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Jul 19 at 12:15 comment added Mario Potschatski @VerbalKint Alright, thank you very much!
Jul 18 at 8:11 comment added Verbal Kint Yes, this is how it has to be done: the input caps are part of the EMI filter while the buck \$Z_{in}\$ is determined without them. Then you can then compare \$Z_{out}\$ of the filter and \$Z_{in}\$ of the buck (or another dc-dc converter).
Jul 18 at 8:00 comment added Mario Potschatski @VerbalKint To understand this correctly from your seminar. Make it simple and pull the 2 input capacitors from the buck converter to the filter side so that they are taken into account in the output impedance of the filter. Correct?
Jul 18 at 5:36 comment added Verbal Kint Oh, sure, if this is \$Z_{in}\$ that you want, they surely do but for this exercise, they are usually not accounted for. Check my APEC 2017 seminar, I derived the input impedance in open- and closed-loop conditions.
Jul 18 at 5:05 comment added Mario Potschatski So do the two input capacitors have no influence on the input impedance of the buck converter?
Jul 16 at 18:49 comment added Verbal Kint You can get rid of the two input capacitors, they are useless for the analysis as the voltage source is perfect. They would have a role if a) the voltage source is affected by an output resistance (or impedance) and b) if they add an equivalent series resistance or ESR. If you want the control-to-output transfer function, you consider \$v_{in}\$ equal to 0 V in ac. I recommend you have a look at my APEC 2013 seminar, it covers small-signal analysis using the PWM switch from Vatché Vorpérian published in 1986.
S Jul 16 at 7:52 review First questions
Jul 16 at 13:14
S Jul 16 at 7:52 history asked Mario Potschatski CC BY-SA 4.0