Optoelectronics Lecture notes

1. Chapter 7: Photovoltaic
Devices
Solar Energy Spectrum
Photovoltaic Device Principles
Solar Cells Materials, Devices & Efficiency

2. Introduction
• Photovoltaic devices or solar cells
– Convert the incident solar radiation energy into
electrical energy
• Incident photons are absorbed to photo-
generate charge carriers
– then pass through an external load to do electrical
work

3. Photovoltaic Device Applications
• Cover a wider range
– From small consumer electronics such as a solar
cell calculator using less than a few mW to
photovoltaic power generation by a central power
plant (generating a few MW)
• There are several MW photovoltaic power
plants and tens of thousands of small 1 kW
scale photovoltaic generation systems
currently in use.

4. Solar Energy Spectrum
• The intensity of radiation emitted from the sun
has a spectrum that resembles a black body
radiation at a temperature of 6000K
• The actual intensity spectrum on Earth’s surface
depends on
– the absorption and scattering effects of the
atmosphere.
– the atmosphere composition and radiation path length
through the atmosphere

5. Solar Energy Spectrum, cont
• Clouds increase the absorption and scattering
of sunlight and hence substantially reduce the
incident intensity.
• On a clear sunny day, the light intensity
arriving on the Earth’s surface is roughly 70%
of the intensity above the atmosphere.

6. 0
Black body radiation at 6000 K
AM0
AM1.5
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
0
0.5
1.0
1.5
2.0
2.5
Wavelength (m)
Spectral
Intensity
W cm-2 (m)-1
The spectrum of the solar energy represented as spectral
intensity (I) vs wavelength above the earth's atmosphere
(AM0 radiation) and at the earth's surface (AM1.5
radiation). Black body radiation at 6000 K is shown for
comparison (After H.J. Möller, Semiconductors for Solar
Cells, Artech House Press, Boston, 1993, p.10)
© 1999 S.O. Kasap,Optoelectronics(Prentice Hall)
Fig. 1

7. Definition of light intensity
• Light intensity (W/m2) variation with
wavelength is typically represented by
intensity per unit wavelength
– Called spectral intensity I so that I  is the
intensity in a small interval .
• Integration of I over the whole spectrum
gives the integrated or total intensity I.

8. AM0
• The integrated intensity above Earth’s
atmosphere gives the total power flow
through a unit area perpendicular to the
direction of the sun.
• This quantity is called solar constant or air-
mass zero (AM0) radiation
• It is approximately constant at a value of 1.353
kW m–2.

9. AM1
• Absorption and scattering effects increase
with the sun beam’s path through the
atmosphere.
• The shortest path through the atmosphere is
when the sun is directly above that location
and the received spectrum is called air mass
one (AM1)

10. AMm
• All other angles of incidence ( 90°) increase
the optical path through the atmosphere and
hence the atmospheric losses.
• Air mass m (AMm) is defined as the ratio of
the actual radiation path h to the shortest
path h0 that is m = h/h0.
• Since h = h0 sec, AMm is AM sec.

11. AM1.5
• The spectral distribution for AM1.5 is shown in Fig.1
– There are several sharp absorption peaks at certain
wavelengths due to various molecules such as ozone, air,
water vapor etc.
– Atmospheric molecules & dust particles scatter the
sunlight.
• This spectrum refers to incident energy on a unit
area normal to sun rays.
– Which have to travel the atmospheric length h as shown in
Fig. 2.

12. Terrestrial sunlight
• The terrestrial light has a diffuse component and
direct component
• Diffuse component increases with cloudiness and
sun’s position and has a spectrum shifted toward the
blue light
– The scattering of light increases with decreasing
wavelength so that shorter wavelengths in the original sun
beam experience more scattering than longer wavelength
– On the clear day, the diffuse component can be roughly
20% of the total radiation and significantly higher on
cloudy days.

