All this would be true but one must consider automated insertion
equipment requirements. Most equipment has an absolute minimum hole size
requirement and then a relationship requirement as leads get larger in
diameter.

> -----Original Message-----
> From: TechNet [mailto:[log in to unmask]] On Behalf Of Donald Kyle
> Sent: Wednesday, March 31, 2004 8:12 AM
> To: [log in to unmask]
> Subject: Re: [TN] Finished Hole Size vs Lead diameter
>
>
> This might be what you are looking for.  Please excuse me if
> there is an error in this text as I'm still half asleep.
>
>
> Refer to IPC2221, section 9.0 &  IPC-2222 section 9.2 for
> complete information.
>
> To find the minimum through-hole land size, a worst-case
> land-to hole relationship must be established.  The following
> equation from IPC-2221 will do that.
>
> Where:
> Copper weight = 1 oz and number of layers are 8 or less.
> (see table 9-1 in
> IPC-2221)
>
> For internal layers:
> a = the drilled hole diameter ( see table 9-3 in IPC-2222)
> For external layers:
> a = maximum diameter of the finished hole (see table 9-3 in IPC-2222)
>
> b = minimum annular ring requirements = 2 mils
>
> c = standard fabrication allowance = 8-16 mils (see table 9-1
> in IPC-2221)
>
> For example, take a 20 AWG wire that is 38 mils in diameter
> at its maximum, 8 layer PWB.
>
> The maximum diameter of the finished hole would be 38 + 6
> (hole to lead
> space) + 3 (tolerance)  = 47 mils.  The nominal diameter
> would be 38 + 6 = 44 mils.
>
> The drilled hole diameter is 44 + 6  =  50 mils.
>
> Minimum through-hole land size = a + 2*b + c
> Internal land = 50 + 4 + 10 = 64 mils
> External land = 47 + 4 + 10 = 61 mils
> Remember, these are the minimum land sizes and they can be
> made larger.
>
> In IPC-2222, section 9.2.2 Plated-Through Holes, the minimum
> and maximum finished hole size is discussed.  It states that
> the maximum hole size is less than or equal to the minimum
> lead diameter plus 28 mils (level A) and that the minimum
> finished hole size is greater than or equal to the maximum
> lead diameter plus 10 mils (level A).
>
> The following methods calculate the minimum land size for a
> typical DIP-I.C. with leads as follows.  Width = (15 mils  23
> mils), Thickness = (8 mils  15 mils), Diagonal = (17 mils - 27 mils).
>
> Using the IPC-2221 formula
>
> For internal lands
> Minimum land size = a + 2*b + c
>
> Where:
> a = drilled hole diameter
> b = minimum annular ring
> c = fabrication allowance
>
> drilled hole diameter = maximum diameter of finished hole +6
> mils (depends on type of finish).
>
> a =   51 mils
> b = 2 mil (internal)
> c = 14 mils (8 layers or less)
> c = 16 mils (10 layers or more)
> c = add 2 mils/oz of Cu for Cu > 1oz
>
> Number of Layers 10 8
> a42 mils42 mils
> 2*24 mils4 mils
> c16 mils14 mils
> Min land size =62 mils60 mils
>
> Using the IPC-2221 formula
>
> For external lands
> Minimum land size = a + 2*b + c
>
> Where:
> a = maximum diameter finished hole
> b = minimum annular ring
> c = fabrication allowance
>
> maximum diameter finished hole = minimum lead diameter (17
> mils) + 28 mils = 45 mils.
>
> a = 45 mils
> b = 2 mil (external)
> c = 14 mils (8 layers or less)
> c = 16 mils (10 layers or more)
> c = add 2 mils/oz of Cu for Cu > 1 oz.
>
> Numbers Layers 10 8
> a45 mils45 mils
> 2*24 mils4 mils
> c16 mils14 mils
> Min land size =65 mils63 mils
>
>
>
>
>
>
>
> At 09:26 AM 3/29/2004 -0600, you wrote:
> >Fellow TechNetters,
> >
> >     Can someone guide me to an IPC document that make reference to
> >Finish Hole Size versus lead acceptable lead diameter with
> respect to
> >ROUND, RECTANGULAR and SQUARE
> >Leads.   I have a case in which the square lead deformed the PTH and
> >minimal solder
> >fillet is exhibited inside the barrel.   At least in my opinion.
> >
> >In advance I thank you for your timely response in this matter.
> >
> >Victor,

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