LEARNING EXAMPLE DESIGN CAMERA FLASH CIRCUIT

DESIGN CONDITIONS50 70

80CF

B

V V

R

_

_1

1

100 (peak power)FLASH CIRCUIT

R

ms

P mW

1FIND , ,

S FV C R

380 10FC B F F

R C C s 12.5C F

worst case(max voltage drop)

The constraint in V_CF sets the range for feasible batteries

60 (splits the range)S

V V

2

_1_

1

0.1S

R PEAK

VP W

R

136R k

WHAT IS THE CHARGING TIME?4 6

13.6 10 12.5 10 0.45

CHARGE FRC s

0067.0

0183.00498.0135.0

5

432

368.0

t

et

50 0.83 60

(within 17% of final value

REACHES FIRING VOLTAGE WITHIN 2 TIME CONSTANTS!

LEARNING EXAMPLEALTERNATOR CIRCUIT TO GENERATE HIGH VOLTAGE PULSESFROM A SMALL DC VOLTAGE SOURCE

Single pole-double throw (SPDT) switch

1T

Connected to battery for T1 seconds

1 1

0

1pos 1: ( )

T IN

IN

V Ti t V dt

L L

Current through inductor when switchmoves to pos2

1

1( ) IN

PEAK

RV Tv v T

L DESIGN EQUATION

1

1

1pos 2: ( ) ( ) ( )

t TIN

O

V Tt T i t e v t Ri t

L

L

R

DESIGN SPEC

500PEAKv V

1

13

100 5500 1

10

TT ms

TIME TO DISCHARCHE?310

10100DISCHARGE

s S

In 5 time constants the voltage isbelow 1% of initial value

Put a safety margin and waita bit more (1ms?)

LEARNING APPLICATIONHEART PACEMAKER

SCRv

SCRi

51

Simplified SCR model

SCR “fires”

A50

As soon as the SCR switches off the capacitor starts charging. Hence, assume

2.0)0( Cv

0,)( 21

teKKtvt

C

RRC 61016)( KVvC

212.0)0( KKVvC 8.5

6

2

1

K

K

0,8.56)(

tetv RCt

C

Find R so that the SCR is ready to fireafter one second of capacitor charging

RCC eVv

1

8.565)1(

:Required

758.11

8.51

RC

e RC kR 569569.RC

Charging phase

THE DISCHARGE STAGE

With the chosen resistor discharge startsafter one second and the capacitor voltageis 5V

1,)()1(

21

teKKtvt

C

s569.0VvC 5)1(

)(1050)(10569.066)( 66 ARIvC

45.275

45.22

221

1

KKK

K

145.2745.22)( 569.0)1(

tetvt

C

For SCR turn off 2.0)1( offC Tv

sTe off

Toff

11.065.22

45.27569.0

%example6p12%visualizes one cycle of pacemaker%charge cycletau=0.569;tc=linspace(0,1,200);vc=6-5.8*exp(-tc/tau);%discharge cycle. SCR ontd=linspace(1,1.11,25);vcd=-22.45+27.45*exp(-(td-1)/tau);plot(tc,vc,'bd',td,vcd,'ro'),grid, title('PACEMAKER CYCLE')xlabel('time(s)'), ylabel('voltage(V)')legend('SCR off', 'SCR on')

LEARNING EXAMPLEANALYSIS AND CONTROL OF “INDUCTIVE KICK”

DETERMINE PEAK VOLTAGES ACROSS INDUCTOR AND SWITCH.Current in steady state is 1A before switching

1 2( ) , 0

t

Li t K K e t

1 2

2

(0 ) 1

( ) 0

( ) , 0

L

L

t

L

K K i

K i

i t e t

0 (switch open )L

RR

1( ) ( ) ; 0

t

L

div t L t e t

dt

Circuit to control kick

!! ( )Lv inductive kick

Trying to make discontinuous theinductor current!!!

!( )SWITCHv KVL must hold

(0 ) (0 ) 0

(0 ) (0 ) 1

(0 ) 1 [ ]

C C

L L

SWITCH

v v

i i A

v R V

circuit is now second order and may oscillate

2 1 1Ch. Eq.: 0

Rs s

L LC

Select R, C for adequate damping andnatural frequency; 6. ., 1, 10

ne g

21 12 ;

n n

R

L LC

DESIGN EQS.

