Blog for students taking ECEN5817 Resonant and Soft Switching Techniques in Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2012
Wednesday, February 1, 2012
H(s) for 19.1
I am getting something really ugly for H(s) for the resonant tank...
I have something similar for Zo, but am pushing C and Re left through the inductor rather than Cb and L, like you are doing. How did you get Ls/n and n/Cbs? I thought impedances were multiplied by n^2 when they were pushed through transformers. So, if I was pushing Cb and L to the right of the transformer, they would become Cb/n^2 and L/n^2. See L in pages 251-252 for a reference.
Olga, I think I took nearly the same direction as you did. H(s) = Z_out/Z_1 (where Z_1 does not mean Z_in). I first moved all components from primary to secondary of transformer. If you did the same, I think you are missing a multiple of n with your sL term: sL/n^2 rather than sL/n. Also, it is important to note that C_b is not part of the resonant tank (as far as I understand things). It is a DC blocking capacitor used to create an AC signal into the tank network. If it is considered as part of your impedance calculations, I believe it can be eliminated anyway because the problem statement mentions that C_b is of large value. This means that the series combination of the Z_1 impedance, 1/(s*n^2*C_b) + sL/n^2 is dominated by inductance because 1/(s*n^2*C_b) approaches zero for large C_b (and frequencies approaching resonance) while the inductive impedance increases. H(s) then becomes much easier to work with and turns out to be similar to textbook examples.
Thanks for the comments. I did leave out the extra n term, good catch. Keith, did you just leave out Cb all together then? I definitely agree that it is not part of the resonant tank, but was unsure if its impedance played a significant role in the transfer function. It would certainly simplify things a lot if it can be neglected.
My understanding is that it is going to remove the DC offset that is present in the switching section, but maybe not.
Olga, Glad I was able to help. Regarding your question, I did not discount the contribution of C_b entirely. As you mentioned, its function is to remove the DC component of the resonant tank input current; giving us an AC waveform to work with our sinusoidal approximation. In determining the H(s) transfer function the C_b contribution was neglected though. As I mentioned above, because C_b is large the combination series impedance Z_1 will be dominated by the inductor L.
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4 comments:
Hi Olga,
I have something similar for Zo, but am pushing C and Re left through the inductor rather than Cb and L, like you are doing. How did you get Ls/n and n/Cbs? I thought impedances were multiplied by n^2 when they were pushed through transformers. So, if I was pushing Cb and L to the right of the transformer, they would become Cb/n^2 and L/n^2. See L in pages 251-252 for a reference.
Olga,
I think I took nearly the same direction as you did. H(s) = Z_out/Z_1 (where Z_1 does not mean Z_in). I first moved all components from primary to secondary of transformer. If you did the same, I think you are missing a multiple of n with your sL term: sL/n^2 rather than sL/n. Also, it is important to note that C_b is not part of the resonant tank (as far as I understand things). It is a DC blocking capacitor used to create an AC signal into the tank network. If it is considered as part of your impedance calculations, I believe it can be eliminated anyway because the problem statement mentions that C_b is of large value. This means that the series combination of the Z_1 impedance, 1/(s*n^2*C_b) + sL/n^2 is dominated by inductance because 1/(s*n^2*C_b) approaches zero for large C_b (and frequencies approaching resonance) while the inductive impedance increases. H(s) then becomes much easier to work with and turns out to be similar to textbook examples.
Thanks for the comments. I did leave out the extra n term, good catch. Keith, did you just leave out Cb all together then? I definitely agree that it is not part of the resonant tank, but was unsure if its impedance played a significant role in the transfer function. It would certainly simplify things a lot if it can be neglected.
My understanding is that it is going to remove the DC offset that is present in the switching section, but maybe not.
Olga,
Glad I was able to help. Regarding your question, I did not discount the contribution of C_b entirely. As you mentioned, its function is to remove the DC component of the resonant tank input current; giving us an AC waveform to work with our sinusoidal approximation. In determining the H(s) transfer function the C_b contribution was neglected though. As I mentioned above, because C_b is large the combination series impedance Z_1 will be dominated by the inductor L.
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