Packaging and wire bonding - flip-chip: Visual walkthrough

WHAT this page does. The parent note handed you two headline formulas: $N_{wire} = \frac{4 L}{p} N_{flip} = (\frac{L}{p})^{2}$ This page builds both from absolutely nothing — no formula assumed, every letter earned — and shows in pictures why one grows in a straight line and the other explodes. By the end you will see why the whole industry flipped its chips over. Along the way we are honest about the rounding — real pad counts are whole numbers, and we say exactly when the clean formula is off, why, and how to force a valid (non-negative, integer) answer.

Step 0 — The three symbols we are allowed to use

Before a single formula, let us agree what every letter means, anchored to a picture.

Figure — Packaging and wire bonding - flip-chip

Figure s01 — what to look for: the lavender square is the die (side $L$ , both red arrows). Along the top edge, mint dots are pads; the short slate arrow between two dot centres is one pitch $p$ . The empty middle labels $N$ , the count we are hunting.

The question the whole page answers:

If both methods use the same pitch $p$ on the same die of side $L$ , how many connections does each give — and how does that number grow as the die grows?

Step 1 — Counting along ONE edge (the atom of the argument)

WHAT. Take a single straight edge of the die, of length $L$ . Lay down connection points spaced $p$ apart, stop-when-full (our fixed convention). How many fit?

WHY. Wire bonding can only reach pads that sit on the rim of the die (the wire is sewn in from outside), so an edge is the natural unit to count first. Master one edge and the rest is bookkeeping.

PICTURE. Look at the red dots marching along the top edge, and notice the leftover grey stub at the right end where a whole pitch no longer fits.

Figure s02 — what to look for: the slate line is one edge of length $L$ (lavender arrow). Coral dots are pads spaced by one pitch $p$ (mint arrow between the first two). The dashed grey stub on the right is the leftover space — smaller than $p$ , so no extra pad fits there. That stub is exactly why the real count is rounded down.

Step 2 — Wire bonding uses all 4 edges (the perimeter formula)

WHAT. A square has four edges. Wire bonding can place pads on every one of them, each edge counted stop-when-full. Add them up — but the four corners are shared, so subtract the overlap.

WHY. The die is face-up and glued down; wires arrive from outside and can only grab the border. So the total wire-bond budget is the sum over the border — the perimeter — minus the corners we'd otherwise count twice.

PICTURE. Four edges, each carrying its own row of $⌊ L / p ⌋$ dots. The middle of the die is empty — no wire reaches there. The four corner dots (circled) each belong to two edges at once.

Figure s03 — what to look for: the butter square is the die; coral dots line all four edges. The four circled dots at the corners are each counted by two edges — that is the double-count. The centre text marks the wasted interior real estate that flip-chip will reclaim.

Formula The perimeter (wire-bond) count — integer form, then scaling form

Exact integer form (the one you actually build a chip with): $N_{wire} = max (4 ⌊ \frac{L}{p} ⌋ - 4, 0)$

$⌊ L / p ⌋$ — pads on one edge, stop-when-full (from Step 1).
$\times 4$ — four edges of equal length $L$ .
$- 4$ — the four corner pads are each shared by two edges, so summing all four edges counts each corner twice; we subtract the four duplicates once.
$max (\cdot, 0)$ — a clamp to zero. If the die is so small that each edge holds one pad or fewer, $4 ⌊ L / p ⌋ - 4$ could go negative, which is nonsense — you cannot have $- 4$ connections. The clamp forces the physically-real answer $0$ . (Step 6 walks exactly when this bites.)

Clean scaling form (used only for the growth argument, never for a real count): $N_{wire} \approx \frac{4 L}{p}$ This drops two approximations at once, and here is precisely why each is allowed when $L ≫ p$ :

dropping the floor on each edge loses $< 1$ pad per edge ⇒ $< 4$ pads total;
dropping the $- 4$ loses the four corners. Both are $O (1)$ constants; the leading term $4 L / p$ grows without bound, so their relative weight $\to 0$ . For a small die they do not vanish — see the next box.

Formula Exactly when the

$- 4$ (and the floor) matter — annotated Relative size of the corner correction against the leading term: $\frac{correction}{leading term} = \frac{4}{4 L / p} = \frac{p}{L}$

Numerator $4$ — the four dropped corners.
Denominator $4 L / p$ — the clean scaling count.
Result $p / L$ — the fraction of the answer you throw away by using the clean form. Big when $L$ is close to $p$ (small die), tiny when $L ≫ p$ .

Die size	scaling $4 L / p$	floored $4 ⌊ L / p ⌋$	integer $max (4 ⌊ L / p ⌋ - 4, 0)$	error $= p / L$
$L = p$	$4$	$4$	$max (0, 0) = 0$	$100%$
$L = 2 p$	$8$	$8$	$4$	$50%$
$L = 10 p$	$40$	$40$	$36$	$10%$
$L = 100 p$	$400$	$400$	$396$	$1%$
$L = 1000 p$	$4000$	$4000$	$3996$	$0.1%$
Read the $L = p$ row carefully: the naive $4 L / p$ says $4$ , but the honest integer form clamps to 0 — a $100%$ error. For a real chip ( $L ≫ p$ ) the columns collapse together and the correction is a rounding whisper.

