Yes you are right Brian.The current out put is very less.It will be the typical leakage current only.So assuming a 5mA worst case and as per your calculations ,Output power=100V*5mA=0.5W @100% efficiency it will require 150mA from 3.3V so by putting margin the input supply required will be 3.3V,300mA worst.
So the requirement translates to
Input -3.3V,300mA
Output-100V,5mA.
I was checking different dc dc converter solutions, im supposing that simple boost converter is not easy here since the voltage is very high( correct me if im wrong)
Then remains flyback,push pull,half bridge,full bridge etc..
Can some body give me a suggestion??
Regards