13. Direct Diffuse
(a) Illustration of the effect of the angle of incidence  on the ray path length and the
definitions of AM0, AM1 and AM(sec). The angle  between the sun beam and the horizon
is the solar latitude (b) Scattering reduces the intensity and gives rise to a diffused radiation
Atmosphere
AM0
AM1

AM(sec)
h0
h

(a) (b)

Tilted PV device
Earth
© 1999 S.O. Kasap,Optoelectronics(Prentice Hall)
Fig. 2

14. Example:
Solar Energy Conversion
• Suppose that a particular family house in a
sunny geographic location over a year
consumes a daily average electrical power of
500W. If the annual average solar intensity
incident per day is about 6 kWhm–2, and a
photovoltaic device that converts solar energy
to electrical energy has an efficiency of 15%,
what is the required device area.

15. Solution
• Since we know the average light intensity incident,
Total energy available for 1 day = Incident solar energy in 1 day
per unit area x Area x Efficiency,
Which must equal to the average energy consumed per
house in 1 day. Thus,
 
3.6m.3.6mpanelaor3.13
15.0min/60min/60106
24min/60min/60500
EfficiencyareaunitperenergysolarIncident
houseperEnery
Area
2
126







m
hrsdaymhrW
hrshrsW

16. Photovoltaic Device Principles
• Consider a pn-junction with a very narrow & heavily doped n-
region
– Illumination through the thin n-side
• The depletion region (W) extends primarily into p-side
– Built-in field Eo in the depletion layer
• Electrode attached to the n-side must allow illumination to enter
the device & result in a small series resistance.
– They deposited onto n-side to form an array of finger electrodes on the
surface.
• A thin anti-reflection coating on the surface reduces reflections
and allows more light to enter the device

17. Fig. 3

18. Finger electrodes
p
n
Bus electrode
for current collection
Finger electrodes on the surface of a solar cell
reduce the series resistance
© 1999 S.O. Kasap,Optoelectronics(Prentice Hall)
Fig. 4

19. Photovoltaic Device Principles, cont
• As the n-side is very narrow,
– Most of the photons are absorbed within the depletion
region and within the neutral p-side and photo-generate
EHPs.
– EHPs in the depletion region are immediately separated by
the built-in field Eo
– The e drifts and reaches the neutral n+ side whereupon it
makes this region negative by an amount of charge – e.
– The hole drifts and reaches the neutral p-side and thereby
makes this side positive.

20. Photovoltaic Device Principles, cont
• An open circuit voltage develops between the
terminals of the device with the p-side positive with
respect to the n-side
• If external load is connected then the excess electron
in the n-side can travel around the external circuit,
do work & reach the p-side to recombine with the
excess hole there.

21. Minority carrier diffusion length
• The EHPs photo-generated by long wavelength
photons that are not absorbed in the neutral p-
side can only diffuse in this region as there is no
electric field.
• If the recombination life time of the electron is e,
it diffuses a mean distance Le given by Le = (2 De
e)½
– where De is its diffusion coefficient in the p-side

22. Le
Lh W
Iph
x
EHPs
exp(x)
Photogenerated carriers within the volume Lh + W +Le give rise to a photocurrent Iph. The
variation in the photegenerated EHP concentration with distance is also shown where  is the
absorption coefficient at the wavelength of interest.
© 1999 S.O. Kasap,Optoelectronics(Prentice Hall)
Fig. 5

23. Minority carrier diffusion length
• Those electrons within a distance Le to the depletion
region can readily diffuse & reach this region
whereupon they become drifted by Eo to the n-side.
• Consequently, only those EHPs photo-generated
within the minority carrier diffusion length Le to the
built-in field Eo can contribute to the photovoltaic
effect.
• Those photo-generated EHPs further away from the
depletion region than Le are lost lost by
recombination.
• It is therefore important to have the minority carrier
diffusion length Le as long as possible.

24. Minority carrier diffusion length
• The reason for choosing the p-type in the bottom
layer of Si pn-junction is to make the electrons to be
minority carriers
– Electron diffusion length in Si is longer than the hole
diffusion length.
• The same idea also apply to EHPs photo-generated
by short-wavelength photons absorbed in the n-side
• Those holes photo-generated within a diffusion
length Lh can reach the depletion layer and become
swept across to the p-side

25. Photocurrent
• The photo-generation of EHPs that contribute to the
photovoltaic effect therefore occurs in a volume
covering Lh + W + Le.
• If the terminals of the device are shorted then the
excess electron in the n-side can flow through the
external circuit to neutralize the excess hole in the p-
side.
• This current due to the flow of the photo-generated
carriers is called photocurrent.

