Conductivity and Current in Extrinsic Semiconductors
Electronic
Physics
Dr. Ghusoon Mohsin
Ali
Si
n
ce
(n
n
)
>>
( p
n
)
Example
P-type sample
with
l
=6mm,
A=
0.5mm
2
,
R=120Ω.
Calculate majority
and
minority
carriers if
intrinsic
carrier
density
is
2.5×10
19
/
m
3
.
Given
μ
n
=0.38m
2
/Vs, μ
p
=0.18m
2
/Vs
.
Solution
Conductivity of Extrinsic
Semiconductor
.
n
e
p
e
i
n
i
p
For
n-type
n
e
p
e
n
n
n
n
p
n
e
n
n
n
and
if
n
n
=
N
D
then
n
N
D
e
n
For
p-type
n
e
p
e
p
p
n
p
p
Since
(p
p
)
>>
(n
p
)
p
e
p
p
p
and
if
p
p
=
N
A
then
N
e
p
A
p
p
e
p
p
p
l
73
A
A
R
l
Electronic
Physics
Dr. Ghusoon Mohsin
Ali
2
n
i
=n
p
p
p
Drift Current
Density
An electric field is applied
to a
semiconductor will produce
force
on
electrons
and
holes so that they will experience
a
net acceleration and net
movement.
This
net
movement
of
charge
due
to
an
electric
field
is
called
drift
. The net drift
of charge
gives rise to
a
drift current
.
The current density
due
to electrons
drift
The current density
due
to electrons
drift
J
ne
E
n
n
J
pe
E
p
p
The total current density
due to
electrons and holes
drift
J
J
J
ne
E
pe
E
n
p
n
p
Example
Calculate the drift current density
in a
gallium
arsenide
sample
at
T=300
K, with doping concentration
N
a
=0,
N
d
=10
22
/m
3
,
µ
n
=0.85m
2
/V.s,
µ
p
=0.04m
2
/V.s, E=10 V/cm,
n
i
=2.2×10
17
/m
3
.
Solution
p
74
p
p
100
3.46
10
21
/
m
3
e
1.6
10
19
0.18
p
p
n
2
p
p
n
i
1.9
10
17
/
m
3
Electronic
Physics
Dr. Ghusoon Mohsin
Ali
Since N
D
>>N
a
then it
is
n-type
Current density
due
to hole diffusion
is
Where
D
n
, D
p
(cm
2
/s)
are
electron and hole diffusion constants,
respectively.
Density gradient
of
electrons
Density gradient
of
holes
The minority
hole
concentration
is
n
2
p
i
3.24
10
2
/
m
3
n
J
1
.
6
1
0
19
0
.
8
5
1
0
22
1
0
3
1
3
6
1
0
4
A
/
m
2
Diffusion Current Density
It
is gradual flow
of
charge from
a region
of high density
to a
region
of
low
density.
Current
density
due to
electron
diffusion
is
n
dx
n
J
eD
dn
p
dx
p
J
e
D
dp
dn
dx
dp
dx
75
A
n
N
N
n
D
n
10
22
/
m
3
Electronic
Physics
Dr. Ghusoon Mohsin
Ali
Total Current
Density
The total current density
is
the
sum of
these
four
components
Einstein
Relation
This relation between
the
mobility
and
diffusion
coefficient.
Where
L the
distance
traveled by charge carrier before,
recombination
τ
life
time.
n
dx
p
dx
n
p
J
e
n
E
e
p
E
e
D
d
n
e
D
dp
e
kT
D
p
D
p
n
n
D
kT
e
39
Example
Determine the diffusion coefficient, assume
the
mobility
is
1000cm
2
/V.s
at
T=300K.
Solution
e
D
kT
0.0259
1000
25.9
cm
2
/
s
n
n
p
p
76
Rcombination
Recombination
that
result
from
the
collision
of
an
electron
with
a
hole.
This process is the return
of a
free electrons in the
conduction
band
to
valence band, there is
a
net recombination
rate by
difference between the
recombination and generation
rates.
L
D
L
D
the
current density
J
x
is
given
by:
V
H
Fig. 5.7.
Hall measurement
situation
77
x
y
z
Electronic
Physics
Dr. Ghusoon Mohsin
Ali
Hall
Effect:
As shown there is
a
current
I
x
resulting from
an
applied electric field
E
x
in
x-
direction
an
electrons will drift
v
x
.
