> can
-
not move (fixed)
>
-
Once
acceptor impurity (like Boron) is put into a
en
silicon
crystal
' than
s from
the
an
· 32 in
⑥
·
?
mor
valence
shell
⑳
④
0
Si
⑨
Si
.
^
Si
·
Immobile/fixed
@
pt
0005 ì
P
∅
5 é in
valence shell
un
pt
⑳
positive charge
.
∅
B
Si
si
⑩
(net + charge)
i
Free é
Mobile (Net
>
-
Negative chargel
Type
←
P
→ pt
Hole
-
ㅱ
type of charge
carrier
of material
concentration in p-type material
equilibrium
Dt
P = NA
Po
=
ND
No Electron
conc
.
in
n-type material
⇌
P= #
Pro & Hole conc
no
2
=
Is Electron concentration
material
in
p-type
leaves
-
electrically negative
:
Hole , that eventually
negatively charged
be ionized and
t
cherge
-Si
a
,
velence
ㆁ
rises
starting freely moving around the
it
,
> When
crystal happens the region around the Boron
atom is no longer
electrically neutral > Boron becomes to
imoree shell
⑬
ㆁ
it
,
atom
the boron
.
in
n-type material
under
一
Diffusion
Because
holes
difference
of
fromp-side
will
in
n-type
→
charges
·
11 ※
Guolleess
(
-
BB
·
in
holes
gonne , remaining
are
charged Boron ions
G Fixed charge
.
3
of free and Holes
free e
holes
of
a
depletion
e
free é or holes
region
diffuse onto p-side because of difference
concentration
㉚
pt
Fixed net
charges
·
·
Space-Charge Region
charge
few holes
rich
in
free é
as
well
holes from p-side
as
n-side will
find
↳ As
more
free and
free
e
that
diffuse to
recombine
and
holes
.
diffuses
and
starting
recombine , the DEPETION REGION
grows ,
from the interface
till after
and keep
growing
time
some
to an
comes
and
,
,
,
stop
equilibrium
,
growing .
we
got two fixed charges combined in
distance (two
depletion Region separated by
distance set up
opposite charges separated by
↳
the
a
,
a
Electric field
an
south
pole of two magnetics create a magnetic field
creating an Electric field built into the device
,
such
as
a
north
pole
and
a
& Built-in
Electric Field
Vbi
Voltage (potencial difference)
=
Built-in
n-side
P-side
neutral
丹
⇌
Region
neutral
Region
←
W
~
in
willfree fromsidediffusedto psea
Bt
are
Depletion Region
Recombination of the diffused
formation
Pt
gonmes remaing
Tefreecharged phosphoros
↳ =
=
and
ㅗ
P
pt ·
> No
.
2
py
%
very few free é
Diffusion
will
@
θ
-
1-
>
-
tile
Holes
rich
é
immo
the E
·
Free
pt
BB
:
side
D…
P-typeSide
immobile
Pnjunation Diode
concentration
onto n-side
,
difuse
SmallParticlesmorefromwer
:
P-type
n-type
℃
C
Efn
Efi
Efi
Efp
Ev
Ev
P-type
n
℃
-
typetype
℃
Efn
Eai
Ei
Efp
品
Ev
Ev
Ev
P-type
It
can
together .
its
not have two different
As the
conduct band
fermi
fermi
level
of
and
valence
band
levels
n-type
moves
,
so
side
down
2-Hype
they aligna
moves
with
it
down ,
Etp
Ev
neutral
.
region
Deplotion
Region
P-side
ω
glfp
EFi -EFp = KTAn()
G↓
Ef.
-
Ean
-
「
ti
=
め「 &n
gQfn
Efp
=
KTen
)
( 器)
℃
Efn
neutral region
n-side
Ev
Holes
↳
>
Free é
Q
=· · · · · Region
⑳
㉚
Neutral
⑳
⑩
Region
0
p-side
M- Side
min
t
#
C
Energy Barrier Height
~
Efi
g9-p
quei
=
qfp-tn s
Free é
世 ·
EF
ょ
Ev
gPfp
EF:
Efn
-
-
:
EFp
far
=
\
RTln()
KTln
glfn
(|
q φtn
Efi
Ev
V5
o
=
qVbi
. 796 et
0
=
Vbi
0 796 V
=
.
Vbi =
qUbi
=
=
&V
g φfp + glfn
0 347
+
.
