Cmos Vlsi Design 4th Ed Solution Manual
Solution of CMOS VLSI Design 4th (Odd). 1. SolutionsSolutions for CMOS VLSI Design 4th Edition. Last updated 12 May 2010.Chapter 11.1 Starting with 100,000,000 transistors in 2004 and doubling every 26 months for 12 ⎛ 12 ⋅ 12⎞ -8 ⎝ 26 ⎠ years gives 10. 2 ≈ 4.6B transistors.1.3 Let your imagination soar!1.5 A B C D Y1.7 A Y A Y (a) (b) B A A B Y Y (c) B C (d) 1. 2 SOLUTIONS 1.9 A1A1 A0A0 Y0 Y1 A0 A1 Y0 A2 Y2 A1 Y3 Y1 A0 (a) (b) 1.11 The minimum area is 5 tracks by 5 tracks (40 λ x 40 λ = 1600 λ2).
1.13 B A GND Y VDD n+ n+ n+ p+ p+ n well p substrate 1.15 This latch is nearly identical save that the inverter and transmission gate feedback. CHAPTER 2 SOLUTIONS 3 has been replaced by a tristate feedaback gate. CLK D Y CLK CLK CLK1.17 VDD A B C D A D B C F F C D A B GND (a) (b) (c) 5 x 6 tracks = 40 λ x 48 λ = 1920 λ2. (with a bit of care) (d-e) The layout should be similar to the stick diagram.1.19 20 transistors, vs. Mazda protege 2018 service manual. 10 in 1.16(a).
A B A C Y B C1.21 The Electric lab solutions are available to instructors on the web. The Cadence labs include walking you through the steps.Chapter 2. 4 SOLUTIONS 2.1 W ⎛ 3.9. 8.85 ⋅ 10−14 ⎞ ⎛ W ⎞ W β = μCox = ( 350 ) ⎜ −8 ⎟⎜ ⎟ = 120 μ A / V 2 L ⎝ 100 ⋅ 10 ⎠⎝ L ⎠ L 2.5 Vgs = 5 2 1.5 Vgs = 4 Ids (mA) 1 Vgs = 3 0.5 Vgs = 2 Vgs = 1 0 0 1 2 3 4 5 Vds 2.3 The body effect does not change (a) because Vsb = 0. The body effect raises the threshold of the top transistor in (b) because Vsb 0. This lowers the current through the series transistors, so IDS1 IDS2. 2.5 The minimum size diffusion contact is 4 x 5 λ, or 1.2 x 1.5 μm.
Cmos Vlsi Design 4th Edition Solution Manual
The area is 1.8 μm2 and perimeter is 5.4 μm. Hence the total capacitance is C db ( 0V ) = ( 1.8 ) ( 0.42 ) + ( 5.4 ) ( 0.33 ) = 2.54fF At a drain voltage of VDD, the capacitance reduces to – 0.44 – 0.12 C db ( 5V ) = ( 1.8 ) ( 0.42 ) ⎛ 1 + - ⎞ + ( 5.4 ) ( 0.33 ) ⎛ 1 + - ⎞ 5 5 -= 1.78fF ⎝ 0.98⎠ ⎝ 0.98⎠ 2.7 The new threshold voltage is found as 2.
1017 φs = 2(0.026) ln = 0.85V 1.45. 1010 100. 10−8 γ= −14 2 (1.6.
10−19 ) (11.7. 8.85. 10−14 ) ( 2. 1017 ) = 0.75V 1/ 2 3.9. 8.85. 10 Vt = 0.7 + γ ( φs + 4 − φs = 1.66V ) The threshold increases by 0.96 V.
Cmos Vlsi Design Solutions
CHAPTER 3 SOLUTIONS 52.9 The threshold is increased by applying a negative body voltage so Vsb 0.2.11 The nMOS will be OFF and will see Vds = VDD, so its leakage is −Vt I leak = I dsn = β v e e 2 1.8 nvT T = 69 pA2.13 Assume VDD = 1.8 V. For a single transistor with n = 1.4, −Vt +ηVDD I leak = I dsn = β v e e2 1.8 T nvT = 499 pA For two transistors in series, the intermediate voltage x and leakage current are found as: −Vt +η x η (VDD − x ) −Vt − x ⎛ −x ⎞ I leak = β v e e2 1.8 nvT ⎜ 1 − e ⎟ = β vT e e vT 2 1.8 nvT T ⎜ ⎟ ⎝ ⎠ −Vt +η x ⎛ −x ⎞ η (VDD nvx )−Vt − x − ⎜1 − e ⎟ = e nvT vT e T ⎜ ⎟ ⎝ ⎠ x = 69 mV; I leak = 69 pA In summary, accounting for DIBL leads to more overall leakage in both cases.
Cmos Vlsi Design Weste
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.