Chap6 photodetectors

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Transcript of Chap6 photodetectors
PhotodetectorsPrinciple of the pn junction Photodiode
Schematic diagram of a reverse biased pn junction photodiode
SiO2Electrode
ρ net
–eN a
eN d x
x
E (x )
R
E max
e–h+
Iph
hv > Eg
W
E
n
Depletionregion
ARcoating
Vr
Electrode
Vout
Net space charge across the diode in the depletion region. Nd and Na are the donor and acceptor concentrations in the p and n sides.
The field in the depletion region.
p+
Photocurrent is depend on number of EHP and drift velocity.
The electrode do not inject carriers but allow excess carriers in the sample to leave and become collected by the battery.
6
Principle of pn junction photodiode
• (a) Reversed biased p+n junction photodiode.
• Annular electrode to allow photon to enter the device.
• Antireflection coating (Si3N4) to reduce the reflection.
• The p+side thickness < 1 μm.• (b) Net space charge distribution,
within SCL.• (c) The E field across depletion
region.
PhotodetectorsPrinciple of the pn junction Photodiode
(b) Energy band diagram under reverse bias.
(a) Crosssection view of a pin photodiode.
(c) Carrier absorption characteristics.
Operation of a pin photodiode.
Variation of photon flux with distance.
A physical diagram showing the depletion region.
A plot of the the flux as a function of distance.
There is a loss due to Fresnel reflection at the surface, followed by the decaying exponential loss due to absorption.
The photon penetration depth x0 is defined as the depth at which the photon flux is reduced to e1 of its surface value.
PhotodetectorsPrinciple of the pn junction Photodiode
PhotodetectorsRAMO’s Theorem and External Photocurrent
t t
eh+
Iphoto(t)
Semiconductor
V
x
l L − l
t
vhole
0 Ll
e–h+
0t
0
Area = Charge = e
evh/L eve /L
An EHP is photogenerated at x = l. The electron and the hole drift in opposite directions with drift velocities vh and ve.
The electron arrives at time telectron = (Ll )/ve and the hole arrives at time thole = l/vh.
photocurrent
ielectron(t)
E
Le hv
+
Le
Le eh vv
thole
telectron
tholeihole(t)
i (t)
telectron
tholevelectron
iphoto(t)
As the electron and hole drift, each generates ielectron(t) and ihole(t).
The total photocurrent is the sum of hole and electron photocurrents each lasting a duration th and te respectively.
PhotodetectorsRAMO’s Theorem and External Photocurrent
( ) ee
e ttL
eti <= ;
v ( ) hh
h ttL
eti <= ;
v
( ) ( ) edttidttiQhe t
h
t
ecollected =+= ∫∫ 00
( )transit
d ttL
teti <= ;
v)(
( ) ( )h
he
e
lttand
lLtt
vv=−= Transit time
( ) dttiVdxEedoneWork e⋅=⋅=dt
dxv
L
VE e ==
Ramo’s Theorem
Photocurrent
The collected charge is not 2e but just “one electron”.
If a charge q is being drifted with a velocity vd(t) by a field between two biased electrodes separated by L, the motion of q generates an external current given by
PhotodetectorsAbsorption Coefficient and Photodiode Materials
][24.1
][eVE
mg
g =µλ
Absorbed Photon create ElectronHole Pair.
Cutoff wavelength vs. Energy bandgap
xeIxI α−⋅= 0)( Absorption coefficient
Incident photons become absorbed as they travel in the semiconductor and light intensity decays exponentially with distance into the semiconductor.
13
Absorption Coefficient• Absorption
coefficient α is a material property.
• Most of the photon absorption (63%) occurs over a distance 1/α (it is called penetration depth δ)
0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8
Wavelength (µm)
In0.53Ga0.47As
Ge
Si
In0.7Ga0.3As0.64P0.36
InP
GaAs
aSi:H
12345 0.9 0.8 0.7
1×103
1×104
1×105
1×106
1×107
1×108
Photon energy (eV)
Absorption coefficient ( α) vs. wavelength (λ) for various semiconductors(Data selectively collected and combined from various sources.)
