Lecture 8: Solar Cell, LED, Metal/Semiconductor Junction and Heterojunction. Requirement: understand and explain in word. * Some of the content from C. Hu : “Modern Semiconductor devices for Integrated Circuits”. 5.7 Solar Cells. Solar Cells is also known as photovoltaic cells .
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Requirement: understand and explain in word.
* Some of the content from C. Hu : “Modern Semiconductor devices for Integrated Circuits”
5.7 Solar Cells
I
Dark IV
light
N
Eq.(4.9.4)
P
I
0.7 V
sc

0
V
E
Solar Cell
c
IV
Eq.(4.12.1)
Maximum
E
–
I
poweroutput
v
sc
+
(a)
Solar Cell Basics
directgap semiconductor
indirectgap semiconductor
Directgap semiconductor: Absorption coefficient is larger .
Si is most prevalent for solar cell because of low cost.
Light Absorption
α(1/cm): absorption coefficient
1/α: light penetration depth
A thinner layer of directgap semiconductor can absorb most of solar radiation than indirectgap semiconductor.
Si solar cell > 50 um in thickness to absorb most of the photons because of low α
If light shines on the Ntype semiconductor and generates holes (and electrons) at the rate of G s1cm3 ,
If the sample is uniform (no PN junction), d2p’/dx2 = 0 p’ = GLp2/Dp= Gtp
Assume very thin P+ layer and carrier generation in N region only.
G is really not uniform. Lp needs be larger than the light penetration depth to collect most of the generated carriers.
How to raise Voc ?
A particular operating point on the solar cell IV curve maximizes the output power (I V).
Output Power
NRL’s new triplejunction solar cells could achieve 50 percent efficiency
E percent efficiencyc
Nonradiative recombination through traps
Radiative recombination
Ev
5.8 Light Emitting Diodes and SolidState Lighting
Trap
Indirect band gap
Example: Si
Direct recombination is rare as k conservation is not satisfied
Direct band gap
Example: GaAs
Direct recombination is efficient as k conservation is satisfied.
4.13.1 LED Materials and Structure percent efficiency
LED percent efficiencyMaterials and Structure
compound semiconductors
binary semiconductors:
 Ex: GaAs, efficient emitter
ternary semiconductor :
 Ex: GaAs1xPx , tunable Eg (to vary the color)
quaternary semiconductors:
 Ex: AlInGaP , tunable Eg and lattice constant (for growing high quality epitaxial films on inexpensive substrates)
Eg(eV)
red
yellow
blue
Red
Yellow
Green
Blue
Lightemitting diode materials
AlInGaP Quantun Well percent efficiency
Common LEDs
5.9 MetalSemiconductor percent efficiencyJunction
Vacuum level, percent efficiency
E
0
= 4.05 eV
c
c
Si
Si
q
y
y
M
M
q
f
E
Bn
c
E
f
E
v
fBn Increases with Increasing Metal Work Function
: Work Function of metal
: Electron Affinity of Si
Theoretically,
fBn= yM – cSi
Depletion percent efficiency
Metal
Neutral region
layer
qfBn
Ec
Ef
NSi
Ev
Ec
PSi
Ef
Ev
qfBp
SchottkyBarriers
Energy Band Diagram of Schottky Contact
Schottky barrier heights for electrons and holes percent efficiency
fBn + fBp Eg
fBnincreases with increasing metal work function
Vacuum level, percent efficiency
E
0
c
= 4.05 eV
Si
q
y
M
q
f
E
Bn
c
+

E
f
E
v
Fermi Level Pinning (Schottky barrier lowering)
q percent efficiencyfbi
qfBn
Ec
Ef
Ev
qfBn
q(fbi + V)
qV
Ec
Ef
Ev
Using CV Data to Determine fB
Question:
How should we plot the CV data to extract fbi?
q percent efficiencyfbi
2
qfBn
1
/C
Ec
Ef
Ev
V

f
bi
Using CV Data to Determine fB
Oncefbi is known, fB can be determined using
Thermionic percent efficiencyEmission Theory
v
thx

E
q(
f

V)
B
c
q
f
Ntype
B
E
qV
E
fn
V
Metal
Silicon
fm
E
v
x
I percent efficiency
Schottky
Schottky diode
I
f
f
PN junction
PN junction
B
B
diode
V
V
Applications of Schottly Diodes
5.10 percent efficiencyHeterojunction
Heterojunction gives us additional parameters to manipulate the ratio of electron/hole current
More will be discussed in ECE684: HEMT
What is the energy band diagram at thermal equilibrium? percent efficiency
What is Vbi