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ECE 250  Electronic Device Modeling  RoseHulman Study Resources

ECE250 HW2S
School: RoseHulman
Course: Electronic Device Modeling
ECE 250 HW #2 Solution Define units and constants 19 6 eV eV := 1.602 10 coul volt KB := 86 10 K Specify some constants for silicon Eg_Si := 1.12 eV 19 BSi := 5.23 10 15 1 cm K 3 1.5 q := 1.602 10 coul Problem 3.109 Length := 1

ECE_250_HW1P
School: RoseHulman
ECE 250 Homework #2 Due 9/15/2008 (Monday). Problem 1: A silicon semiconductor is to be designed such that the majority carrier electron concentration is no=71015 cm3. a) Should donor or acceptor impurity atoms be added to intrinsic silicon to achie

Problem 5Answers
School: RoseHulman
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Lecture1
School: RoseHulman
Course: Introduction To NMOSFET
3.2 See Sedra and Smith Text Eqn 3.2 and Eqn 3.48

Lecture3
School: RoseHulman
Course: Introduction To NMOSFET
Equation 3.35 in Sedra & Smith Text (Note Vm is "Vp" in the textbook.) This equation is NOT in the Sedra & Smith textbook. The textbook's "Idav" expression is only for the average diode current during the charging interval, which is not useful.

Lecture2
School: RoseHulman
Course: Introduction To NMOSFET
Sedra and Smith Figure 3.12 "Minority Carrier Concentration Gradients on both sides of a forwardbiased pn junction." Sedra and Smith Eqn 3.47 Cj=Cjo/(1+Vr/Vo)m Where m = 1/2 to 1/3, depending on junction grading profile. Note: Vr = Vd = Cathode to anode

Lecture4
School: RoseHulman
Course: Introduction To NMOSFET
12/14/2009 ECE250 (KEH) 1 Introduction to Bipolar Junction Transistors Transistors (Read Chapter 3 of Text) 12/14/2009 ECE250 (KEH) 2 Bipolar Junction Transistor Bipolar Junction Transistor Currentcontrolled current source current source Made by sandwi

ECE_250_HW1S
School: RoseHulman
Course: Electronic Device Modeling
9/2/07 8:55 AM % % % % % Problem 1 C:\Documents.\ECE250_HW1_Problem1.m ECE250 HW1 Solutions 1 of 1 VD = 10:.1:10; rho = 100e6; ID=zeros(1,length(VD); for i=1:length(VD) if VD(i)>0 ID(i) = rho*VD(i).*VD(i); else ID(i)=0; end end plot(VD,1000*ID);

ECE 250 Homework 4  Solutions
School: RoseHulman
Course: Electronic Device Modeling
Problem 1 5 4 3 2 1 D D1 Breakdown_Voltage = 5 Forward_Voltage = 0.7 Vin R1 Vo D + 1k + V1 FREQUENCY = 1 AMPLITUDE = 15 + R2 1k C C 0 B B A ECE Department 5500 Wabash Avenue Terre Haute, IN 47803 Ph: (812) 8778512 FAX: (253) 3699

ECE250_F02_HW3_S
School: RoseHulman
Course: Electronic Device Modeling
ECE 250 HW #2 Solution Define units and constants 19 6 eV eV := 1.602 10 coul volt KB := 86 10 K Specify some constants for silicon Eg_GaAs := 1.4 eV 19 B GaAs := 2.1 10 14 1 cm K 3 1.5 q := 1.602 10 coul Problem 1: Nd := 10 cm

ECE 250 Homework 6  Solutions
School: RoseHulman
Course: Electronic Device Modeling
982.9uA R3 6.2k + R1 1.548Meg Q1 + 5.906V + 773.8mV 6.553uA R2 1.106Meg QB150  V1 DC = 12 + 0 * 04/23/02 16:37:00 * PSpice Lite (Mar 2000) * * Profile: "SCHEMATIC1Bias" [ C:\WEBSITE\ROSE_CLASSES\ECE250\HOMEWORK\SPRING_2002\HW6\ece 250 hw6

ECE 250 Exam 1 Fall 2007Solution
School: RoseHulman
Course: Electronic Device Modeling
ECE 250 HW #2 Solution Define units and constants eV := 1.602 10 19 6 eV coul volt KB := 86 10 K Specify some constants for silicon Eg_Si := 1.12 eV 19 BSi := 5.23 10 15 1 cm K 3 1.5 q := 1.602 10 coul A := 1 m E := 2 Voltage

ECE 250 Final Exam Fall 2007  Solution
School: RoseHulman
Course: Electronic Device Modeling
VG := 15 V 62 k 62 k + 91 k VP := 2 V VG = 6.078 V IDSS := 2 mA VGS := 0 V ID := 1 mA Given VGS ID = IDSS 1 VP 2 VG = VGS + ID 12 k VGS := Find V , I ( GS D) ID VGS = 0.92 ID = 0.583mA IDSS := 20 mA VGS := 0 V VP := 8 V