I didn't realize that was in ieee286 as edison mentioned, but there it is. Here is a piece of that discussion:
"Hence,on the assumption that tan ä ,which is determined by the dielectric losses,remains unchanged with
voltage,the overall tan ä value of the bar insulation will vary with the second term on the right hand side of
Equation (3),which represents the discharge power loss contribution to the dissipation factor.
As long as an increasingly larger number of voids begin to undergo discharge with rising applied voltage,
the value of tan ä will continue to increase.Once all voids become ionized and are discharging,the tan ä
value,after attaining a maximum,will commence decreasing with voltage.This behavior is manifested
when the power loss due to all the partial discharges is increasing at a lower rate than the square of the
applied voltage term,V 2 ,in the denominator of the second term of the right hand side of Equation (3).
Consequently,a negative tip-up value of tan ä (if it is caused by partial discharge losses)occurs when all the
existing voids become ionized and begin discharging at some lower voltage,and a further rise in applied
voltage does not result in any additional discharging voids."
Sorry, it loses a little bit without the equations which I can't text copy in here.
What they're saying kind of makes sense.
Each void can discharge only once.
IF the pd at the low votlage has pretty well got all of the voids discharging, then increasing the voltage results in increased capacitive current but no further resistive (arcing) component, so overall decrease in power factor.
Now trying to remember the theory... which type of void arcs at a lower voltage: small or large? That's the one you've got (according to the theory which is a little dubious imho). I have to go back and look at some references to figure that out.
I'm sure doble can provide plenty of info on the subject.