HW2 - HW2 Solution 1. IS = 2. 27 9 9V = 1.20mA I L < 1.20...

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HW2 Solution 1. I S = 27 9 15 k Ω = 1.20 mA I L < 1.20 mA | R L > 9 V 1.2 mA = 7.50 k Ω 2. I Z = V S V Z R S V Z R L = V S R S V Z 1 R S + 1 R L P Z = V Z I Z I Z nom = 60 15 () V 150 Ω 15 V 100 Ω = 150 mA P Z nom = 15 V 150 mA = 2.25 W I Z max = 60 V 1.1 150 Ω 0.90 15 V 0.90 1 150 Ω 0.90 + 1 100 Ω (1.1) = 266 mA P Z max = 15 V 0.90 266 mA = 3.59 W I Z min = 60 V 0.90 150 Ω 1.1 15 V 1.1 1 150 Ω 1.1 + 1 100 Ω (0.9) = 43.9 mA P Z min = 15 V 1.1 43.9 mA = 0.724 W 3. V 1 = V P - V on = 49.3 V and V 2 = -(V P -V on ) = -49.3V. 4. i D 0 + = 5 V 1 k Ω = 5 mA I F = 5 V D 1 k Ω = 5 0.6 1 k Ω = 4.4 mA I r = 3 0.6 1 k Ω =− 3.6 mA τ S = 7 ns ln 1 4.4 mA 3.6 mA = 5.59 ns 5. i D 0 + = 5 V 5 Ω = 1 A I F = 5 V on 5 Ω = 5 0.6 1 Ω = 0.880 A I R = 3 0.6 5 Ω 0.720 A S = 250 ns ln 1 0.880 A 0.720 A = 200 ns
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Additional Problems 1. (a) LC 1 = ω j j diode V A C C φ = 1 0 Change in voltage results in variation of depletion width. This results in change in capacitance which changes frequency as a result. Thus, output frequency can be modulated by changing the input voltage.
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This note was uploaded on 03/02/2009 for the course ECE 137a taught by Professor Staff during the Winter '08 term at UCSB.

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HW2 - HW2 Solution 1. IS = 2. 27 9 9V = 1.20mA I L < 1.20...

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