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Chapter 1 1.4

Antireflection coating

a Consider three dielectric media with flat and parallel boundaries with refractive indices n1, n2 and n3. Show that for normal incidence the reflection coefficient between layers 1 and 2 is the same as that between layers 2 and 3 if n2 = √[n1n3]. What is the significance of this? b Consider a Si photodiode that is designed for operation at 900 nm. Given a choice of two possible antireflection coatings, SiO2 with a refractive index of 1.5 and TiO2 with a refractive index of 2.3 which would you use and what would be the thickness of the antireflection coating? The refractive index of Si is 3.5. Explain your decision. 1.8

Thin film coating and multiple reflections: A1

A0

A2

A3

n1 B1

B3 B2

B5 B4

n2 B6

n3 C1

C2

C3

Thin film coating of refractive index n2 on a semiconductor device

Figure 1Q8 Consider a thin film coating on an object as in Figure 1Q8. Suppose that the incident wave has an amplitude of A0, then there are various transmitted and reflected waves as shown in Figure 1Q8. We then have the following amplitudes based on the definitions of the reflection and transmission coefficients, A1 = A0r12

B1 = A0t12

B2 = A0t12r23

A2 = A0t12r23t21

B3 = A0t12r23r21 B4 = A0t12r23r21r23

A3 = A0t12r23r21r23t21

B5 = A0t12r23r21r23r21

C1 = A0t12t23 C2 = A0t12r23r21t23

B6 = A0t12r23r21r23r21r23

C3 = A0t12r23r21r23r21t23

and so on. Assume that n1 < n2 < n3 and that the thickness of the coating is d. For simplicity, we will assume normal incidence. The phase change in traversing the coating thickness d is φ = (2π/λ)n2d where λ is the free space wavelength. The wave has to be multiplied by exp(−jφ) to account for this phase difference. The reflection and transmission coefficients are given by, n −n n − n3 r1 = r12 = 1 2 = −r21 , r2 = r23 = 2 n1 + n2 n2 + n3 t 1 = t 12 =

and a

2n1 , n1 + n2

Show that 1 − t1t 2 = r1

2

b

Show that the reflection coefficient is

t 2 = t 21 =

2n2 , n1 + n2

t 23 =

2n3 , n2 + n 3

r=

Areflected tt = r1 − 1 2 A0 r1

∞

∑(−r r e k =1

)

− j2 φ k

1 2

which can be summed to r=

c

r1 + r2 e − j 2 φ − j 2φ 1+ r1r2 e

Show that the transmission coefficient is

Ctransmitted t1t 23e jφ ∞ − j 2φ k t= =− −r1 r2e ) ∑ ( A0 r1r2 k=1

t=

which can be summed to

 t1t 23e jφ  r1r2e − j 2φ  = − j 2φ  r1r2  1+ r1 r2 e

t1t 23e − jφ − j 2φ 1+ r1r2 e

1.9 Antireflection coating: Consider a thin film coating on an object as shown in Figure 1Q9. Using the transmission coefficient in Question 1.8 show that the a normally incident light beam has maximum transmission into medium 3 when

d=

mλ 4n2

where m is an odd-integer and λ is the free space wavelength. Show, in addition, that we need r1r2 = 1. which requires choosing n2 = (n1n3)1/2. Derive the same result using the reflection coefficient in Question 1.8. A1

A0

A2

A3

n1 B3

B1 B2

B5 B4

n2 B6

n3 C1

C2

C3

Thin film coating of refractive index n2 on a semiconductor device

Figure 1Q9 1.16 Diffraction by a lens Any lens in practice is an aperture and the image of a point is therefore a diffraction pattern. Suppose a lens with a diameter of 2 cm has a focal length of 40 cm. Suppose that it is illuminated with a plane wave, a collimated beam of light, of wavelength 590 nm. What is the diameter of the Airy disk at the focal point? What is your conclusion?

