EE_331F_2008_-_HW5S - EE 331 Devices and Circuits I Problem...

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Unformatted text preview: EE 331 Devices and Circuits I Problem #5 Solution 4.4 (a) ' n " ox Autumn 2008 14 3.9o cm 2 3.9 8.854 x10 F / cm K nC n n 500 9 Tox Tox 100 V sec 50 x10 m cm / m ox F A A 34.5 x 106 2 34.5 2 V sec V V (b) & (c) Scaling the result from part (a) yields ' Kn 34.5x106 ' Kn 34.5 A 50nm V 20nm 2 86.3 A V 2 ' | Kn 34.5 A 50nm V 10nm 2 173 A V 2 ' | Kn 34.5 A 50nm V 2 5nm 345 A V2 4.8 a 0 0.8V I D = 0 ( b ) VGS - VTN = 0.2V , VDS = 0.25V Saturation region 200 A 5m 2 K ' W V I D n GS VTN DS DS V V 1 0.8 40.0 A 2 2 L 2 2 V 0.5m ( c ) VGS - VTN = 1.2V , VDS = 0.25V triode region A 5m V 0.25 'W I D Kn GS VTN DS DS V V 200 2 2 0.8 0.25 538 A L 2 2 V 0.5m ( d ) VGS - VTN = 2.2V , VDS = 0.25V triode region A 5m V 0.25 'W I D Kn GS VTN DS DS V V 200 2 3 0.8 0.25 1.04 mA L 2 2 V 0.5m W A 5m mA e Kn Kn' L 200 V 2 0.5m 2.00 V 2 4.10 a 0 1V cutoff region, I D = 0 b 1V = 1V cutoff region, ID = 0 (c) VGS - VTN = 1V , VDS = 0.1V triode region A 10m V 0.1 'W I D Kn GS VTN DS DS V V 250 2 2 1 0.1 231 A L 2 2 V 1m (d) VGS - VTN = 2V , VDS = 0.1V triode region A 10m V 0.1 'W I D Kn GS VTN DS DS V V 250 2 3 1 0.1 488 A L 2 2 V 1m W A 10m mA e Kn Kn' L 250 V 2 1m 2.50 V 2 1 4.11 Identify the source, drain, gate and bulk terminals and find the current I in the transistors in Fig. P-4.3. W = L 10 1 D G +5 + V (a) S (b) GS +0.2 V I B V + V DS GS - 0.2 V S I V B D + DS W = L 10 1 + +5 G - (a) VGS VG VS 5V VDS VD VS 0.2V Triode region operation A 10 V 0.2 'W I = I D Kn GS VTN DS DS V V 100 2 5 0.70 0 .2 840 A L 2 2 V 1 (b) VGS VG VS 5 0.2 = 5.2V VDS VD VS 0 0.2 0.2V A 10 0.2 I = I S 100 2 5.2 0.70 0 .2 880 A 2 V 1 4.15 a R b R on ' Kn 1 W VGS VTN L 1 23.0 6 100 100 x10 5 0.65 1 on 1 35.7 6 100 100 x10 3.3 0.50 1 4.20 a For VGS 0, VGS VTN and ID 0 b For VGS 1 V , VGS VTN and ID 0 c VGS VTN ID = 2 - 1 = 1V and VDS 3.3 | VDS VGS VTN so the saturation region is correct 375 A 5m A 5m mA 22 'W 375 2 3.75 2 2 1 V 1.88 mA | K n K n 2 2 V 0.5um L V 0.5um V = 3 - 1 = 2V and VDS 3.3 | VDS VGS VTN so the saturation region is correct 375 A 5m 22 3 1 V 7.50 mA 2 2 V 0.5m d VGS VTN ID 4.29 VDS > VGS - VTN so the transistor is saturated. 2 Kn 250 A 2 2 5 0.75 0.0256 2.60 mA 1 VGS VTN 1 VDS 2 2 2V K 250 A 2 2 (b) ID n VGS VTN 5 0.75 2.26 mA 2 2 2V (a) ID 4.32 (a) The transistor is saturated by connection. ID VGS 12 VGS 100 20 ( )(VGS 0.75)2 100k 2 1 1.08V 12 1.08 109.2 A 100k ID (b) ID 12 VGS 100 20 ( )(VGS 0.75)2 (1 0.02VGS ) 100k 2 1 Starting with the solution from part (a) and solving iteratively yields VGS = 1.08 V and IDS = 109.2 A, essentially no change 4.42 (a) VTN 1.5 0.5 4 0.75 0.75 2.16V | VGS < VTN Cutoff & ID 0 (b) ID = 0. The result is independent of VDS . 4.49 ( a ) VGS VTP 1.1 0.75 0.35V | VDS 0.2V Triode region 0.20.2 40.0 A 40A 20 ID 1.1 0.75 2 V 2 1 ( b ) VGS VTP 1.3 0.75 0.55V | VDS 0.2V Triode region 0.20.2 72.0 A 40A 20 1.3 0.75 ID 2 V 2 1 (c ) VTP 0.75 .5 1 .6 6 0.995V VGS VTP 1.1 0.995 0.105V | VDS 0.2V saturation region 1 40A 20 2 ID 2 1.1 0.995 4.41 A 2 V 1 (d ) VGS VTP 1.3 0.995 0.305V | VDS 0.2V triode region 10A 10 0.20.2 32.8 A ID 1.3 0.995 V 2 1 2 3 ...
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This document was uploaded on 01/31/2011.

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