Explanation Given Information: In the MOSFET L , V D D and t o x are scaled by the same factor in order to keep the magnitudes of all electrical fields unchanged. Similarly, V t cannot be scaled. The given table with missing entries is shown below. Calculation: From row 1, scaling factors are determined as follows: 0.5 μm s 1 = 0.25 μm s 1 = 0.5 0.25 s 1 = 2 0.25 μm s 2 = 0.18 μm s 2 = 0.25 0.18 s 2 = 1.388 0.18 μm s 3 = 0.13 μm s 3 = 0.18 0.13 s 3 = 1.385 Row 2: The parameters L and t o x are scaled by the same factor. The values of t o x after scaling is calculated as shown below. ( t o x ) 2 = ( t o x ) 1 s 1 = 10 nm 2 = 5 nm ( t o x ) 3 = ( t o x ) 2 s 2 = 5 nm 1.388 = 3.6 nm ( t o x ) 4 = ( t o x ) 3 s 3 = 3.6 nm 1.385 = 2.6 nm Row 3: The formula for oxide capacitance, C o x is: C o x = ε o x t o x ε o x = 3.45 × 10 − 11 F / m ( Silican dioxide ) t o x = 10 nm C o x = 3.45 × 10 − 11 10 × 10 − 9 = 3.45 × 10 − 3 F / m 2 = 3.45 × 10 − 3 F ( 10 6 μm ) 2 = 3.45 × 10 − 3 × 10 − 12 F μm 2 = 3.45 × 10 − 15 F μm 2 = 3.45 fF μm 2 The values of C o x is inversely proportional to t o x . Thus, it is directly proportional to scaling factor. ( C o x ) 2 = s 1 ( C o x ) 1 = 2 × 3.45 fF μm 2 = 6.9 fF μm 2 ( C o x ) 3 = s 2 ( C o x ) 2 = 1.388 × 6.9 fF μm 2 = 9.58 fF μm 2 ( C o x ) 4 = s 3 ( C o x ) 3 = 1.385 × 9.58 fF μm 2 = 13.27 fF μm 2 Row 4: The process transconductance parameters ( k n ' ) for ( μ n = 500 cm 2 V .s ) is determined as follows: Row 5: The transconductance parameters ( k n ) for ( W L = 10 ) is determined as follows: Row 6: The device area, A ( μm 2 ) after scaling is determined as follows: W L = 10 W 1 L 1 = 10 W 1 0.5 = 10 W 1 = 5 μm A 1 = W 1 × L 1 = ( 5 μm ) × ( 0.5 μm ) = 2.5 μm 2 W 2 L 2 = 10 W 2 0.25 = 10 W 2 = 2.5 μm A 2 = W 2 × L 2 = ( 2.5 μm ) × ( 0.25 μm ) = 0.625 μm 2 W 3 L 3 = 10 W 3 0.18 = 10 W 3 = 1.8 μm A 3 = W 3 × L 3 = ( 1.8 μm ) × ( 0.18 μm ) = 0.324 μm 2 W 4 L 4 = 10 W 4 0.13 = 10 W 4 = 1.3 μm A 4 = W 4 × L 4 = ( 1.3 μm ) × ( 0.13 μm ) = 0.169 μm 2 Row 7: The value of V D D after scaling is determined as follows: ( V D D ) 1 = 5 V ( V D D ) 2 = ( V D D ) 1 s 1 = 5 2 = 2.5 V ( V D D ) 3 = ( V D D ) 2 s 2 = 2.5 1.388 = 1.8 V ( V D D ) 4 = ( V D D ) 3 s 3 = 1.8 1.385 = 1.3 V Row 9: The drain current I D in every case is determined as follows:For first case: V D D = V G S = V D S V G S > V t 5 V > 0.7 V Hence, transistor would not be in cutoff region. It would be either in triode or saturation region. V G D = V G S − V D S = 0 V G D < V t 0 < 0.7 Hence, transistor is in saturation region: The formula for drain current I D is: I D = 1 2 k n ( V G S − V t ) 2 = 1 2 × 1.725 mA V 2 × ( 5 − 0.7 ) 2 = 0.8625 × 18.49 mA = 15.9 mA For second case: V D D = V G S = V D S V G S > V t 2

Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition

7th Edition

ISBN: 9780199339136

Author: Adel S. Sedra, Kenneth C. Smith

Publisher: Oxford University Press

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In the context of MOS devices, state briefly the advantages of:(a) constant field scaling(b) high-κ dielectrics(c) mid-gap metals

Very large-scale integration is the process of creating an integrated circuit by combining millions of MOS transistors or millions BJT transistors onto a single chip . In recent, statistics shows that electronic industries prefer MOSFET then BJT. In your own understanding could you please analyze why VLSI circuits prefer MOSFETs than BJTs?