Chapter 2 2.1

Dielectric slab waveguide

a

Consider the rays 1 and 2 in Figure 2.3. Derive the waveguide condition.

b C

Consider the two rays 1 and 2 in Figure 2.4. Show that the phase difference when they meet at at a distance y above the guide center is

Φm = k12(a − y)cosθm − φm

c

Using the waveguide condition show that y Φ m = Φ m (y) = mπ − (mπ + φm ) a n2

A

B′

1

θ

θ

π−2θ

E k1 2

A′

y 2 θ−π/2

C

2a

θ

n1 n2

n2

A

x 1

B

Figure 2.3 Two arbitrary waves 1 and 2 that are initially in phase must remain in phase after reflections. Otherwise the two will interfere destructively and cancel each other.

C

1

θ

θ

a−y π−2θ

E

z

k Guide center

A′

a

y

y

2 z x

Figure 2.4 Interference of waves such as 1 and 2 leads to a standing wave pattern along the y-direction which propagates along z.

2.5 Dielectric slab waveguide Consider a dielectric slab waveguide which has a thin GaAs layer of thickness 0.2 µm between two AlGaAs layers. The refractive index of GaAs is 3.66 and that of the AlGaAs layers is 3.40. What is the cut-off wavelength beyond which only a single mode can propagate in the waveguide assuming that the refractive index does not vary greatly with the wavelength? If a radiation of wavelength 870 nm (corresponding to bandgap radiation) is propagating in the GaAs layer, what is the penetration of the evanescent wave into the AlGaAs layers? What is the mode field distance of this radiation? 2.8 A multimode fiber Consider a multimode fiber with a core diameter of 100 µm, core refractive index of 1.475 and a cladding refractive index of 1.455 both at 850 nm. Consider operating this fiber at λ = 850 nm. a

Calculate the V-number for the fiber and estimate the number of modes.

b

Calculate the wavelength beyond which the fiber becomes single mode.

c

Calculate the numerical aperture.

d

Calculate the maximum acceptance angle.

e Calculate the modal dispersion ∆τ and hence the bit rate × distance product given that rms dispersion σ ≈ 0.29∆τ where ∆τ is the full spread. 2.9 A single mode fiber Consider a fiber with a SiO2-13.5%GeO2 core of diameter of 8 µm and refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm where the fiber is to be operated using a laser source with a half maximum width of 2 nm. a

Calculate the V-number for the fiber. Is this a single mode fiber?

b

Calculate the wavelength below which the fiber becomes multimode.

c

Calculate the numerical aperture.

d

Calculate the maximum acceptance angle.

e Obtain the material dispersion and waveguide dispersion and hence estimate the bit rate × distance product (B×L) of the fiber. 2.18 Microbending loss It is found that for a single mode fiber with a cut-off wavelength λc = 1180 nm, operating at 1300 nm, the microbending loss reaches 1 dB m-1 when the radius of curvature of the bend is roughly 6 mm for ∆ = 0.00825, 12 mm for ∆ = 0.00550 and 35 mm for ∆ = 0.00275. Explain these findings.

Chapter 3 3.2 GaAs GaAs has an effective density of states at the conduction CB Nc of 4.7×1017 cm-3 and an effective density of states at the VB edge Nv of 7×1018 cm-3. Given its bandgap Eg of 1.42 eV calculate the intrinsic concentration and the intrinsic resistivity at room temperature (take as 300 K). Where is the Fermi level? Assuming that Nc and Nv scale as T3/2, what would be the intrinsic concentration at 100 °C? If this GaAs crystal is doped with 1018 donors cm-3 (such as Te), where is the new Fermi level and what is the resistivity of the sample? The drift mobilities in GaAs are shown in Table 3Q2 Table 3Q2 Dopant

impurities scatter carriers and reduce the drift mobility (µe for electrons and µh for holes).

