learning application
DESCRIPTION
LEARNING APPLICATION. HEART PACEMAKER. Simplified SCR model. SCR “fires”. Charging phase. As soon as the SCR switches off the capacitor starts charging. Hence, assume. Find R so that the SCR is ready to fire after one second of capacitor charging. %example6p12 - PowerPoint PPT PresentationTRANSCRIPT
LEARNING APPLICATIONHEART PACEMAKER
SCRv
SCRi
51
Simplified SCR model
SCR “fires”
A50
Charging phase
As soon as the SCR switches off the capacitor starts charging. Hence, assume
2.0)0( Cv
0,)( 21 teKKtv
t
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
THE DISCHARGE STAGE
With the chosen resistor discharge startsafter one second and the capacitor voltageis 5V
1,)()1(
21 teKKtv
t
C
s569.0VvC 5)1(
)(1050)(10569.066)( 66 ARIvC
45.275
45.22
221
1
KKK
K
145.2745.22)( 569.0)1(
tetv
t
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')
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 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(
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')
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