r. w. ericksonecee.colorado.edu/~ecen5797/course_material/lecture28.pdftransistor gate driver b(t) i...

R. W. EricksonDepartment of Electrical, Computer, and Energy Engineering

University of Colorado, Boulder

Fundamentals of Power Electronics Chapter 9: Controller design1

Chapter 9. Controller Design

9.1. Introduction9.2. Effect of negative feedback on the network transfer

functions9.2.1. Feedback reduces the transfer function from disturbances

to the output9.2.2. Feedback causes the transfer function from the reference

input to the output to be insensitive to variations in the gainsin the forward path of the loop

9.3. Construction of the important quantities 1/(1+T) andT/(1+T) and the closed-loop transfer functions


Controller design

9.4. Stability9.4.1. The phase margin test9.4.2. The relation between phase margin and closed-loop

damping factor9.4.3. Transient response vs. damping factor

9.5. Regulator design9.5.1. Lead (PD) compensator9.5.2. Lag (PI) compensator9.5.3. Combined (PID) compensator9.5.4. Design example


9.1. Introduction

Output voltage of aswitching converterdepends on duty cycled, input voltage vg, andload current iload.

+–

+

v(t)

–

vg(t)

Switching converter Load

Pulse-widthmodulator

vc(t)

Transistorgate driver

b(t)

iload(t)

b(t)

TsdTs t v(t)vg(t)

iload(t)

d(t)

Switching converter

Disturbances

Control input

}}

v(t) = f(vg, iload, d )


The dc regulator application

Objective: maintain constantoutput voltage v(t) = V, in spiteof disturbances in vg(t) andiload(t).Typical variation in vg(t): 100Hzor 120Hz ripple, produced byrectifier circuit.

Load current variations: a significant step-change in load current, suchas from 50% to 100% of rated value, may be applied.A typical output voltage regulation specification: 5V ± 0.1V.

Circuit elements are constructed to some specified tolerance. In highvolume manufacturing of converters, all output voltages must meetspecifications.

v(t)vg(t)

iload(t)

d(t)

Switching converter

Disturbances

Control input

}}



The dc regulator application

So we cannot expect to set the duty cycle to a single value, and obtaina given constant output voltage under all conditions.Negative feedback: build a circuit that automatically adjusts the dutycycle as necessary, to obtain the specified output voltage with highaccuracy, regardless of disturbances or component tolerances.


Negative feedback:a switching regulator system

+–

+

v

–

vg

Switching converterPowerinput

Load–+

Compensator

vrefReference

input

HvPulse-widthmodulator

vc

Transistorgate driver

b Gc(s)

H(s)

ve

Errorsignal

Sensorgain

iload


Negative feedback

vref

Referenceinput

vcve(t)

Errorsignal

Sensorgain

v(t)vg(t)

iload(t)

d(t)

Switching converter

Disturbances

Control input

}}+– Pulse-width

modulatorCompensator



9.2. Effect of negative feedback on thenetwork transfer functions

Small signal model: open-loop converter

Output voltage can be expressed as

wherev(s) = Gvd(s) d(s) + Gvg(s) vg(s) – Zout(s) i load(s)

Gvd(s) = v(s)d(s) vg = 0

i load = 0

Gvg(s) = v(s)vg(s) d = 0

i load = 0

Zout(s) = – v(s)i load(s) d = 0

vg = 0

+–

+– 1 : M(D) Le

C Rvg(s) j(s)d(s)

e(s)d(s)

iload (s)

+

v(s)

–


Voltage regulator system small-signal model

• Use small-signalconverter model

• Perturb andlinearize remainderof feedback loop:

vref(t) = Vref + vref(t)

ve(t) = Ve + ve(t)

etc.Reference

input

Errorsignal

+–

Pulse-widthmodulator

Compensator

Gc(s)

Sensorgain

H(s)

1VM

+–

+– 1 : M(D) Le

C Rvg(s) j(s)d(s)

e(s)d(s)

iload (s)

+

v(s)

–

d(s)

vref (s)

H(s)v(s)

ve(s) vc(s)


Regulator system small-signal block diagram

Referenceinput

Errorsignal

+–

Pulse-widthmodulatorCompensator

Sensorgain

H(s)

1VM Duty cycle

variation

Gc(s) Gvd(s)

Gvg(s)Zout(s)

ac linevariation

Load currentvariation

+

–+

Output voltagevariation

Converter power stage

vref (s) ve(s) vc(s) d(s)

vg(s)

iload(s)

v(s)

H(s)v(s)


Solution of block diagram

v = vrefGcGvd / VM

1 + HGcGvd / VM+ vg

Gvg

1 + HGcGvd / VM– i load

Zout

1 + HGcGvd / VM

Manipulate block diagram to solve for . Result isv(s)

which is of the form

v = vref1H

T1 + T + vg

Gvg

1 + T – i loadZout

1 + T

with T(s) = H(s) Gc(s) Gvd(s) / VM = "loop gain"

Loop gain T(s) = products of the gains around the negativefeedback loop.


