ECE 441
Problem Set #2
Due: Wednesday, February 4
Professor E. Rosenbaum
Spring Semester 2015
Reading Assignment: Sections 4.1, 3.1-3.2
1.
Calculate the magnitude of the built-in field in the quasi-neutral region of an exponential impurity
distribution:
e
SECONDARY
AND PRIMARY
DATA
Sanjay Kalidindi
SECONDARY AND PRIMARY
DATA COLLECTION
Secondary
Published
information available from other
sources that has already been gathered.
This info is relevant to the problem at
hand. Either internal or external to an
ECE 441 Spring 2015 TCAD Assignment 1 Solution
1.
(a)
As explained in the tutorial, you should define a fine mesh near the junction and a coarse
mesh for the rest of the device. Also, since the device is uniform in the y-direction, only
the mesh size in t
ECE 441
Midterm Exam #2
April 6, 2015
Name: :3 Ol 00S
Problem 1 (30)
Problem 2 (20)
Problem 3 (20)
Problem / (30)
Total (100 points)
Read careﬂlly:
This is a 50 minute exam. You must stop work and turn in your exam when the instructor
announces that the
ECE 441 Spring 2015 - Homework 3 Solution
1.
The exponential impurity distribution is
() = 0 ( )
Use the quasi neutral approximation, it can be derived:
|()| = |
1
()
()
25.9
| = | | = 0.32
= 8.09 102
For the abrupt junction with Nd = 1017 3, Na = 5 101
ECE 441 Spring 2015 Homework 4 Solution
1.
(a)
From = and = we have =
=
=
. Therefore, from =
d
From = dt , we can derive
d
= dt
The general solution for this differential equation is
() = C exp ( )
Assume the initial charge distribution is (, , , 0),
ECE 441 Spring 2015 - HW1 Solution
1.
(a)
Since total charge in the region < 2 is zero, from Gausss Law it is obvious that
= 0 for > 2 . Similarly, we also have = 0 for < 0.
The negative sheet charge located on the metal side at = 0 is equal to the total
1. In the circuit shown below, an unspecied semiconductor device is represented by a box.
The applied voltage is unknown; it may be zero or non-zero, positive or negative.
However, the band diagram (E vs. x) is available, and it is also shown below.
Bas
ECE 441 Spring 2015 - HW1 Solution
1.
(a)
Since total charge in the region < 2 is zero, from Gausss Law it is obvious that
= 0 for > 2 . Similarly, we also have = 0 for < 0.
The negative sheet charge located on the metal side at = 0 is equal to the total
ECE 441 Spring 2015 - HW #14 Solution
1.
a)
The short channel PMOS drain current equations are:
For VGS < VTP, VDS > VDSAT:
ID =
W
L
eff Cox (VGS VT )VDS 2 VDS 2 )
1
V
1 DS
Esat L
For VGS < VTP, VDS VDSAT:
ID =
Weff Esat Cox
(E
(VGS VT )2
) (1
VDS VDSAT
ECE 441 Spring 2015 - HW #13 Solution
1. (postponed from HW #12)
The off-state leakage current of a long channel MOSFET can be expressed as
IOFF = exp (
),
Cd
where n = 1 +
In order to reduce the leakage current by one order of magnitude, we need to i
ECE 441
Problem Set #13
Due: Wednesday, May 6
Professor E. Rosenbaum
Spring Semester 2015
Reading Assignment: Sections 10.1-10.2
1.
The ID model equations for an n-channel MOSFET are listed below. You are to write the
corresponding set of equations for a
ECE 441 Spring 2015 - Homework 12 Solution
1. (P 8.4); postponed from HW #11.
Since Si = 2 cm, from Figure 1.6 we get Na = 7.5 1015 3.
In depletion, the gate voltage and surface potential is related by
1
VG = VFB + (s p ) + C 2qNa s (s p )
ox
2s (s p )
Al
ECE 441
Problem Set #13
Due: Wednesday, April 29
Professor E. Rosenbaum
Spring Semester 2015
Reading Assignment: Section 9.2
1.
