Solutions_3 - EE 310 Problem Set No. 3 Solutions 2.50 (a)...

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Unformatted text preview: EE 310 Problem Set No. 3 Solutions 2.50 (a) (i) I 5 V, D1 and D 2 on 5 0.6 5 O 0.6 O O O 5 5 0.5 0.5 0.88 1.0 1.2 0.20 0.20 2.0 2.0 O I O 0.7 V (ii) O 5V 0.5 0.5 5 I 0.455 V (b) (i) (ii) I I 5 V, 5 V, O O 4.4 V 0.6 V 2.59 (a) For (b) For I D1 O I I 0.5 V, I D1 I D2 I D3 I D3 0, 0 O 0.5 V 1.5 V, D1 on; I D 2 1.5 0.7 0.0667 mA 48 0.7 0.0667 8 1.23 V I (c) For 3 3 V, D1 and D 2 conducting, I D 3 0 0.7 1.7 O O O 4 8 6 0.75 0.0875 0.2833 O 0.25 0.125 0.1667 2.069 0.7 8 2.069 1.7 6 0.171 mA 0.0615 mA O 2.069 V Then I D1 I D2 (d) For 5 I O 5 V, O all diodes conducting 0.7 O 1.7 O 2.7 4 0.25 O 0.25 0.125 0.1667 4 8 6 1.25 0.0875 0.2833 0.675 So O 2.90 V 2.90 0.7 0.275 mA 8 2.90 1.7 0.20 mA 6 2.90 2.7 0.05 mA 4 Then I D1 I D2 I D3 EE 310 Problem Set No. 3 Solutions 3.3 (a) Enhancement-mode (b) From Graph VT = 1.5 V Now 0.03 0.15 0.39 0.77 K n 2 1.5 K n 3 1.5 K n 4 1.5 K n 5 1.5 2 2 2 2 0.25 K n 2.25 K n 6.25 K n 12.25 K n Kn Kn Kn Kn iD (sat) iD (sat) 0.12 0.0666 0.0624 0.0629 2 From last three, K n (Avg) 0.0640 mA/V (c) iD (sat) iD (sat) 0.0640(3.5 1.5) 2 0.0640(4.5 1.5) 2 0.256 mA for VGS 0.576 mA for VGS 3.5 V 4.5 V 3.5 (a) V DS sat 2.2 V DS VGS V TN 2.2 0.4 1.8 V 1.8 Saturation (b) V DS sat VGS VTN 1 0.4 0.6 V V DS 0.6 1 0.4 V V DS sat 0.6 V (c) VGS 1 1 0 Cutoff V DS sat Nonsaturation 3.10 (a) I D 0.6 kn W VGS 2L 0.12 2 VTN 2 W 1.4 0.8 L 2 W L 27.8 Or W (b) I D 27.8 0.8 22.2 m 0 .4 2 0.12 27.8 2 1.4 0.8 0.4 2 0.534 mA (c) V DS sat VGS V TN 1.4 0.8 0.6 V EE 310 Problem Set No. 3 Solutions 3.17 Kp kp W 2L 50 12 2 0.8 Kp 0.375 mA/V 2 (a) Nonsaturation ID 0.375 2 2 0.5 0.2 0.2 2 0.21 mA (b) Nonsaturation ID ID 0.375 2 2 0.5 0.8 0.375 2 2 0.5 1.2 0.8 1.2 2 0.66 mA 0.81 mA (c) Nonsaturation 2 (d) Saturation ID ID 0.375 2 0.5 0.375 2 0.5 2 0.844 mA 0.844 mA (e) Saturation 2 3.20 VGS r0 VGS r0 VA 2 V, I D 1 ID 0.2 2 1.2 r0 2 2 0.128 mA 781 k 1.57 mA 1 0.01 0.128 4 V, I D 1 0.01 1.57 1 1 0.01 0.2 4 1.2 r0 VA 63.7 k 100 V 3.28 0.8 0.12 80 VGS 2 0.4 2 V GS 0.808 V VGS 1 Rin V DD R1 1 200 1.8 R1 R1 R2 445 k 363 k 0.80825 R1 R 2 Rin 200 k EE 310 Problem Set No. 3 Solutions 3.30 VG 22 6 22 8 3 1.40 V 3 K p R S V SG 2 SG VTP 2 V SG VG 3 ID V SD 2 0.5 0.5 V SG 1.6V SG 0.64 V SG 1.40 0.25V 0.6V SG 1.44 2 0 V SG 4.72 V 1.483 V 0.5 1.483 0.8 6 0.2332 mA 0.2332 0.5 5 3.33 ID V DS 0.2 0.2 mA 1 1.8 0.2 4 1 0.8 0.4 VTN VGS 0.8 V 0.4 V 0.4 VGS VTN 2 Now V DS sat VDS sat ID 0.2 0.4 VGS 0.8 V kn W VGS 2L 0.12 2 W L 0.8 0.4 2 W L 20.8 Now VG VGS I D R S VG 1 1 Rin V DD R1 0.8 0.2 1 1.0 V R1 360 k 1 200 1.8 R1 R2 450 k R1 R 2 Rin 200 k 3.37 IQ 50 5 1 5 500 VGS 1.2 1.516 2 VGS 2 1.516 V (a) (i) VDS IQ VDS 6.516 V VGS 7.61 V 2.61 V 0.5 VGS 1.2 2.61 VDS (ii) VDS (b) (i) Same as (a) (ii) VGS VDS VGS VDS 1.516 V 2.61 V EE 310 Problem Set No. 3 Solutions 3.38 ID 0.25 VGS VD K n VGS VTN 0.2 VGS 0.25 0.6 0.2 9 0.25 24 2 0.6 2 VGS VD 1.72 V 3V VS 1.72 V ...
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This note was uploaded on 04/26/2010 for the course E E 310 taught by Professor Wharton,markjedwards,perrysto during the Spring '10 term at Pennsylvania State University, University Park.

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