Power Electronics of Piezoelectric ElementsPower Electronics Laboratory ... – Sonar transducer ... Prof. S. Ben-Yaakov, Power Electronics of Piezoelectric Elements, June 2006 ...medinid/energy-gordon/Piezoelectric/Slides... ·

1

Prof. S. Ben-Yaakov, Power Electronics of Piezoelectric Elements, June 2006 [1]

Power Electronics LaboratoryDepartment of Electrical and Computer Engineering

Ben-Gurion University of the NegevP.O. Box 653, Beer-Sheva 84105, ISRAEL

Phone: +972-8-646-1561; Fax: +972-8-647-2949;Email: sby@ee. bgu.ac.il; Website: www.ee.bgu.ac.il/~pel

Gordon Seminar, Tel-Aviv University, June 2006

Shmuel (Sam) Ben-Yaakov

Power Electronics of Piezoelectric Elements


1. Introduction• Piezoelectricity• Brief overview of Piezoelectric devices

• Actuators• Vibrating vans• Motors • Micro-PowerGenerators/Dampers• Transformers• Miscellaneous

OUTLINE

2


OUTLINE (Cont.)

2. Models of Piezoelectric devices3. Drivers4. Rectifiers5. PT based CCFL Ballasts

• The stability issue• Envelope Simulation• Thermal effects


Piezoelectricity

Discovery-1880 by Pierre and Jacques Curie– Sonar transducer– Pickup and microphone– High frequency quartz resonators

1. Introduction

3


Piezoelectricity (Cont.)1940- Piezoelectric Ceramics

e.g. lead-zirconate-titanate (PZT), lead-titanate (PbTiO2), lead-zirconate (PbZrO3), and barium-titanate (BaTiO3)Plastic Piezo material, PVDF (Polyvinylidene fluoride)

–Powerful “sonars”–Systems of piezo-ignition –Ceramic tone-transducers, Buzzers, Speakers –Piezoelectric motors an actuators –Piezoelectric transformers–“Exotic” devices: damper, power sources…


Range of Applications

TechOnLine –Applications for Piezoelectric Ceramics.htm

4


This seminar relates to modern piezoelectric devices with particular emphasis on piezoelectric transformers.

Overview of Piezoelectric devices and associated electronics from the PowerElectronics point of view.

OBJECTIVE


V

V

• Mechanical electrical interaction• Electrical field Mechanical Stress

Piezoelectric material

Electrode

1. Brief overview of piezoelectric devicesThe piezoelectric effect

5


where

61valuestakesq,p31valuestakesk,j,i

−−[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]kTikqiqi

kkpqEpqp

ETdD

EdTsS

⋅ε+⋅=

⋅−⋅=

ntdisplacemeelectricDfieldelectricE

companentStressT

companentStrainS

i

k

q

p

==

=

=

stressttanconsatttanconstypermittivi

ttanconsricPiezoelectd

fieldelectricttanconsatttanconscomplianceTS

s

Tik

kp

constEq

pEpq

=ε

=

=∂

∂=

=

Compressed notation and matrix arrays:


Constants – IEEE Standard

table_piezo_nomenklature_01.jpg table_piezo_nomenklature_02.jpg

6


Features and Applications:

• Small deflection (µm range)• Static and dynamic applications• Light deflection • Positioning, no friction or backlash • Valve control

L

LStack of piezoelectric elements

Actuators


XY Positioning

200 nm span

7


Serial BimorphParallel Bimorph

•Same idea as bi-metal•Large deflection, mm range

Bimorph Benders


Electricalterminals

ppx −∆

VanBimorphs Piezoelectric element

Base

Vibrating Van

8


Bi-Morph Actuators and Vibrating Vans

• Large deflection • Light Choppers• Remote operation• Valve control • Fan



• Nanomotion Ltd. Israel

piezo ceramicelectrode A

electrode B

electrode C B

A

common

Ellipticmovement

S1 S2

stator

piezo actuatordriver

VAC

stage

Piezoelectric motors

9


Operation Demo

Nano_motor.avi


Nanomotion’s NanoLens

10



• Linear motion• Circular motion• Sub-micron motion and positioning• Small size• Vacuum compatible• Camera lenses• HD drive• Microelectronics manipulators

Piezoelectric Motors


MASS

Mechanical Vibration

Micro-Power Generators (Mechanical to Electrical energy harvesting)

11


Silicon beamPiezoelectric

element

Vout

63Ni radioisotopeemitter

[12]

Micro-Power Generator


MIT’s Piezo Tennis Shoe

12


Shaking table

Piezoelectric element

Beam

Vin

Dampers


Sports Active Damping Patent

13


Ski DampingCEDRAT TECHNOLOGIES & SKI ROSSIGNOL


Dampers experimenthttp://live.pege.org/2005-material/oscillation-damping.htm

14



• Active suspension• Skis• Motorcycles• Remote energy sources• Tennis shoes• Structures

Micro-Power Generators and Dampers


Piezoelectric Transformers

Vin VoPT RL

Vin

Vo

[15]

15


Transformer examples

P

P

T

Vin

Vout

• Radial mode[16]


VP

VS

primarypart

secondarypart

x

xdisplacement

potentialVS

supportpoint

poling

• High voltage gain

[15-18, 70]

Rosen Type Piezoelectric Transformer

16


• Higher voltage transfer ratio

Vin

A

A

Rosen Type Piezoelectric TransformerMultilayer


Step Down

17


Characteristics of Piezo transformersAdvantages– Potentially low costs– Compact size– High efficiency – Ability to work at high frequency– Good insulation capability– No windings, i.e. no magnetic fields

Disadvantages– Resonant device (frequency and load dependent)– Low Power– Cost


1. Fluorescent lamp driver for laptop backlight (commercial)

2. Fluorescent lamp driver for LCD monitor (TV) backlight3. Ionizer (commercial)4. Fluorescent lamp ballast (high power) 5. Cell phone battery charger 6. Laptop battery charger 7. Isolated gate driver

The main reason for commercial holdup: price

Piezoelectric Transformer Applications

18


• Welders • Ultrasonic Cleaners• Humidifiers• Nebulizers• Massage and skin scrubbers• Ozonator (high voltage)

Miscellaneous devices and applications


Purpose: •Analytical derivations•Simulation

Approach:• Mechanical-Electrical analogy• Equivalent circuit (based on Mason’s Model)

2. Models of piezoelectric devicesModeling of Piezoelectric Elements

devices

19


Electrical-Mechanical Relationships

TechOnLine - Piezoelectric Sidebar 2Bridging the Mechanical and Electrical Worlds.htm


Mechanical-Electrical Analogyc

rm

Electronicsystem

MechanicalSystemm-mass L

r-losses R

c=1/stiffness C

v-velocity i

F-force u

CrLrRm

20


Electrical Coupling

• Electrical connection by plated electrodes


Equivalent Circuit

LrRm CrVin

1:n

Cin

• The transformer emulates the coupling between the electrical-mechanical energies

Original

Reflected to PrimaryLrRm CrVin

Cin

21


Resonant modes

Longitudal

Shear

Flexural


Standing-Waves Wavelengths

displacement

Z

λ/2(a)

Half wavelength

22


λ(b)

displacement

Full wavelength

Standing-Waves Wavelengths


Many resonant modesModel

1 nn

LSn CSn RSnCin

1 n2

LS2 CS2 RS2

1 n1

LS1 CS1 RS1

Vin

23


• Usually operated below resonance• Mass includes the work piece • For practical actuators ZCm<< Rm• Cm in the µF range• Highly capacitive