P-Side
Example 2
n-
Neutral
:
NA
Ppo
Na = 4 x 10 <m
PPo
a find free e
PPo
=
"po
NA
=
Na
nno
3
hole (ht
NiD
Pno
N ) = 9 x 1000cm-
-
3
4 × 1016 cm
concentration
in
3
9 x 18
Who = ND =
3
4 × 1016 ( m
Level
Fermi
q4fp
Ny
=
9x 10 6cm
%
=
-
3
q4fp
q φ 5p
Draw
=
=
0
0
.
the
.
g φ tn
0259h)4x100)
qVbi
Diagram
. side
p
Es
1 .1 eV
.
Efi
KT \n
=
9 &Fn
383 eV
Band
Ss
2Son
=
both sides
on
kTln()
=
m
both sides
Mno
positions
.
Side
562Scm- Pro
=
=i2 ==
b) Find
2)
and
=
Depletion region
-
=
eV
0 449
Neutral
ผ
Upo
%
=
=
(‰)
=
q4fp + q(fn
Ev
=
. 383
0
. 167 eV
8
~
+ 0
.
404
=
0 787 eV
.
n-side
:
.
. 55 ev
5
( 00)
. 404 eV
0
0 383 eU
Etp
\n
002s 9
qVbi
=
q0fp + qPfn
* 787 eV
.
Es
0 146 eV
.
E
0 484 eV
Fn
.
Efi
Ev
长
d) Find
Vbi
w
=
bi
depletion
e) Total
Voltage
Built-in
the
0 787
=
.
at
width
bias ?
zero
=
under
equilibrium
Vr = 0
>
-
2Es (Vbi-30
=
N'
qN'
NAND
=
Na + ND
N'
NaNp
=
(4x 1010)(9x1816)
=
permisivity
Es
.
.
.
1)1 . 602 + 10 19)
-
of the semiconductor
defines how much
permisivity
electrical charge the material can
(8 8 sux 1 o 19 x 11 A) ( 0 A 82)
α
1016
(凶 10 ) ( 9 × 10 )
Na + ND
ω=
. 77x
2
=
℃
o
.
(2 77 x 1016)
.
=
=
hold.
W = 1,917
x
10 -
5
0.917
cm =
Capacitors high permittivity
↑M = 192 mm
would be able to store
f find the peak È field at
Emax
2
=
Emax
Es
(Vbi-Vr)
10
☆=∴ 008
cm
s
=
82. 1
So
=
Relative
Er
.
a 1 xl
09
v
/ cm
8 854x10-14 E/am
.
=8
kV/cm
for
Si
)
=
Es = Go + Ep/si)
흙
8 854 41 D
.
9) Electric field at
w
Wat-IOV
=
-10 Volts
Bias
.
Bias :
2Es (Vbi
-
Vr)
=
qN'
ω=
α
(8 - 8"
1 , 600
↳
Emax
" )(
;階 ; (
's
2
o . a8 t
. =
7
× 10
t
.
l xlo -5
30 4 × 10器
.
cm
o
==f)
max =
permittivity
permittivity of free space
=
-
charge
more
(Vascum)
x
2
Emax
the material
Bias.
zero
W
=
-
-
= 304 KV
Im
=
3.0x
Yen
×
H
.7
charge density (n-sidel
p+
B-X
-
+
P(p )
B
>
-
Nij
NyX( q)
f(b-)
ES
·
:
-
+
=
+
p-side
(charge)
q
~→+
Nat
charge density
1
=
=
q
NaX( q)
-
x = E
‰
c
p Electric field (E)
=
.
㉚
↑
peak value =
↳
Emax
NT
CONSTANT
Xp
In
)
most
cases
Xn
ㅇ
p-side
n-side
these
two et
same value !
NA
ND
required to solve the
problems
XDXn
W
tpt th
ω=
6606
×
뮤
=
더
☆
Ex:
p-side
WA =
1 1 &
Yp
Xn
ㅇ
p
No =
3x
181 = =
n-side
Xp
- The
>
Xn
dopping concentration
depletion width is
NA
ND
W
is
going
less
on
3
=
higher
on
to
bemore
the
heavily doped
on
Xn
higher
is
3
on
the n-side ; the
the p-side - depletion Width
the
lightly
side .
doped
side
and
Depletion
side is
lightly doped side.
region is wider on the
very
depletion width on the heavily doped
The
narrow .
