ECE1750, Spring 2017

Week 11 – Power Electronics ControlControl

1

Power Electronic Circuits Control

• In most power electronic applications we need to control some variable, such as the output voltage of a dc-dc converter, automatically so we can compensate for potential deviationsautomatically so we can compensate for potential deviations.

• When we control a given variable so that it reaches a given reference value automatically, it is said that we are regulating such variable.

• Some reasons for regulating a variable, such as the output voltage of a dc-dc converter, are:

• Deviations from ideal (expected) behavior to real behavior• Deviations from ideal (expected) behavior to real behavior (e.g. account for losses).• Deviations in input variables (e.g. different input voltage from th t d)the one expected).

Buck converter modeling (review)

Consider a buck converter:• Consider a buck converter:

• State variables (those associated to energy storage):

• Capacitor voltage

3

• Inductor current

Buck converter modeling (review)

In continuous conduction we have two states:• In continuous conduction we have two states:

• State 1, main switch is ON • State 0, main switch is OFF

( ) 1 ( )Lin C

di t V v tdt L

( ) 1 ( )L

Cdi t v tdt L

4

( ) ( )1 ( )C CL

dt Ldv t v ti tdt C R

( ) ( )1 ( )C CL

dv t v ti tdt C R

Buck converter modeling (review)• In continuous conduction we have two states:In continuous conduction we have two states:

• State 1, main switch is ON • State 0, main switch is OFF

( ) 1( ) 1 ( )L

in C

q tdi t V v tdt L

( ) 0( ) 1 ( )L

C

q tdi t v tdt L

( ) ( )1 ( )C C

Ldv t v ti tdt C R

( ) ( )1 ( )C C

Ldv t v ti tdt C R

( ) 1 ( ) ( )

( ) ( )1

Lin C

C C

di t q t V v tdt Ldv t v t

• Switched model of a dc-dc buck converter

5

( ) ( )1 ( )C CL

dv t v ti tdt C R

q(t) is the switching function

• MOSFETS are usually controlled with a pulse-width modulation (PWM) strategy.

• In essence the PWM process involves comparing a sawtooth (or any other

Switching Function Realization

• In essence, the PWM process involves comparing a sawtooth (or any other periodic function with linear transitions) with a voltage level. If the voltage level is constant, it will tend to produce a constant (dc) output.

Sawtooth or triangleadjustable analog input (duty cycle control) ---- vcont(t)

Sawtooth or triangle

Linear portions of sawtooth allows to directly translate voltage levels into time intervals (on-times)

12V

Gate Driver

( TC1427)(e.g. TC1427)

6Comparator

(e.g., LM393)

D = 8/12 = 0 75

• PWM Switching

Function realization D 8/12 0.75

Gate Driver

( TC1427)(e.g. TC1427)

7Comparator

(e.g., LM393)

D = 6/12 = 0 5

• PWM Switching

Function realization D 6/12 0. 5

Gate Driver

( TC1427)(e.g. TC1427)

8Comparator

(e.g., LM393)

D = 4/12 = 1/3

• PWM Switching

Function realization D 4/12 1/3

Where is coming from?

What if it can be adj sted so the o tp t

Gate Driver

( TC1427)

adjusted so the output voltage remains constant regardless of potential deviations?

(e.g. TC1427)p

9Comparator

(e.g., LM393)

Buck converter modeling – Average model• Let’s go back to the switched model of a buck converter:Let s go back to the switched model of a buck converter:

( ) 1 ( ) ( )Lin C

di t q t V v tdt L

( ) ( )1 ( )C C

Ldv t v ti tdt C R

Let’s apply the average

1 T

Let s apply the average operator on both sides of both equations

1 ( ) 1 1 ( ) ( )T T

Ldi t dt q t V v t dt

0

1 dtT

0 0

0 0

( ) ( )

( ) ( )1 1 1 ( )

in C

T TC C

L

dt q t V v t dtT dt T L

dv t v tdt i t dtT dt T C R

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Buck converter modeling – Average model

0 0

1 ( ) 1 1 ( ) ( )

