final exam review
DESCRIPTION
Final Exam Review. word0 is high if A2 A1 A0 = 000. logical effort of each input is (1+3.5)/3 per wordline output. word7 is high if A2 A1 A0 = 111. 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. - PowerPoint PPT PresentationTRANSCRIPT
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Final Exam Review
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word7 is high if
A2 A1 A0 = 111
word0 is high if
A2 A1 A0 = 000
logical effortof each inputis (1+3.5)/3per wordline output
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Skewed gates
HI-skew gu = 2.5/3 = 5/6
gd = 2.5/1.5 = 5/3
gavg = 5/4
p = 2.5/3 = 5/6
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LO-skew gu = 2/1.5 = 4/3 (unskewed inverter with equal rise time pMOS size 1, nMOS size 0.5
gd = 2/3 (unskewed inverter with equal fall time pMOS size 2, nMOS size 1
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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
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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
• gA = 10/9
• gB = 2
• gtotal = gA + gB = 28/9
• Asymmetric gate approaches g = 1 on critical input• But total logical effort goes up
Areset
Y
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4/3
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reset
A
Y
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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
6C
2C2
2
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B
Ax
Y
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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 gd = (4/3)/3 = 4/9 parasitic delay pd = (6/3)/3 = 6/9
logical effort gu = gd x 1/3 = 4/3 parasitic delay pu = pd x 3 = 18/9 ( only I/3)
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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
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8-input footed domino AND gate
For H > 2.9 4-stage is better
electrical effort
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Chapter 1:
Chapter 2:
Chapter 3:
Chapter 4:
Chapter 6:
Chapter 7:
Chapter 11:
Flash Memory