ECE/Uconn ENGR-ECE4243-6243 HW #4: Absorption and Emission, Excitons, F.C. Jain 09/20/16

ECE/UConn ENGR-ECE4243-6243 HW #4: Absorption and Emission, Excitons, F.C. Jain 09/20/16 (due 09/27/2016)

Assigned HOMEWORK

Q.1(a) What is an exciton?

Q1(b) if the photon energy is 1.4eV what is the energy of the band gap to form an exciton. Given the exciton binding energy is 0.004eV.

Q.2(a) What is your understanding of quantum confined Stark effect in a quantum well?

Q2(b) Write an expression for photon energy needed to form an exciton in a quantum well having electron level 0.054eV and heavy hole level as 0.020eV and the binding energy of exciton is 0.008eV.

Q3(a). What determines the photon absorption shift to longer wavelength (red Stark shift) in the presence of electric field?

HINT: (i) The binding energy of excitons decreases when an electric field is applied across the quantum wells.

(ii) The energy level separation (Ee1 and Ehh1) also lowers little bit in magnitude as an external electric field is applied.

That is, Ee1 +Ehh1 (when no field or lower field) > Ee1 +Ehh1 (when higher field is applied)

Q.3(b) Is electro-absorption higher in multiple quantum wires as compared to quantum wells? Explain why.

Q.4(a). Describe in few lines the operation of an electro-absorptive MQW modulator such as shown on page 187 (Fig. 2).

Q.4(b) Is the index of refraction of a multiple quantum well (MQW) layer a function of external RF or DC electric field?

Q.4(c) what optical modulators can you think use this property?

REVIEW Questions not to be submitted

Q1 Figure 1 shows two 10.0 mm thick Si and GaAs samples illuminated by a 10 mW photon source. Assume the source to be monochromatic and emitting photons with energy hn=1.9 eV.

(a) Given the absorption coefficient a(hn=1.9eV of l =0.65 mm) = 3000 cm-1 in crystalline Si and index of refraction nr=»3.45, find the power absorbed in Si in one pass.

Fig. 1a Photon absorption in Si. / Fig. 1b Photon absorption in GaAs. / Fig. 1c Absorption coefficient vs l.

(b) If the crystalline Si is replaced by amorphous Si (a-Si), find the absorbed power in film of 1 mm in thickness. Given a (hn=1.9eV of l =0.65 mm) = 10,000 cm-1 in a-Si sample. Assume the same index of refraction.

(c) Find the power absorbed in GaAs. Use the plot of absorption coefficient in Fig. 1c for GaAs at 1.9eV. For index of refraction of GaAs, use home work #6 Q2 on laser diode where AlGaAs-GaAs information is provided. For AlxGa1-xAs e = 13.18 –3.12x; for GaAs x =0. Index = (e)1/2.

(d) Find the excess energy per photon in c-Si, a-Si and GaAs. HINT: The excess energy is (hn-Eg) per photon.

(e) Find out the portion of absorbed energy that is not used up in electron-hole pair (EHP) generation in crystalline Si and GaAs. Hint: It is wasted in the form of heat. Multiplying this by the number of photons will give the power that is wasted.

Q2. Fig 2 shows an n+-p Si diode with following device/material parameters.

Fig. 2. An abrupt n+-p Si solar cell (the light is incident from the left or n+ side).

Given: n+-side: Donor concentration ND=1020 cm-3, minority hole lifetime tp=2x10-6 sec.

Minority hole diffusion coefficient Dp=12.5 cm2/sec.

p-side: Acceptor concentration NA=5x1017cm-3, tn=10-5sec. Dn=40 cm2/sec.

Junction area A=1 cm2, nI (at 300K)=1.5x1010cm-3, er (Si)=11.8, e0=8.85x10-14 F/cm, e=e0er.

Effective mass: electrons me=mn=0.26mo, holes mh=mp= 0.64 mo,

Assume all donors and acceptors to be ionized at T=300K.

(a) Find reverse saturation current Is. Determine the open circuit voltage Voc, if the light generated current ISC= IL = 27mA= the short circuit current.

(b) Determine the maximum output power Pmp or Pm =Vm*Im.

(d) What is the effect on Voc and fill factor FF if the doping of p-Si is raised to 1018cm-3.

Voc will increase or Decrease

FF will increase or Decrease

(e) What happens to power output if the operating temperature is raised from 300K to 500K?

Circle one: Increases OR Decreases.

Q.3(a) Show the polarity of output voltage and direction of current in load RL of all solar cells in Fig. 3.

