ECE 630: Physics and Models of Semiconductor Devices
Course Outline
Fall 2013
Course description
Semiconductor physics. Metal-semiconductor contacts. Physical principles and operation of
pn-junctions, MOS capacitors, MOS field-effect transistors, and rela
ECE630
Week 3 (i) Finish semiconductor physics
(ii) Start pn-junctions
Last week:
Doping
Density of states and the Fermi function
Equations for the concentrations of e-s & h+s
Mobility, drift, conductivity, diffusion
Today:
Generation/recombination
Semiconductor Physics
ECE630
Electrical resistivity of materials
insulators 1010 1018 cm
semiconductors 10-4 108 cm
conductors 10-6 10-4 cm
what is so special about semiconductors?
we can control and engineer the electronic properties of
semiconductors ov
ECE630
Week 5 Metal/semiconductor contacts
Last week:
- Finished pn-junctions
- Brief introduction to metal/semiconductor contacts (work
functions, Schottky barriers)
This week:
- Ohmic vs. Schottky and how to achieve each
- I-V of a Schottky diode
- Cont
ECE730T17- Midterm 2012 Solutions
Question 1
(a) Sulphur. Although nickel will more easily capture holes, for a hole to recombine both a hole and
an electron are needed and the location of sulphurs energy state is the best for capturing both.
(Explanation
ECE730T17- Midterm 2012
Oct 23, 11:30am. Time: 80 minutes
Answer all 4 questions. Total marks: 100 points
Equation/constants sheet included. Calculators permitted. No other aids allowed.
Clearly mark your NAME and ID NUMBER in your answer-book.
Please don
ECE630
Week 4 pn junctions part 2
Last week:
- Shockley-Read-Hall theory of recombination
- Continuity equations
- Built-in voltage, electric-field, depletion region width, etc.
of a step-junction pn-junction
- Qualitative discussion of I-V characteristic
ECE730T17 HW#2 Solutions
(1) When discussing SRH recombination is class, we derived U which is the recombination rate of
excess carriers. This recombination rate is equal to p/, so the higher U is, the lower the excess
carrier lifetime. The energy levels
Mass action law
n0 = ni e
E F E Fi
kT
p0 = ni e
E Fi E F
kT
Multiply these together and you get:
n0 p0 = n
2
i
this very useful relation is called
the mass action law.
This tells us that the product of n0 and p0 is always a constant for a
given semic
(4)
5 1015
= 0.3294 V
10
1.5 10
The surface potential is
s = 2 fp = 2(0.3294 ) = 0.659 V
We have
Q
V FB = ms ss = 0.90 V
C ox
Now
Q SD (max )
VT =
+ s + V FB
C ox
We obtain
fp = (0.0259 ) ln
x dT
4 s fp
=
eN a
1/ 2
(
)
4(11.7 ) 8.85 10 14 (0.
ECE 630, Homework #5: MOS Capacitors
(1) A MOS capacitor is constructed using a TiN (titanium nitride) gate, silicon oxide, and n-type
silicon. The work function of TiN is 4.78 V, the electron affinity of silicon is 4.01 V, and the donor
dopant concentrat
System of Units,
Conversion Factors,
and General Constants
Table H.1 I International system of units*
Quantity
Length
Mass
Time
Temperature
Current
Frequency
Force
Pressure
Energy
Power
Electric charge
Potential
Conductance
Resistance
Capacitance
Magnetic
ECE630, Homework #2
This is not to be handed in. Solutions will be provided. If additional constants or semiconductor
properties are required to answer the question, see the Constants & semiconductor properties pdf on
the course website
(1) In silicon, zi
ECE630, Homework #3
This is not to be handed in. Solutions will be provided. If additional constants or semiconductor
properties are required to answer the question, see the Constants & semiconductor properties pdf on
the course website
(1) The total junc
ECE630, Homework #1 Semiconductor Physics
This is not to be handed in. Solutions will be provided. If additional constants or semiconductor properties are required
to answer the question, see the Constants & semiconductor properties pdf on the course webs
ECE630, Homework #4: Metal/Semiconductor contacts SOLUTIONS
0 = ( ) = 5.01 4.01 = 1.09 V
Question 1
=
= + ( )/
( ) = kT ln (NC/ND) = .0259*ln(2.8 x 1019/1016) = 0.2056 eV so
= 0.8844 V
= 4.22
B will remain the same with increased doping. i however
ECE630, Homework #4: Metal/Semiconductor contacts
This is not to be handed in. Solutions will be provided. If additional constants or semiconductor
properties are required to answer the question, see the Constants & semiconductor properties pdf on
the cou
ECE730T17, Homework #3 solutions
(1)
For a silicon p + n junction with N d = 5 1015 cm 3 , from the PN Junction Breakdown Voltage graph from
lecture, V B 100 V
2 ( )
1
=
+
Since Vbi is typically 1V, ( Vbi - Va) VB. Also, Na > Nd, so equation becomes
Homework #1 solutions
1(a) At energy E1 , we want
1
1
E1 E F
E EF
exp 1
kT 1 + exp kT
1
E EF
1 + exp 1
kT
= 0.01
This expression can be written as
E EF
1 + exp 1
kT
1 = 0.01
E1 E F
exp
kT
or
E EF
1 = (0.01) exp 1
kT
Then
E1 = E F + kT
V
V
depletion layer
inversion layer
(electrons)
semiconductor (p)
NA-
x
air
-Q1
charge density
+Q1
metal
semiconductor (p)
NA-
air
charge density
+Q1 -Q1
metal
metal
air
metal
charge density
the field effect
.
+Q2
-Q2
x
x
large V (>Vth)
the field effect f