Chapter 4 PN Junctionsinst.eecs.berkeley.edu/~ee130/sp06/chp4.pdf · 2005-01-20 · Semiconductor...
Transcript of Chapter 4 PN Junctionsinst.eecs.berkeley.edu/~ee130/sp06/chp4.pdf · 2005-01-20 · Semiconductor...
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-1
Chapter 4 PN Junctions
A PN junction is present in every semiconductor device.
N-type
P-type
Donors
V
I
Reverse bias Forward bias
N P
VI
diodesymbol
– +
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-2
N-region P-region(a)
(b)
(c)
(d)
Depletionlayer
NeutralP-region
NeutralN-region
Ef
4.1.1 Energy Band Diagram and Depletion Layer of a PN Junction
A depletion layer exists at the PN junction. n ≈ 0 and p ≈ 0 in the depletion layer.
Ec
EfEv
Ec
Ev
Ef
Ec
Ev
Ef
Ev
Ec
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-3
4.1.2 Built-in Potential
Can the built-in potential be measured with a voltmeter?
(b)
(c)
Ef
Ec
Ev
N-typeN d
P-typeN a
(a)
qφbi
–xn xpx0
V
Nd Na
φbi
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-4
Thermal Couple and Thermoelectric GeneratorThe total built-in voltage in a closed circuit is zero, therefore it cannot be read.
Thermal Couple:When the junction of two wires, such as W and Pt, is placed in a furnace, a non-zero voltage can be read with a voltmeter between the two cold ends.
Thermoelectric Generator:In addition to voltage, significant electric power can also be extracted. Heat-to-electricity conversion efficiency can be optimized by using exotic semiconductors of P and N types.
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-5
4.1.2 Built-in Potential
d
ckTAqcd N
Nq
kTAeNNn ln=⇒== −
2
2
lni
ackTBqc
a
i
nNN
qkTBeN
Nnn =⇒== −
−=−=
d
c
i
acbi N
Nn
NNq
kTAB lnln 2φ
N-region
Ef
Ec
Ev
qφbi qB
qA
2lni
adbi n
NNq
kT=φ
P-region
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-6
4.1.3 Poisson’s Equation
Gauss’s Law:
εs: semiconductor permittivity (~ 12 for Si)ρ: charge density (C/cm3)
What can Poisson’s equation tell us?
∆x
ρ(x) (x + ∆x)
x
areaA
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-7
4.2.1 Field and Potential in the Depletion Layer
On the P-side of the depletion layer, ρ = –qNa
On the N-side, ρ = qNd
4.2 Depletion-Layer Model
N P D eple tion Layer N e utral R egi on
–xn 0 xp
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-8
4.2.1 Field and Potential in the Depletion Layer
The electric field is continuous at x = 0.
Naxp = Ndxn
Which side of the junction is depleted more?
A one-sided junction is called a N+P junction or P+N junction
N P D eple tion La yer N e utral R egi on
–xn 0 xp
N eut ra l Re gion
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-9
4.2.1 Field and Potential in the Depletion Layer
On the P-side,
Arbitrarily choose the voltage at x = xp as V = 0.
On the N-side,
2)(2
)( xxqNxV ps
a −=ε
2)(2
)( ns
d xxqNDxV +−=ε
2)(2 n
s
dbi xxqN
+−=ε
φ
x
E
–xn xp
Ec
Ef Ev
φbi , built-in potential
0
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-10
4.2.2 Depletion-Layer Width
V is continuous at x = 0
If Na >> Nd , as in a P+N junction,
What about a N+P junction?
where density dopant lighterNNN ad
1111≈+=
N P D eple tion La yer N e utral R egi on
–xn 0 xp
N eut ra l Re gion
+==+
da
bisdeppn NNq
Wxx 112 φε
nd
bisdep x
qNW ≈=
φε2
0≅= danp NNxx
qNW bisdep φε2=
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-11
EXAMPLE: A P+N junction has Na=1020 cm-3 and Nd=1017cm-3. What is a) its built in potential, b)Wdep , c)xn , and d) xp ?