Postponed from Problem Set #12. An NMOS transistor has body doping NA = 1017cm-3 and oxide
thickness Tox = 10nm. The desired th
ECE 441
Problem Set #12
Due: Wednesday, April 22
Professor E. Rosenbaum
Spring Semester 2015
Reading Assignment: Sections 8.4, 9.1
1.
Postponed from problem set #11: P 8.4, with the following clarifications and changes. You are to
construct an accurate pl
ECE 441 Spring 2016 - HW1 Solution
1.
According to Eqn. (1.1.23) and (1.1.24), the effective density of states is
3/2
,
2,
= 2(
)
2
From Table 1.3, the density of states effective mass for electrons and holes are
= 1.080 ; = 0.810
Thus
(a)
At 77K,
2 3/
ECE 441 Spring 2016 HW2 Solution
1.
(a)
Since the only current carrier in a metal is electron, the conductivity is
=
Thus
n =
= 0.375 2 1 1
This number is much smaller compared with the electron mobility in Si and Ge.
(b)
The drift velocity is related w
Sanjay Kalidindi
CH.4 NOTES
Suppose you have manufactured a new
product and you want to sell it in a
foreign country. Three things you would
need to know to accomplish your goal.
Customs
Languages
Times
How are you going to get involved
Import
Export
Lice
ECE 441
TCAD Assignment #2
Due: May 1, 2015
In this exercise, you will use TCAD to simulate an NMOS transistor. You will learn how to
simulate a quasi-static C-V characteristic and an I-V characteristic. You will investigate the
subthreshold behavior of a
ECE 441
Formula Sheet: Midterm #2
Monday, April 6, 2015
Professor E. Rosenbaum
Spring Semester 2015
1
1
exp
,
,
exp
exp
Abrupt PN junction:
,
ln
Avalanche multiplication:
, VJ is the reverse bias on the junction
Debye length:
, where N is the net doping c
ECE 441
TCAD Assignment #1
Due: Friday, March 13*
In this exercise, you will use TCAD to simulate a silicon PN junction diode. You will learn how
to (i) create a device structure using the TCAD script, (ii) simulate the (static) I-V characteristic,
and (i
ECE 441
Formula Sheet: Midterm #1
Friday, February 20, 2015
Professor E. Rosenbaum
Spring Semester 2015
1
1
exp
,
,
exp
exp
Abrupt PN junction:
,
ln
Avalanche multiplication:
, VJ is the reverse bias on the junction
Debye length:
, where N is the net dopi
ECE 441 Spring 2016 HW6 Solution
1. M&K P 5.12
Assume the diode has a long base on both the p-side and the n-side. The minority carrier
currents can be expressed as
( < ) = (
( > ) = (
+
)
) = 2 (
)
Since the net recombination rate in the space ch
ECE 441 Spring 2016 HW4 Solution
1.
The E-field is lowest near the center of the device. This is because the doping level near the
center of the device is highest. Since the resistor is biased at Vab=10V, there is significant current
flowing through the r
ECE 441 Spring 2016 HW5 Solution
1. M&K P 4.9
Assume we have an abrupt p+n diode. The capacitance of the diode can be expressed as
Cd =
=
2( )
Where is the junction area. Thus,
1
2( )
=
2
2
a)
The slope of the curve for 1V < Va < 0 is
d 1
2
1.5 1026
ECE 441 Spring 2016 HW3 Solution
1. (M&K P. A1.2 with modifications)
(a)
Since total integrated charge in the region < 2 is zero, from Gausss Law it is
obvious that = 0 for > 2 . Similarly, we also have = 0 for < 0.
The negative sheet charge located on th
ECE441
Spring 2016
Assignment # 6
Date: February 26, 2016
Due: March 4, 2016
1.)
M&K P 5.12
2.)
Derive equation (5.2.9a) from (5.2.8) using (5.2.1), (5.2.2), (5.2.5), and (5.2.6).
3.)
An n-type silicon sample contains recombination centers with their ener
ECE441
Spring 2016
Assignment # 4
Date: February 12, 2016
Due: February 19, 2016
1.)
For the silicon slab resistor we simulated in the TCAD tutorial, plot the E-field
along the x-axis when Vab=10V. Explain the physical reason for the particular
shape of t