The KLC gives 0.15mm for the minimum annular ring width for drilled through-holes, but nothing about the width itself. How to determine the width of the annular ring?
For example TO-220-3_Horizontal_TabDown.kicad_mod (https://www.vishay.com/docs/66542/to-220-1.pdf) the (circular) hole size is 1.1 mm and the (oval) pad size 1.905 x 2 mm.
Annular ring is smallest copper dimension minus by drill diameter divided by two.
So in your case (1.905 - 1.1)/2
So minimum width is drill diameter plus two times minimum annular ring
How is the 1.905 or 2 mm arrived at?
Experience. And the 1.905 probably to achieve some pad to pad clearance.
As @Rene_Poschl said, experience is the rule.
From “PCB Design tutorial” by David L. Jones:
“There is an important parameter known as the pad/hole ratio. This is the ratio of the pad size to the hole size. Each manufacturer will have their own minimum specification for this. As a simple rule of thumb, the pad should be at least 1.8 times the diameter of the hole, or at least 0.5mm larger. This is to allow for alignment tolerances on the drill and the artwork on top and bottom layers. This ratio gets more important the smaller the pad and hole become, and is particularly relevant to vias.” Copyright© 2004 David L. Jones
PCB Libraries has an informative file of tables for proportional through-hole padstack dimensions.
The best is to get your PCB manufacturer guidelines…
here some for reference, including annular tips
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