L

L

Actuators


Vibrating Van Model

inC

1rC

1rL

1mR

P.R.B

)j(Yin ω 2rC 3rC

VanBimorphs Piezoelectric element

Base

• Low frequency • Operation at resonance for maximum

displacement

[2]

24


Piezoelectric Motor Model m

R2

Rs

Vs

0Vdc

V4

TD = 0

TF = 100nPW = 1/(2*f req)PER = 1/f req

V1 = 0

TR = 100n

V2 = Vdc

C2

C_in

in

V3

FREQ = f reqVAMPL = amplVOFF = 0

R1

Rm

C1

Cr

L1

Lr

1

2

V11Vac0Vdc

PARAMETERS:Lr = 0.00019836C_in = 39nCr = 3.5382e-8f req = 45kampl = 27.4*1.41Vdc = 0Rs = 1uRm = 155.37

0

• Operation near resonance


Ti me

300us 320us 340us 350usV( M)

0V

50V75V

SEL>>

I ( Vs )- 2. 0A

0A

2. 0AV( I N)

0V

50V

100V

Measurement

ModelSimulation

Voltage

Current

Voltage

Current

Drive

Drive

25


Ti me

300us 320us 340us 350usI ( Vs )

- 1. 0A

0A

1. 0AV( I N)

- 40V

0V

40V

SEL>>

Voltage

Current

Voltage

Current

Measurement

ModelSimulation


os)opt(L C

Rω

=1

RL

PoutoCini RL

( )22

1 Los

L)rms(inout

RC

RIP

ω+=

Ropt

Piezo Source Load

[82]

Basic Generator/Damper Model

26


Piezoelectric Transformer (PT) Model (for One Resonant Mode, Close to Resonant Frequency)

CrLr RmCin Co

Energy-Couplinginput

Energy-Couplingoutput

1:n1 1:n2

PT RL

Vin

Vo


Equivalent Circuits of a PTCrLr RmCin Co

CrLr RmCin CoVin Vo

ir

nir

nVo

1:n

A

B

• Model B is preferred for simulation[22]

27


Reflecting the output side to the primary

oo CR ′′′′ andofconnectionSeries

2o

o nRR =′

o2

o CnC =′

oo CR ′′ andofconnectionParallel

2oo

oo )RC(1

RR′′ω+

′=′′

2oo

2oo

oo )RC()RC(1CC

′′ω′′ω+′=′′

Lr

Cin

Rm Cr

Vin V'o

R''o

C''o

Lr

Cin C'o

Rm Cr R'oVin V'o

[22]


Input output voltage transfer ratioRosen Type; Cr=37.234pF, Lr=155.3mH, Rm=136.1157 W,

n=4.4899, Cin=720.32pF, Co=19.404pF, frs=66.191kHz.

28


Generic Characteristics of PTsDefinitions:

maxinVoutV

m21k ⎟⎟⎠

⎞⎜⎜⎝

⎛ ′=

)rLrC(2inV

oP

basPoP*

oP ==

100inPoP×=η

rCrLoRoC

oRoCrsQ =ω=

mRrCrs

1mQ

ω=

rCoC2n

rCoCc =′

=[22]

- output capacitance times the stiffness of the ceramic

- maximum value of output to input voltage ratio

- output power per unit system

- efficiency

- normalized load

- mechanical quality factor


ηη= 0.5

ηm

(k21)ηm

[Q]

k21m

Po m*

Po*

(Po)ηm*

[k21m]

[η][Po]*

cQm

Qmc + 1

1

1 + 12c

Generic Characteristics

[22]

29


0

0.5

1

1.5

2

2.5

3

10 10

5

10

15

20

25

30

VoutV in m

))

R o [Ω ]

PoVin

2mWV 2

10 2 10 3 10 4 10 5 10 6

P oV in

2

V outV in )

)

mm

Experimental Results

circles - experimental results; lines - theoretical prediction. [22]


Model Parameters Extraction• Measurements (Network Analyzer)• Fitting

[2, 15]

Problems• Model is drive-level dependent• Model is load dependent• Model is non-linear• Some have a low Q

Most published parameters are based on low voltage measurements for one load

30


Drive Dependence & NonlinearityInput admittance ,magnitude

150 160 170 180 190 200 210 220 230 240 2500

0.51

1.52

2.5

Input admittance ,phase [degrees]

3x 10-4

Vin=5(Vrms)Vin=25(Vrms)

150 160 170 180 190 200 210 220 230 240 250-20

020406080

100

frequency [Hz]

Vin=5(Vrms)Vin=25(Vrms)

non-linear region

[2]


Red- V0=500V, green – 400V, blue – 100Vrms

Vo/Vin

Rosen type single layer

Model dependence on output voltage

31


Rosentype Multilayer. ELECERAM Ltd.

Model dependence on load


Rosen type single layer

Fitting range

Simulation/Experimental agreement

32


Ω50output input

RA

A×

B.R.P,inY

Network Analyzer

inY

P.R.B

.Amp,outV

B.R.PI

AmplifierRF

Ω50

RShunt

RAtten

• Connection for 50Ω input resistance analyzer

DUT

[2]

Measurements under high power excitation


150 160 170 180 190 200 210 220 230 240 2500

0.5

1

1.5

2 x 10-4 Input admittance magnitude

measuredcalculated

150 160 170 180 190 200 210 220 230 240 2500

20

4060

80100

frequency[Hz]

Input admittance phase[degrees]

measuredcalculated

Fitting area

Fitting area

f1=195[Hz]f2=200.5[Hz]

f > frC=67.8nF Lr=84.551H Cr=8.6184nF Rm=6508Ω

[2]

Results of Least square fitting

33


Transformer Model- Parameter fitting

• Can follow the impedance measurements procedure by shorting the output

• Shorting the output may lead to erroneous results

• Proposed method: Fitting under nominal voltage/power conditions by a Forward-Backward method – under loadedcondition


EXPERIMENTAL SETUP

Ω50output input

RA

A×

Network Analyzer

AmplifierRF

Ω50

Rd1

PT

Rd2

VoutVin

• Forward connection

34


FORWARD

⎥⎦

⎤⎢⎣

⎡

ω−⎟⎟

⎠

⎞⎜⎜⎝

⎛+ω+ω−++

==

from

f

ror

2

r

o

f

min

oo

RCnCnR

RnLjCnL

CnC

RnRn

1VV

k

L r Cr Rm

Cin Co+-Vo

n

I r

IrnVin Rf

T1 T2

Vo

L r Cr Rm

Cin

I r

Vin Con2 Rf

n2

Von

( ) ( )

( ) ( )( )f

ff

fffout

inf

kRekImtan

kImjkReVV

k

=ϕ

+=⎟⎟⎠

⎞⎜⎜⎝

⎛=


BACKWARD

⎥⎦

⎤⎢⎣

⎡

ω−⎟⎟

⎠

⎞⎜⎜⎝

⎛+ω+ω−++

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

rrinm

r

rinr

2

r

in

r

min

oi

RCnCnR

RnLjCnL

CnC

RnR

n1

1VV

k

L r Cr Rm

Cin+-Vo

n

I rT2

inVCoIrn

T1

RrVo

L r Cr Rm

Cin

I r

Vinn

RrVo

( ) ( )

