sorry for that.
assume iL1 equal to iL2 may not be helpful for me,maybe I didn't find that out.idk
but before diode turns off Vc=Vs-Vo=L2*di/dt may be something important to solve the problem.
How to solve Cuk converter look alike circuit?
- Thread starter kunx2
- Start date
assume iL1 equal to iL2 may not be helpful for me,maybe I didn't find that out.idk
but before diode turns off Vc=Vs-Vo=L2*di/dt may be something important to solve the problem.
Hello again,
Well another fact is that the average voltage across C1 is equal to the negative average voltage of the output cap C2. So if vC2=5v then vC1=-5v, in the average that is.
That means when the switch closes, the output circuit with the diode, L2, and C2 and load, are like a buck circuit with input voltage Vin-vC2 and since vC2 is negative of the output, that means that the 'buck' input would be Vin+Vout.
That is also the time when L2 charges up.
That's very interesting so see if that helps.
[LATER]
I took another look at this and it appears that there is a method to the madness. The idea is to discover the basic operation and then the solution becomes more apparent. Understanding the basic relationships means we avoid going to the time domain solution which turns this into a messy set of partial differential equations which are not easy to solve. I might fool around with that method anyway at some point if i get in the right mood.
Well another fact is that the average voltage across C1 is equal to the negative average voltage of the output cap C2. So if vC2=5v then vC1=-5v, in the average that is.
That means when the switch closes, the output circuit with the diode, L2, and C2 and load, are like a buck circuit with input voltage Vin-vC2 and since vC2 is negative of the output, that means that the 'buck' input would be Vin+Vout.
That is also the time when L2 charges up.
That's very interesting so see if that helps.
[LATER]
I took another look at this and it appears that there is a method to the madness. The idea is to discover the basic operation and then the solution becomes more apparent. Understanding the basic relationships means we avoid going to the time domain solution which turns this into a messy set of partial differential equations which are not easy to solve. I might fool around with that method anyway at some point if i get in the right mood.
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Yes!! That 's help me a lot .
with that equation I can tell that output voltage will be around 2v to 2.35v!is that right?
thanks a lot.
Hello again,
Well, i get a bit higher than that (around 25 percent higher) and verified this with a simulation based on a set of four ODE's solving them numerically while finding the optimum diode 'off' duty cycle. I then changed the parameters of the problem several times and was able to predict those outputs also with the technique, then derived a formula for the duty cycle and verified that, then derived a formula for the max duty cycle. The max duty cycle then led me to a formula for the max Vout given any load or input voltage, and that is a consequence of the circuit that there is a theoretical maximum output voltage for any given load and input voltage. That's something that should be calculated also in order to specify the input/output relationship in its entirety.
So yes the average capacitor voltages are the negative of each other and that is a key point, but another hint is that the inductors energy vs the required output energy seems to have solved the problem without having to calculate the diode turn off time. I think we would still have to calculate the diode turn off time in order to calculate the output ripple voltage though but i havent gotten to that yet ... just the average output voltage and theoretical max output voltage.
[LATER]
Ok calculating the diode 'on' time is easy considering the energy in L2.
Results match with the ODE set calculations so now i dont have to find the optimum diode 'on' time just calculate it outright. The diode off time interval would then be easy also if needed because then that's just the remaining time in the cycle.
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thank you again.
I might do something wrong with my calculation. By the way,my friend told me that circuit called "Zeta".
Hello again,
Ok thanks i might read up on that.
But basically all you have to do is calculate the average power in both inductors and equate that to the average output power, as we can assume that both inductors dump all their energy into the output for each cycle. Knowing that vC1=-vC2 means both inductors are energized by Vin so we know the energy each inductor receives each cycle.
Looks like there might be a state space solution on the web somewhere too so we can look into that at some point also perhaps.
[LATER]
I checked around on the web a little and found that the relationships they are giving for the input/output all seem to assume something about the choice of components. Many seem to select L2=L1 for example. Another assumes that L2 is magnetically coupled to L1.
Then some are stating that the output voltage Vout follows this rule:
Vout=Vin*D/(1-D)
but clearly that can not always be the case unless there are some other assumptions (maybe like continuous inductor operation).
So the idea then is to read each article carefully and dont assume anything unless they explain how and why they got there in full.
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Hello again,
Ok i verified the result Vout=Vin*D/(1-D) using state space averaging but that is ONLY for continuous inductor current operation. For your circuit you have discontinuous inductor current so the result is somewhat different. You'll have to work out the details based on inductor energy.
I am sure you can work this out as you did the other simpler topologies.
Ok i verified the result Vout=Vin*D/(1-D) using state space averaging but that is ONLY for continuous inductor current operation. For your circuit you have discontinuous inductor current so the result is somewhat different. You'll have to work out the details based on inductor energy.
I am sure you can work this out as you did the other simpler topologies.