Dopant concentration (cm-3)

0

1015

1016

1017

1018

µe (cm2 V-1 s-1)

8500

8000

7000

4000

2400

µh (cm2 V-1 s-1)

400

380

310

220

160

3.3 Direct bandgap pn junction Consider a GaAs pn junction which has the following properties. Na = 1016 cm-3 (p-side), Nd = 1016 cm-3 (n-side), B = 7.21×10-16 m3 s-1, cross sectional area A = 0.1 mm2. What is the diode current due to diffusion in the neutral regions at 300 K when the forward voltage across the diode is 1 V? See Question 3.2 and Table 3Q2 for GaAs properties.

3.4

The Si pn junction

Consider a long pn junction diode with an acceptor doping, Na, of 1018 cm-3 on the p-side and donor concentration of Nd on the n-side. The diode is forward biased and has a voltage of 0.6 V across it. The diode cross-sectional area is 1 mm2. The minority carrier recombination time, τ, depends on the dopant concentration, Ndopant (cm-3), through the following approximate relation

τ=

5 × 10 −7 (1+ 2 × 10 −17 N dopant )

a Suppose that Nd = 1015 cm-3. Then the depletion layer extends essentially into the n-side and we have to consider minority carrier recombination time, τh, in this region. Calculate the diffusion and recombination contributions to the total diode current given that when Na = 1018 cm-3, µe ≈ 250 cm2 V-1 s-1, and when Nd = 1015 cm-3, µh ≈ 450 cm2 V-1 s-1. What is your conclusion? b Suppose that Nd = Na. Then W extends equally to both sides and, further, τe = τh. Calculate the diffusion and recombination contributions to the diode current given that when Na = 1018 cm-3, µe ≈ 250 cm2 V-1 s-1, and when Nd = 1018 cm-3, µh ≈ 130 cm2 V-1 s-1. What is your conclusion?

3.5 AlGaAs LED emitter An AlGaAs LED emitter for use in a local optical fiber network has the output spectrum shown in Figure 3Q5. It is designed for peak emission at 820 nm at 25°C. a What is the linewidth ∆λ between half power points at temperatures −40°C, 25°C and 85°C? What is the empirical relationship between ∆λ and T given three temperatures and how does this compare with ∆(hυ) ≈ 2.5kBT − 3kBT? b

Why does the peak emission wavelength increase with temperature?

c

Why does the peak intensity decrease with temperature?

d

What is the bandgap of AlGaAs in this LED?

e

The bandgap, Eg, of the ternary alloys AlxGa1-xAs follows the empirical expression, Eg(eV) = 1.424 + 1.266x + 0.266x2.

What is the composition of the AlGaAs in this LED? f When the forward current is 40 mA, the voltage across the LED is 1.5V and the optical power that is coupled into a multimode fiber through a lens is 25 µW. What is the efficiency? Relative spectral output power ?0蚓 1

25蚓

85蚓

0 740

800

840

The output spectrum from GaAlAs LED. Values normalized to peak emission at 25蚓.

880 900

Wavelength (nm)

Figure 3Q5

3.7 as

External conversion efficiency

ηext =

The external power or conversion efficiency ηext is defined

Optical power output P = o Electrical power input IV

One of the major factors reducing the external power efficiency is the loss of photons in extracting the emitted photons which suffer reabsorption in the pn junction materials, absorption outside the semiconductors and various reflections at interfaces. The total light output power from a particular AlGaAs red LED is 2.5 mW when the current is 50 mA and the voltage is 1.6 V. Calculate its external conversion efficiency.

3.8 Linewidth of LEDs Experiments carried out on various direct bandgap semiconductor LEDs give the output spectral linewidth (between half intensity points) listed in Table 3Q8-1. We know that a spread in the wavelength is related to a spread in the photon energy, hc ∆ λ ≈ 2 ∆E ph (1) E ph Suppose that we write Eph = hc/λ and ∆Eph = ∆(hυ) ≈ mkBT where m is a numerical constant. Show that,

∆ λ ≈ λ2

mkB T hc

(2)

and by appropriately plotting the data in Table 3Q8-1 find m. Table 3Q8-1 shows the linewidth ∆λ1/2 for various visible LEDs. Radiative recombination is obtained by appropriately doping the material. Using m ≈ 3, T = 300 K, in Eq. (2) calculate the expected spectral width for each and compare with the experimental value. What is your conclusion? Do you think EN in Figure 3.24 (b) is a discrete level?