( ) ( )( )r

rr

rrrout

inr

kRekImtan

kImjkReVV

k

=ϕ

+=⎟⎟⎠

⎞⎜⎜⎝

⎛=

35


Solving for ϕ

( ) ( )( )

or2

r

o

f

m

from

f

r

f

ff

CnLC

nCR

nRn

RCnCnR

RnL

kRekIm

tanω−++

ω−⎟⎟

⎠

⎞⎜⎜⎝

⎛+ω

==ϕ

( ) ( )( )

inr2

r

in

r

m

rrinm

r

r

r

rr

CnLC

nCR

nRn1

RCnCnR

RnL

kRekImtan

ω−++

ω−⎟⎟

⎠

⎞⎜⎜⎝

⎛+ω

==ϕ


Expanding

( ) ( )( ) fror

3formfr

fromrr2

fRCCLRCCRRC

1RCCRCLtan

ω−++ω

−+ω=ϕ

( ) ( )

rrinr3

rinrmrr2

rrinmrr2

rRCCLRCCRRC

n1

1RCCRCLtanω−⎟

⎠⎞

⎜⎝⎛ ++ω

−+ω=ϕ

36


FITTING

( ) ( )( ) ( )⎩

⎨⎧

=ϕω−ω+ϕω=ϕω−ω+ϕω

1tanbbtanb1tanaatana

r32

2r3

1

f32

2f3

1 ai-bi are found by mean square error method

( ) ( )( ) fror

3formfr

fromrr2

fRCCLRCCRRC

1RCCRCLtan

ω−++ω

−+ω=ϕ

( ) ( )

rrinr3

rinrmrr2

rrinmrr2

rRCCLRCCRRC

n1

1RCCRCLtan

ω−⎟⎠⎞

⎜⎝⎛ ++ω

−+ω=ϕ


Least-Square Fitting

rinrmrr23 RCCRRCn1b ++=

rrinr1 RCCLb =

rrinmrr2 RCCRCLb +=

fror1 RCCLa =

fromrr2 RCCRCLa +=

formfr3 RCCRRCa ++=

Initial estimation of PT equivalent parameters:Lr(ini),Cr(ini), Rm(ini), Cin(ini), Co(ini), n(ini)

( ) ( )( ) ( )⎩

⎨⎧

=ϕω−ω+ϕω=ϕω−ω+ϕω

1tanbbtanb1tanaatana

r32

2r3

1

f32

2f3

1 ai-bi are found by mean square error method

37


Calculations and Experimental Results


Phase

38


Load Dependence of Parameters0 – 600 KOhm

nCr

Rm Fr


DRIVERS

Main Issue:

• High input capacitance • Need for nearly sinusoidal drive• Fast response

3. Drivers

39


controlGSV

SV SVSI

dJswitchingP

t

t

t

Switching losses due to overlap Pd linear with fS !

Switching lossesHard Switching


Soft Switching Real and pseudo

snubberdtdV

DVDI

dtdVD

DV

t

switchingsoft"True"

switchingsoft"Pseudo"

snubberdtdI

40


Inverter

PTInputMatchingNetwork

Vo

Vin

VDS

ControlIDS

VDS

IDS

Hard switching

Achieving ZVS of the inverter switches

VDS IDS

Soft switching

[23, 24]

Input Matching Network


Main problem:Extremely high input capacitance

Requirements:• DC to low frequency• High accuracy• Low frequency ripple• Voltage range: 100V-1000V• Charge recovery method

Actuators Drivers

41


Class ABVCC

-VEE

Vin

Q1

Q2VCC

Q3

Q4

• “Charge recovery”

[26, 27, 58]

Class D

• High Losses


Commercial Amplifier

42


Stability Criterion

)f(LG1KHA 1

CL +=

The system is unstable if 1+LG(f) has roots in the right half of the complex plane.


Nyquist

))f(LG(gIm

))f(LGRe(1−

mΦ

Nyquist criterion can be used to test for the location of 1+LG(f) roots.

43


Capacitive Load - Stability Issue

LCOR

fRinR

inV]Hz[f

1f 2f

Pf

LO C,R]Hz[f

]Hz[f

LG

]dB[A

]DB[LG dec/dB20−

dec/dB40−

dec/dB60−

combined

1f 2fPf

dec/dB20−

dec/dB40−

]DB[LG

LOP CR2

1fπ

=


Decoupling ResistorOne possible solution

]dB[A

]Hz[f1f 2f

Pf

LO C,R]Hz[f

]Hz[f

OLA

]dB[A

]dB[A dec/dB20−dec/dB40−

dec/dB20−combined

1f 2fPf

Zf

dec/dB40−

Zf

dec/dB20−

dec/dB40−

LCOR

fR

inR

inV SR

LSOP C)RR(2

1f+π

=

LSZ CR2

1fπ

=

Signal attenuation at f > fP

44


Demo Circuit


High Power Vibrating Devices Drivers(welders, atomizers, etc.)

• Resonant drivers• Class D

Ls Cs Vp

1:n

VinLs

Vp

1:n

Vin

LLCC

LC and PWM

Q1

Q2

CinVCC

D1

D2

1. LC

D3

D4

Q3

Q4

LrRm Cr

Outputfilter

2. LCC3. PWM

[26]

45


Converter topology

LC LLCC PWM

]A[i ]mH[L ]A[i ]mH[L ]A[i ]mH[L Series

inductance SL 0.01 8.8 0.01 6.37 0.01 0.258

Parallel inductance PL - - 0.0029 6.37 - -

PWM: Most compact but higher losses due to hard switching at 250kHz

(After [26])

Operating frequency ~ 20kHz


Vibrating Van Drivers

Requirements:

• Low frequency• Low power• Constant frequency• Locking to resonant frequency

46


Vin

Iin

Squarewave Sinusoidal Trapezoidal

Vin

Iin

[2]

Vin

Iin

Vin

Iin

Vin

IinSeries inductance (Lseries) that needs to be placed in series with the PRB to achieve ZVS:

C)f2(1L 2r

seriesπ

>

For commercial PRB, Lseries>10H !

Vibrating Van Drive


Motors Drivers

Requirements:Relatively high frequencySinusoidal driveConstant frequency operationVariable amplitudeFast response

[6, 7, 13, 14, 46]

At operating point:

ZCm<< Rm Very high reactive current

47


• Soft switching• Requires high Q• Sensitivity to capacitance variations• High circulating current • High switch current

[3, 4]

Resonant inverter


LinRLCLC2

Vin

OSCILLATOR + DRIVERS

Lmn2

T1 ZL

V

n1

n1

VtapC1

DQ2

VGS2

VD2

2D1

Q1VGS1

D1

time

V

Ts

Ts 2

VD1VD2 VD2

Vtap

VGS1

VGS2

Vtap

Vtap

time

time

time

time

The Current-Fed Push-Pull Parallel-Resonant Inverter (CFPPRI)

rs ff =

rs ff <

rs ff >

Frequency deviation will cause:

Efficiency reduction.

Output signal distortion.

