Hi,
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A while back, I noticed the familiar double-pi filter circuit that¡¯s often used at output of transmitters. One day, I did some calculations and realized the connection between the values of L and C¡¯s (each ¡°pi¡± filter has an inductance L in series and two C¡¯s in shunt at each end). For each pi filter, at the frequency at which a tank of L and C (mounted in parallel) meaning L*C*(w^2) = 1, the impedance as seen at the input is equal to the inverse of the impedance of the load at the output, times L divided by C. Stacking two of these pi filters in series, the input impedance is equal to the load impedance (again, this is at the frequency which L and C resonates).
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I made a video (poor quality, sorry about it), it describes the gists of the above explanation:
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When an inductor L is substituted by a parallel L¡¯/C¡¯, as long as the impedance of the L¡¯/C¡¯ is equal to that of the original L, the transformation of impedance (from load to input) at the frequency of interest still stays the same. However, the parallel L¡¯/C¡¯ would block the frequency at which they resonate at. This can be used to reduce substantially a harmonic of the frequency of interest.
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Values of L and C are chosen with considerations of maximum of voltage across the capacitor at the middle (2 of C¡¯s) and the currents that flow through the inductors L¡¯s and C¡¯s. Higher currents would make the filter having larger ohmic loss. Too high of a voltage mike blow capacitors.
I simulated in LTSpice and described this half-wave filter here (the article is in Vietnamese but you can ask Safari to auto-translate to English, it¡¯s not 100% but close enough):?
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Thang Le
AA6SV