Solution:a)
b)
c)
d)
V 1cm10
cm1010lnV026.0ln 620
61720
2 ≈×
== −
−
i
adbi n
NNq
kTφ
µm 12.010106.1
11085.812222/1
1719
14
=
××
××××=≈ −
−
d
bisdep qN
W φε
µm 12.0=≈ 8
10V
2.1cm10
12.(14)Tj/TT11 1 Tf0 2.05924 -2305924 08270.23 516804 Tm
<007c>Tj18397338 0 TD<0020>Tj-391725 0 TD
<0075>Tj-.5026 0 TD<0020>Tj-979026 0 TD
<0020>Tf0 10359240 1035924 0 268.638840.5 TD<0010>Tj/TT6 1 Tf34764.16 -1.188 TD
(a)Tj-953399 0 TDd(1)Tj56276159 0 TD
nep(N)Tj-.07799 0 TD
(N)Tj-042399 0 TD
x(N)Tj-206705 0 TD
x)
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-12
4.3 Reverse-Biased PN Junction
densitydopantlighterNNN ad 1111
≈+=
• Does the depletion layerwiden or shrink withincreasing reverse bias?
+ –V
N P(a)
(b) V = 0
(c) reverse-biased
qV
qNbarrier potential
qNVW srbis
dep⋅
=+
=εφε 2|)|(2
Ec
Ef
EvEf
Ev
qφbi
Ec
EcEfn
Ev
qφbi + qV
Ec
EfpEv
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-13
4.4 Capacitance-Voltage Characteristics
• Is Cdep a good thing?• What are three ways to reduce Cdep?
N P
Nd Na
Conductor Insulator Conductor
Wde p
dep
sdep W
AC ε=
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-14
4.4 Capacitance-Voltage Characteristics
• Is there a way to obtain both Na and Nd of a P+N junction?
222
2
2)(21
AqNV
AW
C S
bi
s
dep
dep εφ
ε+
==
Vr
1/Cdep2
Increasing reverse bias
Slope = 2/qNεsA2
– φbi
Capacitance data
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-15
EXAMPLE: If the slope of the line in the previous slide is 2x1023 F-2 V-1, the intercept is 0.84V, and A is 1 µm2, find the lighter and heavier doping concentrations Nl and Nh .
Solution:
⇒= ln 2i
lhbi n
NNq
kTφ 318026.084.0
15
202
cm 108.1106
10 −×=×
== eeNnN kT
q
l
ih
biφ
• Is this an accurate way to determine Nl ? Nh ?
315
28141923
2
cm 106 )cm101085.812106.1102/(2
)/(2
−
−−−
×=
×××××××=
×= AqslopeN sl ε
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-16
4.5 Junction Breakdown
A Zener diode is designed to operate in the breakdown mode.
V
I
VB, breakdown
P N
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-17
4.5.1 Peak Electric Field
2/1
|)|(2)0(
+== rbi
sp VqN φ
ε
• What kind of junctionis shown in the figure?
N+ PNa
Neutral Region
0 xp
xxp
(a)
(b)
increasingreverse bias
increasing reverse bias
p
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-18
4.5.2 Tunneling Breakdown
Dominant breakdown cause when both sides of a junction are very heavily doped.
bicrits
B qNV φε
−=2
2
V/cm106≈= critp
(a)
(b)Empty StatesFilled States
V
I
(c)
Breakdown
-
Ev
Ec
Ev
EcEf
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-19
4.5.3 Avalanche Breakdown
impact ionization
qNV crits
B 2
2ε=
avalanche breakdown
EcEfn
Ec
Ev
Efp
originalelectron
electron-holepair generation
daB N
1N1
N1V +=∝
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-20
4.6 Carrier Injection Under Forward Bias–Quasi-equilibrium Boundary Condition
minority carrier injection
– +V
N PEc
EfEv
Ec
Efp
Ev
V = 0
Ef n
Forward biased
qφbi
qV
-
+
qφbi–qV
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-21
4.6 Carrier Injection Under Forward Bias–Quasi-equilibrium Boundary Condition
kTEEkTEEc
kTEEc
fpfnfpcfnc eeNeNn /)(/)(/)(P)0( −−−−− ==
kTqVP
kTEEP enen fpfn /
0/)(
0 == −
• The minority carrierdensities are raisedby eqV/kT
• Which side gets morecarrier injection?