ZVS

Boost period

Hard switching

[3]

48


Self-Adjusting CFPPRI

L inCL

ZL

RL

VD2

V in

D2

C2

C3

Ibias

D1

Q1 Q2

VD1

R in2

n

T1

Lrn2

n11

12

n3R2

Phasecomparator

LPFFin Rin1R1C1

Fβ

D3

Q3Vin

+-

PWMModulator

R f

Vref

A1

Vtap

VGS2VGS1

COMP1

Vref

Q QTDR2DR1

FF

phas

e fe

edba

ck

Soft Switching Controller (SSC)

Current feedback

Controlledinductor

[28]• Reactive power locked in resonant tank


Transformers Drivers

Requirements:

• Relatively high frequency• Near Sinusoidal waveform • Soft switching• Gain• Power range (DC-DC, Ballast)• Ignition voltage (Ballast)

49


PTControl InputMatchingNetwork

Iin

Vin Vout

Hard switching Soft switching

VDS

IDS

VDS IDS

[29]

Half-Bridge Inverter Topology


Inductor-less Half Bridge drive

PT

• Simplest and most elegant

[30-32]

50


Steady-State Current and Voltage Waveformsof the ZVS PT Inverter

)tsin(I)t(i m ψ−ω=

i(t)

VinD2

D1

Iin

Cin

Q1

Q2

VDC

Res.tank

VGS1

VGS2

Im and ψ are the current peak and the initial phase

VGS2 VGS1

Vin

iin

i(t)

VGS

t

t

t

t

t0 t1 t2 t3 t4 t5

t0-t1 – charging time

t0-t2 – dead time

t2-t3 – Q1-ON

t3-t4 – discharging time

t3-t5 – dead time

[32]


Normalized load factor Q and frequency range

k

∆κ

∆κ∆κ

r∆

1.0151.005 1.01 1.02 1.025

0.05

0.15

0.25%)97=(ηPT

37.0Q =

25.0Q =

%)5.90(ηPT=13.0Q =

%)5.94(ηPT=

∆r is the normalized charging time

k is the normalized operation frequency

Q is the normalized load factor

∆k is the frequency range for soft switching

Limitations:Qmax at ∆r=0.25

Qmin at ∆PD=10%

0.13<Q<0.37 [32]

51


f=120kHz, RL=130Ω (Q=0.15)

tr

[32]

Experimental voltage curves


Advantages:•Simple •Low cost

Disadvantages:•Small operational range•Non optimal operation•Not applicable to all transformers•Trapezoidal waveform• No voltage gain

[29-32]

Inductor-less Half Bridge drive

52


Vin LpInverter PTCb

Vin

LpInverter PT

Cb

A B

A - DC on PTB – Only AC on PT

Voltage on Cb = ½ Vin [21, 29, 50]

Parallel Inductor


Q1

Q2Cin

Vin

D1

D2

CDS1

CDS2

PT

VGS2

VGS1

ControlLP

CB

r

CP C2

DT)D1(TL −=

2DS1DSinr CCCC ++=

cycledutyD −

periodT−T1.0TC ≈

rCoftimeingargchthe

• Advantages: ZVS, lower EMI, constant voltage• Disadvantages: Non sinusoidal waveform, higher

conduction losses, no voltage gain

• Trading switching losses with conduction losses[21, 29]

Voltage-Fed Half Bridge Inverter with a Parallel Matching Inductor LP

53


Series Inductor

PTLin

• Simple method for obtaining soft switching• May attenuate or boost PT input voltage• May change overall frequency response• Best dealt by simulation • A coupling capacitor will eliminate DC on PT

[25, 29, 48, 50, 62, 63, 69, 74]


Input Impedance of the PT

Fr equency

100KHz 150KHzp( I ( L2) ) - p( V( PTI NAC) )

0d

- 100d

100d

SEL>>

V( OUTAC) / V( PTI NAC)

0. 50

1. 00

0. 05

• Not always capacitive around the operating frequency

inductive Fr equency

100. 0KHz43. 6KHzp( I ( L4) ) - p( V( PTTAC) )

0d

50d

100dV( out ) / V( PTTAC)

0

2. 5

5. 0

7. 0

SEL>>

Vout/Vin

Phase

PT “A” PT “B”

54


Fr equency

100KHz 120KHz 140KHz 150KHzV( OUTAC) / V( I NPUTAC)

0

1. 0

2. 0

P( - I ( V6) )- 100d

0d

100d

SEL>>

V( OUTAC) / V( PTI NAC)0

1. 0

2. 0

• Examination by small signal (AC) simulation

Overall Vout/Vin

PT response

Phase of series inductor current

1mH 1.6mH

2.4mH

PT Operating Region


Ti me

920. 0us 930. 0us914. 3us 935. 0usV( I NPUTTRAN)

50V

- 10VSEL>>

V( PTTRANS)0V

25V

50VI ( L4)

- 40mA

0A

40mA

Inductor current

PT voltage

HB commutation

• Large-signal time-domain (TRAN) simulation

55


Ti me

672. 00us 674. 00us 676. 00usV( I NPUTTRAN)

0V

20V

40V

52V

Green=1mHRed=1.6mHBlue= 2.4mH

• HB commutation


Diode Clamped Resonant Snubber

Q1

Q2

Lr

Cin

VCC

DQ1

DQ2

PTC1

C2

D1

D2 Cex

• Forces soft commutation [39, 44]

56


Lin=10m; Cex=1n

Single layer, RosenType, ELS-60 Eleceram Co.

Ti me

700. 0us 710. 0us 720. 0us 730. 0usV( I NPUTTRAN)

0V

200VV( PTTRANS)

- 200V

0V

200V

SEL>>

I ( L104)- 100mA

0A

100mAV( OUTTRAN)

- 500V

0V

500V

Output voltage

Inductor current

PT input voltage

HB commutation


0Itt0Itt

Q43

D31

>−

>−0Vtt

offQtt

inC10

20

>−

−−

Advantages:1. Only one switch2. Low cost3. Better EMI suppression

Disadvantage:1. Small operational range2. High voltage stress

L Lr

Cin C'o

Rm

Cr R'oVin V'oDQ

VCin

IDIQ

IIN PT

Control

[33-35, 54]

Class E Inverter

57


Dual inductor Current-Fed Push-Pull Inverter (PPRI)

VinVoPT

Q1CDS1

D1 Q2

CDS2D2

L2L1IL1 IL2

RoPTCVDS1

VDS2

D2D1Q2Q1Interval

1o tt −

21 tt −

32 tt −

43 tt −

on off off off

off

off

off off

off off

off

on

on

on on

on

Advantages: Nearly sinusoidal waveformDisadvantages: Narrow operational range, no voltage gain

[36]


Single Inductor PPRI

• Extra voltage gain by transformer• Good for fixed power level

LinC2

Vin

OSCILLATOR + DRIVERS

Lmn2

T1

PT

V

n1

n1

VtapC1

D

Q2VGS2

VD2

2D1

Q1VGS1

D1

[28]

58


Resonant Forward-Flyback Inverter

Vin VoPTRo

PT

QC D

Lm

Llkg2

CIN

IME

IME – Integrated Magnetic ElementLm – magnetization inductanceLlkg2 – leakage inductance reflected to the secondary

[47]


Resonant forward-flyback self-oscillating inverter(magnetic element on pot core P14/8) and Rosen type PT (PXE43 48 x 8 x 2.2 mm)

[47]

A Piezoelectric Cold Cathode Fluorescent Lamp Driver Operating from a 5 Volt Bus

59


[47]

Experimental Results


Power Control

• Frequency shift• PWM

• Resonant converters• Asymmetrical (half bridge) • Low frequency PWM (ON-OFF)

• PFM Combined PWM and PFM [37, 66]

60


Half Bridge Driver with Active-Clamp

VinDC

PT

Q1

LS

LrCr

VLrQ2

VGS1 VGS2

VDS1

VLr

t

t

t

VGS

)Dsin(D

VV inLr π

−π=

12

• ZVS• Voltage gain• Squarewave drive [37, 38]

0 0.2 0.4 0.6 0.8 1

in

oVV

Active clamp

Asymmetrical

D


Lo

PTinV 2D

1D fL

fC LR

PTinV

2D1D fC

LR

L1

L2

Current Doubler

Half Wave

[21, 38, 43]

