Homework 1_soln

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

This preview shows pages 1–7. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

## 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.

### Page1 / 15

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

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online