0000
Triangle
area
h
Vbi
=_ b
=
.
h
= WxEmax
.W
. Emax
f
Valid ONLY under equilibrium
0
Emax
Vbi
=
=
no
( Ur
If
external
=
external voltage
an
voltage (vr)
)
O
is
applied
:
vrl
Emax =
Vb
:
-
℃
Xp8
Pside
Xnn side
∅
W
Xp
×
× A
,
뮤
←
xn =
>
-
at least
a
p
t
-
n
thesmallecentration
>
-
n
p +
-
n
smalter
ND Na
dopping conc.
100
3× 6 D
-
n
times
higher to be highly
1⑯
4 xlO97
X
saxhuxrox
Highly
I
+
ptn
(砦
ノー
symmetrical function
doppedNA
ND
Monte
http
is
p
XPA ) Xn
1
.
7x1016
,
}
assymmet
ro
yonction
ol
Vbiz
행
↓ε
>
The amount
30949
/
+
the
↳Since
has reduced
region
charges have
strength
reduced
of it the
and because
W
the
positive
region reduces
negative
and
electric field become
associated with it
voltage
,
(Buit-in electric field
.
of
the
,
depletion
within the
charge
of
this
since
(depletion width) reduces
W
reduces
,
built-in
weaker
also
,
reduces
voltage also reduces
↳ Effective voltage drop across
↳ Barrier
the p-n
Height
will
qUbi
-
also
reduce
qUf
neutral
neutral
BavrielHeight
P
same
level
Efn
P
พ
^
모
not in the
Ec
function
Forward
P side
n- side
.
ㄴ
ω
>
ω
Bias
él
Ev
V
+
>
-
lots
of free e will be
n-side (4 number of
the
é
h+
So
going
poured
free e)
on
Energy ↑
Energy t
the
to
average
be
energy
upward
,
level
because
M-Side
¤
e
'
H
*
e
π
:
Effective
Barries
poured into n-side and they are
going
injected onto the p-side
to be injected from the prside to the n-side
the holes are
going
Free e
to
be
In
the
free
I
are
neutral region of p-side is rich in holes so
injected there it's recombined very quickly.
,
is
,
is
,
Vf
Height ↓
Width/Depletion Width t
Barrier
is
As this depletion width W and the effective barrier height is
reduced, more and more free electrons would be able to jump across
this and will be able to go to the p side.
e
Bias
energy
So we see that this depletion width has also reduced. So two things
have reduced as a result of this applied forward bias. The depletion
width has reduced. That means the width over which this jump has
to be made. Also, the height of this barrier has reduced.
e
스
Forward
free
what results
higher upward push Ef upward
in
reducing the barrier height (qVbi-qVf)
-
P-side
the inside
on
when
a
When you apply sufficiently high forward bias voltage, then lots of free electrons
would be able to easily cross this thin junction because at that condition, the
depletion width would be even narrower, and they would be able to go from n side
to p side.
So once electrons starts flowing from n side to p side, that means the current is
actually going to be from p side to n side, because electrons carry negative charge.
So the current, positive current, would be in the opposite direction. So in the
external circuit, we see a current in the p side. Whereas in reality, electrons are
going from the negative terminal of the power supply and going into the n side.
P-side
M-Side
¤
current
→
H
Hlow
e
e
→
1)
e
…
Minority
∅
e
e
injected
_
since
ㅓ
Forward
Bias
:
Effective
Barries
Height ↓
peside is
rity
Similarly
·
injected that
Holes
P- side
want to diffuse
e
‰
←
<
∵
Highof treecentration
that
were
injected and
want to diffuse
Important the increase
:
current with
is
NOT
the
in
voltage
LINEAR !
,
minority
n-side
m
“品
Injection
from n-side
carriers
Width/Depletion Width t
Barrier
carrier
rich
to p-side become minority
in
holes .
So
,
e
in
peside
to
n-side
=
are
mino-
'
holes
injected
from pside
become
carriers
2)
Minority
Carrier
3)
Minority
carriers
Diffusion
recombination
Both recombination of electrons and holes will give rise to the net electric
current, I, or ID.
The forward current increases when you increase the forward voltage, simply
because as we put that higher the forward voltage Vf, smaller is going to be this
barrier height, and as a result, there would be a higher probability that more and
more free electrons would be able to jump across n-side to p-side. Similarly,
holes would be able to cross this barrier from p-seed to n-side.
In a PN junction, because the current that will increase as a result of increase of voltage would be exponential
increase, not a linear increase. So you increase the voltage a little bit, and the current will
Increase much higher, Exponential increase.
Semiconductor devices
carriers
non-linear devices