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On 27/05/2025 11:46 pm, john larkin wrote:On Tue, 27 May 2025 14:16:11 +0100, liz@poppyrecords.invalid.invalid>
(Liz Tuddenham) wrote:
Bill Sloman <bill.sloman@ieee.org> wrote:
>On 27/05/2025 9:09 am, john larkin wrote:>
[...]>The specs, as far as I can tell, suggest 1KV at 1 ma in and 3.3v at 3>
ma out. The required efficiency is then 1%.
Actually 1kV at 10uA in.
Good grief! A 10-megohm quarter watt resistor with a zener diode is
the obvious answer.
A 3.3v real zener will probably leak more than 10uA. Low voltage
zeners are awful.
And he wants 3 mA out... I think.
The guy I was talking to had 10uA at 1kV, and wanted a 3.3V output. 1mW
at 3.3V is 3mA. The 3ma is the upper limit the circuit could supply.
>>
[...]Misunderstanding the constraints can lead people to propose>
inappropriate solutions.
How can anyone misunderstand something they have never been told?
You managed it.
>Bill was the confused person.>
John Larkin does like to think that.
>The problem gets much more interesting with 10 uA in and 99%>
efficiency requirement.
There never was an efficiency requirement. 50% would have been nice.
>A Baxandall isn't going pull microamps at 1 KV.>
Keeping a resonant tank resonating at a couple of kV is going to use up
power in the winding resistance. I think I can get 245H out of RM14
core. If we can keep the parallel capacitance down to 25pF that would
resonate at 2kHz and the reactive impedance would be 3 Mohm, giving a
peak current 0f 1mA. The resistance of the winding is going to be about
480R, so I^2.R is 500uW, or 0.5uA at 1kV.
>
A properly designed Baxandall might well pull less than 1uA if lightly
loaded.
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