Chapter 7a

Chapter 7: Photovoltaic Devices

Solar Energy Spectrum

Photovoltaic Device Principles

Solar Cells Materials, Devices & Efficiency

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

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.

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

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.

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 2.00

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 spectralintensity (I) vs wavelength above the earth's atmosphere

(AM0 radiation) and at the earth's surface (AM1.5radiation). 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

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.

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.

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)

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.

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.

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.

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)

h0h

(a) (b)

Tilted PV deviceEarth


Fig. 2

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.

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.3.6m panel aor 3.13

15.0min/60min/60106

24min/60min/60500

Efficiencyareaunit per energy solar Incident

houseper Enery Area

2

126

m

hrsdaymhrW

hrshrsW

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

Fig. 3

Finger electrodes

p

n

Bus electrode

for current collection

Finger electrodes on the surface of a solar cellreduce the series resistance


Fig. 4

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.

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.

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

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 theabsorption coefficient at the wavelength of interest.


Fig. 5


• 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.


• 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

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.

• 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.

• At long wavelength, around 1-1.2m, the absorption coefficient of Si is small and the absorption depth (1/) is typically > 100m

• 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 200-500m and Le tends to be shorter than this.

• Crystalline Si has a bandgap of 1.1eV, which correspond to a threshold wavelength of 1.1m.

• The incident energy in the wavelength region >1.1m 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

• 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.

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)

Solution

The EHP photo-generation rate from the illuminated crystal surface followsGoexp(–x)

The total number of EHP generated per unit time in a small volume Ax is Gph(Ax). Thus:

The total number EHP generated per unit time in

enoEHP

LWl

xoen

LWlAG

dt

dN

dxxGALWlen

exp1or

exp0

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

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=11018cm–3s–1 in Eq.(1), we find Iph20mA whereas Eq.(2) gives 21mA.

On the other hand for strong absorption at 0.83μm, = 105m–1

(absorption depth = 1/ =10μm) Eq.(1) gives Iph40mA. 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.

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

Iph

R

I

V V = 0

Iph

I = Id I

ph

V

Id

Isc

= –Iph

R

(a) (b) (c)

(a) The solar cell connected to an external load R and the convention for the definitions ofpositive voltage and positive current. (b) The solar cell in short circuit. The current is thephotocurrent, Iph. (c) The solar cell driving an external load R. There is a voltage V and current

I in the circuit.

Light


Fig. 6

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.

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.

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.

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.

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 operation is always in thenegative current region.


Fig. 7

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

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-V for a solar cell under an

illumination of 600 Wm-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 cellbut the current through it is in the opposite direction to the convention thatcurrent flows from high to low potential. (b) The current I and voltage V inthe circuit of (a) can be found from a load line construction. Point P is theoperating point (I, V). The load line is for R = 30 ž .

LightI

R

V

I

(a) (b)


Fig. 8

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.

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.

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.

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%

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?

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-V for a solar cell under an

illumination of 600 Wm-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 cellbut the current through it is in the opposite direction to the convention thatcurrent flows from high to low potential. (b) The current I and voltage V inthe circuit of (a) can be found from a load line construction. Point P is theoperating point (I, V). The load line is for R = 30 ž .

LightI

R

V

I

(a) (b)


Fig. 80.425V

-14.2mA= -1/30

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’ = 14.210–30.425= 6.035 mW

• This is not necessarily the maximum power available from the solar cell. The input sunlight power isPin = (Light intensity) (Surface area)

= (600Wm–2)(0.01m)2= 0.060WEfficiency, = 100Pout/ Pin= 100(0.006035/0.060)=10.06%

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?

Solution

• The general I-V characteristic under illumination is given by Eq(2). Setting I =0 for open circuit we haveI = –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)

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?

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Chapter 7a

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