According to David Harris's presentation for eve224a course: (slides 6-11 and 47)
Delay d = f+p = g*h+p
Where d is process-independent delay, f is effort delay (stage effect), p is parasitic delay, g is logical effort, h is electrical effort (fanout; h = C_out/C_in)
In the Wikipedia article "Logical Effort" there are some examples too:
Delay in an inverter. By definition, the logical effort g of an inverter is 1
Delay in NAND and NOR gates. The logical effort of a two-input NAND gate is calculated to be g = 4/3
For NOT gate with FO1 (driving the same NOT gate):
g=1; h=1; p=1; so d = 1*1 + 1 = 2
For NOT gate with FO4 (the FO4 metric itself):
g=1; h=4 (Cout is 4 times more than Cin); p=1 so d = 1*4+1 =5 (the same result is at page 20 of books "Logical Effort: Designing Fast CMOS Circuits", draft from 1998)
1 FO4 delay is equal to 5 process-independent units (defined by harris, slide 6)
For NAND gate with two inputs (p=2) which drives the same:
g=4/3; h=1; p=2; d= 4/3 * 1 + 2 = 10/3 = 3,3 (a 1.5 times slower than NOT with FO1, but faster than NOT FO4)
For NAND gate asked by me - 2 inputs which drives 3 same NANDs:
g=4/3; h=3; p=2; d= (some magic inside) 4/3 * 3 + 2 = 6
So
Delay of 1 FO4 gate is equal to 5/6 delay of NAND (2-in, 3 FO).
The last problem is to convert chain delay of 18 NANDs to chain delay of FO4. (slide 41 of harris)
Hmm.. seems I need only to multiply 18 NANDs delay with 6/5... 21,6 FO4.
Thanks!
The NMOS FETs out of order. In the schematic gate "b" is closer to ground but in the layout gate "a" is closer to ground.
The mismatched number of terminals indicates there may be some additional substrate (bulk) and nwell terminals assumed in the schematic devices. If you query the FETs in the schematic, is there an entry for the well or bulk node? There probably is, and it's probably for some default node names other than gnd! and vdd!. (I never use the three-terminal device symbols for this reason.) You'll want to use a dedicated pin for supply and one for ground, as well, eventually, e.g. vdd and vss, rather than the global vdd! and vss!. Global nets cause all sorts of headaches later (at least in big chips).
Good luck with it.
Best Answer
This depends on the process and area vs. speed trade-offs. Clearly, making a NAND5 using 5 p-FETs and 5 n-FETs would be the simplest in theory, but it might not be the most effective. For one thing, the pull-down network resistance begins to get large, which would slow down the fall time if a large capacitance is being driven.
Synthesis will instead probably realize a NAND5 as multiple gates using standard cells, rather than using the fewest-transistors solution. As it is, typical libraries don't have bigger fan-ins than 4.
Example: http://www.vlsitechnology.org/html/cells/wsclib013/lib_gif_index.html
Another: https://www.cl.cam.ac.uk/teaching/0910/SysOnChip/tanner_ami.pdf