> Vo = 15 V
=> IL (avg) = 5,01 A
6.20 Design a boost converter to provide an output of 18 V from a 12 V source. The load is 20 W.
The output voltage ripple must be less then 0,5 percent. Specify the duty ratio, the switching
frequency, the inductor size and rms current rating, and the capacitor size and rms current
rating.
Design for continuous current, assume ideal components.
Diket : Vo = 18 V, Vs = 12 V, Po = 20 W, rVo < 0,5%
jawab :
Vo = Vs / (1-D)
D = 1 - Vs/Vo
D = 1 - 12/18
= 0,333
f > 20 kHz => 40 kHz
R = Vo^2 / Po
R = 18^2 / 20
= 16,2 Ω
IL = Is
IL = Vs / [R(1-D)^2]
IL = 12 / [16,2(1-0,333)^2]
= 1,66 A
Metode 1
Lmin = D x R(1-D)^2 / [2 x f]
Lmin = 0,333 x 16,2(1-0,333)^2 / [2 x 40k]
= 29 uH
L = Lmin x 10
= 29u x 10
= 290 uH
r = D /[R x C x f]
C = D / [R x r x f]
C = 0,333 / [16,2 x 0,5% x 40k]
= 101,8 uF
C > 101,8 uF
Ic = IL -(Vo/R)
Ic = 1,66 -(18/16,2)
= 0,55 A
PSIM simulation view
- Circuit diagram
- Output gelombang
=>Vo = 18 V
=> IL (avg) = 1,66 A
=> IC (avg) = 0,55 A