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### Solutions_3

Course: E E 310, Spring 2011
School: Penn State
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Word Count: 516

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310 EE Problem Set No. 3 Solutions 2.50 (a) (i) I = 5 V, D1 and D 2 on 5 - ( O + 0.6 ) 5 - O O O - 0.6 + = + 5 5 0.5 0.5 0.88 + 1.0 + 1.2 = O (0.20 + 0.20 + 2.0 + 2.0 ) O = 0.7 0.5 I = -0.455 V 0.5 + 5 V (ii) I = -5 V O = (b) (i) I = 5 V, O = 4.4 V (ii) I = -5 V, O = -0.6 V 2.59 (a) For I = 0.5 V, I D1 = I D 2 = I D 3 = 0 , O = 0.5 V (b) For I = 1.5 V, D1 on; I D 2 = I D 3 = 0 1.5 - 0.7 =...

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