α (m1)
1.0
Figure 5.3
PhotodetectorsAbsorption Coefficient and Photodiode Materials
0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8
Wavelength (mm)
Ge
Si
In0.7Ga0.3As0.64P0.36
InP
GaAs
aSi:H
123450.9 0.8 0.7
1×103
1×104
1×105
1×106
1×107
1×108
Photon energy (eV)
Ab
sorp
tion
Coe
ffic
ien
t α
(m
1)
1.0
In0.53Ga0.47As
AbsorptionThe indirectgap materials are shown with a broken line.
15
Absorption Coefficient• Direct bandgap semiconductors
(GaAs, InAs, InP, GaSb, InGaAs, GaAsSb), the photon absorption does not require assistant from lattice vibrations. The photon is absorbed and the electron is excited directly from the VB to CB without a change in its kvector (crystal momentum ħk), since photon momentum is very small.
E
CB
VB
k–k
Direct Bandgap Eg Photon
Ec
Ev
(a) GaAs (Direct bandgap)
E
k–k
(b) Si (Indirect bandgap)
VB
CB
Ec
Ev
Indirect Bandgap, Eg
Photon
Phonon
(a) Photon absorption in a direct bandgap semiconductor. (b) Photon absorptionin an indirect bandgap semiconductor (VB, valence band; CB, conduction band)
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
0 momentumphoton VBCB ≈=− kk
Absorption coefficient α for direct bandgap semiconductors rise sharply with decreasing wavelength from λg (GaAs and InP).
16
Absorption Coefficient• Indirect bandgap
semiconductors (Si and Ge), the photon absorption requires assistant from lattice vibrations (phonon). If K is wave vector of lattice wave, then ħK represents the momentum associated with lattice vibration ħK is a phonon momentum.
E
CB
VB
k–k
Direct Bandgap Eg Photon
Ec
Ev
(a) GaAs (Direct bandgap)
E
k–k
(b) Si (Indirect bandgap)
VB
CB
Ec
Ev
Indirect Bandgap, Eg
Photon
Phonon
(a) Photon absorption in a direct bandgap semiconductor. (b) Photon absorptionin an indirect bandgap semiconductor (VB, valence band; CB, conduction band)
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Kkk ==− momentumphonon VBCB
Thus the probability of photon absorption is not as high as in a direct transition and the λg is not as sharp as for direct bandgap semiconductors.
PhotodetectorsAbsorption Coefficient and Photodiode Materials
E
CB
VB
k–k
Direct Bandgap Eg
E
k–k
VB
CBIndirect Bandgap
Photon
Phonons
Photon absorption in an indirect bandgap semiconductor
EgEC
EV
EC
EV
Photon absorption in a direct bandgap semiconductor.
Photon
PhotodetectorsQuantum Efficiency and Responsivity
νηhP
eI
photonsincidnetofNumbercollectedandgeberatedEHPofNumber ph
0
==
External Quantum Efficiency
0(W)
(A)
P
I
PowerOpticalIncident
ntPhotocurreR ph==
Responsivity
ch
e
h
eR
λην
η == Spectral Responsivity
Responsivity vs. wavelength for a typical Si photodiode.
0 200 400 600 800 1000 12000
0.10.20.30.40.50.60.70.80.9
1
Wavelength (nm)
Si Photodiode
λ g
Res
pons
ivit
y (A
/W)
Ideal PhotodiodeQE = 100% ( η = 1)
Photodetectors
20
The pin Photodiode• The pn junction photodiode has two
drawbacks:– Depletion layer capacitance is not
sufficiently small to allow photodetection at high modulation frequencies (RC time constant limitation).
– Narrow SCL (at most a few microns) long wavelengths incident photons are absorbed outside SCL low QE
• The pin photodiode can significantly reduce these problems.
Intrinsic layer has less doping and wider region (5 – 50 μm).
Reversebiased pin photodiode
PhotodetectorsThe pin Photodiode
pin energyband diagram
pin photodiode circuit
SiO2
Schematic diagram of pin photodiode
p+
iSi n+
Electrode
ρnet
–eNa
eNd
x
x
E(x)
R
E0
e–h+
Iph
hυ > Eg
W
Vr
Vout
Electrode
E
PhotodetectorsThe pin Photodiode
Small depletion layer capacitance gives high modulation frequencies.
High Quantum efficiency.
In contrast to pn junction builtinfield is uniform
A reverse biased pin photodiode is illuminated with a short wavelength photon that is absorbed very near the surface.
