HW6 - 4.28 760 − 140 µA ∆iD = = 310 µS | As a check,...

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Unformatted text preview: 4.28 760 − 140 µA ∆iD = = 310 µS | As a check, we can use the results from Problem 4.22. 5−3 V ∆vGS µA gm = K n (VGS − VTN ) = 125 2 (4 − 1.5)V = 313 µS V ∆i 390 − 15 µA µA = 188 µS | Checking : gm = 125 2 (3 − 1.5)V = 188 µS (b) gm = D = V ∆vGS 4−2 V (a) gm = 5.74 V IC = β F I B = β FO 1 + CE I B | We need two Q - points from the output characteristics. VA For example : ( mA, 14 V) and 10 (5 mA, 5 V) 14 5 10mA = β FO 1 + 0.1mA and 5mA = β FO 1 + 0.06 mA yields VA VA 14 5 100 = β FO 1 + and 83.3 = β FO 1 + . Solving these two equations yields VA VA β FO = 72.9 and VA = 37.6 V . 13.15 NPN Common-Emitter Amplifier +18 V 270 k Ω 360 kΩ = 5.84 V 360 kΩ + 750 kΩ = R1 R2 = 360 kΩ 750 kΩ = 243 kΩ VEQ = 18V REQ R EQ Q ( 1 V I B = 0.245 µA | IC = 90 I B = 22.0 µA EQ 5.84 V ) 5.84 = 243 x103 I B + 0.7 + 91 228.2 x103 I B 243 k Ω 228.2 k Ω VCE = 18 − 2.7 x105 IC − 2.28 x105 I E = 6.99 V Q − point : (22.0 µA, 6.99 V ) 13.31 PMOS Common-Gate Amplifier 12 + VGS 200 x10−6 2 = (VGS − 1) 33000 2 = −0.84V | ID = 338 µA ID = VGS VDS = −(12 − 33000 ID − 22000 ID + 12) M1 +12 V 33 k Ω 22 k Ω -12 V VDS = −5.41 V Q − point : (338 µA, - 5.41 V ) 13.63 g m = 40( µA)= 40µS | rπ = 1 40 50 + 1.5 V = 1 MΩ | ro = = 51.5 MΩ 40µS 1 µA RBB = RB rπ = 5 MΩ 1 MΩ = 833kΩ 833kΩ Av = (−40µS ) 51.5 MΩ 1.5 MΩ 3.3 MΩ = −40.0 10 kΩ + 833kΩ ( ) 13.98 For the bias network : VEQ = 10V 430 kΩ = 4.343V | REQ = 430 kΩ 560 kΩ = 243kΩ 430 kΩ + 560 kΩ 2 5 x10−4 (VGS − 1) | VGS = 4.343 − 2 x104 I D → VGS = 1.72 V | I D = 131 µA 2 = 10 − 63kΩ( µA)= 1.75V ≥ VGS − VTN so active region assumption is ok. 131 ID = VDS 1 + 1.75 0.0133 131 = 586 kΩ g m = 2 5 x10−4 ( µA) = 362µS | ro = 131µA 243kΩ Av = − (362µS ) 586kΩ 43kΩ 100kΩ = −10.3 243kΩ + 1kΩ ( ) ( ) ...
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HW6 - 4.28 760 − 140 µA ∆iD = = 310 µS | As a check,...

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