9.2.1. Feedback reduces the transfer functionsfrom disturbances to the output

Original (open-loop) line-to-output transfer function:

Gvg(s) = v(s)vg(s) d = 0

i load = 0

With addition of negative feedback, the line-to-output transfer functionbecomes:

v(s)vg(s) vref = 0

i load = 0

=Gvg(s)

1 + T(s)

Feedback reduces the line-to-output transfer function by a factor of1

1 + T(s)

If T(s) is large in magnitude, then the line-to-output transfer functionbecomes small.


Closed-loop output impedance

Original (open-loop) output impedance:

With addition of negative feedback, the output impedance becomes:

Feedback reduces the output impedance by a factor of1

1 + T(s)

If T(s) is large in magnitude, then the output impedance is greatlyreduced in magnitude.

Zout(s) = – v(s)i load(s) d = 0

vg = 0

v(s)– i load(s) vref = 0

vg = 0

= Zout(s)1 + T(s)


9.2.2. Feedback causes the transfer function from thereference input to the output to be insensitive to

variations in the gains in the forward path of the loop

Closed-loop transfer function from to is:

which is independent of the gains in the forward path of the loop.This result applies equally well to dc values:

v(s)vref

v(s)vref(s) vg = 0

i load = 0

= 1H(s)

T(s)1 + T(s)

If the loop gain is large in magnitude, i.e., || T || >> 1, then (1+T) ! T andT/(1+T) ! T/T = 1. The transfer function then becomes

v(s)vref(s) 5

1H(s)

VVref

= 1H(0)

T(0)1 + T(0) 5

1H(0)


9.3. Construction of the important quantities1/(1+T) and T/(1+T)

Example

T(s) = T0

1 + stz

1 + sQtp1

+ stp1

2 1 + stp2

At the crossover frequency fc, || T || = 1

fp1

QdB

– 40 dB/decade

| T0 |dB

fz

fc fp2

– 20 dB/decade

Crossoverfrequency

f

|| T ||

0 dB

–20 dB

–40 dB

20 dB

40 dB

60 dB

80 dB

– 40 dB/decade

1 Hz 10 Hz 100 Hz 1 kHz 10 kHz 100 kHz


Approximating 1/(1+T) and T/(1+T)

T1 + T 5

1 for || T || >> 1T for || T || << 1

11+T(s) 5

1T(s) for || T || >> 1

1 for || T || << 1


Example: construction of T/(1+T)

T1 + T 5

1 for || T || >> 1T for || T || << 1

fp1

fz fcfp2

– 20 dB/decade

– 40 dB/decade

Crossoverfrequency

f

|| T ||

0 dB

–20 dB

–40 dB

20 dB

40 dB

60 dB

80 dB

T1 + T



Example: analytical expressions for approximatereference to output transfer function

v(s)vref(s) = 1

H(s)T(s)

1 + T(s) 51

H(s)

v(s)vref(s) = 1

H(s)T(s)

1 + T(s) 5T(s)H(s) = Gc(s)Gvd(s)

VM

At frequencies sufficiently less that the crossover frequency, the loopgain T(s) has large magnitude. The transfer function from the referenceto the output becomes

This is the desired behavior: the output follows the referenceaccording to the ideal gain 1/H(s). The feedback loop works well atfrequencies where the loop gain T(s) has large magnitude.At frequencies above the crossover frequency, || T || < 1. The quantityT/(1+T) then has magnitude approximately equal to 1, and we obtain

This coincides with the open-loop transfer function from the referenceto the output. At frequencies where || T || < 1, the loop has essentiallyno effect on the transfer function from the reference to the output.


Same example: construction of 1/(1+T)

11+T(s) 5

1T(s) for || T || >> 1

1 for || T || << 1fp1

QdB

– 40 dB/decade

| T0 |dB

fz

fc fp2Crossoverfrequency

|| T ||

0 dB

–20 dB

–40 dB

20 dB

40 dB

60 dB

80 dB

–60 dB

–80 dB

f1 Hz 10 Hz 100 Hz 1 kHz 10 kHz 100 kHz

QdB

– | T0 |dBfp1

fz

11 + T

– 40 dB/decade+ 40 dB/decade

+ 20 dB/decade

– 20 dB/decade


Interpretation: how the loop rejects disturbances

Below the crossover frequency: f < fcand || T || > 1

Then 1/(1+T) ! 1/T, anddisturbances are reduced inmagnitude by 1/ || T ||

Above the crossover frequency: f > fcand || T || < 1

Then 1/(1+T) ! 1, and thefeedback loop has essentiallyno effect on disturbances

11+T(s) 5

1T(s) for || T || >> 1

1 for || T || << 1


Terminology: open-loop vs. closed-loop

Original transfer functions, before introduction of feedback (“open-looptransfer functions”):