The photogenerated electron has to diffuse to the depletion region where it is swept into the i layer and drifted across.
hυ > Eg
p+ iSi
e–
h+
W
Drift
Diffusion
Vr
E
PhotodetectorsThe pin Photodiode
pin diode
(a) The structure;
(b) equilibrium energy band diagram;
(c) energy band diagram under reverse bias.
PhotodetectorsThe pin Photodiode
Quantum efficiency versus wavelength for various photodetectors.
PhotodetectorsPhotoconductive Detectors and Photoconductive gain
W
AC r
dep
εε 0=
Electric field of biased pin
Junction capacitance of pin
PhotodetectorsThe pin Photodiode
Small capacitance: High modulation frequency
RCdep time constant is ∼ 50 psec.
WV
WV
EE rr ≈+= 0
Response time
ddrift
Wt
v=
Edd µ=v
The speed of pin photodiodes are invariably limited by the transit time of photogenerated carriers across the iSi layer.
For iSi layer of width 10 µm, the drift time is about is about 0.1 nsec.
Drift velocity vs. electric field for holes and electrons in Silicon.
102
103
104
105
107106105104
Electric field (V m1)
Electron
Hole
Dri
ft v
eloc
ity
(m s
ec1)
PhotodetectorsThe pin Photodiode
Example
Bandgap and photodetection
(a) Determine the maximum value of the energy gap which a semiconductor, used as a
photoconductor, can have if it is to be sensitive to yellow light (600 nm).
(b) A photodetector whose area is 5×102 cm2 is irradiated with yellow light whose
intensity is 20 mW cm−2. Assuming that each photon generates one electronhole
pair, calculate the number of pairs generated per second.
Solution
(a) Given, λ = 600 nm, we need Eph = hυ = Eg so that,
Eg = hc/λ = (6.626×1034 J s)(3×108 m s1)/(600×109 m) = 2.07 eV
(b) Area = 5×102 cm2 and Ilight = 20×103 W/cm2.
The received power is
P = Area ×Ilight = (5×102 cm2)(20×103 W/cm2) = 103 W
Nph = number of photons arriving per second = P/Eph
= (103 W)/(2.059×1.60218×1019 J/eV)= 2.9787×1015 photons s1 = 2.9787×1015 EHP s1.
(c) For GaAs, Eg = 1.42 eV and the corresponding wavelength is
λ = hc/ Eg = (6.626×1034 J s)(3×108 m s1)/(1.42 eV×1.6×1019 J/eV)
= 873 nm (invisible IR)∴The wavelength of emitted radiation due to EHP recombination is 873 nm.
(d) For Si, Eg = 1.1 eV and the corresponding cutoff wavelength is,
λ g = hc/ Eg = (6.626×1034 J s)(3×108 m s1)/(1.1 eV×1.6×1019 J/eV)
= 1120 nmSince the 873 nm wavelength is shorter than the cutoff wavelength of 1120 nm, the Si photodetector can detect the 873 nm radiation (Put differently, the photon energy corresponding to 873 nm, 1.42 eV, is larger than the Eg, 1.1 eV, of Si which mean
that the Si photodetector can indeed detect the 873 nm radiation)
Example
Bandgap and Photodetection
(c) From the known energy gap of the semiconductor GaAs (Eg = 1.42 eV), calculate the
primary wavelength of photons emitted from this crystal as a result of electronhole
recombination. Is this wavelength in the visible?
(d) Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why?
Solution
Example
Absorption coefficient
(a)If d is the thickness of a photodetector material, Io is the intensity of the incoming
radiation, the number of photons absorbed per unit volume of sample is
[ ]υ
αhd
dInph
)exp(10 ⋅−−=
(a) If I0 is the intensity of incoming radiation (energy flowing per unit area per second), I0 exp(−α d ) is the transmitted intensity through the specimen with thickness d and thus I0 exp(−α d ) is the “absorbed” intensity
Solution
Example
(b) What is the thickness of a Ge and In0.53Ga0.47As crystal layer that is needed for
absorbing 90% of the incident radiation at 1.5 µm?