Table 3Q8-1 Linewidth ∆λ1/2 between half points in the output spectrum (Intensity vs wavelength) of GaAs and AlGaAs LEDs. Peak wavelength of emission (λ) nm

650

810

820

890

950

1150

1270

1500

∆λ1/2 nm

22

36

40

50

55

90

110

150

Material (Direct Eg)

AlGaAs

AlGaAs

AlGaAs

GaAs

GaAs

InGaAsP

InGaAsP

InGaAsP

Table 3Q8-2 Linewidth ∆λ1/2 between half points in the output spectrum (Intensity vs wavelength) of four various visible LEDs using SiC and GaAsP materials. Peak wavelength of emission (λ) nm

468

565

583

600

635

∆λ1/2 nm

66

28

36

40

45

Color

Blue

Green

Yellow

Orange

Red

Material

SiC (Al)

GaP (N)

GaAsP (N)

GaAsP (N)

GaAsP

3.10

LED-Fiber coupling Efficiency

a It is found that approximately 200 µW is coupled into a multimode step index fiber from a surface emitting LED when the current is 75 mA and the voltage across the LED is about 1.5 V. What is the overall efficiency of operation? b Experiments are carried out on coupling light from a 1310 nm ELED (edge emitting LED) in multimode and single mode fibers. (i) At room temperature, when the ELED current is 120 mA, the voltage is 1.3 V and light power coupled into a 50 µm multimode fiber with NA (numerical aperture) = 0.2 is 48 µW. What is the overall efficiency? (ii) At room temperature, when the ELED current is 120 mA, the voltage is 1.3 V and light power coupled into a 9 µm single mode fiber is 7 µW. What is the overall efficiency?

Chapter 4 4.8

Fabry-Perot optical resonator

a. Consider an idealized He-Ne laser optical cavity. Taking L = 0.5 m, R = 0.99, calculate the separation of the modes and the spectral width following Example 1.7.1.

b. Consider a semiconductor Fabry-Perot optical cavity of length 200 micron with end-mirrors that have a reflectance of 0.8. If the semiconductor refractive index is 3.7, calculate the cavity mode nearest to the free space wavelength of 1300 nm. Calculate the separation of the modes and the spectral width following Example 1.7.1. 4.10

Threshold current and power output from a laser diode

a Consider the rate equations and their results in Section 4.10 It takes ∆t = nL/c second for photons to cross the laser cavity length L, where n is the refractive index. If Nph is the coherent radiation photon concentration, then only half of the photons, 1/2Nph, in the cavity would be moving towards the output face of the crystal at any instant. Given that the active layer has a length L, width W and thickness d, show that the coherent optical output power and intensity are  hc N phdW  Po = (1 − R)  2nλ 

 hc N ph  I= (1 − R)  2nλ 

2

2

and

where R is the reflectance of the semiconductor crystal face.

b If α is the attenuation coefficient for the coherent radiation within the semiconductor active layer due to various loss processes such as scattering and R is the reflectance of the crystal ends then the total attenuation coefficient αt is, 1  1  ln 2 2L  R 

αt = α +

Consider a double heterostructure InGaAsP semiconductor laser operating at 1310 nm. The cavity length L ≈ 60 µm, width W ≈ 10 µm, and d ≈ 0.25 µm. The refractive index n ≈ 3.5. The loss coefficient α ≈ 10 cm-1. Find αt, τph.

c For the above device, threshold current density Jth ≈ 500 A cm-2 and τsp ≈ 10 ps. What is the threshold electron concentration? Calculate the lasing optical power and intensity when the current is 5 mA. 4.12

