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Photodetectors

Photodiodes

2Slide Set 6ELEC4105


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.

6Slide Set 6ELEC4105 6

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.


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


A generic photo-diode

Photodetectors: Principle of the p-n junction Photo-diode


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


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)


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


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.


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


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.



• 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).



• 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.



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: 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


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


The pin Photo-diode

•

•

Intrinsic layer has less doping and wider region (5 – 50 μm).


Reverse-biased p-i-n photodiode

Photo-detectors: PIN Photo-diode

pin energy-band diagram

pin photodiode circuit


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



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



p-i-n diode

(a) The structure;

(b) equilibrium energy band diagram;

(c) energy band diagram under reverse bias.



The responsivity of PIN photodiodes



Quantum efficiency versus wavelength for various photo-detectors

Photo-detectors: Photo-conductive Detectors and Gain


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.



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

)



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.


(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


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


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)


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?


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


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 ==⋅=


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.


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)

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