{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EE_331F_2008_-_HW5S

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online