Power Electronics

http://eebookforengineer.files.wordpress.com/2013/02/power-electronics-by-daniel-hart.pdf

http://www.browndoggadgets.com/collections/charge-controllers http://www.instructables.com/id/Solar-USB-Charger-20-21/step1/Parts-and-Tools/

((D/(1-D))-Vd/Vg)*1/(1+(D*Ron+RL)/((1-D)^2*R))

(1-(1-D)/D*Vd/Vg)*1/(1+(D*Ron+RL)/((1-D)^2*R))

hw 4

1.<math>Vpv-IL*RL -IL*Ron*D-Vd*(1-D)-Vbus*(1-D)</math>

2.<math>IL*(1-D) - IL*tr/Ts - Qr/Ts -Ibus</math>

13.<math>(1-Vd/Vg + sqrt(Vd^2/Vg^2-2*Vd/Vg+1+4*D1^2/K))/2</math>

hw 5

5.

((1-D)/(n*D*Vg))*(Vg*n*D/(1-D)-Vd)*(R/(D*(n/(1-D))^2*(Rp+Ron)+Rs/(1-D)+R))

maxima mathcad clone

hw6

1. vghat - dhat/(1-D)*Ron/R*V - D*ihat*Ron - (1-D)*vhat + dhat*V 2. -vhat/R +(1-D)*ihat -dhat*V/(R*(1-D))

I'm kind of where Hamza is. When you pass the current source through the D*Ron resistance and find the thevenin equivalent by replacing the current source with a voltage source = to IR where I is that $$-\frac{V*dhat}{R*(1-D)^2}$$ and R = D*Ron that seems to cover it. The current you have left $$-\frac{V*dhat}{R*(1-D)^2}$$ then passes through the inductor just like creating a second voltage term $$-\frac{V*dhat}{R*(1-D)^2}*s*L$$ which has no Ron in it. How is my thinking messed up.