ch04_solutions

ch04_solutions - CHAPTER 4 4.1 (a) VG > VTN corresponds...

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49 CHAPTER 4 4.1 (a) V G > V TN corresponds to the inversion region (b) V G << V TN corresponds to the accumulation region (c) V G < V TN corresponds to the depletion region 4.2 (a) C ox " = e ox T ox = 3.9 e o T ox = 3.9 8.854 x 10 - 14 F / cm ( 29 50 x 10 - 9 m 100 cm / m ( 29 = 6.91 x 10 - 8 F cm 2 = 69.1 nF cm 2 (b), (c) & (d): Scaling the result from part (a) yields C ox " = 69.1 nF cm 2 50 nm 20 nm = 173 nF cm 2 | C ox " = 69.1 nF cm 2 50 nm 10 nm = 346 nF cm 2 | C ox " = 69.1 nF cm 2 50 nm 20 nm = 691 nF cm 2 4.3 C d = e s 2 e s qN B 0.75 V ( 29 = 11.7 8.854 x 10 - 14 F / cm ( 29 2 11.7 ( 29 8.854 x 10 - 14 F / cm ( 29 1.602 x 10 - 19 10 15 / cm 3 ( 29 0.75 V ( 29 = 10.5 x 10 - 9 F / cm 2 4.4 (a) K n ' = m n C ox " = m n e ox T ox = m n 3.9 e o T ox = 500 cm 2 V - sec 3.9 8.854 x 10 - 14 F / cm ( 29 50 x 10 - 9 m 100 cm / m ( 29 K n ' = 34.5 x 10 - 6 F V - sec = 34.5 x 10 - 6 A V 2 = 34.5 m A V 2 (b) & (c) Scaling the result from part (a) yields K n ' = 34.5 m A V 2 50 nm 20 nm = 86.3 m A V 2 | K n ' = 34.5 m A V 2 50 nm 10 nm = 173 m A V 2 | K n ' = 34.5 m A V 2 50 nm 5 nm = 345 m A V 2 4.5 (a) a ( 29 Q " = C ox " V GS - V TN ( 29 = e ox T ox V GS - V TN ( 29 1 V ( 29 = 3.9 8.854 x 10 - 14 F / cm ( 29 25 x 10 - 9 m 100 cm / m ( 29 1 V ( 29 = 1.38 x 10 - 7 C cm 2 b ( 29 Q " = C ox " V GS - V TN ( 29 = e ox T ox V GS - V TN ( 29 1 V ( 29 = 3.9 8.854 x 10 - 14 F / cm ( 29 10 x 10 - 9 m 100 cm / m ( 29 2 V ( 29 = 6.91 x 10 - 7 C cm 2
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50 4.6 a ( 29 v n = - m n E = - 500 cm 2 V - s 1000 V cm = - 5.00 x 10 5 cm s a ( 29 v n = - m n E = - 400 cm 2 V - s 3000 V cm = - 1.20 x 10 6 cm s 4.7 The carrier velocity must increase as the carriers travel down the channel to compensate for the decrease in carrier density 4.8 a ( 29 K n = K n ' W L = 200 m A V 2 60 m m 3 m m = 4.00 mA V 2 b ( 29 K n = 200 m A V 2 3 m m 0.15 m m = 4.00 mA V 2 | c ( 29 K n = 200 m A V 2 10 m m 0.25 m m = 8.00 mA V 2 4.9 a ( 29 0 < 1.5 V I D = 0 b ( 29 1V <1.5V I D = 0 (c) V GS - V TN = 0.5 V , V DS = 0.1 V triode region I D = K n ' W L V GS - V TN - V DS 2 V DS = 250 m A V 2 10 m m 1 m m 2 - 1.5 - 0.1 2 0.1 ( 29 = + 113 m A (d) V GS - V TN =1.5 V , V DS = 0.1 V triode region I D = K n ' W L V GS - V TN - V DS 2 V DS = 250 m A V 2 10 m m 1 m m 3 - 1.5 - 0.1 2 0.1 ( 29 = + 363 m A e ( 29 K n = K n ' W L = 250 m A V 2 10 m m 1 m m = 2.50 mA V 2 4.10 a ( 29 0 < 1.0 V I D = 0 b ( 29 1V =1V I D = 0 (c) V GS - V TN =1 V , V DS = 0.1 V triode region I D = K n ' W L V GS - V TN - V DS 2 V DS = 250 m A V 2 5 m m 0.5 m m 2
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This note was uploaded on 05/02/2008 for the course ECE 220 taught by Professor Cockrum during the Spring '08 term at Cal Poly Pomona.

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ch04_solutions - CHAPTER 4 4.1 (a) VG &gt; VTN corresponds...

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