Annular Ring Size

[Ranayna]

Hi,

i have just created my first, rather complex, custom pattern (Footprint) in Diptrace.
It is a Molex RJ45 Jack. Here is the datasheet, which I also used for all measurements: https://www.molex.com/pdm_docs/sd/857891001_sd.pdf

The process of creating the pattern was straight-forward, if somewhat tedious. I set the Grid to 0.01 mm and plastered my sheet with measurements 
Everything seems to be working out, at least according to the 3D model. After properly offsetting the model, everything fits 
 
I have now a question though that is not specific about Diptrace. I want to know about the required minimum size of annular rings for the two pairs of power pins (P1/P3 and P2/P4 on the datasheet)
Those two pins are only 1.47 mm apart, and therefore do not work with the same ring size of the other pins. Ironically, these are the power pins, which can have almost 60 volts on them at about 500 mA I think.

Currently, I have these set up with 0.95mm hole diameter and 1.2mm pad diameter. With that, the clearance between the pads is just 0.27mm...
Is that still ok? Those seem awfully close together...

Greetings,

Ranayna

[Deridex]

I can only recommend a bigger annular-ring-size

I'm not 100% sure, but i think 0,35mm(As example:Hole 0,25mm; Diameter 0,6mm)should be minimum on basic layouts.
Remember that you need to be able to work with that pads 

[Ranayna]

Yeah, the usability of that small ring is one of the things I worry about. But now that I think about it, that should be easily solvable by using rectangular pads. With those I think I could get away with making the pads just 1.1mm wide, increasing the clearance between the pads to 0.37.
That setup would result in just 0.07mm copper at the narrowest place, will that become an issue for the PCB maker?

[tautech]

0.95mm hole size is massive and probably doesn't need to be so large. Try 0.8mm

[Ranayna]

Hmm, the datasheet says 0.89mm +0.05/-0.03mm. I chose 0.95 to make sure the part fits.

But... following your comment I measured one pin of the jack with a (admittedly cheap chinesium) digital caliper, and got 0.48mm...
It seems I misinterpreted the drawing to mean pin size, whereas it seems to suggest hole sizes.