2 Replies Latest reply on Dec 13, 2010 8:46 AM by jmatthews

    Layout problem regarding OAR

    dhaval.shah

      hi everyone.

      i am new in PCB Design..


      I have made one design and when i make order for manufacturing they have some queries like.
               " Minimum OAR (Outer Annular Ring) in our  service is 0.125mm.Your design has OAR of 0.0mm for drill size of 0.30mm end size."         


      i don't know how to correct it..

      In my design drill size of the pad is 1.0mm and at TOP and BOTTOM it is 1.20mm as well as in PLANE its size is 1.40mm.
      but i think this one is incorrect..

       

       

      And they have suggest me to calculate OAR as per following formulas.. But I am too much confused from this.

       

      Calculation of OAR:OAR = (Cupper pad diameter – TOOLSIZE)/2

       

      TOOLSIZE = ENDSIZE + 0.10mm for all PTH holes of 0.45mm(ENDSIZE) and smaller

                                    + 0.15mm for all PTH holes bigger then 0.45mm(ENDSIZE)

                                             + 0.05mm for all NPTH holes.

       

      Example calculation for OAR in your design on the 0.30mm ENDsize drills:

          OAR = (0.4191mm – (0.30mm + 0.10mm))/2

                   = (0.4191mm – 0.40mm)/2

      = 0.0191mm/2

                   = 0.0096mm

       

       

      If Any one knows the formula to calculate OAR for different Drill sizes??

       

      Please let me know if anybody knows solution.

       

      They have provide me one documents showing my errors which i have attached.

       

      So you can find that attached file more comfort.

       



      Thanks

        • 1. Re: Layout problem regarding OAR
          jmatthews

          That is a confusing explanation.

           

          First off, understand that I don't think in metric, so bear with me if there are any conversion errors.

           

          One thing I do is specify in my fab drawing notes is that the listed sizes in the drill table are the finished hole sizes. I don't worry about how they (the fab shop) get there.

           

          In fabrication, the actual hole size drilled for a finished plated hole is a bit larger than what you specify. The oversize I'm used to seeing is usually 3 mils (or ~0.076mm) +/-3 mils (or ~0.076mm) for plated holes, and 0.0 +/-2 mils (or ~0.05mm) for non-plated holes. The finished hole size obviously carries the same +/- tolerance (2 or 3 mils) as the drill. So your pad sizes also need to take that plus any layer-to-layer offset tolerances (ask your fabricator what tolerance they meet) into account.

           

          But your shop oversizes their drills for plated holes quite a bit more than what I'm used to seeing, 4 mils (0.1mm)  for 18-mil (0.45mm) and down plated finish holes, 6 mils (0.15mm) for holes larger than the 18-mil (0.45mm) plated finish. And they do list a 2-mil (.05mm) oversize for the NP holes, which meets the tolerance I'm used to seeing.

           

          The minimum annluar ring is the minimum amount they require the pad to be over the hole size on any one edge, their MAR is apparently 0.125mm

           

          Anyway, what they are saying is for a finished plated hole size of 0.3mm size you specified:

           

          Add 0.1mm (because it's smaller than 0.45mm) = 0.4mm

          Then add the MAR: (0.125mm x 2 = 0.25mm)

          Then add the numbers (0.25mm + 0.4mm)

           

          So, the minimum pad diameter for a 0.3mm (~12-mil) hole is, per their specs, 0.65mm (~26 mils). which sounds about right. I'm guessing this is a via?

           

          They don't say what size pad you're using, but I'm guessing it is smaller than 0.65mm?

          • 2. Re: Layout problem regarding OAR
            jmatthews

            BTW, I don't know about that shop, but you can often waive requirements, as long as continuity is not compromised.