199 , 10R C nF

199 !SWITCHv V

snubbercircuit

Inductor current at the beginning of ON period MUST be the same than the current at the end of OFF period

LEARNING EXAMPLE BOOSTER CONVERTER

STANDARD DC POWER SUPPLY

BOOSTER “ON” PERIODEnergy is stored in inductor.Capacitor discharges

BOOSTER “OFF” PERIODInductor releases energy.Capacitor charges

e.g. booster

THE “ON” CYCLE

onin

t

LonL tL

VIdxxv

Liti

on

00

)(1

)0()(

THE “OFF” CYCLE ontt

off

on

t

tLonoffonL dxxv

Ltitti )(

1)()(

SIMPLIFYING ASSUMPTION: THE OUTPUTVOLTAGE (Vo) IS CONSTANT

0VVv inL

offin

on tL

VVtiI 0

0 )(

offin

onin

o tL

VVt

L

VII 0

0

inoff

offono V

t

ttV

booster) (hence inVV 0

T

tD

ttT

on

offon

:cycleDuty

:Period

DVV in

1

10

By adjusting the duty cycle one canadjust the output voltage level

LEARNING BY DESIGNDESIGN OF ELECTRIC HEATER USING A 24V SOURCE AND1 OHM HEATING ELEMENT

100 400W P W Solution one

Too much power lost in rheostat

SWITCHED INDUCTORALTERNATIVE

Pos 1pos 2

Controlling switching frequency one controls I_peak and average power … And no power loss!

LEARNING EXAMPLE DESIGN DECOUPLING CAPACITOR TO ISOLATE LOAD FROM VARIATION IN SUPPLY VOLTAGE

decouplingcapacitor

(0)O

VO

V

DESIGN EQUATION FOR DECOUPLING CAPACITOR

model forsupply variation

EXPECTED SPIKE

'EXPECTED DURATION

:

:S

V

t

Qualitative operation

acceptable

CIRCUIT AT t=0+ STEADY STATE AFTER SWITCHING

LEARNING BY DESIGN

1.5s and 1s between1A above Stays (2)

100ms; within1A Reaches (1)

:SATISFIES AND,OVERDAMPED IS THAT SUCH FIND )(tiC

AFTER SWITCHING WE HAVE RLC SERIES

0)(1

)()(2

2

tiLC

tdt

di

L

Rt

dt

idL

LL

0))((/520 212 ssssCss :Eq. Ch.

0;)( 2121 teKeKti tsts:RESPONSE DESIRED

Cssss /5;20 2121

12)0()0(;0)0( dt

diLvi L

LL

:CONDITIONSINITIAL

60

0

2211

21

KsKs

KK

For the initial conditions analyze circuit at t=0+. Assume the circuit was in steadystate prior to the switching

+ -

)0()0(dt

diLvL

)0(Cv Circuit at t=0+

VvC 12)0(

http://www.wiley.com/college/irwin/0470128690/animations/swf/7-21.swf

tstsL ee

ssti 21

12

60)(

NOW ONE CAN USE TRIAL AND ERROROR CAN ATTEMPT TO ESTIMATE THEREQUIRED CAPACITANCE

Mesh plot obtained withMATLAB

IF FEASIBLE, GET AN IDEA OF THE FAMILY OF SOLUTIONS

» s=[[1:9]';[11:19]'];» mesh(t,s,ils')» view([37.5,30])» xlabel('time(s)'),ylabel('s_1(sec^{-1})')» title('CURRENT AS FUNCTION OF MODES')

Ils is a matrix that contains allthe computed responses, oneper column

%example6p14.m%displays current as function of roots in characteristic equation% il(t)=(60/(s2-s1))*(exp(-s1*t)-exp(s2*t));% with restriction s1+s2=20, s1~=s2.t=linspace(0,5,500)'; %set display interval as a column vectorils=[]; %reserve space to store curvesfor s1=1:19 s2=20-s1; if s1~=s2 il=(60/(s2-s1))*(exp(-s1*t)-exp(-s2*t)); ils=[ils il]; %save new trace as a column in matrix endend%now with one command we plot all the columns as functions of timeplot(t,ils), grid, xlabel('Time(s)'),ylabel('i(A)')title('CURRENT AS FUNCTION OF MODES')

Estimate charge by estimating area under the curve

For this curve the area is approx. 12 squares

CsAQ 3][5.05.012

mFFV

QC 25025.0

12

3

20/5

20

21

21

Css

ss

056.1

944.18

2

1

s

s

%verification s1=18.944; s2=20-s1; il=(60/(s2-s1))*(exp(-s1*t)-exp(-s2*t)); plot(t,il,'rd',t,il,'b'), grid, xlabel('time(s)'), ylabel('i(A)') title('VERIFICATION OF DESIGN')

Applications