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On Tue, 27 May 2025 01:07:36 +1000, Bill Sloman <bill.sloman@ieee.org>It took me a while to wake up to that. I think that there's an option where you'd use four switching transistors, two to alternately ground either end of the main inductor and two more to alternately switch the feed inductor into the other end of the main inductor, but I haven't worked it out in any detail. It's very much in the brainstorm state at the moment
wrote:
On 26/05/2025 4:20 am, JM wrote:Yes, but that's the centre tap voltage. Due to the autotransformerOn Mon, 19 May 2025 12:23:54 +1000, Bill Sloman <bill.sloman@ieee.org>>
wrote:
>On 19/05/2025 12:15 am, john larkin wrote:On Sun, 18 May 2025 18:11:58 +1000, Bill Sloman <bill.sloman@ieee.org>
wrote:>The drain swing is actually 1.67 times the supply voltage, but it does>
Where does 1.67 come from?
Integrate a series of half-sine peaks that get to 1.67V and the voltage
averages to about 1V.
>
You can do it as purely mathematical exercise, and I did it years ago,
and that's roughly the result I got.
>
There's a voltage drop across the switching FET that's on and the
Baxandall circuit doesn't product perfect half-sine waves, so it hasn't
got a lot to do with precise reality, but it's good enough for
preliminary design.
>
If senile dementia hasn't set in too far I could probably do it again.
action the stress across the off transistor will be twice that, or
1000*PI in this application. The 1700 volt device under consideration
isn't up to the task.
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