Chapter04

Microelectronic Circuit Design

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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 " = ε ox T ox = 3.9 o T ox = 3.9 8.854 x 10 14 F / cm () 50 x 10 9 m 100 cm / m = 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 = s 2 s qN B 0.75 V = 11.7 8.854 x 10 14 F / cm 2 11.7 8.854 x 10 14 F / cm 1.602 x 10 19 10 15 / cm 3 0.75 V = 10.5 x 10 9 F / cm 2 4.4 (a) K n ' = μ n C ox " = n ox T ox = n 3.9 o T ox = 500 cm 2 V sec 3.9 8.854 x 10 14 F / cm ( ) 50 x 10 9 m 100 cm / m K n ' = 34.5 x 10 6 F V sec = 34.5 x 10 6 A V 2 = 34.5 A V 2 (b) & (c) Scaling the result from part (a) yields K n ' = 34.5 A V 2 50 nm 20 nm = 86.3 A V 2 K n ' = 34.5 A V 2 50 nm 10 nm = 173 A V 2 K n ' = 34.5 A V 2 50 nm 5 nm = 345 A V 2 4.5 a Q " = C ox " V GS V TN = ox T ox V GS V TN = 3.9 8.854 x 10 14 F / cm ( ) 25 x 10 9 m 100 cm / m 1 V = 1.38 x 10 7 C cm 2 b Q " = C ox " V GS V TN = ox T ox V GS V TN = 3.9 8.854 x 10 14 F / cm 10 x 10 9 m 100 cm / m 2 V = 6.91 x 10 7 C cm 2 75
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4.6 a () v n =− μ n E 500 cm 2 V s 2000 V cm 1.00 x 10 6 cm s a v n n E 400 cm 2 V s 4000 V cm 1.60 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 0 < 0.8 V I D =0 ( b ) V GS - V TN .2 V , V DS .25 V Saturation region I D = K n ' 2 W L V GS V TN V DS 2 V DS = 200 2 A V 2 5 m 0.5 m 1 0.8 2 = 40.0 A ( c ) V GS - V TN =1.2 V V DS V triode region I D = K n ' W L V GS V TN V DS 2 V DS = 200 A V 2 5 m 0.5 m 2 0.8 0.25 2 0.25 = 538 A ( d ) V GS - V TN =2 V V DS V triode region I D = K n ' W L V GS V TN V DS 2 V DS = 200 A V 2 5 m 0.5 m 3 0.8 0.25 2 0.25 = 1.04 mA e K n = K n ' W L = 200 A V 2 5 m 0.5 m = 2.00 mA V 2 4.9 a K n = K n ' W L = 200 A V 2 60 m 3 m = 4.00 mA V 2 b K n = 200 A V 2 3 m 0.15 m = 4.00 mA V 2 | c K n = 200 A V 2 10 m 0.25 m = 8.00 mA V 2 4.10 a 0 < 1 V cutoff region, I D = 0 b ( ) 1V = 1V cutoff region, I D (c) V GS - V TN =1 V V DS .1 V triode region I D = K n ' W L V GS V TN V DS 2 V DS = 250 A V 2 10 m 1 m 2 1 0.1 2 0.1 =+ 231 A 76
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(d) V GS - V TN =2 V , V DS =0 .1 V triode region I D = K n ' W L V GS V TN V DS 2 V DS = 250 μ A V 2 10 m 1 m 3 1 0.1 2 0.1 () =+ 488 A e K n = K n ' W L = 250 A V 2 10 m 1 m = 2.50 mA V 2 4.11 Identify the source, drain, gate and bulk terminals and find the current I in the transistors in Fig.
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This note was uploaded on 03/05/2008 for the course EE 307 taught by Professor Braun during the Spring '08 term at Cal Poly.

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Chapter04 - CHAPTER 4 4.1 (a) VG > VTN corresponds to the...

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