Upon introduction of feedback, these transfer functions become(“closed-loop transfer functions”):

The loop gain:

Gvd(s) Gvg(s) Zout(s)

1H(s)

T(s)1 + T(s)

Gvg(s)1 + T(s)

Zout(s)1 + T(s)

T(s)


9.3. Construction of the important quantities1/(1+T) and T/(1+T)

Example

T(s) = T0

1 + stz

1 + sQtp1

+ stp1

2 1 + stp2

At the crossover frequency fc, || T || = 1

fp1

QdB

– 40 dB/decade

| T0 |dB

fz

fc fp2

– 20 dB/decade

Crossoverfrequency

f

|| T ||

0 dB

–20 dB

–40 dB

20 dB

40 dB

60 dB

80 dB

– 40 dB/decade



Transient response vs. damping factor

0

0.5

1

1.5

2

0 5 10 15tct, radians

Q = 10

Q = 50

Q = 4

Q = 2

Q = 1

Q = 0.75

Q = 0.5Q = 0.3Q = 0.2

Q = 0.1

Q = 0.05Q = 0.01

v(t)


9.4. Stability

Even though the original open-loop system is stable, the closed-looptransfer functions can be unstable and contain right half-plane poles. Evenwhen the closed-loop system is stable, the transient response can exhibitundesirable ringing and overshoot, due to the high Q -factor of the closed-loop poles in the vicinity of the crossover frequency.When feedback destabilizes the system, the denominator (1+T(s)) terms inthe closed-loop transfer functions contain roots in the right half-plane (i.e.,with positive real parts). If T(s) is a rational fraction of the form N(s) / D(s),where N(s) and D(s) are polynomials, then we can write

T(s)1 + T(s) =

N(s)D(s)

1 + N(s)D(s)

= N(s)N(s) + D(s)

11 + T(s) = 1

1 + N(s)D(s)

= D(s)N(s) + D(s)

• Could evaluate stability byevaluating N(s) + D(s), thenfactoring to evaluate roots.This is a lot of work, and isnot very illuminating.


9.4. Stability

Even though the original open-loop system is stable, the closed-looptransfer functions can be unstable and contain right half-plane poles. Evenwhen the closed-loop system is stable, the transient response can exhibitundesirable ringing and overshoot, due to the high Q -factor of the closed-loop poles in the vicinity of the crossover frequency.When feedback destabilizes the system, the denominator (1+T(s)) terms inthe closed-loop transfer functions contain roots in the right half-plane (i.e.,with positive real parts). If T(s) is a rational fraction of the form N(s) / D(s),where N(s) and D(s) are polynomials, then we can write

T(s)1 + T(s) =

N(s)D(s)

1 + N(s)D(s)

= N(s)N(s) + D(s)

11 + T(s) =

11 + N(s)D(s)

= D(s)N(s) + D(s)

• Could evaluate stability byevaluating N(s) + D(s), thenfactoring to evaluate roots.This is a lot of work, and isnot very illuminating.

Fundamentals of Power Electronics

Effect of feedback on transfer function poles

Feedback moves the poles of the system transfer functions• Good news: we can use feedback to alter the poles and

improve the frequency response• Bad news: if you re not careful, feedback can move the poles

into the right half of the complex s-plane (poles have positivereal parts), leading to an unstable system

G(s)

+–

H(s)

T(s)

vout(s)ve(s) G(s)vin(s)

Open loop Closed loop


Example

G(s) = 1001 + s 3

The gain G(s) below has three poles at s = – 1

Re(s)

Im(s)

–1xxx

Complex s-plane

+—

T(s)

vout(s)ve(s)G(s)

vin(s)

G(s) = 1001 + s 3

H(s) = 1

Add a simple feedback loop:

How does thefeedbackchange thepoles?


Exact closed-loop transfer function

For our simple example, the closed-loop transfer function is

Factor denominator numerically:

which has poles at s = – 5.64 (LHP)

and at s = +1.32 ± j4.07 (RHP)

The RHP poles indicate that theclosed-loop system is unstable.