Ec
Efp
Ev
Efn
0N 0P
x
Ec
Efn
Efp
Ev
x
Efn
0N 0P
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-22
4.6 Carrier Injection Under Forward Bias–Quasi-equilibrium Boundary Condition
kTVq
a
ikTVqP e
Nnenn
2
0)0( ==
kTVq
d
ikTVqN e
Nnepp
2
0)0( ==
)1()0()0( 00 −=−≡′ kTqVPP ennnn
)1()0()0( 00 −=−≡′ kTqVNN epppp
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-23
EXAMPLE: Carrier InjectionA PN junction has Na=1019cm-3 and Nd=1016cm-3. The applied voltage is 0.6 V.
Question: What are the minority carrier concentrations at the depletion-region edges?
Solution:
Question: What are the excess minority carrier concentrations?
Solution:
-311026.06.00 cm 1010)0( =×== eenn kTVq
P-314026.06.04
0 cm 1010)0( =×== eepp kTVqN
-311110 cm 101010)0()0( =−=−=′ Pnnn
-3144140 cm 101010)0()0( =−=−=′ Nppp
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-24
4.7 Current Continuity Equation
τpq
xxJxxJ pp ′
=∆
−∆+−
)()(
τpxA
qxxJ
Aq
xJA pp ′
⋅∆⋅+∆+
⋅=⋅)()(
τpq
dxdJ p ′
=−
∆ x
pJp(x + ∆x)
x
area A
Jp(x)
Volume = A·∆x
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-25
4.7 Current Continuity Equation
Minority drift current is negligible;Jp= –qDpdp/dx∴
Lp and Ln are the diffusion lengths
τpq
dxdJ p ′
=−
pp
pqdx
pdqDτ
′=2
2
22
2
ppp Lp
Dp
dxpd ′
=′
=′
τ
ppp DL τ≡
22
2
nLn
dxnd ′
=′
nnn DL τ≡
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-26
4.8 Excess Carrier Distribution in Biased PN Junction
22
2
pLp
dxpd ′
=′
0)( =∞′p
)1()0( /0 −=′ kTqV
N epppp LxLx BeAexp //)( −+=′
0 ,)1()( //0 >−=′ − xeepxp pLxkTqV
N
P N +
0P 0N 0
x
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-27
0 ,)1()( //0 <−=′ xeenxn nLxkTqV
P
0 ,)1()( //0 >−=′ − xeepxp pLxkTqV
N
4.8 Excess Carrier Distribution in Biased PN Junction
0.5
1.0
–3Ln –2Ln –Ln 0 Lp 2Lp 3Lp 4Lp
N-side
Nd = 2×1017 cm-3
pN' ∝ e–x /L
P-side
nP'∝ ex /L
Na = 1017cm -3
n p
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-28
EXAMPLE: Carrier Distribution in Forward-biased PN Diode
• Sketch n'(x) on the P-side.313026.06.0
17
20/
2
0 cm 101010)1()1()0( −==−=−=′ ee
Nnenn kTqV
a
ikTqVP
N-typeNd = 5×1017 cm-3
Dp =12 cm2/sτp = 1 µs
P-typeNa = 1017 cm-3
Dn=36.4 cm2/sτn = 2 µs
x
N-side P-side1013cm-3
2×1012
n’ ( = p’ )
p´ ( = n’ )
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-29
• How does Ln compare with a typical device size?
µm 8510236 6 =××== −nnn DL τ
• What is p'(x) on the P- side?