( ) ( )1 1 1

T TL

in C

T TC C

di t dt q t V v t dtT dt T L

dv t v t

0 0

( ) ( )1 1 1 ( )C CL

dv t v tdt i t dtT dt T C R

Now, let’s change the order of some operators (this can be done due to linearity of integrals and differentials)y g )

0

0 0

1 ( )1 1 1( ) ( )

TL T T

in C

d i t dtT V q t dt v t dt

dt L T T

0 00

1 1( ) ( )1 1 ( )

T TC T C

L

dt L T T

d v t dt v t dtT Ti t dtdt C T R

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0dt C T R

Buck converter modeling – Average model1 ( )

Td i d

0

0 0

1 ( )1 1 1( ) ( )

1 1

L T Tin C

T T

d i t dtT V q t dt v t dt

dt L T T

0 00

1 1( ) ( )1 1 ( )

TC T C

L

d v t dt v t dtT Ti t dtdt C T R

Let’s define the following average variables:

1( ) ( )T

i t i t dt 1( ) ( )T

v t v t dt Also,

0

1( ) ( )T

d t q t dtT

0

( ) ( )L Li t i t dtT

0( ) ( ) ,C Cv t v t dt

T

( ) 1 ( ) ( )

( ) ( )1

Lin C

d i t d t V v tdt L

d t t

0T

12

( ) ( )1 ( )C CL

d v t v ti tdt C R

Buck converter modeling – Average model• Comparison between average and switched model:Comparison between average and switched model:

( ) 1 ( ) ( )Lin C

d i t d t V v tdt L

( ) 1 ( ) ( )

( ) ( )1

Lin C

di t q t V v tdt Ld

( ) ( )1 ( )C C

Ld v t v ti t

dt C R

( ) ( )1 ( )C CL

dv t v ti tdt C R

• Application to the previously modeled and simulated buck converter:( )Cv t

( )Li t( )Cv t

( )Li t

13• Notice that the average operator works like a low pass filter by filtering

out all of the ripple caused by switching.

PI Controller for DC-DC Buck Converter Output Voltage

Open Loop, DC-DC Converter Process

Voltage

PWM mod DC-DCVout

PWM mod. and MOSFET

driver

DC DC conv.vcont

q(t) has a fixed out inV DV

( ) 1 ( ) ( )Li C

d i t d t V v t

duty cycleD

out in

q(t) ( ) ( )

( ) ( )1 ( )

in C

C CL

d t V v tdt L

d v t v ti tdt C R

q(t)

I V t i t t ll d t ti ll14

• Issue: Vout is not controlled automatically

( )

PI Controllerq(t) has a variable

duty cycle( ) ( )set oute t V v t

Vset VoutPI

controllerPWM mod.

and MOSFET driver

DC-DC conv.

error e(t)

+

vcont d(t)

driver–

1( ) ( ) ( ) ( ) ( ) ( )d K d K K d ( ) ( ) ( ) ( ) ( ) ( )cont P P ii

v t d t K e t e t dt K e t K e t dtT

• The switching function has a duty cycle that changes from cycle to cycle

til th t h t d t t tiuntil the converter reaches steady state operation.

• Proportional term: Immediate correction but steady state error (vcont equals zero when there is no error (that is when Vset = Vout)).

• Integral term: Gradual correction

Consider the integral as a continuous sum (Riemman’s sum)

15Thank you to the sum action, vcont is not zero when the e = 0

Has some “memory”

E.g. Buck converter

• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

( ) 1 ( ) ( )Lin C

d i t d t V v tdt L

( ) ( )1 ( )C C

Ld v t v ti t

dt C R

1

1( ) ( ) ( )Pi

d t K e t e t dtT

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E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

