Control Buck

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Model Derivation and Control Design of a

Buck Converter using

Ansoft SimplorerTM

By:Marius Marita



Overview

• Problem Statement

• Buck Theory• Derive the Transfer Function

• Investigate the Control Methods

• Simulation Results• Conclusion

• Application: Induction Motor Control

• Questions?



Problem Statement

• Obtain the

Mathematical Model

of a Buck Converter

• Find a controller that

covers the whole

range of operation of a

Buck Converterincluding Continuous

and Discontinuous

Mode

R1.V[

1 .1 24 7 7 7 .RA ND .

t[s]

14

-2

0

2

4

6

8

10

12

0 20m2m 4m 6m 8m 10m 12m 14m 16m 18m

R1 R : = 4

C1 C := 1500u

L1

L := 56u

E1

EMF := 27

C4 C := 1500u

S1

D1

R2 R := 0.02

R3

R := 0.017

R4 R := 0.02

SUM1

NEG

NEG1

GAIN

GAIN1

PWM

pwm1

S& H

SAH1

LIMIT

LIMIT1

R5 R := 1.5

S2

STEP1 STEP2

TS1

SAH2.VAL

t[s]

1.2

-0.2

0

0.2

0.4

0.6

0.8

1

0 20m2m 4m 6m 8m 10m 12m 14m 16m 18m

R6

R := 0.001

STEP3

G(s)

GS1

LIMIT

LIMIT2

SUM2

I

b

SUM4

D

DIFF1

S& H

SAH2

RANDOM

RANDOM1

SUM5

+

V

VM1



Buck Theory

• Step down the voltage

• Control the Output

Voltage while the

Load Resistance or the

Input Voltage Changes

• Provide Isolation from

the Line

R R :=C C := 150

L

L := 56

E

EMF :=

S

D



Buck Theory

Continuous

Conduction Mode

• Inductor Current Staysabove zero value for all

time

Discontinuous

Conduction Mode

• Inductor Current reacheszero value for a while

during a one period.

L1.I [A]

t [s]

Defaul

6.

0

1

2

3

4

5

20 20.220.02 20.04 20.06 20.08 20.1 20.12 20.14 20.16 20.18

L1.I [A]

t [s]

3.8

0

0.5

1

1.5

2

2.5

3

6.3m 6.5m6.4m 6.4m 6.4m 6.5m 6.5m 6.5m



Buck Transfer Function

• TF derived from

changing the On Time

of the Switch resulting

in changing the output

voltage (CCM)

• Load Resistance

changes from 1O, to

1.5O, to 2O, and 3O in

CCM

PWM

R1 R := 1

C1 C := 1500

L1

L := 56

E1

EMF := 2

S1

D1

pwm

CONSTCONS

SUM

STEP

AMPL := 0.

0.44

Continuous Conduction Mode

Simulation Setup




• TF derived from Changing

the Load Resistance whilekeeping the On Time of

the PWM Generator

Constant (DCM)

• Load changes from 5O,

7O, 10O, and 12O in

DCM

Discontinuous Conduction Mode

Simulation Setup

PWM

R R :=

C C := 150

L

L := 5

E

EMF :=

S1

D1

pwm

CONSTCONS

0.4

S2

TS

R R := 2

STE

AMPL :

STE

AMPL :




• Second Order

Response for CCM

• First Order Response

for DCM

R1.V [R1.I [A

[s]

Defa

15

10.1

1

1

1

1

20.12 57.1024 26 28 30 32 34 36 38 40 42 44 46 48 50 52




• Second Order CCMVOut(IN)ss= Output Voltage before adding

the change in Duty Ratio

VOut(FV)ss= Output Voltage Steady State

value after adding the change in the

Duty Ratio

V(PEAK)= Peak Value of the Overshoot

Voltage

T(FV First Time)= Time to the first reach of the

Steady State Voltage

T(PEAK)= Time to the Peak Voltage

Calculate:

• DC Gain:

• Ratio of Peak Value to Final Value:

• Damping Ratio from Table:

ζ=0.55

• Normalized Time to Peak Value: t *from

Table

• Calculated Natural Frequency:

• Transfer Function:

( )

( )

PEAK

Out ss

V V

*

( )

n

PEAK

t

T ω =

( )

2

22

22

22 2

( )2

2

(2 )( )

2 2 (2 )

DC n

n n

n n

DC n

n n

K G s

s s

f

K f G s

s f s f

ω

ζω ω

ω π

π

ζ π π

=+ +

= ⋅ ⋅

⋅ ⋅=

+ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅

g

g




• First Order DCM

V1(in)- Voltage Value before

Load Change OccursV1(out)-Voltage Value after

load change occurred and

reached steady state

T(63%)-Time it takes for thesystem to reach 63% of its

final value

• K - DC Gain

• t - 63.2% of the final

value

( )

1

H s

sτ

=

+



Buck Transfer Function• Continuous Conduction Mode

(CCM)• Discontinuous Conduction

Mode (DCM)

( ) ( )

( ) ( ) ( ) ( )

2 2

2 22

27 2 633( )

2 0.50 2 633 2 633Gs

s s

π

π π

⋅ ⋅=

+ ⋅ ⋅ ⋅ + ⋅

.