26. • EHPs photo-generated by energetic photons
absorbed in the n-side near the surface region or
outside the diffusion length Lh to the depletion layer
are lost by recombination as the lifetime in the n-side
is generally very short
• The n-side is therefore made very thin, typically < 0.2
m or less.
• The EHP photo-generated very near the surface of
the n-side however disappear by recombination due
to various surface defects acting as recombination
centre.

27. • At long wavelength, around 1-1.2m, the absorption
coefficient  of Si is small and the absorption depth
(1/) is typically > 100m
• To capture these long wavelength photons we
therefore need a thick p-side and a long minority
carrier diffusion length Le.
– Typically, the p-side is 200-500m and Le tends to be
shorter than this.

28. • Crystalline Si has a bandgap of 1.1eV, which
correspond to a threshold wavelength of 1.1m.
• The incident energy in the wavelength region >1.1m
is then wasted
– This is not negligible amount ~25%
• The worst part of the efficiency limitation however
comes from the high energy photons becoming
absorbed near the crystal surface and being loss by
recombination in the surface region

29. • Crystal surface and interfaces contain a high concentration of
recombination centers which facilitate the recombination of
photo-generated EHP near the surface
– Losses due to EHP recombination near or at the surface can be as high
as 40%
– These combined effects bring the efficiency down to ~45%
• In addition, the anti-reflection coating is not perfect which
reduces the total collected photons by a factor of ~0.8-0.9.
• When we also include the limitations of the photovoltaic
action itself, the upper limit to a photovoltaic device that uses
a single crystal of Si is ~ 24-26% at room temperature.

30. Example
Consider a particular photovoltaic device that is illuminated with
light of such wavelength that photo-generation occurs over the
device thickness and the EHP photo-generation rate Gph, number of
EHPs photo-generated per unit volume per unit time, decays as
Goexp(–x) where Go is the photo-generation rate at the surface
and  is the absorption coefficient. Suppose that the device is
shorted to allow all the photo-generated carriers to flow around the
external circuit (only electron). Suppose that Lh> ln (the n-layer
thickness) so that all the EHPs so that all the EHPs generated within
the volume (ln+W+ Le) contribute to the photocurrent. Further,
assume that EHP recombination near the crystal surface is negligible.
Show that the photocurrent Iph is then
Iph = e GoA/  {1 – exp [–(ln+W+ Le)]} (1)
Where A is the device surface area under illumination (not blocked
by finger electrodes)

31. Solution
The EHP photo-generation rate from the illuminated crystal
surface follows
Goexp(–x)
The total number of EHP generated per unit time in a small
volume Ax is Gph(Ax). Thus:
The total number EHP generated per unit time in
 
   en
oEHP
LWl
x
oen
LWl
AG
dt
dN
dxxGALWl
en

 





exp1or
exp
0

32. Solution, cont
Since the photo-generated electrons flow through the
external circuit, the photocurrent Iph is then
e(dNEHP/dt)
Iph = e GoA/  {1 – exp [–(ln+W+ Le)]}
For long wavelengths,  will be small. Expanding the
exponential we find,
Iph = e GoA (ln+W+ Le) (2)
Which applies under nearly uniform photo-generation
conditions

33. Solution, cont
Taking a crystalline Si device that has A= 5cm  5cm, ln = 0.5μm, W
= 2μm, Le = 50μm, small  such as = 2000m–1 (absorption
depth = 1/ =500μm) for Si at  1.1μm and using
Go=11018cm–3s–1 in Eq.(1), we find Iph20mA whereas Eq.(2)
gives 21mA.
On the other hand for strong absorption at  0.83μm, = 105m–1
(absorption depth = 1/ =10μm) Eq.(1) gives Iph40mA. The
current is doubled simply because more photons are now
absorbed in the volume (ln+W+ Le).
Further increase in  with decreasing wavelength will eventually
(when  < 450nm) constrict the photo-generation to the
surface region where the surface defects will facilitate EHP
recombination and thereby diminish the photocurrent.

