Friday, February 5, 2010

Problem 19.9

Can someone help me compare the Voc value for part A. I used Voc=(4*Vg/pi)*n*||H_inf(s)|| and got Voc=138V. This value seems too big for Vg=12V. Maybe my formula for Voc is wrong! Anybody with the same result?

6 comments:

Vojkan said...

19.9a
Voc(peak)=154V
Your formula is correct.
H_inf(s)=sLp/(s(Ls+Lp)+1/(sC))

Yusuf said...

I am having some doubts about my value of Isc. I get Isc =6uA. The way I did it was to find
mag(Zo_0) and then use Isc = Voc/(n*mag(Zo_0).
I get Zo_0 as 3.36M!!!

Anonymous said...

Zoo = (Xp*Xs)/(Xp+Xs) = 3.2345 (on primary side)

Voc/Zoo = 154/3.2345/(1/N)^2 = 0.85A

Note: R1/R2 = (N1/N2)^2

Vojkan, do you agree?

Also, I think wo and woo are independent of N. If you transform the load to the primary side (R/N^2)....this variable gets shorted and opened....so you just get what you calculated in the other problem. Does this seem right?

Vojkan said...

I agree

Yusuf said...

Shouldn't it be OK to also reflect the short to the primary. In that case we get
nIsc=Voc/mag(Zo_0) giving the Isc formula I mentioned above.

Audrey said...

Yusuf,
What are you calling Z_o0 in your formula? I think there is a Z_o0p found with the shorted input voltage for the primary side of the transformer that needs to be reflected to the secondary to find Z_o0=n^2 Z_o0p.

I guess I don't know what you mean by "reflect the short to the primary" since the only short should be on the input voltage on the primary.