Hello MrAL..
Thanks for your reply.. I am doing a project on boost converter based modular inverter. In that the boost converter has to be operated at no-load condition. So before operating the experimental setup i want to know the time domain solution of it. Please help in this.
Hello again,
First, because you have no load at all you can expect the voltage to go up higher than you want it to unless you have feedback, a constant duty cycle is not good.
The equations for each cycle after the switch opens are as follows...
I=(sin(t*w)*sqrt(C)*E)/sqrt(L)-(v1*sin(t*w)*sqrt(C))/sqrt(L)+i1*cos(t*w)
Vout=(i1*sin(t*w)*sqrt(L))/sqrt(C)-cos(t*w)*E+E+v1*cos(t*w)
where
I is the inductor current after the switch opens, we set I=0 and solve for t,
Vout is the capacitor voltage after the switch opens, we use t from above,
i1 is the initial inductor currrent,
v1 is the initial capacitor voltage,
w=1/sqrt(LC),
i1=dt*E/L,
dt=switch 'on' time period.
i1 always equals dt*E/L because that's the current at the end of the switch 'on' cycle.
Now we can run those equations a couple tens of thousands of times, or we can simplify based on the energy transfer between the inductor and capacitor. This leads to:
Vout=sqrt(N)*K
where
N is the number of switch on/off cycles since startup with zero capacitor voltage, and
K=dt*E/sqrt(L*C), where dt is the switch 'on' time period, and E is the battery supply voltage.
From the above simple equation for Vout, we see that because K is a constant no matter how many cycles we allow we always see an increasing voltage, but we do see a decrease in the amount of voltage increase per cycle as the voltage builds up.
For example, with E=10v and dt=1ms and L=500uH and C=5000uf after 40 cycles we see a capacitor voltage Vout of 40 volts and after one more cycle we see Vout=40.6969 which is an increase of 0.6969 volts, but after 400 cycles we see Vout=126.4911 volts and after one more cycle we see 126.6491 which is an increase of only 0.1580 volts, about 1/4 of the previous change, and after 4000 cycles we see Vout=400v and after 4001 cycles Vout=400.049996875391 which is only a change of 0.05 volts now from the previous cycle. After 400000 cycles we see Vout=4000 volts and after 400001 cycles we see Vout=4000.005 volts, a change that is 10 times less than that at around 4000 cycles. Looking at this in time, we see the following:
cycles=40, t=0.080, Vout=40v, change=0.7v
cycles=4000,t=8.0, Vout=400v, change=0.05v,
cycles=400000,t=800,Vout=4000v,change=0.005v (more then 10 minutes)
cycles=430000,t=1660,Vout=4147v,change=0.0048v (another full minute later)
cycles=40000000,t=80000,Vout=40000v, change=0.0005v (more than 22 hours operation)
cycles=40030000,t=80060,Vout=40014.9971885542v,change=0.0005v (another minute later)
So we see as time goes by it takes longer and longer to get a higher voltage of the same increment. At 10 minutes operation we get around 150v more after another minute, but after 22 hours of operation we only get another 15 volts after another minute of operation.
After three months operation being on 24 hours a day 7 days a week, we see around 400000 volts output with only a change of 5e-5 volts per cycle now. But even though that is low, if we keep it on for another three months we STILL see an increase, to 565685 volts.
So although the slope decreases given enough time the voltage always climbs. The limiting factor is how long the unit is run for before it is turned off.
Now of course that is the super ideal case which will never be seen in real life because even the tiny tiniest power loss will limit the output voltage after some number of cycles. This could be a 1 megohm resistor, but even that tiny loss of power will limit the output.
So you can see two things here:
1. You can not run the unit with no load unless it has feedback to limit the voltage, and even then the feedback might not be able to cut the pulse width back enough to limit the output anyway.
2. You need feedback and at least some load