• Author / Uploaded
• Aditya Kumar

Citation preview

ELEC425/1-2012 1 Assignment 4

Assignment #4. 5.1 (1.0 %), 5.2 (0.6 %), 5.3 (1.0 %), 5.5 (0.4 %), 5.8 (1.0 %), 5.9 (0.4 %), 5.10 (0.6 %) 5.1. Band gap and photodetection. a) Determine the maximum value of the energy gap (bandgap) which a semiconductor, used as a photodetector, 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 2 mW cm⁻². Assuming that each photon generates one electron-hole pair (EHP), calculate the number of EHPs generated per second. 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. a) E=

hc

λ

6.626 × 10 −34 × 3 × 108 E= = 3.313 × 10 −19 J or 2.07 eV −9 600 × 10

b) # of EHP generated per second = # of incident photons per second =

I ph e

=

P0 hν

where Iph/e is # of electrons per second and P₀/hν is # photons per second P0 (2 ×10−3 × 5 ×10−2 )W = = 3.02 ×1014 s −1 Therefore, # of EHP generated per second = −19 hν (3.313 ×10 ) J

ELEC425/1-2012 2 Assignment 4

c) λ=

hc 1.24 = ≈ 0.873 µm this is infrared light and it is not in visible region. E 1.42

d) for Si, Eg = 1.12 eV and for GaAs, Eg = 1.42 eV corresponding cut-off wavelength for Si is λcut−off =

hc 1.24 = ≈ 1.107 µm Eg 1.12

The cut-off wavelength of GaAs is shorter (873 nm) than cut-off wavelength of Si (1.107 μm); thus, Si photodetector will be sensitive to the radiation from a GaAs laser. In other words, the bandgap of the silicon is smaller than it is in GaAs and emitted photons from laser (GaAs) will have higher energy than silicon’s energy bandgap. Consequently, photons will result in generation of EHP in silicon photodetector. 5.2. Absorption coefficient. a) If d is the thickness of a photodetector material, Iₒ is the intensity of the incoming radiation, show that the number of photons absorbed per unit volume of sample is n ph =

I 0 [1 − exp(− αd )] dhv

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? c) Suppose that each absorbed photon liberates one electron (or electron hole pair) in a unity quantum efficiency photodetector and that the photogenerated electrons are immediately collected. Thus, the rate of charge collection is limited by rate of photogeneration. What is the external photocurrent density for the photodetectors in (b) if the incident radiation is 100 μW mm⁻²?

ELEC425/1-2012 3 Assignment 4

Photon energy (eV) 5 4

3

1

2

0.9

0.8

0.7

1×108

Ge

1×107

In0.7Ga0.3As0.64P0.36 In0.53Ga0.47As

Si 1×106 GaAs

α (m-1)

InP

1×105 a-Si:H 1×104

1×103 0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

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

Figure 5.3. S. O. Kasap “Optoelectronics and Photonics” Solution. a) let I₀ is incoming radiation which is represented by energy flowing per unit area per second and I₀[1-exp(-αd)] is the absorbed intensity. Photon flux is the number of photons arriving per unit area per unit second, P₀/(hνArea)=I₀/hν. So, absorbed photon flux per unit area is # of photons absorbed per volume =

I 0 [1 − exp(− αd )] . hν

I 0 [1 − exp(− αd )] dhν

ELEC425/1-2012 4 Assignment 4

b) from Fig. 5.3 α ≈ 6.0 × 10⁵ m⁻¹ is the absorption coefficient for Ge at 1.5 μm incident radiation I (d ) = I 0 exp(− αd ) = 0.1I 0 exp(− αd ) = 0.1 ln 0.1 d =−

α

d =−

ln 0.1 ≈ 3.84 µm 6.0 × 105

from Fig. 5.3 α ≈ 7.5 × 10⁵ m⁻¹ is the absorption coefficient for In0.53Ga0.47As at 1.5 μm incident radiation I (d ) = I 0 exp(− αd ) = 0.1I 0 exp(− αd ) = 0.1 ln 0.1 d =−

α

d =−

ln 0.1 ≈ 3.07 µm 7.5 × 105

c) given ηQE = 1, the rate of charge collection is limited by rate of photogeneration, and I₀ = 100 μW/mm² = 100 W/m² I ph e

=

e × I 0 [1 − exp(− αd )] e × 0.9 I 0 × λ = area hν hc −19 −6 1.6 × 10 × 100 × 0.9 × 1.5 × 10 = = 108.663 A/m 2 or 10.866 A/cm 2 −34 8 6.626 × 10 × 3 × 10

