EMRFD Message Archive 14192

14192 2017-08-06 18:27:25 mosaicmerc Impedance transformer question Message Date From Subject I was looking at this class E calculator page:http://people.physics.anu.edu.au/~dxt103/calculators/class-e.phpTo obtain about 200W at 80 meters with this MOSFET supplied by 36Vdc:http://www.st.com/content/ccc/resource/technical/document/datasheet/a5/9f/eb/4a/66/42/4b/23/DM00150353.pdf/files/DM00150353.pdf/jcr:content/translations/en.DM00150353.pdfI note the output load is about 3 ohms resistive requiring a 1:4 transformer to take it to about 50 ohm.Does this mean the primary of the transformer needs to have a total impedance of 3Ω at the operating frequency? I believe this is so=> 2.pi.F.L being the dominant factor as copper losses will be minimal. The turns ratio needs to be 1:4 to take a 3 ohm inductance on the 1: side to a 48 ohm inductance on the :4 side.Remember, this is AC and not DC, so "resistance" flies out the window when you start looking at active inductance and reactance values.  Work the numbers at different, randomly spaced phase angles and you'll see what I mean.Fortunately we can simplify this as a resistive equation / transform, where the square of the turns yields the transformation ratio.  1*1=1, and 4*4=16, so 3 ohms*16= your 48 ohms -- close enough to 50 to work FB.If you need more on the theory, I recommend Helge Granberg's seminal works at Motorola.Some things in electronics should be accepted as "it just works" and this is one of them.73Jim N6OTQ Thanks Jim. In adjusting the spice sim, I note that a particular on time of about 95nS for the CLASS E NFET achieves the best efficiency in switching the drain at the lowest voltage in the cycle. Moving away from that on time starts to increase the NFET Miller losses dramatically. I was wondering if there's a feedback loop that could govern this and hold the optimum on time?