Problem 1 (25 points)
Consider a silicon pn junction with a cross section area of 1x10-5 cm2, a forward
bias Va = 0.5V, and the following parameters at T = 300K:
N 1016 cm3 , N 1015 cm3 , 107 s, 106 s
d
a
p
n
400 cm / V .s, 1000 cm / V .s
2
p
2
n
n 1.5 x

Problem 1 (25 points)
Consider a silicon pn junction under thermal equilibrium. Assume
Eg 1.12eV ,
ni 1.5x1010 cm3 , kT = 0.026eV.
a) (10 points) Determine the doping in the p region such that EF = EV +3kT. Determine the
doping in the n region such that E

Problem 1 (25 points)
Si at T = 300K contains donor impurity atoms at a density of 5x 1016 cm3 and acceptor impurity
atoms at a density of 2 x1016 cm3 . Assume ni 1.5x1010 cm3 , kT = 0.026eV.
a) (5 points) Is the semiconductor n type or p type?
b) (10 poi

Problem 1 (25 points)
Consider a silicon semiconductor with Eg = 1.12eV, T = 300K.
a) (10 points) Write the probability of a state at energy EC + E being occupied by an
electron.
b) (10 points) Write the probability of a state at energy EV - E being occup

Name:
Problem 1 (25 points)
Consider a silicon semiconductor with Eg = 1.12eV, nl. =1.5x101°cm3 , kT = 0.026eV,
T = 300K. The semiconductor is doped with ND = 1.5x1016 cm'3.
a) (15 points) Determine the probability of a state at energy EC + kT being occup

ESE 231: INTRODUCTION TO SEMICONDUCTOR DEVICES
Spring 2013
Stony Brook University
Department of Electrical and Computer Engineering
COURSE DESCRIPTION
The course covers physical principles of operation of semiconductor devices.
Energy bands, transport
pro

Problem 1 (25 points)
Consider a silicon semiconductor doped with NA = 2x1014 cm-3. Assume Eg = 1.12eV,
ni 1.5x1010 cm3 , kT = 0.026eV at T=300K.
3
2
3
2
T
T
1350
cm
/
V
.
s
,
480
cm / V .s
300
300
2
n
2
p
a) (10 points) Determine the conductivi

3.6
"(In JEsEc°3l (E-EC)|
«sf mtwngm]
when:
kT
EI=EF+4kT and E:=EC+T
Then
n(Ez) E
3.48
(a) no <n. = 9-91»
(b) no = 45x10 cm" 2 electmns: minority
cutie:
nf (mm 2
Po ' " . 3
n 45x10
I)
A7 = 5x10 an" => holes: majon'ty carrier
(C)
p0 = N - N
a I
so
5x10

4.15
(a) O- : eu'o + cyppo and 0 =
Po
Then
gun:
a = + eyrlpu
p0
To nd the minimum conductivity,
do 1 e I :12
_ = 0 : U% + ep :
dp Pf;
which yields
1 2
115.,
Pa = ;[] (Answer 10 Dan (M)
Hf,
Substituting into the conductivity expression
eplni I1
020mm z+

Problem 1 (25 points)
Consider a silicon pn junction with a cross section area of 1x10 -5 cm2, a forward
bias Va = 0.5V, and the following parameters at T = 300K:
N 1016 cm3, N 1015 cm3 , 107 s, 106 s
d
a
p
n
400 cm / V .s, 1000 cm / V .s
2
p
2
n
n 1.5 x