J ph = J ph

P I / area I 0 (1 − exp(− αd )) = = hν hν hν × area I ph

=

The reflection of the light from the surface of the photodetector is neglected. Assumed that the anti-reflective coating has efficiency 100%. 5.3. Ge Photodiode. Consider a commercial Ge pn junction photodiode which has the responsivity shown in Figure 5.20. Its photosensitive area is 0.008 mm². It is used under a

ELEC425/1-2012 5 Assignment 4

reverse bias of 10 V when the dark current is 0.3 μA and the junction capacitance is 4 pF. The rise time of the photodiode is 0.5 ns. a) Calculate its quantum efficiency at 850, 1300 and 1550 nm. b) What is the intensity of light at 1.55 μm that gives a photocurrent equal to the dark current? c) What would be the effect of lowering the temperature on the responsivity curve? d) Given that the dark current is in the range of microamperes, what would be the advantage in reducing the temperature? e) Suppose that the photodiode is used with a 100 Ω resistance to sample the photocurrent. What limits the speed of response? Responsivity(A/W) 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.5

1 1.5 Wavelength(µm)

2

The responsivity of a commercial Ge pn junction photodiode © 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Figure 5.20. S. O. Kasap “Optoelectronics and Photonics”

ELEC425/1-2012 6 Assignment 4

Solution. a) area = 0.008 mm² = 8.0·10⁻⁹ m² ηQE = R=

η EQ =

Wavelength, nm Responsivity, A/W ηQE, %

850 0.26 38.0

I ph / e P0 / hν

I ph P0

Rhν hcR = e eλ

1300 0.58 55.43

1550 0.72 57.71

b) Iph = Id = 0.3 μA P0 = I ph / R = 0.3 × 10−6 / 0.72 ≈ 0.42 × 10 −6 W or 0.42 µW I0 =

P0 0.42 × 10 −6 = = 52.5 W/m 2 or 5.25 mW/cm2 area 8.0 × 10−9

c) The energy band gap is increasing with the decreasing of the temperature; consequently, the cut-off wavelength is decreasing as temperature is decreasing. So, the higher photon energy is needed to initiate photon absorption. For instance, the curves representing the relationship between absorption coefficient and wavelength demonstrated in Figure 5.3 will be shifted to the left, when temperature is decreased. Hence, the same absorption coefficient for a given semiconductor will be at lower wavelength and at higher photon energy when temperature is decreased. The change in the absorption coefficient attributable to the variations in temperature means that the optical power absorbed in the depletion region and the quantum efficiency vary with temperature. The peak of the responsivity in Figure 5.20 will move to the left, to the lower values of the wavelength, with decreasing

ELEC425/1-2012 7 Assignment 4

temperature, since the amount of the optical power absorbed in depletion region increases with the decreasing in temperature value. d) Dark current is proportional to exp(-Eg/kBT), therefore, it will reduced if the temperature will be decreased. The advantage is the improvement of SNR. e) time constant limitation = RC = 100 Ω·4 pF = 0.4 ns The RC constant is comparable to the rise time, 0.5 ns. Therefore, the speed of the response depends on both the rise time and RC constant 5.5. InGaAs pin Photodiodes. Consider a commercial InGaAs pin photodiode whose responsivity is shown in Figure 5.22. 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?

ELEC425/1-2012 8 Assignment 4

Responsivity(A/W) 1 0.8 0.6 0.4 0.2 0 800 1000 1200 1400 1600 1800 Wavelength(nm) The responsivity of an InGaAs pin photodiode © 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Figure 5.22. S. O. Kasap “Optoelectronics and Photonics” Solution. a) R=

I ph P0

= 0.87 from Figure 5.22

2 × I dark 2 × 5 × 10 −9 P0 = = = ≈ 11.49 nW R R 0.87 I ph

ηQE =

hcR 6.626 ×10−34 × 3 ×108 × 0.87 = ≈ 0.6973 or 70.0 % eλ 1.6 ×10−19 ×1550 ×10−9

From the dimensional identities: [Iph] = A=C/s, [P₀] = W = J/s, [R] = A/W = C/J So, responsivity is the charge collected per unit incident energy b) for λ = 1.3 μm R = 0.82 A/W from Figure 5.22

