ee40sp08hwk5-sol

# ee40sp08hwk5-sol - EE 40 Introduction to Microelectronic...

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Unformatted text preview: EE 40: Introduction to Microelectronic Circuits Spring 2008: HW 5 (due 3/7, 5 pm) Venkat Anantharam March 8, 2008 Referenced problems from Hambley, 4th edition. 1. P4.27 The driving sources are DC, so the circuit will be in DC steady state. Hece, we can replace the capacitor by an open circuit and the inductor by a short circuit, getting the circuit depicted in Figure 5. Applying KCL, we have 10 k Ω i R 2 mA- v C +- + 15 V Figure 1: Circuit 2 i R = 2 mA Applying KVL, we have v c = 2 mA * 10 k Ω + 15 V = 20 V + 15 V = 35 V 2. P4.62 (NOTE: The is a typo in part (b) of the problem P4.61, which should say v ′ (0+) = 10 9 V s .) (a) The differential equation describing the current after the switch opens is (see equation (4.102)): d 2 v ( t ) dt + 1 RC dv ( t ) dt + 1 LC v ( t ) = 0 (1) From equation (4.104), the undamped resonant frequency is ω = 1 √ LC = 10 7 1 s From equation (4.103), the damping coefficient is α = 1 2 RC = 10 7 1 s From equation (4.71), the damping ratio is ζ = α ω = 1 Note that ζ is dimensionless. 1 (b) To solve the differential equation (1), we need the initial conditions v (0+) and dv (0+) dt . We are given that v (0+) = 0 and i L (0+) = 0. (These come from v (0- ) = 0 and i L (0- ) = 0 and physics contraints that v (0- ) = v (0+) and i L (0- ) = i L (0+).) we can find dv (0+) dt by determining i c (0+) (where we use a positive reference). Since v (0+) = 0, no current flows through the resistor at 0+, and likewise since i L (0+) = 0, no current flows through the inductor at 0+. Hence, i c (0+) = 1 A This gives C dv (0+) dt = 1 A Therefore, dv (0+) dt = 10 9 V s (c) Since the equation has zero forcing function, we can simply take v p ( t ) = 0 as a particular solution. (d) Since ζ = 0, we are in the critically damped case. From equation (4.75), we see that the comple- mentary solution has the form v c ( t ) = K 1 e − αt + K 2 te − αt because ω n = radicalbig ω 2 o- α 2 = 0 and so we have s 1 = s 2 =- α (see equations (4.72), (4.73), and (4.76)). Here, K 1 and K 2 are constants that should be chosen so that the total solution v ( t ) = v p ( t ) + v c ( t ) satisfies the initial conditions. This means 0 = v (0+) = K 1 and 10 9 V s = dv (0+) dt =- αK 1 + K 2 This results in the final result v ( t ) = 10 9 te 10 7 s t V s 3. P4.66 Treating i ( t ) as the basic variable, the differential equation describing the circuit after the switch is closed can be written as (see equation (4.59)) d 2 i ( t ) dt 2 + R L di ( t ) dt + 1 LC i ( t ) = 1 L d...
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ee40sp08hwk5-sol - EE 40 Introduction to Microelectronic...

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