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!
Model level 3 definition: "Semi-empirical" - a more qualitative model that uses observed operation to define its equations. This dates from 1980 ... Any sort of decent result (i.e. within 10% of reality) would need to use level 5 models (AKA BSIM3). I'd recommend using BSIM 3V3 which is model level 49 in Star-HSPice parlance.
As long as you going to be using out of date models then you should heed your prof and only look at the trends. I suspect that there probably is a reason he said that.
You're dealing with curve fitted results.
Additionally, unless you have parasitic extraction enabled the rail capacitances as you noted are almost certainly not being extracted. The CJSW means Capacitance, Junction Side Wall and is a computed values based upon the width and S/D sizes (as one example). There are excellent SPICE guides that tell you what all the parameters are, I suggest you find and read them.
Best Answer
The 24/12 means that the PMOS has a width of 24 lambda and the NMOS has a width of 12 lambda. That's a total of 36 lambda of gate width, which is 9 times as much as a 4 lambda transistor. Therefore, the capacitance of the output node of the inverter is also about 9 times bigger than that of the 4-lambda transistor.