I am not able to understand how the professor determined the center of the circle for the buck example by inspection as M. Woudn't we first have to go through the process we went through in the first few slides to first get the equation of the circle?
Thanks
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Hello Yusuf,
Center of circle represents steady-state (ss) point of resonant circuit (in this case voltage of C and current on L), in normalized fashion. From most example given in class, it seems the ss values can be determined from the presence of external voltage or current source in LC circuit. If no external source, ss value will be equal to zero.
For buck converter example, the resonant circuit formed during DCM, has diff equation below:
* C dvs/dt = -iL ss value for iL=0
* L diL/dt = vs - V ss value for vs=V
Then, since prof set Vbase=Vg and Ibase=Vg/Ro, normalize value for vs is vs/Vg (M) and for iL is 0/Ibase. Note: V external voltage during ss
becomes the voltage value for capacitor.
The fact that center of circle is M, I think it is just mere coincident given the Vbase=Vg. For other selection of Vbase, we'll get different answer.
Hi Tanto,
Thanks. Your explanation above has clarified my doubt.
Thanks
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