4. RectifiersPT Rectifiers - Inductive Filter

61


PT Rectifiers - Capacitive Filter

VoltageDoubler

DiodeBridge

CF

Vin

PTRo

D2

D1

D3

D4

VinCF

PTRoD1

D2

[19, 30, 31, 39, 40, 62, 73]


Voltage Doubler Current Doubler

VCo

ir

iD iD1iD2

ϑ

ϑ

ϑ

θ

VCo

ir

iD

ϑ

ϑ

ϑ

iD1 iD2

λ

• Voltage and current may not be in phase• Equivalent AC load is not resistive [39, 40, 43]

The PT Voltage and the Current Waveforms

62


Analytical Method

PTinV LRoC Rect

n

vCoinV inC

rL rC mR

nir

C Reqeq

[39-41]


Equivalent Circuit Parameters

mPTPT

)1(2

L

eqrm QKA

sin

R

R1

ϕ+ω=ω

Voltage Doubler

⎪⎪⎩

⎪⎪⎨

⎧

ω

ϕ=

=

VDeq

)1(VDeq

L2

)1(VDeq

R

tanC

Rk81R

)1(pk)1(Co

Lrect k

2V

Vk ==

Current Doubler

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

ω

ϕ=

⎟⎟⎠

⎞⎜⎜⎝

⎛

λπ

=

CDeq

)1(CDeq

L2

)1(

22CDeq

R

tanC

Rk21R

)1(2pk)1(Co

Lrect

kVV

kπ

λ==

Rectifier Voltage Transfer Ratio

Frequency of the Maximum Output Voltage

[43]

63


100000.1 1 10 100 10000

0.5

1

1.5

2

2.5

3

3.5

4

0.12

0.13

0.14

0.15

0.16

0.17

0.18

0.19

0.2

ω C Ro L[ ]

eqR /RL

adC /Co

[C

/C ]

ado

[R

/R ]

eqL

• Equivalent AC load for voltage doubler[40]

Cad =Ceq-Co

Example


Equivalent Circuit of a PT AC/DC Converter Referred to the Primary

Lr C r R m

Cin

Ir

V in ReqCeq

2eq

'eq

2eq

'eq

nCC

nRR

=

=

)1(2

'eq''

eq

)1(2'

eq''eq

sin

CC

cosRR

ϕ=

ϕ=

L r C r R m

C in

Ir

V ineqC

R eq

[22, 39, 40]

Normalized parameters:• KPT=RL/n2Rm – Normalized load factor• APT=ωrCon2Rm – Normalized PT factor• Qm=1/ωrCrRm – PT mechanical quality factor

64


Voltage Transfer Ratio(Same specific PT)

10-2

10-1

100

100 102 104 KPT

ko

101CurrentDoubler

VoltageDoubler

KPT=RL/n2Rm

Overall voltage gain Vo(DC)/Vin(AC)

[43]

0.2

0.4

0.6

0.8

η

100 102 104 KPT

CurrentDoubler

VoltageDoubler

Efficiency


Power Handling Capability(Output power/PT dissipated power)

0.01

0.02

0.0310

20

30

40

100 101 102 103 104

∆PT

kPT

0.01

0.02

0.03

Qm =1000

APT=0.008

APT=0.008Currentdoubler

Voltagedoubler

KPT=RL/n2Rm

APT=ωrCon2Rm

Qm=1/ωrCrRm

[43]

65


Boundary Conditionsfor Choosing the Rectifier type

VD

CD

Qm=1000500200

50

0.01 0.03 0.05APT

KPT

20

60

100

KPT=RL/n2Rm

APT=ωrCon2Rm

Qm=1/ωrCrRm

[43]


Applications• Ionizers• Ozone generation• Sparkers

• High gain• Good insulation

[39, 40]

Piezoelectric Transformers inHigh Voltage Application

66


S 2

S 1 L bC f

D 3

D 1

D 2

PZT

C s e

R L

D 4

C 1

C 2

+

Inverter

C

[39, 40, 61]

HV DC Output


.

2000

3000

4000

5000

6000

1000

0

VLmax

(V )L fr

[V ,

V]

L

[R , MOhm]0 5 10 15 20

L[39, 40]

67


1.004

1.008

1.012

1.000

1.016

1 10 100 1000 10000[RL, kOhm]

[ω∗]

• Need for frequency tracking[40, 45]

Frequency of Maximum Output


The Proposed Frequency Tracking Method

Current iD2 in part of cycle θ is proportional to irThe phase of the current iD2 is the same as one of irThe trailing edge of the iD2 level detection wave may be used as Phase Reference

Phase Reference

D2

D1

Cf RL

VCo VL

Co

Cr RmLrVin

ir irN

VCoN

iD2 leveldetection

iD2

ir

θ ϑ

ϑ

ϑ

[44, 46]

68


Experimental Setup for Resonant Frequency Tracking

370uH10k

4.7k 4.7k

D1

D2CL RL

PTVin Vout

p(Vin) p(Id)

Phasedetector

39k

1n

VCO

4.2k

1.5k3.3k

1k15v

15v 15v

FFQ

QIR2110driver

0.22u

10.2n

10k4.7

10k4.7

CD4046A

VLF

f

Highvoltage

V1V4 V2

V3

BN

6.6n

[44]


The System With PLL Control

[44]

69


for Micro-Power Generators and Dampers

[82]

Resonant Rectifier


Vibrating Piezo - Electrical Model

os)opt(L C

Rω

=1

RL

PoutoCini RL

( )22

1 Los

L)rms(inout

RC

RIP

ω+=

Ropt

Piezo Source Load

70


Capacitive Source Problem

oCini LC

inVouti

LVLI

1BD 2BD

3BD 4BD

RL

inV

inI

outi

1t 2t 3t

)pk(inV

πω+

πω−

π==Los

DosP

Lout RC

VCI

Ii 21

42

IL

LLos

DosP

LL RRC

VCI

)R(P ⋅⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

πω+

πω−

π=

2

21

42

)VV(C

R

L

Dos

)opt(L 21

12 +ω⋅

π=

DB2 ‘ DB3 DB1 ‘ DB4


What if Co=0 ?

ini LC

inVouti

LVLI

1BD 2BD

3BD 4BD

RL

inV

outiIL

π=

πω

+

πω

−π==

=

P

Los

DosP

LoutI

RC

VCI

Ii

oC

221

42

0

LP

LL RI)R(P ⋅⎟⎠⎞

⎜⎝⎛π

=22

∞≈)opt(LR

ini

71


Earlier solutions and their limitations

oCini RLL

os LC

1=ω

Disadvantage

Large inductance

Passive and active solutions

EmulatorDisadvantages

1.Large in size

2.Difficult to tune

3.High sensitivity

4.External source


Proposed Resonant Rectifier Circuit

oCini LC LR

inVouti

1D 2D

resL

derV

cV

1sw 2sw

COMP.

resi

dtd

LVLI

1BD 2BD

3BD 4BD

Capacitive Source

Differentiator

Comparator

Diode Bridge

Inductor & Switches

72


Principles of operation

cV

derV

resi

inV

outi

1t 2t 3t 4t 5t

oCini LC LR

inVouti

1D 2D

resL

derV

cV

1sw 2sw

COMP.

resi

dtd

LVLI

Why the commutation was not completed during t3~t4 ?