1( ) ( )i

d t e t dtT

• Ki = 40, Kp = 0

i

A large value for Ki yields too many transient oscillations

iL

vC

d

17

e

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

1( ) ( )i

d t e t dtT

A d t l f Ki l t V t ith t h t• Ki = 10, Kp = 0 A moderate value for Ki regulates Vout without overshoot or too many oscillations but it takes some time to converge

iL

vCvC

d

18e

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm( )Pd K e t

A l l f K i ld f t b t th i• Ki = 0, Kp = 1

i

A large value for Kp yields fast convergence but there is a steady state error

iL

vC

Vout = 15V

d

19

e

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm( ) ( )Pd t K e t

A ll l f K i ld l d th• Ki = 0, Kp = 0.1

i

A small value for Kp yields slower convergence and there is still a steady state error

iL

vC

Vout = 11V

d

20

e

E.g. Buck converter• Vin = 24 V

1• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

1( ) ( ) ( )Pi

d t K e t e t dtT

A PI t ll hi f t th i t l• Ki = 10, Kp = 0.1 A PI controller achieves faster convergence than an integral controller without the steady state error of a proportional controller

iL

vC

d

21e

Euler method to numerically solve differential equations

E l (f d) th d l ith• Euler (forward) method algorithm:

• Step 0) We have the value of x(tn) from the previous iteration or from the initial conditions.

where

• Step 1) For time tn we calculate1n nt t t

( ( ), )n nf x t t

• Step 2) For time tn+1 we calculate

1( ) ( ) ( ) ( ( ), )n n n n nx t t x t x t f x t t t

• Now x(tn+1) is the input for the next iteration. To start the new iteration go to Step #1.

22

g p

Euler method applied to buck converters

E l (f d) th d i l ith t l diff ti l ti• Euler (forward) method is an algorithm to solve differential equations based on:

( ) 1 ( ) ( )

( ) ( )1 ( )

Lin C

C C

d i t d t V v tdt L

d v t v ti t

( )Li tdt C R

11( ) ( ) ( ) ( )L L i Ci t i t d t V v t t

1

1

( ) ( ) ( ) ( )

( )1( ) ( ) ( )

L n L n n in C n

C nC n C n L n

i t i t d t V v t tL

v tv t v t i t tC R

23

Euler method applied to buck converters

E l l f PI t ll• Euler solver for PI controller

11( ) ( ) ( ) ( )

( )1

L n L n n in C ni t i t d t V v t tL

1( )1( ) ( ) ( ) C n

C n C n L nv tv t v t i t t

C R

1

( ) ( ) ( )k n

d K K

( ) ( ) ( )P id t K e t K e t dt 1 10

( ) ( ) ( )n P n i kk

d t K e t K e t t

1 1( ) ( )n set C ne t V v t

Consider a Buck converter with Vin = 100 V, Vset=40V, Ki = 15, R=2 ohms, L=300 μH, C=200 μF (horizontal axis is time in seconds).

( )Cv t

( )Li t

24

Integral controller simulation

E l l f PI t ll• Euler solver for PI controller

( ) ( ) ( )P id t K e t K e t dt 1

1 10

( ) ( ) ( )k n

n P n i kk

d t K e t K e t t

0k

1 1( ) ( )n set C ne t V v t

Consider a Buck converter with Vin = 100 V, Vset=40V, Ki = 15, R=2 ohms, L=300 μH, C=200 μF (horizontal axis is time in seconds).

( )Cv t

(%)d

( )e t

25<vC(t)> > Vset=40V so e(t)<0 and d

decreases

E.g. Buck converter• Vin = 24 V

In steady state• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

In steady state d(t)=D=2/3

• Ki = 10, Kp = 0.1

26

E.g. Buck converter• Vin = 30 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 OhmIn steady state d(t)=D=0.53

• Ki = 10, Kp = 0.1

27

E.g. Buck converter• Vin = 20 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm In steady state d(t)=D=0.8

• Ki = 10, Kp = 0.1d(t) D 0.8

28

Op AmpsOp ps

–V−I−

+V+

VoutI+

Assumptions for ideal op amp

++

Vout = K(V+ − V− ), K large (hundreds of thousands, or one million)

I+ = I− = 0

Voltages are with respect to power supply ground (not shown)

Output current is not limited

29

Example 1. Buffer Amplifier( t hi h i d i l t l i d i l)(converts high impedance signal to low impedance signal)

) = K(Vi – V t)( +t VVKV –

+VinVout

) K(Vin Vout)( −+out VVKV

inoutout KVKVV

inout KVKV )1(

KVV inout

1

K

K is large

iVV

30

inout VV

Example 2. Inverting Amplifier( sed for proportional control signal)(used for proportional control signal)

–

Rf

Rin Vin

Vout

KVVKVout )0( , so KV

V out .

