Design Article
Power Tip #4: Damping an Input Filter " Part 2 of 2
Robert Kollman, Texas Instruments
9/29/2008 10:09 AM EDT
The design of an input filter usually begins with selection of an input capacitor (CO of Figure 1) based on ripple current rating or hold-up requirements. The next step usually involves selecting an inductor (LO) based on the system's EMI requirements. As we saw last month, near resonance the source impedance of these two elements can be quite high, leading to an unstable system. Figure 1 presents a method to control this impedance by placing a series resistor (RD) and capacitor (CD) in parallel with the input filter. The filter could be damped with just a resistor across CO. However, in most cases the power loss would be unacceptable. An alternative method is to add a series connection of an inductor and resistor across the filter inductor.
Figure 1: CD and RD damp the output filter source impedance.
(Click this image to view a larger, more detailed version)
Comments
NOPROBLEM
10/2/2008 12:08 PM EDT
Interesting. I believe there are three basic topologies that dampen the input filter. This extra capacitor has the disadvantage that for instant connection of a high input voltage, the resistor needs to absorb 0.5CU^2. I experienced this as a problem. Another solution, used by computer products is a LC-series combination in parallel with LO. The math behind it is the same. The theme has been discussed since the eighties. I remember three articles. Only one of them uses the calculation of pole location to find an optimum. This should also work for two stage input filters. I could not see in the article presented here wether or not the NEGATIVE input impedance has been used in calculating the frequency spectrum. The second chart I find very interesting, but I wonder what kind of mathematical model has been used to derive it. Is this 50% approach a rule of a thumb thing or does it give the optimum ? Apart from the optimum value of R, there are also a minimum and a maximum value beyond which there are no stable solutions for C.
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