5a-75e transceiver pins question

[maciejopalinski]

According to https://github.com/q3k/chubby75/blob/master/5a-75e/hardware_V6.0.md

chubby75/5a-75e/hardware_V6.0.md

Lines 206 to 214 in a1dcd4e

	#### LED Transceivers to J* mapping
	Transceivers U23,U24,U28 completely cover RGB pins on J1,J2,J3,J4
	Transceivers U25,U26,U27 completely cover RGB pins on J5,J6,J7,J8
	Transceivers U14,U16,U18 completely cover RGB pins on J9,J10,J11,J12
	Transceivers U9,U12,U15 completely cover RGB pins on J13,J14,J15,J16

That is 6 RGB pins * 16 HUB75 ports = 96 pins in total
Which is the same as 12 transceivers * 8 pins each = 96 pins in total
That is correct.

I don't understand this line:

chubby75/5a-75e/hardware_V6.0.md

Line 216 in a1dcd4e

Transceivers U10,U11,U13,U17,U19,U20,U21,U22,U30,U38,U39 cover all shared pins

There are only 8 shared pins (A, B, C, D, E, CLK, STB, OE):

chubby75/5a-75e/hardware_V6.0.md

Lines 193 to 204 in a1dcd4e

	For each connector, the address, CLK, STB/LAT, and OE bits are shared:
	| FPGA Pin | HUB75 Pin | Function |
	|-----------|-----------|----------|
	| N5 | 9 | A
	| N3 | 10 | B
	| P3 | 11 | C
	| P4 | 12 | D
	| N4 | 8 | E
	| M3 | 13 | CLK
	| N1 | 14 | STB
	| M4 | 15 | OE

Why does it need 11 transceivers * 8 pins each = 88 pins in total for driving 8 shared pins?

TLDR: Where are the remaining 80 pins connected?