Download 4 - Electrical and Computer Engineering

ECE/UConn ENGR-ECE4243-6243 HW #4: Absorption and Emission, Excitons,
09/20/16 (due 09/27/2016)
Assigned HOMEWORK
Q.1(a) What is an exciton?
F.C. Jain
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 m thick Si and GaAs samples illuminated by a 10 mW photon source.
Assume the source to be monochromatic and emitting photons with energy h=1.9 eV.
(a) Given the absorption coefficient (h=1.9eV of  =0.65 m) = 3000 cm-1 in crystalline Si and
index of refraction nr= 11.8 3.45, find the power absorbed in Si in one pass.
Pin=10mW
(hv=1.9eV)
Pin=10mW
(hv=1.9eV)
Pout
Pout
d
d
Front surface
reflectivity
R1
Back surface reflectivity
R2
Front surface reflectivity
R1
Back surface
reflectivity
R2
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 m in thickness. Given 
(h=1.9eV of  =0.65 m) = 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 AlGa1-As  =
13.18 –3.12; for GaAs Index = (
(d) Find the excess energy per photon in c-Si, a-Si and GaAs. HINT: The excess energy is (h-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.
0
Xp
-xn
photons
P
n+
-
+
X
I
Load RL
+
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 p=2x10-6 sec.
Minority hole diffusion coefficient Dp=12.5 cm2/sec.
p-side: Acceptor concentration NA=5x1017cm-3, n=10-5sec. Dn=40 cm2/sec.
Junction area A=1 cm2, nI (at 300K)=1.5x1010cm-3, r (Si)=11.8, 0=8.85x10-14 F/cm, =0r.
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.
(c) Find the fill factor FF.
(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.
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(b). Order these cells in terms of conversion efficiency, providing qualitative explanation.
(c) Which of the cells shown in Fig. 3 would produce the maximum power output P mp.
-xnJunction xp
-xnJunction x
p
0
0
Photons
+
P
n
1?m
n+
1m
100?m
P
100m
X
X
Vout
Load RL
Load RL
(a)
(b)
Junction
Junction
-xp
Photons
xn
-xp
0
P+
1m
xn
0
Photons
P+
1m
N
100m
N
100m
X
X
Load RL
Load RL
(d)
Fig .3. n+-p and p+-n Si solar cells illuminated from front and back.
(C)
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 J for AM =1 condition (use Fig. 4 plot for AM =1).
1353 W/m2
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
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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?
-xn
-xn
xp
0
Si- n+
xp
0
N-Al0.35Ga0.65As
p-Si
p-GaAs
x
x
Vf
Vf=0.7V
Fig. 5(a) A forward biased n+-p Si diode.
Fig. 5(b). A forward biased nAlGaAs-p GaAs diode.
Energy
Energy
Injected
electrons from
n-AlGaAs
Energy Gap
of Si, Eg=1.1eV
k vector
(a)
Injected electrons and
majority holes in p-Si
GaAs Eg
= 1.424eV
Majority
holes in p-GaAs
k vector
(b)
Injected electrons and
majority holes in p-GaAs
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 q 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.
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(b) Find number of photons generated per second in the p-region.
(c) Determine the wavelength of light generated in the medium and as viewed in air.
(d) Identify the location and length of the region where most of the photons are generated.
(e) Calculate the internal efficiency int. at forward bias Vf of 1.5 volts and 1.55V.
HINT: int = q * inj Given--quantum efficiency in GaAs0.6P0.4 is q= 0.25.
 qA Dn n po  qV f
( Dn n po )

( e kT - 1)
I n( xp )
Ln


Ln
=
=
 inj 
(
)
q
Dn n po ( D p pno )
I n ( x p ) + I p (- xn )  qA Dn n po qA D p p no  V f
+

( e kT - 1)
+


Ln
Lp
Ln
Lp


(f) Find the extraction efficiency extraction.
1


2
1
4 nr
1



 1 - 1 -  
 extraction = 
2 (1 + n r )2   n2r  


The device parameters are: n+ side
p-side
ND = donor concentration=1x1018 cm-3
Acceptor concentration NA=1x1016 cm-3
Minority hole lifetime p =5x10-9 sec
Minority electron lifetime n =1x10-8 sec
2
Hole diffusion coefficient Dp=10 cm /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 r =12.84, Free space permittivity o =8.854x10-14 Farad/cm,
Temperature T =300K unless specified.
Index of refraction, nr, of GaAs0.6P0.4 =3.58 Quantum efficiency q= 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
AIR
xp
-xn
Top Contact
nr3
Epoxy or Dome
Material
nr2
t   4 n r 2 2l  1
l = 1,2,3 ...
n
Junction
Antireflection
Coating
p
nr
nr1
Au wire
0
Layer to
photons
reflect
towards the
traveling
bottom
contact
x
Bottom Contact
VF=1.5 Volt
Leads
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.
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