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Hi Kevin. The standard impedance of a half wave is about 2k5 – 3k Ohm so we need impedance transformation of 1:50 – 1:60. The impedance transformation ratio of an auto transformer is determined by the winding ratio. Actually, it is the winding ratio squared. We require a 1:7 to 1:8 ratio to get 1:49 (with 1:7 winding ratio) or 1:64 (with 1:8 winding ratio) impedance transformation. Regardless of the winding ratio, there’s an additional crucial issue for the auto transformer to work efficiently and that is the major inductance which is dependent on the frequency you want to use. Reduced frequencies generally like to see higher major inductance. There is two way to acomplish that: larger or more (stacked) cores or much more primary (thus secundary, you want the ratio to remain the same!) windings.
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used auto parts near me openThere were concerns that the semantics of the generated operator< and its counterparts, were not a good fit for the low-level model of the language. The details here are a bit sublime (see the proposal wording if you’re interested, particularly around operator<=), but the gist is that the generator operators don’t behave exactly the same as a hand-written function would have in some respects.
Two primary (bilfilar) windings on the smaller FT-140-43 core would be OK for 40 – ten meters. If you want the auto transformer to operate on reduced frequencies (80m_, you need a tiny far more inductance. I know that the industrial HyEndFed antennes by PA3EKE use the exact same quantity of windings (2:14) but use the larger FT-240-43 core. The FT-140-43 has an AL-element of 885, the FT-240-43 has an AL issue of AL1075. That larger AL factor obviously is adequate to cover 80m (study on).
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