EE_331F_2009_-_HW5S

EE_331F_2009_-_HW5S - EE 331 Devices and Circuits I Problem...

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

View Full Document Right Arrow Icon
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 2009 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 1 (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 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 2 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. 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: 12 VGS 100 106 20 ( )(VGS 0.75)2 10k 2 1 2 100VGS 149VGS 44.25 0 VGS 1.08V and I D 1.08 A (b) 12 VGS 100 106 20 ( )(VGS 0.75)2 (1 0.02VGS ) 10k 2 1 Starting with the solution from part (a) and solving iteratively yields VGS 1.075V and I D 1.09 A 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 3 ( a ) VGS VTP 1.1 0.75 0.35V | VDS 0.2V Triode region 0.20.2 40.0 A 40A 20 1.1 0.75 ID 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 2 V 1 2 4 ...
View Full Document

This note was uploaded on 12/13/2009 for the course EE 331 taught by Professor Donthavemyschool during the Spring '09 term at Vermont Tech.

Ask a homework question - tutors are online