For the transistor , IS = 4 × 10−16 μA,βF = 75, and βR = 4. (a) Label the collector, base,and emitter terminals of the transistor. (b) What isthe transistor type? (c) Label the emitter-base andcollector-base voltages, and label the normal direction for IE , IC, and IB. (d) Write the simplified formof the transport model equations that apply to thisparticular circuit configuration. Write an expressionfor IE /IB. Write an expression for IE /IC. (e) Findthe values of IE , IC, IB, VC B, and VE B.

Chapter 5 Solutions

Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition

Ch. 5.1 - Prob. 5.1E Ch. 5.1 - Prob. 5.2E Ch. 5.1 - Prob. D5.3E Ch. 5.2 - Prob. 5.4E Ch. 5.2 - Prob. 5.5E Ch. 5.2 - Prob. 5.6E Ch. 5.2 - Prob. 5.7E Ch. 5.3 - Prob. D5.8E Ch. 5.3 - Prob. D5.9E Ch. 5.3 - Prob. D5.10E

Ch. 5.3 - Prob. 5.11E Ch. 5.3 - Prob. 5.12E Ch. 5.3 - Prob. D5.13E Ch. 5.3 - Prob. D5.14E Ch. 5.3 - Prob. 5.15E Ch. 5.4 - Prob. 5.16E Ch. 5.4 - Prob. 5.17E Ch. 5 - Prob. 5.1P Ch. 5 - Prob. 5.2P Ch. 5 - Prob. 5.3P Ch. 5 - Prob. 5.4P Ch. 5 - Prob. D5.5P Ch. 5 - Prob. 5.6P Ch. 5 - Prob. D5.7P Ch. 5 - Prob. 5.8P Ch. 5 - Prob. 5.9P Ch. 5 - Prob. 5.10P Ch. 5 - Prob. 5.11P Ch. 5 - Prob. 5.12P Ch. 5 - Prob. 5.13P Ch. 5 - Prob. 5.14P Ch. 5 - Prob. 5.15P Ch. 5 - Prob. 5.16P Ch. 5 - Prob. 5.17P Ch. 5 - Prob. 5.18P Ch. 5 - Prob. 5.19P Ch. 5 - Prob. D5.20P Ch. 5 - Prob. 5.21P Ch. 5 - Prob. 5.22P Ch. 5 - Prob. 5.23P Ch. 5 - Prob. 5.24P Ch. 5 - Prob. 5.25P Ch. 5 - Prob. 5.26P Ch. 5 - Prob. 5.27P Ch. 5 - Prob. 5.28P Ch. 5 - Prob. 5.29P Ch. 5 - Prob. 5.30P Ch. 5 - Prob. 5.31P Ch. 5 - Prob. D5.32P Ch. 5 - Prob. D5.33P Ch. 5 - Prob. 5.34P Ch. 5 - Prob. 5.35P Ch. 5 - Prob. D5.36P Ch. 5 - Prob. 5.37P Ch. 5 - Prob. 5.38P Ch. 5 - Prob. 5.39P Ch. 5 - Prob. 5.40P Ch. 5 - Prob. 5.41P Ch. 5 - Prob. 5.42P Ch. 5 - Prob. 5.43P Ch. 5 - Prob. D5.44P Ch. 5 - Prob. 5.45P Ch. 5 - Prob. D5.46P Ch. 5 - Prob. 5.47P Ch. 5 - Prob. D5.48P Ch. 5 - Prob. D5.49P Ch. 5 - Prob. D5.50P Ch. 5 - Prob. D5.51P Ch. 5 - Prob. 5.52P Ch. 5 - Prob. D5.53P Ch. 5 - Prob. 5.54P Ch. 5 - Prob. 5.55P Ch. 5 - Prob. 5.56P Ch. 5 - Prob. 5.57P Ch. 5 - Prob. 5.58P Ch. 5 - Prob. 5.59P Ch. 5 - Prob. 5.60P Ch. 5 - Prob. 5.61P Ch. 5 - Prob. 5.62P Ch. 5 - Prob. 5.63P Ch. 5 - Prob. 5.64P Ch. 5 - Prob. 5.65P Ch. 5 - Prob. 5.66P Ch. 5 - Prob. 5.67P

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