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ee40sp08hwk4-sol

# ee40sp08hwk4-sol - EE 40 Introduction to Microelectronic...

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EE 40: Introduction to Microelectronic Circuits Spring 2008: HW 4 Solution Venkat Anantharam February 29, 2008 Referenced problems from Hambley, 4th edition. 1. P3.39 The current through the capacitor is i ( t ) = C dv C ( t ) dt = 10 - 7 ( - 10 3 sin(100 t )) = - 10 - 4 sin(100 t ) A (where we assume that v C ( t ) is given in volts). The voltage across the resistor is therefore v R ( t ) = Ri ( t ) = - 10 - 3 sin(100 t )V . The concept of 1% accuracy used in the problem statement is not very precise because the voltages are time varying. Interpreting this question as asking whether the peak of the parasitic resistance voltage term is within 1% of the peak voltage across the capacitor, the question may be viewed as asking if 10 - 3 < (0 . 01)10 3 . This is of course true. Suppose now that v C ( t ) = 0 . 1 cos(10 7 t ) (again assumed to be given in volts). Then we have i ( t ) = C dv C ( t ) dt = 10 - 7 ( - 10 6 sin(10 7 t )) = - 0 . 1 sin(10 7 t ) A and v R ( t ) = - sin(10 7 t ) V . In this case the peak of the parasitic resistance voltage is actually 10 times the peak voltage across the capacitor, so it becomes very important to include it in the model. 1

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The message of this problem is that the higher the frequency of the sig- nals the more a capacitor approximates a short circuit (which makes it increasingly inappropriate to neglect parasitic resistances in series). 2. P3.68 The current through the inductor (and resistor) is given as i ( t ) = 0 . 1 cos(10 5 t ) A . Therefore we can find the voltage across the inductor as v L = L di ( t ) dt = 10 * 10 - 3 ( - 0 . 110 5 sin(10 5 t )) = - 100 sin(10 5 t )V . The voltage across the resistor is v R ( t ) = Ri ( t ) = 1 * 0 . 1 cos(10 5 t ) V , and the total voltage is v ( t ) = v L ( t ) + v R ( t ) = - 100 sin(10 5 t ) + 0 . 1 cos(10 5 t ) V The concept of 1% accuracy used in the problem statement is not very precise because the voltages are time varying. Interpreting this question as asking whether the peak of the parasitic resistance voltage term is within 1% of the peak voltage across the capacitor, the question may be viewed as asking if 0 . 1 < (0 . 01)100 . This is of course true. Suppose now that i ( t ) = 0 . 1 cos(10 t ) (again assumed to be given in Amps). Then we have v L = L di ( t ) dt = 10 * 10 - 3 ( - 0 . 1 * 10 sin(10 t )) = - 10 - 2 sin(10 5 t ) V , while v R ( t ) = Ri ( t ) = 0 . 1 cos(10 t ). In this case the peak of the parasitic resistance voltage is actually 10 times the peak voltage across the inductor, so it becomes very important to include it in the model. The message of this problem is that the lower the frequency of the sig- nals the more an inductor approximates a short circuit (which makes it increasingly inappropriate to neglect parasitic resistances in series). 2
3. P3.71 i ( t ) = C dv C ( t ) dt = 5 * 10 - 4 * 1000 * 10 cos(1000 t ) = 5 cos(1000 t ) v L ( t ) = L di L ( t ) dt = - 2 * 10 - 3 * 1000 * 5 sin(1000 t ) = - 10 sin(1000 t ) v ( t ) = v C ( t ) + v L ( t ) = 10 sin(1000 t ) - 10 sin(1000 t ) = 0 This is an idealized LC oscillator with no active source, where energy is sloshing back and forth between the capacitor and the inductor.

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ee40sp08hwk4-sol - EE 40 Introduction to Microelectronic...

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