A
magnetic field
B
y
(wb/m
2
) is
superposed on applied electric field
E
x
,
whereby
the current
I
x
and
the
magnetic flux
B
y
are perpendicular to each other. The electrons will
experience
a
Lorentz
force
F
z
perpendicular
to
I
x
and
B
y
(
in
z
direction)
F
ev
B
z
x
y
Thus
the
electrons
under
the
influence
of
this
force
will
tend
to
crowds
one
face in
the sample. This collection
of
electrons to one side establish
an
electric field in z-direction
is
known
Hall
effect E
H
or
E
z
. Resulting
a
voltage V
H
between
the
upper and the lower faces
of
the sample is
observed:
On
the
other
hand
eE
ev
B
x
y
z
E
v
B
J
x
E
x
J
nev
x
x
Electronic
Physics
Dr. Ghusoon Mohsin
Ali
Hall
coefficient
Where
n
is the total number
of
charge carrier
per
volum
y
z
ne
J
E
x
B
R
78
H
1
ne
J
x
I
I
x
x
A
bd
y
z
x
B
H
bd
I
E
R
y
x
B
H
bd
I
b
V
H
R
y
I
x
B
d
V
R
H
H
y
H
V
ed
I
n
x
B
L
A
I
n
e
V
x
x
Where
all
the quantities in the right-hand side of the equation can
n
measured. Thus the
carrier
concentration and carrier type can
e
obtained
directly
from the
Hall
measurement.
J
E
ne
E
x
x
x
Electronic
Physics
Dr. Ghusoon Mohsin
Ali
Example
Determine
the
majority
carrier
concentration and mobility, sample
10
-1
cm
length and 10
-2
×10
-3
cm
2
cross
section area, given Hall effect
parameters
I
x
=1mA,
V
x
=12.V, B=5×10
-2
tesla,
V
H
=-6.25mV.
Solution
The negative Hall voltage indicate
n-type
semiconductor
neV
A
I
L
x
x
y
H
V
ed
I
n
x
B
2
1
3
(1.6
10
19
)(10
5
)(
6.25
10
3
)
(10
3
)(5
10
2
)
5
1
0
m
n
neV
A
I
L
x
x
2
79
0.1
m
/
V
.
s
(1.6
10
19
)(5
10
21
)(12.5)(10
4
)(10
5
)
(10
3
)(10
3
)
Electronic
Physics
Dr. Ghusoon Mohsin
Ali
Problems
Q1:
Calculate the drift
current
density
in
silicon
sample. If
T=300 K,
N
d
=10
21
/m
3
, N
a
=10
20
/m
3
V, µ
n
=0.85m
2
/V.s, µ
p
=0.04m
2
/V.s,
E=35
V/cm..
(Ans:
6.8×10
4
A/m
2
).
Q2:
A
cubic doped n-type silicon semiconductor sample
at
T=300
K,
µ
n
=0.85m
2
/V.s, R=10kΩ, J=50A/cm
2
when
5V is
applied.
Calculate
80
N
D.
.
Q3:
Determine 10mm×1mm×1mm sample,
a
magnetic field 0.2wb/m
2
is
superposed
on
applied
voltage 1mV.
Calculate
Hall voltage
if
electrons density 7×10
21
/m
3
μ
n
=0.4 m
2
/Vs.
This content delves into the intricacies of conductivity in extrinsic semiconductors, exploring the behavior of carriers in n-type and p-type materials. It covers topics such as drift current density, diffusion current density, the Einstein relation, recombination, and the Hall effect. Practical examples and solutions are provided to aid in comprehending these fundamental concepts in electronic physics.