Laser diode efficiency

a

There are several laser diode efficiency definitions as follows: The external quantum efficiency ηEQE,

ηEQE =

of a laser diode is defined as

Number of output photons from the diode (per unit second ) Number of injected electrons into diode (per unit second )

The external differential quantum efficiency, ηEDQE, of a laser diode is defined as

ηEDQE =

Increase in number of output photons from diode (per unit second ) Increase in number of injected electrons into diode ( per unit second)

The external power efficiency, ηEPE, of the laser diode is defined by Optical output power ηEPE = Electical input power If Po is the emitted optical power, show that eP ηEQE = o Eg I

 e  dPo   Eg  dI

ηEDQE = 

ηEPE = η EQE

 Eg   eV 

b A commercial laser diode with an emission wavelength of 670 nm (red) has the following characteristics. The threshold current at 25°C is 76 mA. At I = 80 mA, the output optical power is 2 mW and the voltage across the diode is 2.3 V. If the diode current is increased to 82 mA, the optical output power increases to 3 mW. Calculate the external QE, external differential QE and the external power efficiency of the laser diode. c Consider an InGaAsP laser diode operating at λ = 1310 nm for optical communications. At I = 40 mA, the output optical power is 3 mW and the voltage across the diode is 1.4 V. If the diode current is increased to 45 mA, the optical output power increases to 4 mW. Calculate external quantum efficiency (QE), external differential QE, external power efficiency of the laser diode. 4.15 The SQW laser Consider a SQW (single quantum well) laser which has an ultrathin active InGaAs of bandgap 0.70 eV and thickness 10 nm between two layers of InAlAs which has a bandgap of 1.45 eV. Effective mass of conduction electrons in InGaAs is about 0.04me and that of the holes in the valence band is 0.44me where me is the mass of the electron in vacuum. Calculate the first and second electron energy levels above Ec and the first hole energy level below Ev in the QW. What is the lasing emission wavelength for this SQW laser? What is this wavelength if the transition were to occur in bulk InGaAs with the same bandgap?

4.16 A GaAs quantum well Effective mass of conduction electrons in GaAs is 0.07me where me is the electron mass in vacuum. Calculate the first three electron energy levels for a quantum well of thickness 8 nm. What is the hole energy below Ev if the effective mass of the hole is 0.47me?What is the ch

Chapter 5 5.1

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 2 mW cm−2. Assuming that each photon generates one electron-hole pair, calculate the number of pairs 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?

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 =

Io [1 − exp(−α d)] dhυ

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 an unity quantum efficient photodetector 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? 5.3 Ge Photodiode Consider a commercial Ge pn junction photodiode which has the responsivity shown in Figure 5Q3. Its photosensitive area is 0.008 mm2. It is used under a reverse bias of 10V when the dark current is 0.3 µA and the junction capacitance is 4 pF. The rise time of the photodiode is 0.5ns. a

Calculate its quantum efficiency at 850, 1300 and 1550nm.

b

What is the light intensity 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 1 1.5 0.5 Wavelength(µm)

The responsivity of a commercial Ge pn junction photodiode

2

Figure 5Q3 5.4 Si pin Photodiodes Consider two commercial Si pin photodiodes, type A and type B, both classified as fast pin photodiodes. They have the responsivity shown Figure 5Q4. Differences in the responsivity are due to the pin device structure. The photosensitive area is 0.125 cm2 (4 mm in diameter). a Calculate the photocurrent from each when they are illuminated with blue light of wavelength 450nm and light intensity 1 µW cm-2. What is the QE of each device? b Calculate the photocurrent from each when they are illuminated with red light of wavelength 700 nm and light intensity 1 µW cm-2. What is the QE of each device? c Calculate the photocurrent from each when they are illuminated with infrared light of wavelength 1000 nm and light intensity 1 µW cm-2. What is the QE of each device? d

What is your conclusion?

