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Photodetectors
Photodiodes
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Photo-detectors: Principle of the P-N junction photo-diode
Schematic diagram of a reverse biased p-n junction photodiode SiO2
Electrode
ρ net
–eNa
eN d x
R
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.
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.
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Principle of pn junction photodiode
(a) Reversed biased pn junction photodiode.
• Annular electrode to allow photon to enter the device.
• Anti-reflection coating (Si3N4) to reduce the reflection.
• The p+-side thickness < 1 μm.
(b) Net space charge distribution, within SCL.
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Photo-detectors: Principle of the p-n junction Photo-diode
(b) Energy band diagram under reverse bias.
(a) Cross-section view of a photo-diode
(c) Carrier absorption characteristics.
Operation of a photo-diode
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A generic photo-diode
Photodetectors: Principle of the p-n junction Photo-diode
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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 e-1 of its surface value.
Photodetectors: Principle of the p-n junction Photo-diode
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Photo-detectors: RAMO’s Theorem and External Photo-current
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 = (L-l )/ve and the hole arrives at time thole = l/vh.
t t
e-h+
Iphoto(t)
Semiconductor
V
x
l L − l
t
vhole
0 Ll
e–h+
0t
0
Area = Charge = e
evh/L eve/L
photocurrent
ielectron(t)
E
Le hv
⎟⎠⎞
⎜⎝⎛ +
Le
Le eh vv
thole
telectron
thole ihole(t)
i (t)
telectron
tholevelectron
iphoto(t)
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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.
( ) ee
e ttL
eti <= ;v ( ) hh
h ttL
eti <= ;v
( ) ( ) edttidttiQ he t
h
t
ecollected =+= ∫∫ 00
( )transit
d ttL
teti <= ;v)(
( ) ( )h
he
elttandlLtt vv =
−= Transit time
( ) dttiVdxEedoneWork e⋅=⋅=dtdxv
LVE 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
Photo-detectors: RAMO’s Theorem and External Photo-current
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Photo-detectors: Absorption Coefficient & Photo-diode Materials
][24.1][eVE
mg
g =μλ
Absorbed Photon create Electron-Hole Pair.
Cut-off 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.
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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
InPGaAs
a-Si:H
12345 0.9 0.8 0.7
1×103
1×104
1×105
1×106
1×107
1×108
Photon energy (eV)
α (m-1)
1.0
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0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8Wavelength (mm)
Ge
Si
In0.7Ga0.3As0.64P0.36
InPGaAs
a-Si:H
12345 0.9 0.8 0.7
1×103
1×104
1×105
1×106
1×107
1×108
Photon energy (eV)
Abs
orpt
ion
Coe
ffic
ient
α (
m-1)
1.0
In0.53Ga0.47As
Absorption The indirect-gap materials are shown with a broken line.
Photo-detectors: Absorption Coefficient & Photo-diode Materials
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• 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 k-vector
ħ(crystal momentum k), since photon momentum is very small.
E
CB
VB
k–k
Direct Bandgap Eg Photon
Ec
Ev
(a) GaAs (Direct bandgap)0 momentumphoton VBCB ≈=− kk hh
Absorption coefficient α for direct band-gap semiconductors rise sharply with decreasing wavelength from λ g (GaAs and InP).
Absorption Coefficient
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• Indirect band-gap semiconductors (Si and Ge), the photon absorption requires assistant from lattice vibrations (phonon). If K is wave vector of
ħlattice wave, then K represents the momentum associated with lattice vibration �K is a phonon momentum.
E
k–k
(b) Si (Indirect bandgap)
VB
CB
Ec
Ev
Indirect Bandgap, Eg
Photon
Phonon
Kkk hhh ==− momentumphonon VBCB
Thus the probability of photon absorption is not as high as in a direct transition and the λ g is not as sharp as for direct band-gap semiconductors.
Absorption Coefficient
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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
Photo-detectors: Absorption Coefficient & Photo-diode Materials
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Photo-detectors: Quantum Efficiency and Responsivity
νη
hPeI
photonsincidnetofNumbercollectedandgeberatedEHPofNumber ph
0==
External Quantum Efficiency
0(W)(A)
PI
PowerOpticalIncidentntPhotocurreR ph==
Responsivity
che
heR λ
ην
η == Spectral Responsivity
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Responsivity vs. wavelength for a typical Si photo-diode
0 200 400 600 800 1000 12000
0.10.20.30.40.50.60.70.80.9
1
Wavelength (nm)
Si Photodiode
λg
Res
pons
ivity
(A/W
)
Ideal PhotodiodeQE = 100% ( η = 1)
Photo-detectors
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The pin Photo-diode
•
•
Intrinsic layer has less doping and wider region (5 – 50 μm).
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Reverse-biased p-i-n photodiode
Photo-detectors: PIN Photo-diode
pin energy-band diagram
pin photodiode circuit
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SiO2
Schematic diagram of pin photodiode
p+
i-Si n+
Electrode
ρnet
–eNa
eNd
x
x
E(x)
R
E0
e–h+
Iph
hυ > Eg
W
Vr
Vout
Electrode
E
Small depletion layer capacitance gives high modulation frequencies.
