How small could Z80 microprocessor made with modern transistors?
Not counting the pad ring, it would use approximately \$\left(\frac{7}{6000}\right)^2 = 1.36 \times 10^{-6}\$ the area.

The original die was 3545&times;3350 &micro;m, or 11.8 mm2. Even if you converted it to CMOS, roughly doubling the number of transistors, the core logic would be about 1/400,000 this size, or about 30 square microns.

If you divide the total chip area of the original by the number of transistors, you get about 1400 square microns, or a box about 37.5×37.5 microns. This accounts for not only the size of the transistor, but also its "share" of the chip infrastructure (power distribution, interconnects, isolation, etc.). I'm making the simplifying assumption that all of this scales linearly with minimum feature size, allocating a box of about 44×44 nm to each transistor.

As a standalone chip, it probably wouldn't be much smaller than the original, because of the perimeter required by the pads. A modern wire bonder can handle pads on a 100 &micro;m pitch, so to get 40 pads (leaving the corners empty) would require a die of about 1200&times;1200 &micro;m, or about 1/8 the area.