Responsivity(A/W) 0.6 0.5 0.4 0.3

B

A

The responsivity of two commercial Si pin photodiodes

0.2 0.1 0 200

5.6

Maximum QE

400

600 800 1000 1200 Wavelength(nm)

Show that maximum QE occurs when

dR R = dλ λ that is, when the tangent at λ passes through the origin (R = 0, λ = 0). Hence determine the wavelengths where the QE is maximum for the InGaAs pin, two Si pin and Ge photodiodes in Figure 5Q6 (Figure 5.22, Figure 5.21 and Figure 5.20 respectively in the textbook).

5.12 The APD and excess avalanche noise APDs exhibit excess avalanche noise which contributes to the shot noise of the diode current. The total noise current in the APD is given by in-APD = [ 2e(Ido + Ipho)M2FB]1/2

(1)

where F is the excess noise factor which depends in a complicated way not only on M but also on the ionization probabilities of the carriers in the device. It is normally taken simply to be Mx where x in an index that depends on the semiconductor material and device structure.

a µm.

Table 5Q12 provides measurements of F vs. M on a Ge APD using photogeneration at 1.55 Find x in F = Mx. How good is the fit?

b The above Ge APD has an unmultiplied dark current of 0.5 µA and an unmultiplied responsivity of 0.8 A W-1 at its peak response at 1.55 µm and is biased to operate at M = 6 in a receiver circuit with a bandwidth of 500MHz. What is the minimum photocurrent that will give a SNR = 1? If the photosensitive area is 0.3 mm in diameter what is the corresponding minimum optical power and light intensity? c

What should be the photocurrent and incident optical power for SNR = 10?

Table 5Q12 Data for excess avalanche noise as F vs M for a Ge APD (from D. Scansen and S.O. Kasap, Cnd. J. Physics. Vol. 70, pp. 1070-1075, 1992) M

1

3

5

7

9

F

1.1

2.8

4.4

5.5

7.5

Chapter 6 6.3

Solar cell driving a load

a A Si solar cell of area 4 cm2 is connected to drive a load R as in Figure 6.8 (a). It has the I-V characteristics in Figure 6.8 (b) under an illumination of 600 W m-2. Suppose that the load is 20 Ω

and it is used under a light intensity of 1 kW m-2. What are the current and voltage in the circuit? What is the power delivered to the load? What is the efficiency of the solar cell in this circuit?

b What should the load be to obtain maximum power transfer from the solar cell to the load at 1 -2 kW m illumination. What is this load at 600 W m-2? c Consider using a number of such a solar cells to drive a calculator that needs a minimum of 3V and draws 3.0 mA at 3 - 4V. It is to be used indoors at a light intensity of about 400 W m-2. How many solar cells would you need and how would you connect them? At what light intensity would the calculator stop working? 6.4 Open circuit voltage A solar cell under an illumination of 100 W m-2 has a short circuit current Isc of 50 mA and an open circuit output voltage Voc, of 0.55V. What are the short circuit current and open circuit voltages when the light intensity is halved? 6.5

Shunt resistance

a

Show that

Consider the equivalent circuit of a solar cell as shown in Figure 6.10.

I = − I ph + Id +

V eV V = − I ph + Io exp( ) − Io + Rp nkB T Rp

b Plot I vs. V for a polycrystalline Si solar cell that has n = 2 and Io = 3×10-4 mA, for an illumination such that Iph = 5 mA. Use Rp = ∞, 1000 Ω and then Rp 100 Ω. What is your conclusion? Iph

