Download e-book for kindle: Complex analysis and geometry : proceedings of the by Vincenzo Ancona; E Ballico; Alessandro Silva; Centro

By Vincenzo Ancona; E Ballico; Alessandro Silva; Centro internazionale per la ricerca matematica (Trento, Italy)

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8 V c. 8 V Logic “1” level degrades as it goes through additional logic gates. 50 a. V01 = V02 = 5 V b. 2 V c. 2 V Logic “0” signal degrades as it goes through additional logic gates. 11 (a) VI = 4 V, Driver in Non ⋅ Sat. 13 (a) For VI = 5 V, Load in saturation and driver in nonsaturation. 16 a. 40 30 V1 = V2 = 5 V I R = I D1 = b. 23 Assume the transistor is biased in the saturation region. 33 Yes, the transistor is biased in the saturation region. 25 VS = −VGS . 5 K nD /K nL = 5 + K nD /K nL b.

0 For vI < 0, both diodes forward biased 0 − vI . At vI = −10 V , i1 = −1 mA 10 v −3 For vI > 3, i1 = I . 5 V For vI = 30 V, i = 10 ␯I(V) b. 29 a. 4 b. 31 One possible example is shown. 7 V. Power ratings depends on number of pulses per second and duration of pulse. 33 C ␯I ␯O ϩ Vx Ϫ a. For Vγ = 0 b. 37 a. 93 V b. 5 ) I D1 = c. d. 5 ) I D1 = I D 2 = b. 38 a. 07 V c. 82 V d. 39 a. 6 mA I= I D1 I D3 b. 4 V c. 4 V d. 6, D1 turns on. 6, D2 turns on. 06 For vI > 0. 43 a. 86 mA I D1 = I D1 b. 47 10 K V1 Va Ϫ VD ϩ 10 K ID 10 K Vb V2 10 K a.

2 V Logic “0” signal degrades as it goes through additional logic gates. 11 (a) VI = 4 V, Driver in Non ⋅ Sat. 13 (a) For VI = 5 V, Load in saturation and driver in nonsaturation. 16 a. 40 30 V1 = V2 = 5 V I R = I D1 = b. 23 Assume the transistor is biased in the saturation region. 33 Yes, the transistor is biased in the saturation region. 25 VS = −VGS . 5 K nD /K nL = 5 + K nD /K nL b. 2 K nD /K nL = For VI = 5, driver in nonsaturated region. 8 V Transistor is biased in the nonsaturation region.

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