VLSI solution - Solutions Solutions for CMOS VLSI Design...

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Solutions 1 Solutions for CMOS VLSI Design 4th Edition. Last updated 12 May 2010. Chapter 1 1.1 Starting with 100,000,000 transistors in 2004 and doubling every 26 months for 12 years gives transistors. 1.3 Let your imagination soar! 1.5 1.7 10 8 2 12 12 26 --------------- ⎝⎠ ⎛⎞ 4.6B A B C D Y AY (a) A B Y (b) A B Y (c) (d) A C B Y
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SOLUTIONS 2 1.9 1.11 The minimum area is 5 tracks by 5 tracks (40 λ x 40 λ = 1600 λ 2 ). 1.13 1.15 This latch is nearly identical save that the inverter and transmission gate feedback A0 A0 A1 A1 Y0 Y1 Y2 Y3 (a) Y1 Y0 A0 A1 A1 A0 A2 (b) n+ n+ p substrate p+ p+ n well A Y VDD n+ GND B
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CHAPTER 2 SOLUTIONS 3 has been replaced by a tristate feedaback gate. 1.17 (c) 5 x 6 tracks = 40 λ x 48 λ = 1920 λ 2 . (with a bit of care) (d-e) The layout should be similar to the stick diagram. 1.19 20 transistors, vs. 10 in 1.16(a). 1.21 The Electric lab solutions are available to instructors on the web. The Cadence labs include walking you through the steps. Chapter 2 Y D CLK CLK CLK CLK (b) AB C A VDD GND BC F D A D (a) D F A Y B A C B C
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SOLUTIONS 4 2.1 2.3 The body effect does not change (a) because V sb = 0. The body effect raises the threshold of the top transistor in (b) because V sb > 0. This lowers the current through the series transistors, so I DS 1 > I DS 2 . 2.5 The minimum size diffusion contact is 4 x 5 λ , or 1.2 x 1.5 μ m. The area is 1.8 μ m 2 and perimeter is 5.4 μ m. Hence the total capacitance is At a drain voltage of VDD, the capacitance reduces to 2.7 The new threshold voltage is found as The threshold increases by 0.96 V. () 14 2 8 3.9 8.85 10 350 120 / 100 10 ox WW W CA V LL L βμ μ ⎛⎞ •⋅ == = ⎜⎟ ⎝⎠ 0 1 2 3 4 5 0 0.5 1 1.5 2 2.5 V ds I ds (mA) V gs = 5 V gs = 4 V gs = 3 V gs = 2 V gs = 1 C db 0V 1.8 0.42 5.4 0.33 + 2.54fF C 5V 1.8 0.42 1 5 0.98 ---------- + 0.44 5.4 0.33 1 5 0.98 + 0.12 + 1.78fF φ γ s V = = = •• 2 0 026 21 0 145 10 085 100 10 39 885 10 17 10 8 14 (. ) l n . . .. 216 10 117 885 10 2 10 075 07 19 14 17 1 2 . . . / = =+ + −− V V ts γφ 4 1 6 6 = s V .
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CHAPTER 3 SOLUTIONS 5 2.9 The threshold is increased by applying a negative body voltage so V sb > 0. 2.11 The nMOS will be OFF and will see V ds = V DD , so its leakage is 2.13 Assume V DD = 1.8 V. For a single transistor with n = 1.4, For two transistors in series, the intermediate voltage x and leakage current are found as: In summary, accounting for DIBL leads to more overall leakage in both cases. However, the leakage through series transistors is much less than half of that through a single transistor because the bottom transistor sees a small Vds and much less DIBL. This is called the stack effect . For n = 1.0, the leakage currents through a single transistor and pair of transistors are 13.5 pA and 0.9 pA, respectively. 2.15 V IL = 0.3; V IH = 1.05; V OL = 0.15; V OH = 1.2; NM H = 0.15; NM L = 0.15 2.17 Either take the grungy derivative for the unity gain point or solve numerically for V IL = 0.46 V, V IH = 0.54 V, V OL = 0.04 V, V OH = 0.96 V, NM H = NM L = 0.42 V. 2.19 Take derivatives or solve numerically for the unity gain points: V IL = 0.43 V, V IH = 0.50 V, V OL = 0.04 V, V OH = 0.97 V, NM H = 0.39, NM L = 0.47 V.
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This note was uploaded on 01/15/2012 for the course ECE260A 660090 taught by Professor Bendak,michaelbeshara during the Fall '09 term at UCSD.

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VLSI solution - Solutions Solutions for CMOS VLSI Design...

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