( ) DC c

c

w

G s s w= +

0.004850.0048410.0034960.002705Time Constant

(Seconds)

18.918.418.4517.9DC Gain K DC

121075Resistance (ς)

Test 4 Test 3 Test 2 Test 1 •Variable Data



Investigate Control Methods

• Proportional Integral Derivative (PID)

• Loop Shaping Design (LSD)

• Active Disturbance Rejection Controller

(ADRC)




• Classical PID

∫ ⋅+⋅⋅+⋅= edt

d k dt ek ek sG d i pc )(




• Loop Shaping Design Method (LSD)

11

2

1 1( ) ( )

1 1

m

c pn

c

sG s G s s s s

ω

ω ω

−+ = ⋅ ⋅ ⋅ + +

2

1

( )1 1cC LG s

s sC R C L

⋅=+ +

⋅ ⋅

Low Pass Filter

Transfer Function p




• Active Disturbance Rejection Controller

3 0

0

z uu

b

− +=

'

0 1 2. . .( ) . p d p d u k e k e k r z k z = + ≈ − −2

2

p c

d c

k w

k w

=

= ⋅

( , ( ), ( ), ( )) ( ) ( ) y f t y t y t w t b t u t = + ⋅&& &




• ADRC State Estimator '

z A z B u

y C z

= + =

0

2

0

3

0

0

2

0 0

3

0

3 * 1 0

3 * 0 1

3 * 0 0

3 *03 *

0

1 0 0

0 1 0

0 0 1

w

A w

w

w B b w

w

C

−

= − −

=

=

z1

ày, z2

ày’, and z3

àƒ




• Evaluate each Control Method based on a

Cost Function of the Form:

2 2

2 2

0 0

(| |) (| |) ( ) ( )

T T

e u e u

max k e max k u k e dt k u dt = ⋅ + ⋅ + ⋅ + ⋅∫ ∫




• PID Implementation

R1 R := 4

C1 C := 1500u

L1

L := 56u

E1

EMF := 27

C4 C := 1500u

S1

D1

R2 R := 0.02R3

R := 0.017

R4 R := 0.02

SUM1

NEG

NEG1

GAIN

GAIN1

PWM

pwm1

S & H

SAH1

LIMIT

LIMIT1

R5 R := 1.5

S2

STEP1 STEP2

TS1

R6

R := 0.001

STEP3

G(s)

GS1

LIMIT

LIMIT2

SUM2

I

INT1

D

DIFF1

S & H

SAH2

RANDOM

RANDOM1

SUM5

+

V

VM1




• LSD and ADRC more complex

– ADRC Require State Estimator

– LSD easier to implement writing an .m file in

Matlab



Simulation Results

LSD Response• Good Load

Disturbance

• Better than PID• Cost Function

Jtotal=3.416Emax=0.2239

Umax=1.123Eint=0.1273Uint=1.942



Simulation Results

ADRC Response• Good Transient

response

• Good DisturbanceRejection

• Cost FunctionJtotal=10.58

Emax=0.8631Umax=1.653Eint=0.5311Uint=7.533



Conclusion

• ADRC is the preferred Control Method

Deals Best with Nonlinearities

• LSD has good load disturbance but can not

make up for Nonlinearities

• PID lags behind ADRC and LSD

Not suitable for a Buck Converter.



Buck Application:

Induction Motor Control

E1R1

E2

E3

R2

R3

L1

L3

D1 D2 D3

D4 D6

IM1.N [rpm]5 * C2.V [V]

5 * SUM4.VAL

t [s]

2k

-0.2k

0

0.25k

0.5k

0.75k

1k

1.3k

1.5k

1.8k

0 20.25 0.5 0.75 1 1.3 1.5 1.8

C1

C2

S7

PWM

PWM1

CLR

Q1

Q2

Q0

Binary

cntb31

AND3

and33

AND3

and34

AND3

and35

AND3

and36

INV

inv7

AND3

and37

AND3

and31

DAC

VAL

[3:0]

INPUT

dac1

DAC

VAL

[3:0]

INPUT

dac2

DAC

VAL

[3:0]

INPUT

dac3

DAC

VAL

[3:0]

INPUT

dac4

DAC

VAL

[3:0]

INPUT

dac5

DAC

VAL

[3:0]

INPUT

dac6

S1

S2

S3

S4

S5

S6

OR3

or31

OR3

or32

OR3

or33

OR3

or34

OR3

or35

OR3

or36

AND3

and32

INV

inv2

INV

inv3

INV

inv8

INV

inv1

INVinv5

INV

inv6

INV

inv4

INV

inv9

PWM

PWM2

+

V

VM1

CONST

CONST

CONST

CONSTxn

POW1

MUL1

COMP1

LIMIT

LIMIT1

SUM1

N E G

NEG1

CONST

CONST1

MUL3

L2

D5

GAIN

GAIN1

I

INTG1SUM2

D7

IM1.N [rpm]0.140222k

MUL3.VAL36.6495m

ADC

[3:0]

INPUT

VAL

adc1

CLK

clk1

M3 ~ B

A

C

IM1

STEP1

STEP3

SUM4

STEP5

L4

R5

S8

COMP2

STEP2

STEP4

D8 D9 D10

D11D12

D13

Vdc



Questions?

Control Buck

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