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Presentation Transcript
Electronic Physics Dr. Ghusoon MohsinAli Conductivity of ExtrinsicSemiconductor. =ne + p e i n i p For n-type = n e + p e n n Since (nn) >>( pn) n n p = n e n and if nn=ND then n = NDe n n n For p-type = n e + p e p p n p p Since (pp) >>(np) = p e p p p and if pp=NAthen = N e p A p Example P-type sample with l=6mm, A=0.5mm2, R=120 . Calculate majority and minority carriers if intrinsic carrier density is 2.5 1019/ m3. Given n=0.38m2/Vs, p=0.18m2/Vs . Solution l R = l= A A = p e p p p 73
Electronic Physics Dr. Ghusoon MohsinAli 2 ni =np pp = p = 100= 3.46 1021/m3 1.6 10 19 0.18 p e p p n2 n =i= 1.9 1017 / m3 p p p Drift Current Density An electric field is applied to a semiconductor will produce forceon electrons and holes so that they will experience a net acceleration and net movement. This net movement of charge due to an electric field is called drift. The net drift of charge gives rise to a drift current. The current density due to electrons drift J = ne E n n The current density due to electrons drift J = pe E p p The total current density due to electrons and holes drift J = J + J = ne E + pe E n p n p Example Calculate the drift current density in a gallium arsenide sample at T=300 K, with doping concentration Na=0, Nd=1022/m3, n=0.85m2/V.s, p=0.04m2/V.s, E=10 V/cm, ni=2.2 1017/m3. Solution 74
Electronic Physics Dr. Ghusoon MohsinAli Since ND>>Na then it isn-type n N N n A D n 1022 /m3 The minority hole concentration is n2 p =i= 3.24 102 / m3 n J =( 1.6 10 19) (0.85)( 1022)( 103)=136 104A/ m2 Diffusion Current Density It is gradual flow of charge from a region of high density to a region of low density. Current density due to electron diffusionis J = eDdn ndx n Current density due to hole diffusion is J = eDdp pdx p Where Dn, Dp (cm2/s) are electron and hole diffusion constants, respectively. dn dx dp dx Density gradient of electrons Density gradient of holes 75
Electronic Physics Dr. Ghusoon MohsinAli Total Current Density The total current density is the sum of these four components J = en E + ep E + eDdn eDdp n dx pdx n p EinsteinRelation This relation between the mobility and diffusion coefficient. D D kT n= p = e n p D =kT = e 39 Example Determine the diffusion coefficient, assume the mobility is 1000cm2/V.s at T=300K. Solution D =kT = 0.0259 1000 = 25.9cm2 /s e Rcombination Recombination that result from the collision of an electron with a hole. This process is the return of a free electrons in the conduction band to valence band, there is a net recombination rate by difference between the recombination and generation rates. D D L = L = n n p p Where L the distance traveled by charge carrier before, recombination life time. 76
Electronic Physics Dr. Ghusoon Mohsin Ali Hall Effect: As shown there is a current Ix resulting from an applied electric field Ex in x- direction an electrons will drift vx. A magnetic field By (wb/m2) is superposed on applied electric field Ex , whereby the current Ix and the magnetic flux By are perpendicular to each other. The electrons will experience a Lorentz force Fzperpendicular to Ixand By( in z direction) F = ev B z x y Thus the electrons under the influence of this force will tend to crowds one face in the sample. This collection of electrons to one side establish an electric field in z-direction is known Hall effect EH or Ez. Resulting a voltage VH between the upper and the lower faces of the sample is observed: On the other hand eE = ev B z x y the current density Jx is givenby: E = v B J = nev x Jx = Ex z x y x VH Fig. 5.7. Hall measurementsituation 77
Electronic Physics Dr. Ghusoon MohsinAli J E =xB ne z y I I = x= A Jx x bd Hall coefficient H=1 R Where n is the total number of charge carrier per volum ne I E = R xB Hbd z y I VH=R xB b Hbd y I V = R H xB d y H I n = B x V ed y H Where all the quantities in the right-hand side of the equation can n measured. Thus the carrier concentration and carrier type can e obtained directly from the Hall measurement. J = E =ne E x x x ne V I x= x A L 78
Electronic Physics Dr. Ghusoon MohsinAli I L = x neV A x Example Determine the majority carrier concentration and mobility, sample 10-1cm length and 10-2 10-3cm2cross section area, given Hall effect parameters Ix=1mA, Vx=12.V, B=5 10-2tesla, VH=-6.25mV. Solution The negative Hall voltage indicate n-type semiconductor I n = B x V ed y H (10 3)(5 10 2) 3 n = = 5 10 m 21 (1.6 10 19)(10 5)( 6.25 10 3) I L = x neV A x (10 3)(10 3) = = 0.1m /V.s 2 (1.6 10 19)(5 1021)(12.5)(10 4)(10 5) 79
Electronic Physics Dr. Ghusoon Mohsin Ali Problems Q1: Calculate the drift current density in silicon sample. If T=300 K, Nd=1021/m3, Na=1020/m3V, n=0.85m2/V.s, p=0.04m2/V.s, E=35 V/cm.. (Ans: 6.8 104A/m2). Q2: A cubic doped n-type silicon semiconductor sample at T=300 K, n=0.85m2/V.s, R=10k , J=50A/cm2 when 5V is applied.Calculate ND. . Q3: Determine 10mm 1mm 1mm sample, a magnetic field 0.2wb/m2 is superposed on applied voltage 1mV. Calculate Hall voltage if electrons density 7 1021/m3 n=0.4 m2/Vs. 80