A′

A

I

Id Iph

Rp V

RL

B

Solar cell

Load

6.6 Series connected solar cells Consider two identical solar cells with the properties Io = -6 25×10 mA, n = 1.5, Rs = 20 Ω, subjected to the same illumination so that Iph = 10 mA. Plot the individual I-V characteristics and the I-V characteristics of the two cells in series. Find the maximum power that can be delivered by one cell and two cells in series. Find the corresponding voltage and current at the maximum power point. 6.7 Series connected solar cells Consider two odd solar cells. Cell 1 has Io1 = 25×10-6 mA, n1 = 1.5, Rs1 = 10 Ω and cell 2 has Io2 = 1×10-7 mA, n2 = 1, Rs2 = 50 Ω. The illumination is such that Iph1 = 10 mA and Iph2 = 15 ma. Plot the individual I-V characteristics and the I-V characteristics of the two cells in series. Find the maximum power that can be delivered by each cell and two cells in series. Find the corresponding voltages and currents at the maximum power point. What is your conclusions?

6.11 Maximum Power and Fill Factor Given the solar cell current equation, and that the power extracted from the solar cell is -IV, show that maximum power occurs when V = Vm and I = Im when, I ph Vm V exp( m ) ≈ nVT nVT Io

and

 nV  Im ≈ − Iph 1 − T   Vm 

where VT = kBT/e is the “thermal voltage”. Given n, Isc ≈ Iph and Voc, suggest ways to estimate Im and Vm and hence the FF.

(

6.12

Energy band diagram

a Explain the photovoltaic action by using an energy band diagram with and without illumination. b Sketch the energy band diagram of an n-type semiconductor that has a decreasing Eg from left to right. What happens to an electron and a hole that are photogenerated?

Chapter 7 7.3 Wire-grid polarizer Figure 7Q3 shows a wire grid-polarizer which consists of closely spaced parallel thin conducting wires. The light beam passing through the wire-grid is observed to be linearly polarized at right angles to the wires. Can you explain the operation of this polarizer? x z Ey Ex

Wire-grid polarizer Ey

y

The wire grid-acts as a polarizer 7.7 Birefringence Consider a negative uniaxial crystal such as calcite (ne < no) plate that has the optic axis (taken along z) parallel to the plate faces. Suppose that a linearly polarized wave is incident at normal incidence on a plate face. If the optical field is at an angle 45° to the optic axis, sketch the wavefronts and the rays through the calcite plate.

7.10 Quartz Wollaston prism Draw a quartz Wollaston prism and clearly show and identify the directions of orthogonally polarized waves traveling through the prisms. How would you test the polarization states of the emerging rays? Consider two identical Wollaston prisms, one from calcite and the other from quartz. Which will have a greater beam splitting ability? (Explain).

7.11 Glan-Foucault prism Figure 7Q11-1 shows the cross section of a Glan-Foucault prism which is made of two right angle calcite prisms with a prism angle of 38.5°. Both have their optic axes parallel to each other and to the block faces as in the figure. Explain the operation of the prisms and show that the o-wave does indeed experience total internal reflection.

Absorber o-ray 38.5° e-ray Calcite Optic axis Air-gap

The Glan-Foucault prism provides linearly polarized light

7.13

Optical activity

a Consider an optically active medium. The experimenter A (Alan) sends a vertically polarized light into the this medium as in Figure 7Q13-1. Light emerging from the back of the crystal and is received by an experimenter B (Barbara). B observes that the optical field E has been rotated to E′ counterclockwise. She reflects the wave back into the medium so that A can receive it. Describe the observations of A and B. What is your conclusion? b Figure 7Q13-2 shows a simplified version of the Fresnel prism that converts an incoming unpolarized light into two divergent beams that have opposite circular polarizations. Explain the principle of operation. θ

E

Levo E′

Dextro

E′ z

E′ z

z

Quartz

L

Optic axis

An optically active material such as quartz rotates the plane of polarization of the incident wave: The optical field E rotated to E′. If we reflect the wave back into the material, E′ rotates back to E.

Figure 7Q13-1 L-polarized R-handed quartz

L-handed quartz R-polarized

The Fresnel prism for separating unpolarized light into two divergent beams with opposite circular polarizations (R = right, L = left; divergence is exaggerated)

Figure 7Q13-2