For Ge, α ≈ 5.2 × 105 m1 at 1.5 µm incident radiation.
mmd
d
µα
α
428.410428.49.01
1ln
102.5
1
9.01
1ln
1
9.0)exp(1
65 =×=
−×=
−=
=⋅−−∴
−
For In0.53Ga0.47As, α ≈ 7.5 × 105 m1 at 1.5 µm incident radiation.
mmd µ07.31007.39.01
1ln
105.7
1 65 =×=
−×= −
For Ge, α ≈ 5.2 × 105 m1 at 1.5 µm incident radiation.For In0.53Ga0.47As, α ≈ 7.5 × 105 m1 at 1.5 µm incident radiation.
(b)
Example
0
0.2
0.4
0.6
0.8
1
800 1000 1200 1400 1600 1800Wavelength (nm)
Responsivity (A/W)
The responsivity of an InGaAs pin photodiode
InGaAs pin Photodiodes Consider a commercial InGaAs pin photodiode whose responsivity is shown in fig.
Its dark current is 5 nA.
(a) What optical power at a wavelength of 1.55 µm would give a photocurrent that is twice the dark current? What is the QE of the photodetector at 1.55 µm?
(b) What would be the photocurrent if the incident power in a was at 1.3 µm? What is the QE at 1.3 µm operation?
Solution
(a) At λ = 1.55×106 m, from the responsivity vs. wavelength curve we have R ≈ 0.87 A/W. From the definition of responsivity,
we have
From the definitions of quantum efficiency η and responsivity, ∴
Note the following dimensional identities: A = C s1 and W = J s1 so that A W1 = C J1. Thus, responsivity in terms of photocurrent per unit incident optical power is also charge collected per unit incident energy.
0)(
)(
P
I
WPowerOpticalIncident
AntPhotocurreR ph==
nWWA
A
R
I
R
IP darkph 5.11
)/87.0
)(10522 9
0 =××===−
hc
e
h
eR
ληυ
η ==
%)70(70.0)1055.1)(106.1(
)/87.0)(/103sec)(1062.6(619
834
=××
×⋅×== −−
−
mcoul
WAsmJ
e
hcR
λη
Solution
(b) At λ = 1.3×106 m, from the responsivity vs. wavelength curve, R = 0.82 A/W.
Since Po is the same and 11.5 nW as in (a),
The QE at λ = 1.3 µm is
%)78(78.0)103.1)(106.1(
)/82.0)(/103sec)(1062.6(619
834
≈××
×⋅×== −−
−
mcoul
WAsmJ
e
hcR
λη
nAnWWAPRI ph 43.9)15.1)(/82.0(0 ==⋅=
π p+
SiO2Electrode
ρnet
x
x
E(x)
R
hυ > Eg
p
Iphoto
e – h+
Absorption
regionAvalanche region
Electrode
n+
E
PhotodetectorsAvalanche Photodiode (APD)
h+
πn+ p
e–
Avalanche region
E
e
h+
E c
E v
Impact ionization processes resulting avalanche multiplication
Impact of an energetic electron's kinetic energy excites VB electron to the CV.
SiO2
Guard ring
ElectrodeAntireflection coating
nn n+
p+
π
p
SubstrateElectrode
n+
p+
π
p
Substrate
Electrode
Avalanche breakdown
Si APD structure without a guard ring
More practical Si APD
PhotodetectorsAvalanche Photodiode (APD)
Schematic diagram of typical Si APD.
Breakdown voltage around periphery is higher and avalanche is confined more to illuminated region (n+p junction).
E
N n
Electrode
x
E (x)
R
hυ
Iph
Absorptionregion
Avalancheregion
InP InGaAs
h+
e–
InP
P + n+
VrVout
E
PhotodetectorsHeterojunction Photodiode
Separate Absorption and Multiplication (SAM) APD
InGaAsInP heterostructure Separate Absorption and Multiplication APD
P and N refer to p and ntype widerbandgap semiconductor.
InP
InGaAs
h+
e–
Ec
Ev
Ec
Ev
InP
InGaAs
Ev
Ev
h+
∆Ev
E
PhotodetectorsHeterojunction Photodiode
Separate Absorption and Multiplication (SAM) APD
InGaAsP grading layer
(a) Energy band diagram for a SAM heterojunction APD where there is a valence band step ∆Ev from InGaAs to InP that slows hole entry into the InP layer.
(b) An interposing grading layer (InGaAsP) with an intermediate bandgap breaks ∆Ev and makes it easier for the hole to pass to the InP
layer.