34. pn junction photovoltaic I-V
characteristics
• Consider an ideal pn junction photovoltaic device
connected to a resistive load R.
– Note that Fig 6(a) define the convention for the direction of
positive current and positive voltage.
• If the load is a short circuit, only current is generated
in the circuit.
– This is called photocurrent Iph, which depends on the
number of EHPs photo-generated within the volume
enclosing the depletion region (W) and the diffusion lengths
to the depletion region
– The greater is the light intensity, the higher is the photo-
generation rate and the larger is Iph

35. Iph
R
I
V V = 0
Iph
I = Id
 Iph
V
Id
Isc
= –Iph
R
(a) (b) (c)
(a) The solar cell connected to an external load R and the convention for the definitions of
positive voltage and positive current. (b) The solar cell in short circuit. The current is the
photocurrent, Iph. (c) The solar cell driving an external load R. There is a voltage V and current
I in the circuit.
Light
© 1999 S.O. Kasap,Optoelectronics(Prentice Hall)
Fig. 6

36. Current & light intensity
• If I is the light intensity then the short circuit current is
Isc = –Iph = –KI (1)
– Where I is a constant that depends on the particular device.
• The photocurrent does not depends on the voltage
across the pn junction because there is always some
internal field to drift the photo-generated EHP.
– We exclude the secondary effect of the voltage modulating
the width of the depletion region.
– The photocurrent therefore flows even when there is not a
voltage across the device.

37. Load
• If R is not a short circuit then a positive
voltage V appears across the pn-junction as a
result of the current passing through it.
– This voltage reduces the built in potential Vo of the
pn junction and hence leads to minority carrier
injection and diffusion just as it would in a normal
diode
– Thus, in addition to Iph there is also a forward
diode current Id in the circuit.

38. pn junction current
• Since Id is due to the normal pn junction behavior, it
is given
Id = Io [exp {eV/(nkBT)} – 1]
where
Io is the reverse saturation current,
n is the ideality factor that depends on the semiconductor
material and fabrication characteristics (n = 1 – 2)
• In open circuit, the net current is zero
– The photocurrent Iph develops just enough photovoltaic
voltage Voc to generate a diode current Id = Iph.

39. Solar cell I-V curve
• The total current through the solar cells is
I = – Iph + Io [exp {eV/(nkBT)} – 1] (2) Solar cell I-V
• The overall I-V characteristics of a typical Si solar cells
is shown in Fig.7
– It corresponds to the normal dark characteristics being
shifted down by the photocurrent Iph, which depends on the
light intensity, I.
– The open circuit output voltage Voc of the solar cells is given
by the point where I-V curve cuts the V-axis
– Voc depends on the light intensity and lies in the range 0.4-
0.6V.

40. V
I (mA)
Dark
Light
Twice the light
0.60.40.2
20
–20
0
Iph
Voc
Typical I-V characteristics of a Si solar cell. The short circuit current is Iph
and the open circuit voltage is Voc. The I-V curves for positive current
requires an external bias voltage. Photovoltaic operationis always in the
negative current region.
© 1999 S.O. Kasap,Optoelectronics(Prentice Hall)
Fig. 7

41. Load Line
• When the solar cell is connected to a load as in
Fig. 8, the load has the same voltage as the solar
cell and carries the same current
– But the current I through R is now in the opposite
direction to the conventional that current flows from
high to low potential.
– Thus I = – V/R (3) the load line

42. V
I (mA)
0.60.40.2
–20
0
Voc
–10
Isc= –Iph
V
The Load Line for R = 30 ž
(I-V for the load)
I-Vforasolarcellunderan
illuminationof600Wm-2.
Operating Point
Slope = – 1/R
P
I
(a) When a solar cell drives a load R, R has the same voltage as the solar cell
but the current through it is in the opposite direction to the convention that
current flows from high to low potential. (b) The current I and voltage V in
the circuit of (a) can be found from a load line construction. Point P is the
operating point (I, V). The load line is for R = 30 ž .
Light
I
R
V
I
(a) (b)
© 1999 S.O. Kasap,Optoelectronics(Prentice Hall)
Fig. 8

43. Actual current & voltage
• The actual current I’ and voltage V’ in the
circuit must satisfy both the I-V
characteristics of the solar cell and that of
the load.
• We can find I’ and V’ by solving eqs (2) &
(3) simultaneously but this is not a trivial
analytical procedure.
– A graphical solution using the solar cell
characteristics however is straightforward.