Step 3 — Flip-chip covers the whole FACE (filling area, not edges)

WHAT. Now flip the die over so its active face points down, and grow solder bumps on the entire face, not just the rim — a full grid of dots covering the square.

WHY. Once you connect straight down through a bump instead of sideways through a wire, the interior of the die becomes reachable. Suddenly the empty middle from Step 2 is usable real estate.

PICTURE. The same square, now peppered edge-to-edge with bumps in a grid.

Figure s04 — what to look for: the mint square is the die face-down. Lavender dots fill it corner-to-corner — the interior that was blank in s03 is now packed with bumps. That reclaimed middle is the whole reason the count is about to square.

Step 4 — Counting a 2-D grid (rows × columns)

WHAT. Count the bumps in the grid. Each row is just an edge — it holds $⌊ L / p ⌋$ bumps (Step 1 again, stop-when-full). Then ask: how many rows are there, stacked from top to bottom?

WHY. A grid is the same edge-count repeated downward. The height of the die is also $L$ (it's square), and rows are spaced by the same pitch $p$ — so the number of rows is also $⌊ L / p ⌋$ .

PICTURE. Highlight one row (across) and one column (down); both hold $⌊ L / p ⌋$ dots.

Figure s05 — what to look for: the top coral row and the left mint column each hold $⌊ L / p ⌋$ dots. The lavender dots are the rest of the grid. Row count × column count = total bumps — that multiplication is where the power-of-2 is born.

Formula The area-array (flip-chip) count — integer form, then scaling form

Exact integer form: $N_{flip} = bumps per row ⌊ \frac{L}{p} ⌋ \times number of rows ⌊ \frac{L}{p} ⌋ = ⌊ \frac{L}{p} ⌋^{2}$

First floor term — how many bumps across one row (horizontal), stop-when-full.
Second floor term — how many rows fit vertically in the height $L$ .
Multiplying rows × columns is how you count any grid — like counting a chessboard as $8 \times 8$ . No clamp is needed: a product of two non-negative floors is already $\geq 0$ .

Clean scaling form: $N_{flip} \approx (\frac{L}{p})^{2}$

The exponent 2 comes directly from two directions: connections live on a 2-dimensional resource.
Dropping the two floors here can lose at most (roughly) one row and one column — an $O (L / p)$ edge strip, not the whole $O ((L / p)^{2})$ area. So the relative error again $\to 0$ as $L ≫ p$ ; for large grids the floored and scaling values are nearly identical.

Step 5 — Racing the two formulas: where does flip-chip win?

WHAT. Put both counts on one graph against die size $L$ (same pitch $p$ ) and find the crossover — the die size where flip-chip overtakes wire bonding.

WHY. Two formulas mean nothing until you compare them. The crossover tells us for which chips the extra complexity of bumping pays off.

PICTURE. A straight line ( $N_{wire}$ ) and an upward-bending curve ( $N_{flip}$ ). They cross once; past that point the curve rockets away.

Figure s06 — what to look for: the coral straight line is $4 L / p$ ; the lavender curve is $(L / p)^{2}$ . They meet at the slate dot ( $L = 4 p$ , both $= 16$ ). The mint shaded wedge to the right is the region where flip-chip's curve has pulled ahead — and it only widens.

Step 6 — The degenerate cases (never leave a scenario unshown)

WHAT. Check the formulas at their extremes, so nothing surprises you later — including the sub-pitch die where fractions must become whole numbers and the clamp must fire.

WHY. A formula you trust must survive its edge cases: sub-pitch die, zero-size die, one-pad-per-edge die, and the tie point. Each is where the floor and the $max (\cdot, 0)$ clamp earn their keep.

PICTURE. Four mini-panels: a die narrower than one pitch, a vanishing die, a single-pad die, and the exact crossover.

Figure s07 — what to look for: panel 1 shows a die narrower than one pitch — not even one pad fits, so both counts are zero. Panel 2 shrinks the die to nothing. Panel 3 ( $L = p$ ) has 4 coral corner-dots (wire) beating a single lavender centre bump (flip). Panel 4 ( $L = 4 p$ ) shows the exact tie, 16 = 16.

