Key F06 Homework 7

Key F06 Homework 7 - ECET 157 F 06 Homework Set 7 Key 14-5...

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Unformatted text preview: ECET 157 F 06 Homework Set 7 Key 14-5 a. VC init = 20 V, b. τ = (1 kO + 1 kO ) × 1 µF = 2 ms, c. VC init = 20 V, IC init = − 20 V = −10 mA (direction opposite reference) with C modeled as E of 20 V, 1 kO + 1 kO d. VC ss = 0 V, I C ss = 0 mA, e. VC (t ) = 0 V + (20 − 0 )V × e − t/2 ms = 20 V·e− t / 2 ms, iC (t ) = 0 mA + (- 10 mA − 0 mA ) × e − t/2 ms = −10 mA·e− t / 2 ms , f. VC (4 ms ) = 20 V × e -4 ms/2 ms = 2.71 V, iC (4 ms ) = −10 mA × e −4 ms/2 ms = −1.35 mA, g. exponential fall 20 V to 0 V in 10 ms, exponential rise −10 mA to 0 mA in 10 ms 14-9 a. τ = 10 kO × 0.1 µF = 1 ms, b. iC (t ) = 0 mA + (6 mA − 0 mA )× e − t/1 ms = 6 mA·e− t / 1 ms, c. I C (τ ) = 6 mA × e −1 = 2.21 mA, I C (2τ ) = 6 mA × e −2 = 812 µA, I C (3τ ) = 6 mA × e −3 = 299 µA, I C 4(τ ) = 6 mA × e −4 = 110 µA, I C (5τ ) = 6 mA × e −5 = 40 µA, 1 mA = 1.79 ms, 6 mA d. iC (500 µs ) = 6 mA × e −500 µs/1 ms = 3.64 mA, e. t = −1 ms × 1n f. exponential fall 6 mA to 0 mA in 5 ms and stays at 0 mA 14-10 a. τ = 100 kO × 0.01 µF = 1 ms, b. VC (t ) = 0 V + (10 − 0 )V × e − t/1 ms = 10 V·e− t / 1 ms, c. VC (τ) = 10 V × e −1 = 3.68 V, VC (2 τ ) = 10 V × e −2 = 1.35 V, VC (3 τ ) = 10 V × e −3 = 498 mV, VC (4 τ ) = 10 V × e −4 = 103 mV, VC (5 τ ) = 10 V × e −5 = 67 mV, 9V = 105 µs, f. 10 V d. VC (2.5 ms ) = 10 V × e -2.5 ms/1 ms = 821 mV, e. t = −1 ms × 1n exponential fall 10 V to 0 V in 5 ms and stays at 0 V 14-11 a. τ = 2.2 kO × 1 µF = 2.2 ms, b. T ≥ 10τ ? T = c. 0-25 ms exponential rise −4 V to +4 V in 11 ms 25-50 ms exponential fall +4 V to −4 V in 11 ms, d. 0-25 ms exponential fall +8 V to 0 V in 11 ms 25-50 ms exponential rise −8 V to 0 V in 11 ms, e. Use Ohm’s Law: iR(t) = vR(t) / R, 0-25 ms exponential fall 1 = 50 ms > 22 ms ? yes,. 20 Hz +3.64 mA to 0 mA in 11 ms 25-50 ms exponential rise −3.64 mA to 0 mA in 11 ms, 2 V-4 V = 3.05 ms -8 V f. t = −2.2 ms × 1n 14-12 a. τ = 1 kO × 1 µF = 1 ms b. T ≥ 10τ ? T = 1 = 10 ms ≥ 10 ms ? yes. 100 Hz c. 0-5 ms exponential rise 0 V to 5 V in 5 ms 5-10 ms exponential fall 5 V to 0 V in 5 ms d. 0-5 ms exponential fall +5 V to 0 V in 5 ms 5-10 ms exponential rise −5 V to 0 V in 5 ms e. Use Ohm’s Law: iR(t) = vR(t) / R 0-5 ms exponential fall +5 mA to 0 mA in 5 ms 5-10 ms exponential rise −5 mA to 0 mA in 5 ms 0.8 V = 1.83 ms 5V f. t = −1 ms × 1n 14-13 t r = 5.065 ms − 232 µs = 4.83 ms, tf = t r = 4.83 ms 14-14 t r = 2.3 ms − 105 µs = 2.20 ms, tf = t r = 2.20 ms, 15-21 a. positive clipper, b. 0 Vp, Vout = d. 0.7 Vp, −4.55 Vmin, e. I = 10 kO × −5 Vp = −4.55 Vmin, c. 0.2 Vp, −4.55 Vmin, 11 kO 5 Vp − 0.7 V 1 kO = 4.3 mAp, f. 70 µAp, −455 µAmin, g. Vdiode reverse = VRL = 4.55 V 15-23 a. rectifier circuit, positive half wave signal out, b. 5.0 Vp, 0 Vmin, c. 4.8 Vp, 0 Vmin, d. 4.3 Vp, 0 Vmin, e. I = 4.3 Vp 1 kO = 4.3 mAp, f. positive half wave, 4.3 mAp, 0 Vmin, g. Vdiode reverse = 5 V 15-24 a. input capacitor filtered rectifier, positive dc output with an ac ripple voltage b. ∼ 5 Vdc c. ∼ 4.8 Vdc d. ∼ 4.3 Vdc ...
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