SOLUTIONS CHAPTER 1
1.1.
Show that Equation (1.6) follows from Equation (1.3).
q2
Solution: Equation (1.3) is dEP =
dr . Integrating both sides we obtain
4 0 r 2
1
q2 1
q2
q2
dEP =
dr =
dr =
+ const = EP
2
2
4 0 r
40 r
4 0 r
To find the constant, we
Physical Operation of BJTs
Structure review
Voltage vs. position in an npn
structure
cutof
active
saturation
Designing for high
The Early efect
Parasitic capacitance (C, C)
From Prof. J. Hopwood
The npn Bipolar Junction Transistor
n-type
p-type
n+-
Metal-OxideSemiconductor
Fields Effect Transistors
(MOSFETs)
From Prof. J. Hopwood
Structure: n-channel MOSFET
(NMOS)
body
B
source
S
gate
G
IG=0
drain
D
ID=IS
IS
metal
oxide
n+
p
L
n+
W
Circuit Symbol (NMOS)
D
G
ID= IS
B
IG= 0
IS
S
(IB=0, should be rever
EECE Tﬁﬂl, Midterm Exam. Name agent/MA '5'" 5
l5
1. {35} a. Silicon is doped with lxlﬂ'ﬁ cm'3 boron and lxllf} cm"1 phosphoms. Find the electron
and hole concentrations at 3313K. [Note that boron is group 3 and phosphorus is group 5, for those
ot‘you who
ECEG201,MIdlerlnEX-IIL Nitrite 331 f; pg
1. (20) a. Calculate the conductivity of silicon doped with phosphorus at oonccmnlions or 1x10”,
1x10“, and lxloz“cm".
b. For a lprrt x lprr. x 1pm piece of silicon doped as above, with contacts on opposite ends, w
Q.
EECE 7201, Midterm Exam. Name ' $1,; ban 5
1. (25) a. Silicon is doped with 1x10” cm'J boron and 1x10” cm'3 phosphorus. Find the electron
and hole concentrations at 300K. (Note that boron is group 3 and phosphorus is group 5, for those
of you who don’t
Q.
EECE 7201, Midterm Exam. Name
1. (30) a. Silicon is doped with 1x1014 cm'3 boron and 1x10” cm"3 phosphorus. Find the electron
and hole concentrations at 300K. (Note that boron is group 3 and phosphorus is group S. for those
of you who don’t have a go
EECE 7201, Midterm Exam. Name ii) {a g ,3 n L,
{c 1. (25) a. Silicon is doped with SKID19 cm'3 phosphorus (group 5 element!) Find the electron and
'2 hole concentrations at 300K.
13. Find the conductivity of this material.
7 c. For a 1m x tum x 1pm piece
ECE G201
Homework 1 Solutions.
Problem 1. This is easy and self-checking.
Problem 2. Solution (from Prof. Hopwood)
The He+ is essentially identical to the H atom, except that the nucleus has a +2 charge.
Therefore, the attraction force between the remaini