Formula Edge-case table — every case walked, with the floor and clamp made explicit

$0 < L < p$ (die narrower than one pitch): $L / p < 1$ , so $⌊ L / p ⌋ = 0$ . Then wire $= max (4 \cdot 0 - 4, 0) = max (- 4, 0) = 0$ — this is exactly where the clamp fires: the raw $4 ⌊ L / p ⌋ - 4 = - 4$ is impossible, and $max (\cdot, 0)$ rescues it to $0$ . Flip $= ⌊ 0 ⌋^{2} = 0$ . So both methods give 0 — the die is simply too small to bond even one pad. The clean forms would return $4 L / p$ and $(L / p)^{2}$ , both non-zero fractions here, which is precisely why the floor and clamp are mandatory.
$L \to 0$ (die shrinks to nothing): $⌊ L / p ⌋ \to 0$ , so wire $\to max (- 4, 0) = 0$ and flip $\to 0$ . No die, no connections. ✅
$L = p$ (die exactly one pitch wide): $⌊ L / p ⌋ = 1$ . Wire (stop-when-full) $= max (4 \cdot 1 - 4, 0) = max (0, 0) = 0$ by the plain formula — because with one pad per edge, all four pads sit at the corners and the corner subtraction removes them all. In practice a designer would place those 4 distinct corner pads and count 4; our stop-when-full-plus-corner-subtraction formula is a lower bound in this tiny regime. Either way, wire's 4 corner slots beat flip's single centre bump ( $⌊ 1 ⌋^{2} = 1$ ): wire bonding wins the small-die regime. This is why tiny chips still use wire bonding.
$L = 4 p$ (the tie): $⌊ L / p ⌋ = 4$ . Wire $= max (4 \cdot 4 - 4, 0) = 12$ (integer form with corner fix); flip $= ⌊ 4 ⌋^{2} = 16$ . The scaling forms both give $16$ (the clean crossover of Step 5); the honest integer forms show flip is already ahead by the tie point once corners are subtracted — the corner penalty nudges the true crossover slightly below $4 p$ .
$L ≫ p$ (real chips): the $L^{2}$ curve buries the $L^{1}$ line; the floor, the $- 4$ , and the clamp all become negligible ( $O (1)$ against a growing $O ((L / p)^{2})$ ). Flip-chip dominates.

The one-picture summary

Figure s08 — what to look for: left = the perimeter method, coral dots on the border only, count $4 ⌊ L / p ⌋ - 4$ , a straight line; the blank middle is wasted. right = the area array, lavender dots filling the whole face, count $⌊ L / p ⌋^{2}$ , an exploding curve. The reclaimed middle on the right is precisely the emptiness on the left — that is why the count squares instead of merely doubling.

Left: the perimeter — dots on the border only, integer count $max (4 ⌊ L / p ⌋ - 4, 0)$ , scaling like $4 L / p$ , a straight line. Right: the area array — dots filling the whole face, integer count $⌊ L / p ⌋^{2}$ , scaling like $(L / p)^{2}$ , an exploding curve. The lonely empty middle on the left is the exact real estate the right-hand picture reclaims — and that reclaimed area is why the count squares instead of doubling.

Recall Feynman retelling — explain the whole walkthrough to a 12-year-old

Imagine a square garden and you want to plant flowers. The pitch $p$ is how far apart flowers must be; the garden's side is $L$ . So along one fence you fit about $L / p$ flowers — rounded down, because you can't plant a fraction of a flower, and if the last stub of fence is too short, it stays bare. (We always use the rule "start at one end, stop when the next won't fit.") Old way (wire bonding): you're only allowed to plant along the four fences. Four fences × $⌊ L / p ⌋$ flowers, minus the 4 corner flowers you'd otherwise count twice = $4 ⌊ L / p ⌋ - 4$ . And if that ever comes out negative (a garden smaller than one spacing), you clamp it to zero — you simply planted nothing. The whole middle stays bare grass. Make the garden twice as big and you get roughly twice as many flowers — a straight line. New way (flip-chip): now you plant the whole garden in neat rows. Each row has $⌊ L / p ⌋$ flowers, and there are $⌊ L / p ⌋$ rows, so $⌊ L / p ⌋ \times ⌊ L / p ⌋ = ⌊ L / p ⌋^{2}$ flowers. Make the garden twice as big and you get four times as many — it curves upward fast. If the garden is smaller than one flower-spacing, no flowers fit at all — zero either way. They roughly tie when the garden is about 4 flowers wide; any bigger and filling the whole area crushes the fence-only method. That "fill the area, don't just line the edge" trick is the entire reason chips got flipped upside-down.

Recall Quick self-test

One edge holds how many pads (our stop-when-full convention)? ::: $⌊ L / p ⌋$ Wire-bond total, exact integer form? ::: $max (4 ⌊ L / p ⌋ - 4, 0)$ Why the $max (\cdot, 0)$ clamp? ::: to stop a tiny die giving a nonsense negative count. When is the $- 4$ correction 50%? ::: when $L = 2 p$ (tiny die). Flip-chip total and why squared? ::: $⌊ L / p ⌋^{2}$ ; rows × columns = 2-D grid. What happens for $0 < L < p$ ? ::: floor is 0, clamp fires — both methods give 0. The scaling crossover die size? ::: $L \approx 4 p$ (slightly below once corners are subtracted). Which method wins for a small die like $L = p$ ? ::: Wire bonding (4 corner pads vs 1 bump).

Visual walkthrough — Packaging and wire bonding - flip-chip