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I have read someone pointing out that 22nm would be the only way Apple could have the required number of transistors on the die. I have not personally checked that information, but it seems a better base to go on than just say "it would be too risky".

Do the math, it isn't hard. 102 sq mm is 102 trillion sq nm. Divide by a billion or so, take the square root, and Apple has about 300x300 nm per transistor. Rule of (very rough) thumb, you need 8x the half-pitch in one dimension for a transistor, and there will be at least two or three different transistor widths that have to be supported. So 28 nm or even 32 nm makes perfect sense.

Note that the Apple is a SOC (system on chip) design, and has some analog areas as well as the CPUs and GPUs. But that is a relatively small portion of the chip.

Also notice that L2 and L3 cache can be laid down with greater density, and most layout programs for 'random' logic areas will leave some transistors sites unused and unpowered. It is much, much easier for the CD layers to have a regular layout than to customize for a hole here and a hole there. Since these transistors are not connected to the higher metal layers, they draw very little power. Some parasitic loads on the adjacent transistors, but it is again easier to have identical for all transistors in an array, than to compute the effects of having a hole.

No, the real surprise here, and kudos to Apple for managing it, is to get an ARMv8 64-bit chip in production--and with an operating system--this fast. Apple only has to support drivers for the devices in the iPhone 5s, But even with a small set of drivers, Apple is so far ahead of Microsoft in being able to 'just' recompile them. (It is never just but that is why you have so many software engineers.)
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