ELEC425/1-2012 9 Assignment 4

P0 =

I ph R

= 11.49 nW from part (a) I ph = P0 R = 11.49 × 10 −9 × 0.82 = 9.422 nA

ηQE =

hcR 6.626 ×10−34 × 3 ×108 × 0.82 = ≈ 0.784 or 78.4 % eλ 1.6 ×10−19 ×1.3 ×10−6

5.8. Transient photocurrents in a pin photodiode. Consider a reverse biased Si pin photodiode as shown in Figure 5.23. It is appropriately reverse biased so that the field in the depletion region (i-Si layer) E=Vr/W is the saturation field. Thus, photogenerated electrons and holes in this layer drift at saturation velocities vde and vdh. Assume that the field E is uniform and that the thickness of the p⁺ is negligible. A very short light pulse (infinitesimally short) photogenerates EHPs in the depletion layer as shown in Figure 5.23 which results in an exponentially decaying EHP concentrations across W. Figure 5.23 shows the photogenerated electron concentration at time t = 0 and also at a later time t when the electrons have drifted a distance ∆x = vde∆t. Those that reach the back electrode B become collected. The electron distribution shifts at a constant velocity until the initial electrons at A reach B which represents the longest transit time τe = W/vde. Similar argument apply to holes but they drift in the opposite direction and their transit time τh = W/vdh where vdh is their saturation velocity. The photocurrent density at any instant is j ph = je (t ) + jh (t ) = eN e vde + eN h vdh

where Ne and Nh are the overall electron and hole concentration in the sample at time t. Assume for convenience that the cross sectional area A = 1 (derivations below are not affected as we are interested in the photocurrent densities). a) Sketch the hole distribution at a time t where τh > t > 0 and τh = hole drift time = W/vdh. b) The electron concentration distribution n(x) at time t corresponds to that at t = 0 shifted by vdet. Thus the total electrons in W is proportional to integrating this distribution n(x) from A at x = vdet to B at x = W.

ELEC425/1-2012 10 Assignment 4

Given n(x) = n₀exp(-αx) at t = 0, where n₀ is the electron concentration at x = 0 at t = 0 we have Total number of electrons at time t =

W

∫

vdet

n0 exp[− α ( x − vdet )]dx

and Ne =

Total number electrons at time t Volume

Then N e (t ) =

1 W

W

∫

vdet

n0 exp[− α ( x − vdet )]dx =

n0 W

   t    1 − exp − 1 − W α    τ    e     

where Ne(0) is the initial overall electron concentration at time t = 0, that is, N e (0 ) =

1 W

W

∫

0

n0 exp(− αx )dx =

n0 [1 − exp(− αW )] Wα

We note that n₀ depends on the intensity I of the light pulse so that n₀ ∝ I. Show that for holes, N h (t ) =

 n0 exp(− αW )   t   exp αW 1 −  − 1 Wα    τ h  

c) Given W = 40 μm, α = 5 × 10⁴ m⁻¹, vde = 10⁵ m/s, vdh = 0.8 × 10⁵ m/s, n₀ = 10¹³ cm⁻³, calculate the electron and hole transit times, sketch the photocurrent densities je(t) and jh(t) and hence jph(t) as a function of time, and calculate the initial photocurrent. What is your conclusion?

ELEC425/1-2012 11 Assignment 4

Photogenerated electron concentration exp(−αx) at time t = 0

v de x A

B

W E

hυ > E g

h+

e–

iph

R

Vr

An infinitesimally short light pulse is absorbed throughout the depletion layer and creates an EHP concentration that decays exponentially © 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Figure 5.23. S. O. Kasap “Optoelectronics and Photonics”

Solution. a) The hole distribution is resemble to the electron distribution as it is shown in Figure 5.23. An infinitesimally short light pulse is absorbed throughout the depletion layer and creates an EHP concentration that decays exponentially.

ELEC425/1-2012 12 Assignment 4

b) 1 W

n0 [exp(− αvdht ) − exp(− αW )] 0 Wα  n exp(− αW )   t   N h (t ) = 0 exp αW 1 −  − 1 Wα    τ h  

N h (t ) =

W −vdht

∫

n0 exp[− α ( x + vdht )]dx =

c) 40 × 10 −6 = 400 ps 105 40 × 10 −6 τ h = W / vdh = = 500 ps 0.8 × 105

τ e = W / vde =

At time t = 0, Ne(0) = Nh(0), n₀ = 10¹³ cm⁻³ = 10⁷ m⁻³ n0 [1 − exp(− αW )] Wα 1019 N e (0 ) = 1 − exp − 5 × 10 4 × 40 × 10 −6 ≈ 4.323 × 1018 m -3 or 4.3 × 1012 cm -3 40 × 10 −6 × 5 × 10 4 N e (0 ) =

[

(

)]

The initial currents are je (0 ) = eN e (0 )vde = 1.6 × 10 −19 × 4.3 × 1018 × 105 ≈ 6.9 × 10 4 A/m 2 or 6.9 A/cm 2 or 69 mA/mm2 jh (0 ) = eN h (0 )vdh = 1.6 × 10 −19 × 4.3 × 1018 × 0.8 × 105 ≈ 5.5 × 10 4 A/m 2 or 5.5 A/cm 2 or 55 mA/mm2

the total initial photocurrent is je(0)+jh(0) = 69+55= 124 mA/mm² The individual transient photocurrents are given by je (t ) = eN e (t )vde =

jh (t ) = eN h (t )vdh =

en0vde Wα

   t   1 − exp − αW 1 −   for t < τ e   τ e   

 en0vdh exp(− αW )   t   exp αW 1 −  − 1 for t < τ h Wα    τ h  

The response is determined by the slowest transient time. There is a kink in the photocurrent waveform when all the electrons have been swept out at τe = 400 ns.

