ejemplo amathcad buck boost

7/23/2019 Ejemplo aMathcad Buck Boost

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6. Buck-boost Converter

Circuit and its parameters:

C

200uIC = 0

V-

V+

PARAMETERS:

Fs = 100kD = 0.6

0

I

p u l s e

outDiode

Dideal

V1

12Vdc

L

100uHIC = 0

R1

3

V2

TD = 0TF = 10nPW = {D/Fs}PER = {1/Fs}V1 = 0TR = 10nV2 = 15

Q1

MOSFET

Basic Formula:

D D

V V g

o

−= 1 RV I I o

Do == L D I D I )1( −= L s g DI I I ==

Pspice Simulation:

alculation:

vout

i L

C

Circuit Parameters:

Vg 12:= Vo 18:=L 100 10

6−⋅:= C 200 10

6−⋅:= f 100 10

3⋅:= R 3:=

Initial guess D 0.5:=

Given

Vo

Vg

D

1 D− D Find D( ):= D 0.6=



icmax 9.36=icmax iLmax Io−:=ic

Io−

Io−

iLmax Io−

iLmin Io−

:=

iL

iLmin

iLmax

iLmax

iLmin

:= iD

0

0

iLmax

iLmin

:=is

iLmin

iLmax

0

0

:=VL

Vg

Vg

Vo−

Vo−

:=t

0

D T⋅D T⋅

T

:=

The waveforms are a piecewise linear. Collect numbers for plotting.

iLmax 15.36=iLmax IL∆iL

2+:=

iLmin 14.64=iLmin IL

∆iL

2−:=

IL 15=IL

Io

1 D−:=Io 6=Io

Vo

R :=

∆iL 0.72=∆iL

1

LVg⋅ D⋅ T⋅:=

T 1

f :=

VL t( ) Vg( ) 0 t< D T⋅<if

Vo−( ) otherwise

:=



0 5 .10 6

1 .10 5

20

13.33

6.67

6.67

13.33

20Inductor Voltage

VL

t

0 2.5 .10 6

5 .10 6

7.5 .10 6

1 .10 5

14.5

14.75

15

15.25

15.5Inductor Current

iL

t

0 5 .10 6

1 .10 5

5

10

15

20Switch current

is

t0 5 .10

61 .10

5

5

10

15

20Diode current

iD

t

Is D IL⋅:= Is 9= ID 1 D−( ) IL⋅:= ID 6=

0 5 .10 6

1 .10 5

10

6.67

3.33

3.33

6.6710

Capacitor current

ic

t

∆vo1

CIo⋅ D⋅ T⋅:= ∆vo 0.18=

Ig Is:= Ig 9= Vg 12= Io 6= Vo 18=

Pg Vg Ig⋅:= Pg 108= Po Vo Io⋅:= Po 108=

Due to ideal components (switch, diode, L and C) that we assume, the circuit is lossless, the

efficiency is 100% and as such Pg = Po.



Lmin 2.4 10 6−

×=∆iL1

LVg⋅ D⋅ T⋅

Lmin1

∆iL

Vg⋅ D⋅ T⋅:=

iL2

iLmin

iLmax

iLmax

iLmin

:=∆iL 30=

iLmax 30=∆iL iLmax:=iLmax 30=iLmax 2 IL⋅:=iLmin 0:=

IL 15=IL

Io

1 D−:=Io 6=Io

Vo

R :=

R 3:=

Case II: R, f, and other circuit parameters are kept constant, L is varied

R 125=R Vo

Io

:=

iL1

iLmin

iLmax

iLmax

iLmin

:=Io 0.144=Io 1 D−( ) IL⋅:=IL 0.36=IL

∆iL2

:=

Case I: L, f, and other parameters are kept constant, R is varied iLmax ∆i:=iLmin 0:=

DCM/CCM Boundary

Case III: R, L, and other circuit parameters are kept constant, f is varied

R 3:=Io

Vo

R := Io 6= IL

Io

1 D−:= IL 15=

iLmin 0:= iLmax 2 IL⋅:= iLmax 30= ∆iL iLmax:= iLmax 30=

∆iL 30=

f min1

∆iL

Vg⋅ D⋅ 1

L⋅:=

∆iL1

LVg⋅ D⋅

1

f ⋅ f min 2.4 10

3×=



Collect numbers for plotting

t1

0

D

f min

D

f min

1

f min

:= iL3

iLmin

iLmax

iLmax

iLmin

:= IL3

15

15

15

15

:= IL2 IL3:=IL1

0.36

0.36

0.36

0.36

:=

Tmax1

f min

:= Tmax 4.167 10 4−

×=

Waveforms for iL and IL:

0 5 .10 6

1 .10 5

0

0.25

0.5

0.75

1Case I

iL1

IL1

t

0 5 .10 6

1 .10 5

0

10

20

30

40Case II

iL2

IL2

t

0 2 .10 4

4 .10 4

6 .10 4

0

20

40Case III

iL3

IL3

t1

-tammat-

ejemplo amathcad buck boost

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