(b). Order these cells in terms of conversion efficiency, providing qualitative explanation.

(a) /
(b)

(C) / (d)

Fig .3. n+-p and p+-n Si solar cells illuminated from front and back.

Q.4. What does Isc and Voc primarily depend on: Circle the right answers

Band gap doping levels thickness of the n-region in a p-n cell incident solar power

Q.5. (a) Show that the long wavelength loss is 176.86 W/m2, as shown below (925 -748.14) W/m2 for Air Mass m = 1 condition.

(b) Calculate the excess energy lost is 298.05 W/m2 (not used in producing electron hole pairs) in the spectral range represented by triangle abJ for AM =1 condition (use Fig. 4 plot for AM =1).

Fig. 4. Solar spectrum under AM O (m=0) and AM1. See Figs. 37 and 78 (Notes pages 448 and 449). The incident power received is 1353 W/m2 for AM=0 and 925 W/m2 for AM=1.

UConn ECE 4211 HW #5 LEDs-I F. Jain 03032015 Due 3/10/15

Q.6 (a). How do the excess carriers recombine in forward biased Si diode Fig .5(a) [which is an indirect gap semiconductor]?

(b) What is the distance from junction edge (xp) in which electron concentration decays by 1/e?

(c) How do excess carriers recombine in AlGaAs-GaAs diode Fig. 5(b)? What is the difference in recombination of injected minority electrons in p-Si and p-GaAs?

Fig. 5(a) A forward biased n+-p Si diode. /
Fig. 5(b). A forward biased nAlGaAs-p GaAs diode.

Fig. 6 shows the energy of injected electrons at a point x on the p-side.

Q.7. (a) Write energy and momentum conservation equation in terms of electron, hole, phonons (in indirect gap Si) and photon energy and momentum.

(b) If the quantum efficiency hq for GaAs is 0.95 and for Si 0.05 how many photons are produced per second for an electron current of 1mA in these devices on the p-side.

Q.8(a). Find the energy of emission for point C for the InGaAs in Fig. 7a.

(b) Find the composition of InGaAs active layer to design LEDs operating at 1.35 and 1.55 microns using band gap-lattice parameter data shown in Figs. 7a and 7b.

/ Fig. 7a (left)

Fig. 7b Lattice constant versus energy gap.

Q.8(c) Name material composition of AlGaInP that will emit at 6500Å, 5500λ, 5850Å. Use the energy gap vs lattice constant given in Fig. 3b. Use vertical dash line (AlxGa1-x)0.5In0.5P.

Q. 9. An n+-p GaAs0.6P0.4 homojunction diode is schematically shown in Fig.8.

(a) Calculate In (x=xp) and Ip (x=-xn) at a forward bias of Vf =1.55 Volt.

(b) Find number of photons generated per second in the p-region.

(d) Identify the location and length of the region where most of the photons are generated.

(e) Calculate the internal efficiency hint. at forward bias Vf of 1.5 volts and 1.55V.

HINT: hint = hq * hinj Given--quantum efficiency in GaAs0.6P0.4 is hq= 0.25.

(f) Find the extraction efficiency hextraction.

The device parameters are: n+ side p-side

ND = donor concentration=1x1018 cm-3 Acceptor concentration NA=1x1016 cm-3

Minority hole lifetime tp =5x10-9 sec Minority electron lifetime tn =1x10-8 sec

Hole diffusion coefficient Dp=10 cm2/s Electron diffusion coefficient Dn=50 cm2/s

Assume all donors and acceptors to be ionized at 300K.

GaAs0.6P0.4 energy gap Eg= 1.85 eV

Band gap type = direct, Intrinsic concentration at 300K ni = 300 cm-3

ni at 500K = 1x109 cm-3, Junction area A = 1x10-3 cm2,

Dielectric constant er =12.84, Free space permittivity eo =8.854x10-14 Farad/cm,

Temperature T =300oK unless specified.

Index of refraction, nr, of GaAs0.6P0.4 =3.58 Quantum efficiency hq= 0.25.

Electron effective mass mn = 0.067 mo, hole effective mass = 0.62mo

Electron charge q=1.6x10-19 Coulombs, Boltzmann constant k=1.38x10-23 J/K

Fig. 8. A forward biased nGaAs0.6P0.4 - pGaAs0.6P0.4 diode. / Fig. 9 An encapsulated LED showing layers reducing losses and improving extraction efficiency.

Q.10. Fig. 9 shows an encapsulated LED with various layers used to reduce losses Identify all layers that improve extraction efficiency, and outline steps to compute the new efficiency.