chap6 photodetectors

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Photodetectors

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Photodetectors

PhotodetectorsPrinciple of the p-n junction Photodiode

Schematic diagram of a reverse biased p-n 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 p-n junction Photodiode

(b) Energy band diagram under reverse bias.

(a) Cross-section view of a p-i-n photodiode.

(c) Carrier absorption characteristics.

Operation of a p-i-n photodiode.

A generic photodiode.

PhotodetectorsPrinciple of the p-n junction 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 e-1 of its surface value.

PhotodetectorsPrinciple of the p-n junction Photodiode

PhotodetectorsRAMO’s Theorem and External Photocurrent

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

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.

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 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.

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

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)

Absorption coefficient ( α) vs. wavelength (λ) for various semiconductors(Data selectively collected and combined from various sources.)

α (m-1)

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

a-Si: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 indirect-gap 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 k-vector (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).

Reverse-biased p-i-n photodiode

PhotodetectorsThe pin Photodiode

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

PhotodetectorsThe pin Photodiode

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+

W

Drift

Diffusion

Vr

E

PhotodetectorsThe pin Photodiode

p-i-n diode

(a) The structure;

(b) equilibrium energy band diagram;

(c) energy band diagram under reverse bias.

PhotodetectorsThe pin Photodiode

PhotodetectorsThe pin Photodiode

The responsivity of pin photodiodes

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 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 s

ec-1)

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 5×10-2 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 is

P = Area ×Ilight = (5×10-2 cm2)(20×10-3 W/cm2) = 10-3 W

Nph = 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,

λ g = 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

[ ]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 m-1 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 m-1 at 1.5 µm incident radiation.

mmd µ07.31007.39.01

1ln

105.7

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

0.8

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 = 1.55×10-6 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)(

)(

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 = 1.3×10-6 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

≈××

×⋅×== −−

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

Iph

Absorptionregion

Avalancheregion

InP InGaAs

h+

e–

InP

P + n+

VrVout

E

PhotodetectorsHeterojunction Photodiode

Separate Absorption and Multiplication (SAM) APD

InGaAs-InP heterostructure Separate Absorption and Multiplication APD

P and N refer to p- and n-type wider-bandgap 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.

x

R

e–h+

iph

hυ > Eg

W

Vr

Photogenerated electron concentration

exp( − αx) at time t = 0

BA

vde

E