1. Co Voltage droped during t2~t3.

2. Power loss during t3~t4.


Losses Calculation

2

r

r

)(2

os2

D)pk(in)loss(R

1

e1Cf)VV(Pr

r

r

⎟⎟⎠

⎞⎜⎜⎝

⎛ωα

+

−⋅⋅⋅−=

ωα

π−

2

r

r

)(

D)pk(inosD)loss(D

1

e1)VV(CfV2Pr

r

⎟⎟⎠

⎞⎜⎜⎝

⎛ωα

+

+⋅−⋅⋅⋅=

ωα

π−

Q)pk(in)loss.(Comp IV2P ⋅≈

s)pk(in)n(gss)pk(in)p(gs)loss(Gate fVQfVQP ⋅⋅+⋅⋅≈

LD)loss(Bridge IV2P ⋅≈Bridge losses

Gate drive losses

Comparator losses

Diode losses

Resistance losses

73


oCexV LR

inV

1D 2D

resL

derV

SV+

SV−

nM pMsensR

hysRderC

derR

COMP. cV

1SD2SD

1SC2SC

LC1BD 2BD

3BD 4BD

inR

LV

+

−

ini

Experiment with dummy current source

• Schottky diodes 1N5817

• Ultra low power IC(MAX921, Maxim, USA)

• MOSFET (VP0104, VN0104)

200V floating source

Rin=100KΩ

Rsens=1KΩ

Co=330nF

Lres=1mH

CL=1µF

fs=185Hz

DifferentiatorCurrent source


LR

inV

1D 2D

resL

derV

SV+

SV−

nM pM

hysRderC

derR

COMP. cV

1SD2SD

1SC2SC

LC1BD 2BD

3BD 4BD

LV

+

−

Actuator Transducer

exV

Longitudinally piezoelectric bimorph van element

RBL-1-006 model, Piezo Systems, Inc, USA

Experiment with Piezoelectric Generator

74


Experimental Circuits

oCini LC

inVouti

LVLI

1BD 2BD

3BD 4BD

RL

2.Proposed rectifier.

3.Proposed rectifier with external supplies.

a. Higher VS reduces Rds(on).b. Supplies the (small) power

consumption of the comparator circuitry.

1.Reference circuit.

oCexV LR

inV

1D 2D

resL

derV

SV+

SV−

nM pMsensR

hysRderC

derR

COMP. cV

LC1BD 2BD

3BD 4BD

inR

LV

+

−

1SD2SD

1SC2SC

ini

External


Results

251%3.16mW3.51KΩ3.34VRESONANT RECTIFIER WITH EXTERNAL SUPPLIES VS=±4V

142%1.79mW2.75KΩ2.2VRESONANT RECTIFIER

-1.26mW2.1KΩ1.6VSTANDARD RECTIFIER

GAIN (%) COMPARED WITH STANDARD

RECTIFIER

OUTPUT POWER

OPTIMAL LOAD

RESISTANCE

OUTPUT VOLTAGE (DC)CIRCUIT TOPOLOGY

230%1.23mW11.43KΩ3.75V

118%0.636mW5.19KΩ1.818V

-0.537mW5.89KΩ1.779V

RESONANT RECTIFIER WITH EXTERNAL SUPPLIES VS=±4V

RESONANT RECTIFIER

STANDARD RECTIFIER

GAIN (%) COMPARED WITH STANDARD

RECTIFIER

OUTPUT POWER

OPTIMAL LOAD

RESISTANCE

OUTPUT VOLTAGE (DC)CIRCUIT TOPOLOGY

Dummy current source

Piezoelectric Generator

75


Experiment Waveforms(with external PS)

inin iV ⋅

ini

inV

1t 2t 3t 4t 5t

inV

derV

5V/div

2mA/div

2V/div

2V/div

Input Voltage

Input Current

Input Voltage

Instantaneous input power

Derivative Signal


The Resonant Rectifier

The operation of the new rectifier is based on self commutation of capacitor voltage

A considerable portion of the losses are due to forward voltage drop of the bridge diodes.

The resonant rectifier exhibits a substantial improvement compared to the conventional rectifier.

Additional improvement could be achieved by replacing the diode bridge by a synchronous rectification scheme.

76


PTInverter VoVin= 5V Lamp

[31, 47-58]

PT Based CCFL Ballasts


Peak voltage

[47]

Cold cathode Fluorescent lamp (CCFL)Drivers

77


Requirements:• Power handling capability• High ignition voltage ~1000V• Sufficient energy to pass the peak voltage• High operating voltage ~ 600V

The Issue of Dynamic Stability

Lamp Current

Lamp Voltage[49, 50, 78]


Dynamic changes of V-I characteristics of a fluorescent lamp operating at high frequency

P1

P2

B

FastCurrent

ChangesSlow Power

Changes

I(lamp) [Arms]

V(lamp) [Vrms] FastCurrent

Changes

AReq1

Req2

Vs

Static V-A line

• Linear approximation of V-I curve [78, 83]

78


eq1 R

)lamp(VG =

)I(fR )rms(lampeq =

RiCi match the dynamic response of lamp

21 )lamp(iE ≡

)p(vE2 ≡

+-

+-

rms

Lamp Model

R

lamp isq p

Ci

Ri

E2E1G1

Lamp

[83]

SPICE Compatible Fluorescent Lamp Model


1fjf

1R

Rfjf

sR)Lf(incZ

0

Ls

eq

0

L

+

−=

eqINCL

sINCLRZfForRZ0fFor

→∞→−→→

• “Right-Half Complex-Plane” Zero [83]

Incremental impedance of fluorescent lamps

79


Frequency100Hz 10KHz 1.0MHz

db

-50

-45

-40

0o

-200o

-100o

| Yinc |

Phase

Negative incremental resistance at low modulating frequency [83]

Incremental Admittance of Experimental Lamp Obtained by Simulation


Modulation frequency 200Hz

Negative incremental resistance

Excitation frequency 50kHz

measuredresponse

simulatedresponse

I

V

I

V

[83]

Response to modulation

80


Modulation frequency 5kHzExcitation frequency 50kHz

Positive incremental resistance

IV

measuredresponse

simulatedresponse

IV

[83]

Response to a modulation


1A

-1A

200V

-200V

measuredresponse

simulatedresponse

I

V

I

V

[83]

Response to a power step

81


Vlamp

Zlamp

Vex +-

IlampZballast

1

Vex

Zballast

Zlamp

VlampIlampBallast

Zballast Zlamp

Vex

lampballast

ZZ

1LoopGain =

Ballast-Lamp InteractionFeedback Model

[78]


Stability Criteria for Carrier Driven SystemsWhat is Z ?

Lamp(Non-Linear

NegativeResistance

Load)

LsCs

Vin fc

ssrc CL2π

1ff =>>

ssrc CL2π

1ff =≈

Stable

Unstable

82


For the Lamp-Ballast system:The relevant impedance is the INCREMENTAL IMPEDANCE under the specific carrier excitation

( )ex

exminc ∆I

∆VfZ =

LsCs

Vex

Iex∆ Iex

∆Vex


How does an incremental impedance (Zinc) behave ?Example & Intuitive Observation

• Yinc(fm) will have a peak when fm = |fr – fc| • Envelope analysis

f r fc

fm fm

( ) ( )( ) ( )t2sint2sinA1tV cmmex ff ππ+=

LsCs

Rs

AM ModulatedSignal

Yinc fm( )

fm sweep

Vex

Y ofResonantNetwork

83


• Fast large signal simulation(as compared to TRAN simulation)

• Very fast small signal simulation(as compared to TRAN simulation)

• Large (TRAN) and Small Signal (AC) compatible

• Can be implemented on any modern circuit simulator

For details see [77-79] and Appendix A

SPICE Compatible Envelope Simulation


[77-79]

