EE_331F_2007_-_HW6S

# EE_331F_2007_-_HW6S - EE 331 Devices and Circuits I Problem...

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EE 331 Devices and Circuits I Autumn 2007 Problem Set #6 Solution 4.85 a () V GG = 100 k Ω 100 k Ω+ 220 k Ω 12 V = 3.75 V | Assume saturation 3.75 = V GS + 24 x 10 3 I D = V GS + 24 x 10 3 100 x 10 6 2 5 1 V GS 1 2 6 V GS 2 11 V GS + 2.25 = 0 V GS = 1.599 V and I D = 89.7 μ A V DS = 12 36 x 10 3 I D = 8.77 V V DS > V GS V TN Saturation is correct. Checking : V GG = 24 x 10 3 I D + V GS = 3.75 V which is correct. Q po int : 89.7 A , 8.77 V 4.90 (a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields I D = 89.6 μ A, V GS = 1.60 V and V DS = 8.77 V 4 (a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields I D = 96.3 μ A, V GS = 1.44 V and V DS = 8.53 V 4 4.97 K n ' = 100 A / V 2 V TN = 0.75 V C hoose V DS = V R D = V R S = 3 V and V GS V TN = 1 V R S = 3 0.25 mA = 12 k Ω R D = 3 0.25 mA = 12 k Ω V GS V TN = 2 I D K n = 1 V and K n = 2 I D 1 V 2 = 500 A / V 2 W L = 5 1 V G = V S + V GS = 3 + 1 + 0.75 = 4.75 V 4.75 V = R 1 R 1 + R 2 9 V 4.75 V = R 1 R 2 R 1 + R 2 9 V R 2 R 2 = 250 k Ω 9 V 4.75 V ⎟ = 473 k Ω⇒ 470 k Ω 4.75 V = R 1 R 1 + 470 k Ω 9 V R 1 = 525 k 510 k Ω R 1 = 510 k Ω , R 2 = 470 k Ω R S = 12 k Ω R D = 12 k Ω W L = 5 1

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4.99 Assume Saturation. For I G =0, V GS =− 10 4 I D 10 4 10 3 2 V GS + 5 () 2 5 V GS 2 + 51 V GS + 125 = 0 V GS 4.10 V and I D = 410 μ A V DS = 15 15000 I D = 8.85 V | V DS > V GS V TN so saturation is ok.
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EE_331F_2007_-_HW6S - EE 331 Devices and Circuits I Problem...

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