44. Operating point of the circuit
• The current and voltage in the solar cell circuit are
most easily found by using a load line construction.
• I-V characteristics of the load in Eq(3) is a straight line
with a negative slope –1/R.
– This is called load line
• The load line cuts the solar cell characteristic at P.
– At P, the load and the solar cell have the same current I’ and
voltage V’.
– Point P satisfies both eqs.(2)&(3) and thus represents the
operating point of the circuit.

45. Power
• The power delivered to the load is Pout= I’V’
– Which is the area of the rectangle bound by I- and V- axes
• Maximum power is delivered to the load when this
rectangular are is maximized when I’=Im & V’=Vm
– By either changing R or the intensity of illumination.
• Since the maximum possible current is Isc and the
maximum possible voltage is Voc, IscVoc represents the
desirable goal in power delivery for a given solar cell.

46. Fill Factor
• To compare the maximum power output ImVm with
IscVoc, the fill factor FF, which is a figure of merit for the
solar cell, is defined as
FF = ImVm/(IscVoc)
– FF is a measure of the closeness of the solar cell I-V curve to
the rectangular shape.
– It is advantageous to have FF as close to unity as possible ut
the exponential pn junction properties prevent this
– Typical FF values are in the range 70-85%

47. Example
• Consider a solar cell driving a 30 resistive load as in
Fig 8(a). Suppose that the cell has an area of 1cmx1cm
and is illuminated with light of intensity 600Wm–2 and
has the I-V characteristics in Fig 8(b).
• 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?

48. V
I (mA)
0.60.40.2
–20
0
Voc
–10
Isc= –Iph
V
The Load Line for R = 30 ž
(I-V for the load)
I-Vforasolarcellunderan
illuminationof600Wm-2.
Operating Point
Slope = – 1/R
P
I
(a) When a solar cell drives a load R, R has the same voltage as the solar cell
but the current through it is in the opposite direction to the convention that
current flows from high to low potential. (b) The current I and voltage V in
the circuit of (a) can be found from a load line construction. Point P is the
operating point (I, V). The load line is for R = 30 ž .
Light
I
R
V
I
(a) (b)
© 1999 S.O. Kasap,Optoelectronics(Prentice Hall)
Fig. 8
0.425V
-14.2mA
= -1/30

49. Solution
• The I-V characteristic of the load is the load line described in
eq.(3), I = –V /30
• The line is drawn in Fig.8(b) with a slope 1/30. It cuts the I-V
characteristics of the solar cell at I’= 14.2mA and V’= 0.425V
which are the current and voltage in the photovoltaic circuit.
• the power deliver to the load is
Pout= I’V’ = 14.210–30.425= 6.035 mW
• This is not necessarily the maximum power available from the
solar cell. The input sunlight power is
Pin = (Light intensity) (Surface area)
= (600Wm–2)(0.01m)2= 0.060W
Efficiency, = 100Pout/ Pin= 100(0.006035/0.060)=10.06%

50. Example
• A solar cell under an illumination of 600Wm–2
has a short circuit current Isc of 16.1mA and an
open circuit output voltage Voc of 0.485V.
What are the short circuit current and open
circuit voltages when the light intensity is
doubled?

51. Solution
• The general I-V characteristic under illumination is given
by Eq(2). Setting I =0 for open circuit we have
I = –Iph + Io [exp(eV/nkBT) – 1] = 0
• Assuming that Voc >>nkBT/e, rearranging the above
equation we can find Voc,
Voc = nkBT/e ln(Iph/Io)
• In Eq.(5), the photocurrent, Iph, depends on the light
intensity I via, Iph=KI. At a given temperature, then the
change in Voc is
Voc2 – Voc1 = nkBT/e ln(Iph2/Iph1) = nkBT/e ln(I2/I1)

52. Solution, cont
The short circuit current is the photocurrent so that
at double the intensity this is
Iph2 = Iph1 (I2/I1) = (16.1 mA) (2) = 32.2 mA
Assuming n = 1, the new open circuit voltage is
Voc2 = Voc1 + nkBT/e ln(I2/I1) = (0.485)+(0.0259)ln(2)
= 0.503V
This is a 3.7% increase compared with the 100%
increase in illumination and the short circuit
current. Ideally do we want Voc to be always the
same?