EMRFD Message Archive 14192

Message Date From Subject
14192 2017-08-06 18:27:25 mosaicmerc Impedance transformer question
I was looking at this class E calculator page:

To obtain about 200W at 80 meters with this MOSFET supplied by 36Vdc:

I 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.
14205 2017-08-08 09:35:23 Jim Strohm Re: Impedance transformer question
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.


14206 2017-08-08 20:40:39 AncelB Re: Impedance transformer question
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

I was wondering if there's a feedback loop that could govern this and
hold the optimum on time?