Envelope Simulation Primer

84


Power System Driven by a Modulated Signal

Modulator-Driver

Reactivenetwork Load

uc(t)

um(t) )t(uout)t(u

The need for Envelope Simulation


( ) ( ) ( )( ) ]etjItIRe[ti tf2j21

cπ+=

A Primer to Envelope Simulation

Any analog modulated signal (AM, FM or PM) can be described by the following expression:

The Current in the network excited by u(t):

( ) ( ) ( ) ( ) ( )( ) ( )( ) ]etjUtURe[

tf2sintUtf2costUtutf2j

21

c2c1

cπ−

=π+π=

85


( ) ( ) ( )

( ) ( ) ( )⎪⎩

⎪⎨

⎧

π+=

π−=

]tLIf2dt

tdIL[j]tV[j

tLIf2dt

tdILtV

1c2

2

2c1

1

( ) ( ) ( )tItIti 22

21 +=

( ) ( ) ( )tVtVtv 22

21 +=

Phasor Analysis

Inductance

LiL

L

LI2

I1

I2(t)ωcL

I1(t)ωcLIm

Re

+ -

+-

V1

V2

( ) ( )( ) ( ) ( )

( ) ( ) ]e)tLIf2jdt

tdIjL

ttIf2dt

tdILRe[(]etjVtVRe[

tf2j1c

2

2c1tf2j

21

c

c

π

π

π+

+π+=−

( ) ( )dt

tdiLtv =


Phasor Analysis

Capacitance

Resistance

VC C

C

CV2

V1ωcCV2ωcC

Im

Re

V1I1

I2

R

R

RIm

Re I1

I2

V1

V2

( ) ( )dt

tdvCti =

( ) ( )tRitv =

86


Splitting the Networkinto Two Cross-Coupled Components -

Imaginary and Real

LoadNetworkSource inV

( )tu

outV


imaginary circuitcomponent

real circuitcomponent

coupling

inre

inim

outre

outim

U1

U2

22 V(outim)V(outre) + outV

Real Load Component

Imaginary Load Component

Splitting the Networkinto Two Cross-Coupled Components -

Imaginary and Real

87


Example: Piezoelectric Transformer Driven by FM Signal (SPICE)

Vin

Excitation

Ro

LoadVoutFMVoVin

Rectifier



Vin

Excitation

Ro

LoadVoutFMRectifier

Co

Lr

+-

Cr Rm1:n

Vo/n

Vo

I(Lr)/n

Ci

Equivalent cirquit of the PiezoelectricTransformer

I(Lr)

88



Vin

Excitation

Ro

LoadVoutFMCo

Lr

+-

Cr Rm1:n

Vo/n

Vo

I(Lr)/n

Ci


I(Lr) ReqCeq

Equivalentreplacementof rectifier



Vin

Excitation

Ro

LoadVoutFMCo

Lr

+-

Cr Rm1:n

Vo/n

Vo

I(Lr)/n

Ci


I(Lr)

( ) ( )( )∫π+π= dttuk2tf2cosAtu mfcc

( )tf2sinA)t(u mmm π= - Harmonic modulating signal

ReqCeq


89


( ) ( )( ) ( )( )( ) ( )tf2sintf2sinsinA

tf2costf2sincosAtu

cmc

cmc

ππβ

−ππβ=


Vin

Excitation

Ro

LoadVoutFMCo

Lr

+-

Cr Rm1:n

Vo/n

Vo

I(Lr)/n

Ci


I(Lr)

( ) ( )( )tf2costf2cosAtu mcc πβ−π=m

mffAkwhere =β

ReqCeq



I(iim)/n

Co

outre

Lr

ire0Vdc

Co

inim

Rm

V(outim)/n

-V(outim)*2*π*fc*Co

Ro

ainre Lr

Cr

V(outre)*2*π*fc*Co

diim0Vdc

c

Ro

Ac*cos((Am*kf/fm)*sin(6.283186*fm*time))

VinputRE

(V(a)-V(b))*2*π*fc*Cr

Crb

Ac*sin((Am*kf/fm)*sin(6.283186*fm*time))

VinputIM

-I(iim)*2*π*fc*Lr

I(ire)*2*π*fc*Lr

V(outre)/n

0

I(ire)/n

outim

Excitation

-(V(c)-V(d))*2*π*fc*Cr

Rm

0

PARAMETERS:Lr = 22.6mCr = 9.83pRm = 1.121kn = 0.647Co = 225pRo = 20k

PARAMETERS:fc = 358k fm = 8k

Am = 1Ac = 1kf = 1000 sqrt(v(outre)**2+v(outim)**2)

out

abs_out

OrCAD Schematics for Envelope Simulation

(Large Signal)

90


Results of Full and Envelope Transient Simulations

The modulating input signal

The Frequency modulated signal

Output signal

Ti me0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms 6. 0ms 7. 0ms

v( out ) v( out put )

- 2. 0V

0V

2. 0V

SEL>>

v( i n) s qr t ( v( a ) *v( a ) +v( b) *v( b) )- 1. 0

0

1. 0v( i nput )

- 1. 0V

0V

1. 0V

Cycle-by-cycle

Cycle-by-cycle

Envelope

Envelope


Small Signal (AC) Envelope Simulation

Amplitude modulation

( )⎩⎨⎧

=

+=

0UtuAkAU

2

mcac1

inre

E1

V(%IN+, %IN-)*ka*AcEVALUE

OUT+OUT-

IN+IN-

V1Ac

inim

um(t)

0 0

inre

0

E1

V(%IN+, %IN-)*ka*AcEVALUE

OUT+OUT-

IN+IN-

V1Ac

inim

VAC

Am

The source is linear and suitable for AC analysis – as is

( ) ( )( ) ( )tf2costuk1Atu cmac π+=

phasorphasorphasor

91


=Ac*kp*u(t)

Small signal

Linearization of Sources for Angle Modulation

Phase Modulation

PM – Nonlinear source

( )( )( )( )⎩

⎨⎧

=

=

tuksinAU

tukcosAU

mpc2

mpc1

( ) ( )( )tuktf2cosAtu mpcc +π=

VAC

inre

0

VDCAcPM

Am

inimkp*Ac

GAIN1

Linear source

=Ac

Small signalinre

inim

Ac*cos(V(%IN) )

Ac*sin(V(%IN))

kp

GAIN1um(t)

phasorphasorphasor


Linear source

VACAm

inre

2*Pi*Ac*kf

INTEG1

inim

0

VDC

0

FM

Ac

Linearization of Sources for Angle Modulation

Frequency Modulation

FM – Nonlinear source

( ) ( )( )( )∫+π= dttuktf2cosAtu mfcc

( )( )( )( )⎪⎩

⎪⎨⎧

=

=

∫∫

dttuksinAU

dttukcosAU

mfc2

mfc1

=Ac

Small signal

=Ac*kp*∫u(t)dt

Small signal

0

2*π*kf

INTEG1

Ac*cos(V(%IN) )inre

Ac*sin(V(%IN))

inim

u(t)

phasorphasorphasor

92


Results: Piezoelectric Transformer Driven by FM signal (AC and Point-by-Point)

for Different Carrier Frequencies

-70

-80

-90

-10 0

-11 0

-12 0

F reque ncy , kH z-36 0

-27 0

-18 0

0 .1 1 10 10 0

G a in , db

P ha se , deg

fc= 3 6 0 kH zfc= 3 5 8 .5 kH zfc= 3 5 7 kH z


ZPT

LinVin

Zlamp

PT

PTs: ELECERAM Co.