High Quantum efficiency.
In contrast to pn junction built-in-field is uniform
Photo-detectors: PIN Photo-diode
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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+ i-Si
e–
h+
Wℓ
Drift
Diffusion
Vr
E
Photo-detectors: PIN Photo-diode
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p-i-n diode
(a) The structure;
(b) equilibrium energy band diagram;
(c) energy band diagram under reverse bias.
Photo-detectors: PIN Photo-diode
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The responsivity of PIN photodiodes
Photo-detectors: PIN Photo-diode
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Quantum efficiency versus wavelength for various photo-detectors
Photo-detectors: Photo-conductive Detectors and Gain
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WAC r
depεε 0=
Electric field of biased pin
Junction capacitance of pin
Small capacitance: High modulation frequency
RCdep time constant is ∼ 50 psec.
WV
WVEE 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 i-Si layer.
For i-Si layer of width 10 μm, the drift time is about is about 0.1 nsec.
Photo-detectors: PIN Photo-diode
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Drift velocity vs. electric field for holes and electrons in Silicon.
102
103
104
105
107106105104
Electric field (V m-1)
Electron
Hole
Dri
ft v
eloc
ity (m
sec-1
)
Photo-detectors: PIN Photo-diode
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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 5×10-2 cm2 is irradiated with yellow light whose
intensity is 20 mW cm−2. Assuming that each photon generates one electron-hole 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 = (6.626×10-34 J s)(3×108 m s-1)/(600×10-9 m) = 2.07 eV
(b) Area = 5×10-2 cm2 and Ilight = 20×10-3 W/cm2. The received power isP = Area ×Ilight = (5×10-2 cm2)(20×10-3 W/cm2) = 10-3 WNph = number of photons arriving per second = P/Eph
= (10-3 W)/(2.059×1.60218×10-19 J/eV)= 2.9787×1015 photons s-1 = 2.9787×1015 EHP s-1.
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(c) For GaAs, Eg = 1.42 eV and the corresponding wavelength is λ = hc/ Eg = (6.626×10-
34 J s)(3×108 m s-1)/(1.42 eV×1.6×10-19 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 cut-off wavelength is λg = hc/ Eg = (6.626×10-34 J s)(3×108 m s-1)/(1.1 eV×1.6×10-19 J/eV) = 1120 nmSince the 873 nm wavelength is shorter than the cut-off 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 electron-hole recombination. Is this wavelength in the visible?
(d) Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why?
Solution
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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[ ]
υα
hddInph
)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
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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 m-1 at 1.5 μm incident radiation.
mmd
d
μα
α
428.410428.49.01
1ln102.5
19.01
1ln1
9.0)exp(1
65 =×=⎟
⎠⎞
⎜⎝⎛
−×=⎟
⎠⎞
⎜⎝⎛
−=
=⋅−−∴
−
For In0.53Ga0.47As, α ≈ 7.5 × 105 m-1 at 1.5 μm incident radiation.
mmd μ07.31007.39.01
1ln105.7
1 65 =×=⎟
⎠⎞
⎜⎝⎛
−×= −
For Ge, α ≈ 5.2 × 105 m-1 at 1.5 μm incident radiation.For In0.53Ga0.47As, α ≈ 7.5 × 105 m-1 at 1.5 μm incident radiation.
(b)
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Example
0
0.2
0.4
0.6
0.81
800 10001200 14001600 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 below. 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?
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Solution
(a) At λ = 1.55×10-6 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 s-1 and W = J s-1 so that A W-1 = C J-1. Thus, responsivity in terms of photocurrent per unit incident optical power is also charge collected per unit incident energy.
0)()(
PI
WPowerOpticalIncidentAntPhotocurreR ph==
nWWA
AR
IR
IP darkph 5.11
)/87.0)(10522 9
0 =××
===−
hce
heR ληυ
η ==
%)70(70.0)1055.1)(106.1(
)/87.0)(/103sec)(1062.6(619
834
=××
×⋅×== −−
−
mcoulWAsmJ
ehcR
λη
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Solution
(b) At λ = 1.3×10-6 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
≈××
×⋅×== −−
−
mcoulWAsmJ
ehcR
λη
nAnWWAPRI ph 43.9)15.1)(/82.0(0 ==⋅=
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π p+
SiO2Electrode
ρnet
x
x
E(x)
R
hυ > Eg
p
Iphoto
e – h+
Absorptionregion
Avalanche region
Electrode
n+
E
Photo-detectors: Avalanche Photo-diode (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.
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SiO2
Guard ring
ElectrodeAnti-reflection coating
nn n+
p+
π
p
SubstrateElectrode
n+
p+
π
p
SubstrateElectrode
Avalanche breakdown
Si APD structure without a guard ring
More practical Si APD
Schematic diagram of typical Si APD.
Breakdown voltage around periphery is higher and avalanche is confined more to illuminated region (n+p junction).
Photo detectors: Avalanche Photo-diode (APD)