Sunday, April 11, 2010

Problem 20.5 Verification

I was wondering if anyone got these numbers:

For 2.38A case: Cr <= 3.8nF and Lr < 6.81e-5 H
For 0.119A case: Cr < = 0.19nF and Lr < 1.37mH

So design: Cr = 0.19nF and Lr < 68.1 microH

n = 0.292

8 comments:

Vojkan said...

Please find: R0, Vbase and J2 for both Imin and Imax.
Verify that both J2s are less than or equal to 1, which is the ZCS boundary condition(L31_annotated slide 17).

Vojkan said...
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Jan Gaciu said...

Hi Vojkan,

Any hints for J2?
I don't see how to be determined based on actual data.

Vojkan said...

Transformer turns ratio is 1:n (1 is referred to the prim side, n is referred to the sec side).
If Vg and Q in forward converter are pushed through the transformer to the secondary side, circuit from L31 slide 12 can be obtained with: V1=n*Vg.
From the circuit: J2=normalized inductor current =normalized output current.

Aaron said...

Hi Brandon,

I took a different approach and was wondering how you come up with different L & C for each of the cases.

--Aaron

Anonymous said...

My answers aren't correct.....

Anonymous said...

But you should get 2 sets of L and C because you have 2 different load currents. Then you select the L and C (from the 2 sets you calculated previously) that will maintain all conditions for both load currents.

Aaron said...

I asked because you should only get one set of L's & C's. If you meet the Js<=1 requirement for the max case, this is one condition. Minizimizing the fet current (J1) means that you pick the max Ro given by the Js<=1 inequality.

Also, you have a max frequency requirement and a u requirement. You know J2 so these two conditions should give you an fo value. This combined with the Ro requirement should give you the L and C value.