Homework 1_soln

Homework 1_soln - ECE 2100 Professor Alyosha Molnar...

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Unformatted text preview: ECE 2100: Professor Alyosha Molnar Homework Assignment #1 Due Feb 6, 2011 Areas covered: Electrical quantities: Charge, Voltage, Current, Power, Energy, Resistance Basic circuit laws: Kirchoff’s current and voltage laws, Ohms law Basic circuits: series, parallel combinations, voltage divider, current divider. Prelab problems: do before coming to lab! 1) Voltage and Current divider. a. Derive an expression for Vout the voltage divider shown in Fig. P1a 2 1 1 1 2 1 R R VR R I Vout R R V I + = ⋅ = + = b. Derive an expression for Iout the current divider shown in Fig. P1b ( 29 2 1 2 1 2 1 2 1 2 || 1 R R IR R V Iout R R R R R R I V + = = + = = Figure p1 . 2) Diode IV relationship: the general diode equation is given by: a. Derive an expression for the dynamic resistance r d = dV/dIin terms of the diode current itself. “modest bias” implies = KT qV Io I D D η exp so D D D D D qI KT dI dV Io I q KT V η η = → = ln b. Derive an expression for the diode voltage as a function of diode current when a series resistance (R s ) is included in the diode model. + = Io I q KT I R V D D s tot ln η c. Derive an expression for the ideality factor ( η ) as a function of the diode current and first derivative of diode current with respect to voltage. D D D D D D D D D dV dI I KT q I KT q KT qV Io KT q dV dI KT qV Io I = = = = η η η η η exp exp Problems from the book: 3) Problem 1.7 I = qAvN where q=charge of electron, A is area, v is velocity and N is electrons/volume (unit check: Coulombs times meters squared time meters per second all divided by meters cubed is Coulombs per second which is Amps) So, v = I/(qAN) = 0.01m/s 4) Problem 1.13 a) As drawn, the element absorbs power, but i is negative, so it will be delivered: p=IV = -400W b) Entering c) Gain : the box is supplying power to them. 5) Problem 1.19 a) -15-10-5 5 10 15 10 20 30 40 50 60 time, s Power, uW b) energy: at 4s, = ½(10uW·4s) = 20 μ J , at 12s = 20 μ J +½(-10uW·4s) = , at 36 s = 0 +½(10uW·4s)+16s(-4uW+ 1/2·6.4uW) = 7.2 μ J at 50s = 7.2 μ J +½(-3.6uW·4s) = 6) Problem 1.29 I just assigned a sign to each device (+1 if current into pos terminal, -1 otherwise) V mA sign mW a 1.6 80 -1 -128 b 2.6 60 -1 -156 c -4.2 -50 1 210 d 1.2 20 -1 -24 e 1.8 30 1 54 f -1.8 -40 -1 -72 g -3.6 -30 1 108 h 3.2 -20 1 -64 j -2.4 30 -1 72 sum: 0 It checks out 7) Problem 2.6 Problem 2....
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This note was uploaded on 06/10/2011 for the course ECE 2100 taught by Professor Kelley/seyler during the Spring '05 term at Cornell.

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Homework 1_soln - ECE 2100 Professor Alyosha Molnar...

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