Re(s)

Im(s)

–5.64

x

x

x

j4.07

+1.32

— j4.07

voutvin

= 1H

T1 + T = G

1 + G =

1001 + s 3

1 + 1001 + s 3

= 100101 + 3s + 3s2 + s3

voutvin

= 100101 + 3s + 3s2 + s3 = 100

s + 5.64 s – 1.32 – j4.07 s – 1.32 + j4.07


Transient response of closed-loop system

One can take the inverse Laplace Transform to find the output waveformvout(t) for a given input. The resulting expression has terms that dependon the poles, of the form

vout(t) = K1e– 5.64t +K2e (1.32 – j4.07)t + K 2*e (1.32 + j4.07)t

The terms with positive real exponents, corresponding to the RHPpoles, lead to growing oscillations that are unstable responses.

Reason: the inverse Laplace transform of K2e (1.32 – j4.07)t + K 2*e (1.32 + j4.07)t is

K2 e1.32t cos 4.07t + K2


Determination of stability directly from T(s)

• Nyquist stability theorem: general result.• A special case of the Nyquist stability theorem: the phase margin test

Allows determination of closed-loop stability (i.e., whether 1/(1+T(s))contains RHP poles) directly from the magnitude and phase of T(s).A good design tool: yields insight into how T(s) should be shaped, toobtain good performance in transfer functions containing 1/(1+T(s))terms.


Determination of stability directly from T(s)

• Nyquist stability theorem: general result.• A special case of the Nyquist stability theorem: the phase margin test

Allows determination of closed-loop stability (i.e., whether 1/(1+T(s))contains RHP poles) directly from the magnitude and phase of T(s).A good design tool: yields insight into how T(s) should be shaped, toobtain good performance in transfer functions containing 1/(1+T(s))terms.


9.4.1. The phase margin test

A test on T(s), to determine whether 1/(1+T(s)) contains RHP poles.The crossover frequency fc is defined as the frequency where

|| T(j2!fc) || = 1 ! 0dB

The phase margin "m is determined from the phase of T(s) at fc , asfollows:

"m = 180˚ + #T(j2!fc)

If there is exactly one crossover frequency, and if T(s) contains noRHP poles, then

the quantities T(s)/(1+T(s)) and 1/(1+T(s)) contain no RHP poleswhenever the phase margin "m is positive.


Example: a loop gain leading toa stable closed-loop system

!T(j2!fc) = – 112˚

"m = 180˚ – 112˚ = + 68˚

fc

Crossoverfrequency

0 dB

–20 dB

–40 dB

20 dB

40 dB

60 dB

f

fp1fz

|| T ||

0˚

–90˚

–180˚

–270˚

m

� T

� T|| T ||


Fundamentals of Power Electronics! Chapter 9: Controller design!27!

Computation of crossover frequency!

fc

Crossoverfrequency

0 dB

–20 dB

–40 dB

20 dB

40 dB

60 dB

f

fp1fz

|| T ||

0˚

–90˚

–180˚

–270˚

m

� T

� T|| T ||


1. The expression for T(s) is:!

T (s) = T0

✓1 +

s

!z

◆

1 +

s

Q!p1+

✓s

!p1

◆2!

2. Write the equation of the asymptote for f > fz:!

kT (j!)k = T0

✓!

!z

◆

✓!

!p1

◆2 = T0f2p1

fzf

3. Equate to 1, and solve for f (= fc):!

fc = T0f2p1

fz


Computation of phase!

fc

Crossoverfrequency

0 dB

–20 dB

–40 dB

20 dB

40 dB

60 dB

f

fp1fz

|| T ||

0˚

–90˚

–180˚

–270˚

m

� T

� T|| T ||


T (s) = T0

✓1 +

s

!z

◆

1 +

s

Q!p1+

✓s

!p1

◆2!

Exact expression for phase:!

\T (j!) = tan�1

✓!

!z

◆� tan�1

2

6664

1

Q

✓!

!p1

◆

1�✓

!

!p1

◆2

3

7775

Expression for phase asymptote over frequency range as illustrated near fc:!

\T (j!) ⇡ (+45

�/dec) log10

✓!

!z/10

◆� 180

�

Evaluate one of the above to find ∠T(jωc), then compute phase margin:!

ϕm = 180˚ + ∠T(jωc)!


Example: a loop gain leading to"an unstable closed-loop system!

∠T(j2πfc) = – 230˚!

ϕm = 180˚ – 230˚ = – 50˚!

fc

Crossoverfrequency

0 dB

–20 dB

–40 dB

20 dB

40 dB

60 dB

f

fp1

fp2

|| T ||

0˚

–90˚

–180˚

–270˚

� T

� T|| T ||

m (< 0)


r. w. ericksonecee.colorado.edu/~ecen5797/course_material/lecture28.pdftransistor gate driver b(t) i...

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