Friday, January 25, 2013
Simplification of circuit for 7.15b
For problem 7.15b, after setting dhat to zero, I end up with the same small-signal equivalent circuit as shown in Fig 7.53 of the text (without the conduction loss terms). I tried to use the transfer function derivation shown in section 8.2.1 of the text, but I ran into difficulty handling the fact that the inductor is connected through the bottom of the ideal transformer and it is not clear how to "push" this impedance through the transformer in order to simply the circuit. Any suggestions?
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What I did was use the definition of Vp (voltage on the primary side of the transformer), Vs (voltage on the secondary side of the transformer) and VL (voltage across the inductor) in terms that simplified in the end...
ReplyDeletei.e. Vout = Vs + VL, Vs = Vp*(-D/D'), Vp = Vg + VL and VL = (ip + is)*Ls where ip is the current through the primary and is is the current through the secondary.
Then you can just use KVL loops and simplify your equations and you will be there... another hint I used is that I kept terms like R||C instead of solving for actual R||C until the end... it makes the algebra easier.
Thanks! You suggestions worked great.
DeleteDrew, See also comments for Alex's 7.15 Transfer function question.
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