Experimental Circuit

93


Vex

Zballast

Zlamp

VlampIlampBallast

Zballast Zlamp

Vex

lampballast

ZZ

1LoopGain =

[78]

Vlamp

Zlamp

Vex +-

IlampZballast

1

Ballast-Lamp InteractionFeedback Model


OUT+OUT-

IN+IN-

E8

sqrt(i(V12)**2+i(V11)**2)EVALUE

tot_cur

R91k

0

0

OUT+OUT-

IN+IN-

E4

Ac*(1+.05*Ka*Am*sin(6.28*fm*time))

EVALUE

V11

0Vdc

V12

0Vdc

OUT+OUT-

IN+IN-

G6

(V(in_re))*6.28*fc*CinGVALUE

OUT+OUT-

IN+IN-

E6

V(out_re)/n_eEVALUE OUT+

OUT-IN+IN-

G3

I(I_re)/n_e+V(out_im)*6.28*fc*Co

GVALUE

C3Co

R5Ro

out_re

R7

1u

R8

1u

OUT+OUT-

IN+IN-

E7

V(out_im)/n_eEVALUE OUT+

OUT-IN+IN-

G4

I(I_im)/n_e-V(out_re)*6.28*fc*Co

GVALUE

C4Co

R6Ro

out_im

0

V2

Am*Ac*KaAc

0

1 2L1

Lr

C1

Cr

R1

RmOUT+OUT-

IN+IN-

E1

-I(I_im)*6.28*fc*LrEVALUE

PARAMETERS:

Lr = 1.125m*(n_a*n_a)Cr = 8.5n/(n_a*n_a)Rm = .67*(n_a*n_a)

Co = 230nRo = 1ufm = 5k

Cin = 32pn_a = 57

n_e = 57/(n_a*n_a)

fc = 50kAc = 1

a

OUT+OUT-

IN+IN-

G1

-(V(c)-V(d))*6.28*fc*CrGVALUE

0

0

I_re

0Vdc

0

ba

1 2L2

Lr

C2

Cr

R2

RmOUT+OUT-

IN+IN-

E2

I(I_re)*6.28*fc*LrEVALUE

c

OUT+OUT-

IN+IN-

G2

(V(a)-V(b))*6.28*fc*CrGVALUE

C5Cin

0

0

I_im

0Vdc

d

0

C6Cin

OUT+OUT-

IN+IN-

E5

sqrt(V(x_re)**2+V(in_im)**2)EVALUE

out

R41k

0

R19

1u

in_im

OUT+OUT-

IN+IN-

G5

-(V(in_im))*6.28*fc*CinGVALUE

x_re

in_re

in_re

Real Part

Imaginary Part

Excitation (AM)

PT envelope simulation - Zinc measurement

94


Frequency

100Hz 1.0KHz 10KHzP(V(tot_cur))

0d

-200d

180dV(tot_cur)

0V

200uV

400uV

fc = 51KHz

fc = 52KHz

fc = 49KHz

fc = 54KHzfc = 53KHz

fc = 50KHz

PT incremental output admittance for several carrier signals, above and below resonance

PT resonance at 51.5 KHz

Magnitude

Phase

PT Envelope Simulation Results


( )rmslamp IfR =

OUT+OUT-

IN+IN-

E3

I(V3)**2EVALUE

OUT+OUT-

IN+IN-

E9

sqrt(V(p))EVALUE

R3

100C101u

IC =

p

0

V3

0Vdc

rms

OUT+OUT-

IN+IN-

G8

V(lamp)/V(Rinc)

GVALUE

lampR11

1meg

V42.5kVdc

0

V51Vac0Vdc

OUT+OUT-

IN+IN-

E10

V(rms)ETABLE

0

Rinc

ETABLE Ilamp

Rlamp

Lamp Model(Orcad 10.3)

95


Thermal dependence of the CCFL

500

520

540

560

580

600

620

640

660

680

700

[mA]Ilamp

[V]Vlamp

0 1 2 3 4 5

(a) Forced air flow

(b) Natural convection

33oC @ 3mA

40oC @ 3mA

Rs1

Rs2


Static Incremental Resistance (Rs)

Forced air flow (mes)

Natural convection (sim)

Natural convection (mes)

Forced air flow (sim)

Natural convection Rs

Forced air flow Rs500

550

600

650

700

750

800

0 0.5 1 1.5 2.5 3.5 4.52 3 4 5[mA]I lamp

[V]Vlamp

-15

-20

-25

-30

-35

-40

-45[KΩ]R s

Rs

V-I

96


Fr equency

10Hz 100Hz 1. 0KHz 10KHz 100KHzV( l amp) / I ( V3)

0

200K

400Kp( V( l amp) / I ( V3) )

0d

90d

180d

SEL>>

Magnitude

Phase

CCFL incremental impedance for lamp currents


Thermal effects on the CCFL’s Zinc

110

120

130

140

150

160

170

180

0.1 1 10

ϕ ZECCFL [deg]

Modulating frequency, fm [KHz]

(a) 33oC

(b) 40oC

86

87

88

89

90

91

92

93ZECCFL [dB]

(a) 33oC

(b) 40oC

97


Stability Criterion

)f(LG1KHA 1

CL +=

The system is unstable if 1+LG(f) has roots in the right half of the complex plane.Nyquist criterion is a test for location of 1+LG(f) roots.


Nyquist

))f(LG(gIm

))f(LGRe(1−

mΦ

lampballast

ZZ

1LoopGain =

98


Possible modes of operation

ZEPT

ZECCFL (T)

TTOP

|Z| , PNominal P

Tmax

P

Mode 1: Carrier frequency is far from resonance –stable mode CCFLPT ZincZinc <

lampballast

ZZ

1LoopGain =


Possible modes of operationMode 2: Carrier frequency is equal to the resonant

frequency – unstable mode

CCFLPT ZincZinc >

Tmax

ZEPT

ZECCFL (T)

T

|Z| , P

Nominal P

P

lampballast

ZZ

1LoopGain =

99


Possible modes of operationMode 3: Carrier frequency is near the resonance –

oscillatory mode CCFLPT ZincZinc ≈

TTOP

|Z| , P

Nominal P

ZECCFL (T)ZEPT

LG < -1 LG > -1

P

UNSTABLESTABLE

lampballast

ZZ

1LoopGain =

Sustained oscillation


-

- 8

- 6

- 4

2

0

2

4

6

8

-2 -1 0 1 2 3 4 5 6 7Re(LG)

Im(LG)

cf = 49 KHz

33oC

40oC

Nyquist Plot lampballast

ZZ

1LoopGain =

Stable

PT-Lamp Envelope Simulation Results

100


Nyquist curves lampballast

ZZ

1LoopGain =

Stable for 40oC but unstable for 33oC

-

4

-

6

-

2

0

2

4

6

- - -3 2 1 0 1 2 3Re(LG)

Im(LG)

cf = 51 KHz33oC40oC

PT-Lamp Envelope Simulation Results


Simulation & Experimental

Time4.00ms 4.25ms 4.50ms 4.75ms 4.95ms

Ilamp

0A

-6mA

6mA

0V

-1.2KV

1.2KV

Vlamp

Vlamp

Ilamp

Cycle-by-cycle(TRAN) Simulation

Experimental

Sustained Oscillations in PT-Lamp System

101


Concluding RemarksOverviewPower Electronics point of viewAs market develops, prices will drop and PE

use will expand

Thank You for Your Attention

The support of theIsraeli Science Foundation

is acknowledged

Documents

Power Electronics of Piezoelectric ElementsPower Electronics Laboratory ... – Sonar transducer ... Prof. S. Ben-Yaakov, Power Electronics of Piezoelectric Elements, June 2006 ...medinid/energy-gordon/Piezoelectric/Slides... ·