Donald,

The "A" choice is the correct interpretation. Total calculated width is 
divided between the planes layers because each plane takes away from the 
thermal relief. Once you have the total thermal web width for each 
layer, you may then calculate the number of connections needed to meet 
minimum line widths for that layer.

Regards,

Gary F.

Donald Kyle wrote:
> All, I would like to open a discussion about Thermal Relief in conductor 
> planes.
> 
> Not whether there should be thermal relieve but how to calculate the web 
> conductor width.
> 
> More precisely, I would like to discuss the calculations in section 
> 9.1.2 of IPC-2222.
> 
> For this discussion, we will be considering a PTH connected to 4 
> internal planes by 2 webs
> on each plane for a total of 8 webs.  (Please, no substitutions.)
> 
> The total calculated thermal width is .80mm and the adjusted thermal 
> width is .60mm as
> shown in IPC-2222, section 9.1.2 below.
> 
> I would like to start by suggesting two interpretations of the Web width.
> 
> a) The total thermal width (.80mm) is divided by 8 and each web on each 
> plane would be .10mm wide.
> 
> b) The total thermal width (.80mm) is divided by 2 and each web on each 
> plane would be .40mm wide.
> 
> c) Your interpretation, opinion.
> 
> Thanks in advance
> Donald Kyle
> 
> 
> 
> 9.1.2 Thermal Relief in Conductor Planes The relationship
> between the hole size, land and web area is critical.
> Typically, divide 60% of the minimum land area diameter
> by the number of webs desired to obtain the width of each
> web in accordance with the following example:
> 
> A. Land Size Calculation
> Maximum hole size = 1.0 mm
> Annular ring      = 2 x 0.05 mm
>                   = 0.10 mm
> 
> Fabrication allowance = 0.25 mm
> 
> Minimum land size = 1.0 mm + 0.10 mm + 0.25 mm
>                   = 1.35 mm diameter
> 
> B. Thermal Relief Calculation
> Total thermal width = 60% of land size
>                             = 0.6 x 1.35 mm
>                     = 0.80 mm
> 
> C. Original Web Size Calculation
> 2-web width = 1/2 of total thermal width
>             = 0.50 x 0.80 mm
>             = 0.40 mm
> 
> 3-web width = 1/3 of total thermal width
>             = 0.33 x 0.80 mm
>             = 0.27 mm
> 
> 4-web width = 1/4 of total thermal width
>             = 0.25 x 0.80 mm
>             = 0.20 mm
> 
> If the actual land diameter chosen is greater than the minimum
> value calculated, the percentage difference between
> the land diameters must be subtracted from the total web
> width calculation, i.e.:
> 
> Minimum land diameter = 1.35 mm
> Actual land diameter  = 1.70 mm
> Percent difference    = (1.70 - 1.35 mm)/1.35 mm
>                       = 25%
> 
> New total web width = total web width percent difference
>                     = 0.80 mm - 25% (0.80 mm)
>                     = 0.60 mm
> 
> D. Adjusted Web Size Calculation
> 
> 2-web width = 1/2 of new total web width
>             = 0.50 x 0.60 mm
>             = 0.30 mm
> 
> 3-web width = 1/3 of new total web width
>             = 0.33 x 0.60 mm
>             = 0.20 mm
> 4-web width = 1/4 of new total web width
>             = 0.25 x 0.60 mm
>             = 0.15 mm
> 
> 
> Total cumulative copper web for all layers in any platedthrough
> hole should not exceed 4.0 mm for 1 oz copper or 2.0 mm for 2 oz copper.
> 
> The total of the thermal relief cross-sectional area divided
> by the number of planes connected to the plated-through
> hole shall not violate current carrying capacity requirements
> for a given hole.
> 
> If the individual web width violates the intended minimum
> conductor width it shall be specified on the master drawing.
> 
> 
> 
> Donald Kyle C.I.D.+ 
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> 

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