1 Final Exam Review. 2 word7 is high if A2 A1 A0 = 111 word0 is high if A2 A1 A0 = 000 logical effort of each input is (1+3.5)/3 per wordline output.

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Presentation transcript:

1 Final Exam Review

2 word7 is high if A2 A1 A0 = 111 word0 is high if A2 A1 A0 = 000 logical effort of each input is (1+3.5)/3 per wordline output

3 Skewed gates HI-skew g u = 2.5/3 = 5/6 g d = 2.5/1.5 = 5/3 g avg = 5/4 p = 2.5/3 = 5/6

4 LO-skew g u = 2/1.5 = 4/3 (unskewed inverter with equal rise time pMOS size 1, nMOS size 0.5 g d = 2/3 (unskewed inverter with equal fall time pMOS size 2, nMOS size 1

5 HI- and LO-Skew Def: Logical effort of a skewed gate for a particular transition is the ratio of the input capacitance of that gate to the input capacitance of an unskewed inverter delivering the same output current for the same transition. Skewed gates reduce size of noncritical transistors –HI-skew gates favor rising output (small nMOS) –LO-skew gates favor falling output (small pMOS) Logical effort is smaller for favored direction But larger for the other direction

6 Asymmetric Gates Asymmetric gates favor one input over another Ex: suppose input A of a NAND gate is most critical –Use smaller transistor on A (less capacitance) –Boost size of noncritical input –So total resistance is same g A = 10/9 g B = 2 g total = g A + g B = 28/9 Asymmetric gate approaches g = 1 on critical input But total logical effort goes up

7 Input Order Our parasitic delay model was too simple –Calculate parasitic delay for Y falling If A arrives latest? 2  If B arrives latest? 2.33 

8 a unit CMOS inverter delivers current I in both rising and falling transitions Pseudo-nMOS inverter: pMOS delivers I/3; nMOS delivers 4I/3 (net pull down current is 4I/3 – I/3 = I logical effort g d = (4/3)/3 = 4/9 parasitic delay p d = (6/3)/3 = 6/9 logical effort g u = g d x 1/3 = 4/3 parasitic delay p u = p d x 3 = 18/9 ( only I/3)

9

10

11 4-input NAND unfooted + Hi INV g = 4/3; g = 5/6 G = 20/18=10/9 p = 5/3; p = 2.5/3 = 5/6 4-input NAND footed + Hi INV g = 5/3; g=5/6 G = (5/3)(5/6) = 25/18 p = 6/3 p = 5/6 P = 6/3 + 5/6 = 17/6

12 8-input footed domino AND gate For H > stage is better electrical effort

13 Chapter 1: Chapter 2: Chapter 3: Chapter 4: Chapter 6: Chapter 7: Chapter 11: Flash Memory