ELEC425/1-2012 13 Assignment 4

14

x 10

Photocurrent density

4

12

Photocurrent density (A/m2)

10 j (t) t

8 j (t) e

Total photocurrent density

6

4

electrons

j (t) h

holes 2

0 0

1

2

3 Time (sec)

4

5

6 x 10

τe

-10

τh

5.9. Fiber attenuation and InGaAs pin Photodiode. Consider the commercial InGaAs pin photodiode whose responsivity is shown in Figure 5.22. This is used in a receiver circuit that needs a minimum of 5 nA photocurrent for a discernible output signal (acceptable signal to noise ratio for the customer). Suppose that the InGaAs pin PD is used at 1.3 μm operation with a single mode fiber whose attenuation is 0.35 dB km⁻¹. If the laser diode emitter can launch at most 2 mW of power into the fiber, what is the maximum distance for the communication without a repeater? Solution. If Iph = 5 nA, R = 0.81 A/W at 1.3 μm wavelength from Figure 5.22 Power absorbed by photodiode = P0 =

I ph R

=

5 ×10−9 = 6.173 × 10−9 W 0.81

ELEC425/1-2012 14 Assignment 4



2 × 10 −3   = 55.105 dB −9   6.173 × 10 

The attenuation loss is 10 log10 (Pin / Pout ) = 10 log10  From attenuation coefficient α = 0.35 dB/km

The maximum distance for the communication without a repeater, L, is given as L=

attenuation loss

α

=

55.105 = 157.444 km 0.35

5.10. Photoconductive detector. An n-type Si photodetector has a length L = 100 μm and a hole lifetime of 1 μs. The applied bias to the photoconductor is 10 V. a) What are the transit times, te and th, of an electron and a hole across L? What is the photoconductive gain? b) It should be apparent that as electrons are much faster than holes, a photogenerated electron leaves the photoconductor very quickly. This leaves behind a drifting hole and therefore a positive charge in the semiconductor. Secondary (i.e. additional electrons) then flow into the photoconductor to maintain neutrality in the sample and the current contributes to flow. These events will continue until the hole has disappeared by recombination, which takes on average a time τ. Thus more charges flow through the contact per unit time than charges actually photogenerated per unit time. What will happen if the contacts are not ohmic, i.e. they are not injecting? c) What can you say about the product ∆σ and the speed of response which is proportional to 1/τ. Solution. a) from given length and applied voltage E=V/L=10 V/100 μm = 10⁵ V/m from the inside cover of the textbook: µe = 1350 cm 2V −1s −1 , µ h = 450 cm 2V −1s −1 The transit times of an electron and a hole across L is given

ELEC425/1-2012 15 Assignment 4

L 100 × 10 −6 te = = = 7.41 ns µ e E 1350 × 10−4 × 105 th =

100 × 10 −6 L = = 22.22 ns µ h E 450 × 10−4 × 105

Another way to solve for transit times of an electron and hole across L in Si is by using Figure 5.7: ve =1.3 × 10⁴ m/s at E = 10⁵ V/m and vh ≈ 4.5 × 10³ m/s L 100 × 10 −4 te = = = 7.69 ns ve 1.3 × 10 4 L 100 × 10 − 4 th = = = 22.22 ns vh 4.5 × 103

Photoconductive gain is G=

Rate of electron flow in external circuit τ (µ e + µ h )E = Rate of electron generation by light absorption L

G=

1× 10 −6 (1350 + 450 ) × 10 −4 × 105 = 180 100 × 10 −6

b) if the contacts are not ohmic, secondary electrons cannot flow into the photoconductor to maintain neutrality. So, only the photogenerated charges can flow through the external circuit; no excess charge can flow and we will not get photoconductive gain. If the contacts cannot inject carriers, then there will be no photocurrent gain, G = 1. c) change in the conductivity or photoconductivity is ∆σ =

eηIλτ (µe + µ h ) hcd

The speed of response is inversely proportional to the recombination time of the minority carriers, τ. For instance, if the light is turned off, it will take τ seconds for the excess carriers to disappear by recombination. Therefore, the product of ∆σ and the speed of response is proportional to ∆σ

1

τ

=

eηIλ (µe + µ h ) hcd

which is constant for a given device geometry and light intensity.