on page 420, we see the steady state model for the buck, and boost converters in DCM.
I do not understand how to solve these for M (M= output Volage/Vg in steady state). The buck-boost on page417 was easy.
Can anyone help me to understand how to solve for these?
thanks
Mark
I am also having a little difficulty with this. On the buck, since the power arrow still goes from Re to the power source, this seems to imply P=V_Re^2/Re. Can anyone verify this?
ReplyDeleteSince the voltage across the power source is V and the current going into it is V_Re/R_e-V/R, I tried to do the math using P=V_Re^2/R2 = V(V_Re/R_e-V/R) and solving for V_Re using the quadratic formula. Then, setting this value equal to V_Re=Vg-V from simple KVL, I was able to start to get something similar looking to the result in the book. Maybe I messed up my algebra somewhere but what I got was V/Vg = 2(R/Re)/(1 +/- sqrt(1-4R/(V Re))).
Like I said - similar but not quite the same. Can anyone verify if I am treating the power source correctly?
Thanks,
Audrey
I believe the approach to solving this is:
ReplyDeleteyou can solve for both the current and voltage across Re (in terms of Vg, or (Vg and V).
You therefore know the power delivered by the power source. You also know the voltage across the power source. B/C you know the voltage across the power source, and the power of it, you know the current it puts out.
Using KCL and the above, you can then write out the KCL equations that only depend on Vg, V and Re.... problem solved (other than lots of algebra).
thanks!
m
Audrey, you're not alone. I also have the problem to solve M for the buck based on figure 11.12 (a). I can't get the M exactly as specified in table 11.1 by solving KCL, KVL equations in figure 11.12 (a).
ReplyDeleteHowever, for Boost, there is no problem to get the M.