EXAMPLE: Carrier Distribution in Forward-biased PN Diode
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-30
4.9 PN Diode IV Characteristics
x0N-side P-side
x0N-side P-side
Jtotal
pLxkTVqN
p
pppN eep
LD
qdx
xpdqDJ )1()(0 −−=
′−=
nLxkTVqP
n
nnnP een
LDq
dxxndqDJ −−−=
′= )1()(
0
xJ
enLDqp
LD
qJJJ kTVqP
n
nN
p
pnPpN
allat
)1()0()0()0( 00
=
−
+=+=
Jtotal
JpNJnP
JpNJnP
Jn = Jtotal – Jp Jp = Jtotal – Jn
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-31
4.9 PN Diode IV Characteristics
dep
depir τ
WqnAII += 0
)1(0 −= kTVqeII
+=
an
n
dp
pi NL
DNL
DAqnI 2
0
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-32
The PN Junction as a Temperature Sensor
What causes the IV curves to shift to lower V at higher T ?
)1(0 −= kTVqeII
+=
an
n
dp
pi NL
DNL
DAqnI 2
0
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-33
4.10 Charge Storage
What is the relationship between τs (charge-storage time)and τ (carrier lifetime)?
x
N-side P-side1013cm-3
2×1012
n'
p’
IQ ∝
sQI τ=
sIQ τ=
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-34
4.11 Small-signal Model of the Diode
kTqVkTqV eIdVdeI
dVd
dVdI
RG /
0/
0 )1(1≈−==≡
qkTIeI
kTq
DCkTqV /)1( /
0 =−=
What is G at 300K and IDC = 1 mA?
Diffusion Capacitance:
qkT/IG
dVdI
dVdQC DCsss τττ ====
Which is larger, diffusion or depletion capacitance?
CRV
I
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-35
4.12 Other PN Junction Devices–From Solar Cells to Laser Diodes
Solar Cells
Also known asphotovoltaic cells, solar cells can convert sunlight to electricity with 15-30% energy efficiency
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-36
4.12.1 Solar Cells
sckTVq IeII −−= )1(0
N P
-
+
short circuit
lightIsc
(a)
V0.7 V
–IscMaximumpower-output
Solar CellIV
IDark IV
0
(b)
Ec
Ev
Eq.(4.9.4)
Eq.(4.12.1)
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-37
4.12.2 Photodiodes and Avalanche Photodiodes
4.12.3 Light Emitting Diodes (LEDs)
• LEDs are made of compound semiconductors such as InPand GaN.
• Terms: direct band-gap, radiative recombination
Ec
Ev
Recombination Recombination centers
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-38
4.12.4 Diode Lasers
• A diode laser must be strongly forward-biased to make Efn – Efp > Eg .
lightout
Clearedcrystalplane
(a)
(b)
• Terms: population inversion, heterojunctions, cleaved mirror, distributed Bragg reflector (DBR)
P+
N+
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-39
4.13 Chapter Summary
The potential barrier increases by 1 V if a 1 V reverse bias is applied
junction capacitance
depletion width
2lni
adbi n
NNq
kT=φ
qNbarrier potentialW s
dep⋅
=ε2
dep
sdep W
AC ε=
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-40
4.13 Chapter Summary
• Under forward bias, there is minority carrier injection.
• The quasi-equilibrium boundary condition of minority carrier densities is:
• Most of the minority carriers are injected into the lighter doped side.
kTVqP enn 0)0( =
kTVqN epp 0)0( =
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-41
4.13 Chapter Summary
• Steady-state continuity equation:
• Minority carriers diffuse outward ∝ e–|x|/Lp
or e–|x|/Ln
• Lp and Ln are the diffusion lengths
22
2
ppp Lp
Dp
dxpd ′
=′
=′
τ
ppp DL τ≡
)1(0 −= kTVqeII
+=
an
n
dp
pi NL
DNL
DAqnI 2
0
Semiconductor Devices for Integrated Circuits (C. Hu) Slide 4-42
4.13 Chapter Summary
qkT/IG DC=
Charge storage:
Diffusion capacitance:
Diode conductance:
GC sτ=
sIQ τ=