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Free tool · DFM
BGA escape planner: traces per channel, fanout and layer count
Check whether a ball pitch can escape at your trace and space, how many traces fit between two lands, and how many layers the fanout will take.
The short answer
A BGA escapes by routing traces out from the pads, either directly between neighbouring balls or through a via in that channel. The room available is the pitch minus the land diameter, and a trace needs its width plus two clearances. On a 0.8 mm pitch with 0.4 mm lands the channel is 0.4 mm: one 0.1 mm trace with 0.1 mm clearance fits and a second does not. On 0.5 mm pitch with 0.075 mm rules nothing fits, so the escape has to go through the pads.
Calculator
BGA escape and fanout planner
Enter the land pattern, the etch rules and the via geometry. The tool returns the channel between two lands, how many traces fit in it, whether a dogbone via fits or the part needs via-in-pad, and a perimeter estimate of the layers the escape takes. It is a feasibility check on a package, not a router: the real fanout decides which ball goes to which layer.
A pitch fills the land and the etch rules it usually implies.
Land pattern and etch rules
Ball pitch of the BGA. Open the button for its units, why it matters and a typical value.
Land (pad) diameter on the board. Open the button for its units, why it matters and a typical value.
Escape trace width. Open the button for its units, why it matters and a typical value.
Trace-to-trace and trace-to-pad clearance. Open the button for its units, why it matters and a typical value.
Balls in the BGA array. Open the button for its units, why it matters and a typical value.pcs
Balls that actually have to escape. Open the button for its units, why it matters and a typical value.pcs
Via and board
Dogbone via pad diameter. Open the button for its units, why it matters and a typical value.
Via drill diameter. Open the button for its units, why it matters and a typical value.
Finished board thickness. Open the button for its units, why it matters and a typical value.
Whether via-in-pad is available. Open the button for its units, why it matters and a typical value.
Filled and plated over: removes the via from the channel entirely.Escape capacity
How many traces leave between the lands, and how many layers that takes.
1 traces per channel, 2 layers needed.
- Traces per channel
- 1
- Channel used
- 0.300mm
- Outer ring
- 76balls
- Per layer
- 152escapes
- Layers needed
- 2
- Via aspect
- 8.0:1
Channel 0.400 mm.
A perimeter model, not a router
The layer count assumes the outer ring escapes outward and each channel carries the balls behind it. A real fanout study decides which balls move to which layer, and it can do better or worse than this number. Use it to check a pinout before the package is chosen.
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Want the derivation? Jump to the math behind the calculator.
The math
The formulas, with the working shown
Each equation is rendered with KaTeX, which also emits a MathML twin for screen readers and answer engines, explained in plain language, and copyable as LaTeX from the button in its corner. The variable list says what every symbol means; the model-limit note says where the formula stops being trustworthy.
The escape channel
The clear width between two adjacent lands, and the only room an escape has.
- p
- ball pitch, mm
- d
- land diameter, mm
Pitch is centre to centre and the land is the copper on the board, so what is left between two lands is where a trace and a via have to live.
A 0.8 mm pitch with a 0.4 mm land leaves 0.4 mm. A 0.5 mm pitch with a 0.3 mm land leaves 0.2 mm, less than one 0.075 mm trace with its clearances.
Model limits
Measured edge to edge along the line between two ball centres. A trace leaving diagonally has more room, which is why real fanouts do better on the corners.
Traces per channel
n traces need n widths and n plus one clearances, so the count is that inequality solved for n.
- w
- trace width, mm
- c
- channel width, mm
Each trace contributes its width, with a clearance to each land and between adjacent traces: n·w + (n+1)·s has to fit inside c.
Rearranging gives the floor expression. Clearance dominates at fine pitch because it is counted one more time than the width.
Model limits
The count assumes straight traces down the channel. A break-out that turns immediately, or one that dodges a via, uses more room.
Outer ring of the array
The perimeter of the square array that holds N balls, and therefore the first layer's direct escapes.
- N
- balls in the array
- N_{\text{ring}}
- balls on the outer ring
A square array of side k holds k squared balls and has 4k − 4 on its perimeter, because the corners are shared.
A 400-ball array is 20 by 20, so 76 balls sit on the ring and the other 324 come through the channels.
Model limits
Real packages are often not square and often have depopulated areas, so the ring can be larger or smaller. A square is the neutral estimate.
Layers the escape needs
The ring escapes one each and each channel carries the balls behind it, so per-layer capacity is that sum.
- N_{\text{sig}}
- balls that need a trace, not a plane
- n
- traces per channel
The first layer carries the ring directly plus n traces through each channel between ring balls. Dividing the signal count by that capacity gives the layers.
A 400-ball part with 260 signal balls at one trace per channel divides by 152, which is two layers of capacity before the routing that follows needs another.
Model limits
A perimeter model, not a router. It gets the order of magnitude right, and a real fanout study can do better or worse depending on where the signals go.
Geometry
The channel is the whole story
Every escape decision comes down to one width: the pitch minus the land diameter. A trace in that channel needs its own width plus clearance on both sides, so the count is the channel minus one clearance, divided by the width plus a clearance. Everything else — array size, via size, layer count — is downstream of that number.
The practice publishes a 2.5 mil (0.063 mm) trace and space capability, which is what makes a 0.5 mm pitch escapable at all. At 0.8 mm an ordinary 0.1 mm rule gives a comfortable single trace; at 1.0 mm it gives two.
- Channel = pitch − land diameter.
- Traces that fit = floor((channel − clearance) / (width + clearance)).
- A dogbone via needs its pad plus clearance to fit in the same channel.
- When nothing fits, via-in-pad is the escape, and that is a fabrication capability.
Fanout
Why the layer count grows so fast
The outer ring of a square array escapes straight outward. Every ball behind it comes out through the channels between two outer balls, so the first layer carries the perimeter plus one trace per channel. The count falls quickly because the perimeter grows more slowly than the area.
That is why a 400-ball 0.8 mm package needs about two or three layers at one trace per channel, and why a 0.5 mm part with no channel capacity cannot be rescued by adding layers: every layer needs the same impossible fanout.
The DFM rules that follow from the escape — annular ring, via-in-pad fill, mask dams — are on our DFM/DFA layout service page, and the routing that follows it on the high-speed PCB layout service page.
FAQ
Common questions
Usually not. A 0.5 mm pitch with a 0.3 mm land leaves a 0.2 mm channel, and one 0.075 mm trace with 0.075 mm clearances needs 0.225 mm. That is why fine-pitch parts are paired with HDI: via-in-pad removes the via from the channel and the trace leaves from the pad itself.
Want this checked on your board?
Send the stackup and interface list. We will check the fanout against the real layout before the quote is fixed.