KCL at the – node is 0

outin

RVV

RVV

. + fin RR

Eliminating V yields

VV

0

f

outout

in

inout

R

VKV

R

VKV

, so

in

in

ffinout R

VRKRKR

V

111 . For large K, then in

in

f

outRV

RV

, so in

finout RR

VV .

31

Example 3. Inverting Difference( d f i l)(used for error signal)

RR

VV

KVVKV b)(

– +

Vout

Va

R

R R

VKVVKV bout 2

)( , so

KVV

V outb 2.

Vb R

R KCL at the – node is 0

RVV

RVV outa , so

0 outa VVVV , yielding 2outa VV

V

.2

Eliminating V yields

outab VVVKV so

about VVK

VKV or

1 abt

VVKKV

22out KV , so

22out KKV , or

221out KV .

For large K then VVV

32

For large K , then baout VVV

Example 4. Inverting Sum( d t ti l d i t l t l i l )(used to sum proportional and integral control signals)

KVVKV )0( VV out

– +

Vout Va

Vb

R

R

R

KVVKVout )0( , so K

V out .

KCL at the – node is

0

RVV

RVV

RVV outba , so

outba VVVV 3 .

Substitutingfor V yields outbaout VVVV

3 , so baout VVV

13 .Substituting for V yields outba VVVK

3 , so baout VVK

V

1 .

Thus, for large K , baout VVV

33

, g , baout

Example 5. Inverting Integrator(used for integral control signal)

Ci

Using phasor analysis, )~0(~ VKVout , so

~

– +

CiRi

Vin Vout

KV

V out~

~ . KCL at the − node is

~~~~ VVVV outin 01

Cj

VVRVV out

i

in

.

~~~

out VV

Eliminating V~ yields 0~

~

out

out

i

inV

KV

CjR

VK . Gathering terms yields

iV~

11~

~11~

i

in

iout R

VK

CjKR

V 111

, or iniout V

KCRj

KV 111

For large K , the

expression reduces to iniout VCRjV ~~ , soCRj

VV inout

~~

(thus, negative integrator action).

34

CRj i

For a given frequency and fixed C , increasing iR reduces the magnitude of outV~ .

Op Amp Implementation of PI Controller

(see slide #14)

– R

Signal flow

PWM mod. and MOSFET

driver

–

+–

+–

–

errorαVout

Rp

15kΩ–

+

++

Summer(Gain = −11)

Proportional(Gain = −Kp)

Buffers(Gain = 1)

VsetVcont

C

Difference(Gain = −1)

–

+

( ) CiRi

35

Inverting Integrator(Time Constant = Ti)

Power Electronic Circuits Control

• Buck converters are relatively simple to control because a PI controller applied to an average model yields a linear system.

( ) 1 1( ( ) ) ( ( ) ) ( )Ld i t K V V d V

( ) ( ( ) ) ( ( ) ) ( )

( ) ( )1 ( )

LP set C set C in C

i

C CL

K V v t V v t dt V v tdt L T

d v t v ti tdt C R

• However, for other converters, PI controllers applied to an average model still yields a non-linear system:

dt C R

( ) 1 1 1( ( ) ) ( ( ) ) 1 ( ( ) ) ( ( ) ) ( )

( ) 1 1

LP set C set C in P set C set C C

i i

d i t K V v t V v t dt V K V v t V v t dt v tdt L T T

d t

( )t

• Moreover, boost and other converters are what are called “non-minimal

( ) 1 1( ( ) ) ( ( ) ) ( )CP set C set C L

i

d v t K V v t V v t dt i tdt C T

( )Cv tR

,phase” converters in which direct regulation of the output voltage likely yields an unstable controller. For boost converters, the variable that can be controlled directly is the inductor current.