Solar photovoltaic cells materials, current ,voltage ,characteristics are explained in this ppt how they work their active region their saturation region their cut-off region

1. PHOTOVOLTAIC MATERIAL
AND ELECTRICAL
CHARACTERISTIC
By
Syed Ajmal

2. Introduction
• A material or device that capable of converting energy in
photons of light into electrical voltage and current –
photovoltaic.
• Photovoltaic use semiconductor materials to convert
sunlight into electricity.
• Basic photovoltaic device is crystalline silicon.
• As long as solar cell is exposed to photons with energies
above the band-gap energy, hole-electron pairs will be
created.
• The problem is electron can fall back into a hole
(recombination) causing both charge carriers to disappear.
• In PV, avoiding the problem by creating electric field within
semiconductor itself that pushes electron in one direction
and holes in the other.

3. P-n junction diode
• P-n junction diode: If we apply voltage, Vd across diode terminal,
forward current flow from p-side to n-side. If diode in reverse
direction, only a very small (≈10-12 A/cm2) reverse saturation
current, I0 will flow.
• V-I characteristic curve for p-n junction diode is described by the
following Shockley diode equation:
• Where Id is diode current in the direction of the arrow, Vd is
voltage across diode terminal from p to n-side, I0 is reverse
saturation current, q is electron charge (1.602 x 10-19C), k is
Blotzmann’s constant (1.381 x 10-23 J/K) and T is the junction
temperature (K).
• A junction temperature of 25oC is often used as a standard, so
diode equation becomes:
)
1
( /
0 
 kT
qV
d
d
e
I
I
C
at
e
I
I o
V
d
d
25
)
1
( 9
.
38
0 


4. • A p-n junction diode allows current to flow
easily from p-side to n-side, but not in reverse.

5. Generic photovoltaic cell
• Simplest equivalent circuit for a photovoltaic cell
– Consists of diode parallel with ideal current source. Ideal current
source delivers current in proportion to solar flux to which it is
exposed.
A simple equivalent circuit for a PV cell consists of a current
source driven by sunlight in parallel with a real diode.

6. • 2 conditions for actual PV and its equivalent circuit:
– Current flows when the terminals are shorted together, short-circuit
current, ISC
– Voltage across terminal when the leads are left open, open-circuit
voltage, VOC.
• Voltage and current equation are:
• When the leads from PV cell are left open, I=0 and we can find VOC:
• At 250C, these above equation becomes:
• In both equations, ISC is directly proportional to solar insolation
(radiation), which means that we can easily plot sets of PV I-V
curves for varying sunlight.
d
SC I
I
I 

)
1
( /
0 

 kT
qV
SC e
I
I
I









 1
0
I
I
In
q
kT
V SC
OC
)
1
( 9
.
38
0 

 V
SC e
I
I
I 








 1
0257
.
0
0
I
I
In
V SC
OC

7. • 2 important parameters for PV are short-circuit
current and open-circuit voltage.

8. Photovoltaic I-V characteristic
curve
• Figure below show I-V relationship for a PV cell when it is dark
(no illumination) and light (illuminated). The dark curve is just
diode curve turned upside-down. The light curve is the dark
curve plus ISC.

9. Example 1
• Consider a 100-cm2 PV cell with reverse
current,I0=10-12 A/cm2. In full sun, it produces ISC
of 40mA/cm2 at 250C. Find VOC at full sun and
again for 50% sunlight. Plot the results.

10. Solution example 1
• The reverse saturation current,
I0=10-12 A/cm2 x 100cm2 = 1x10-10A.
• At full sun, ISC =0.040A/cm2 x 100cm2= 4.0A.
• VOC is:
• Since short-current current is proportional solar
intensity, at half sun ISC=2A and VOC is:
V
In
I
I
In
V SC
OC 627
.
0
1
10
4
0257
.
0
1
0257
.
0 10
0


















 
V
In
I
I
In
V SC
OC 610
.
0
1
10
2
0257
.
0
1
0257
.
0 10
0


















 

11. • Plotting the result

12. More accurate equivalent circuit for a
PV cell
• Consider the impact of shading on a string of cells wired in
series. If any cell in the string is in dark (shaded), it
produces no current and diode is reverse-biased. Thus, no
power will be delivered to the load if any of the cells are
shaded.

13. • Figure below shows a PV equivalent circuit including
parallel leakage resistance, Rp. The ideal ISC deliver
current to diode, Rp and load:
p
d
SC
R
V
I
I
I 

 )
(

14. • From the equation, at any given voltage, the parallel
leakage resistance causes load current for the ideal
model to be decreased by V/Rp as shown in graph
below.

15. • An equivalent circuit with series resistance, Rs shown
below.
• Analysis part of the circuit.
• And then add the impact of Rs,
• To give
)
1
( /
0 



 kT
qV
SC
d
SC
d
e
I
I
I
I
I
S
d IR
V
V 













 

 1
)
(
exp
0
kT
IR
V
q
I
I
I S
SC

16. • Adding series resistance to the PV equivalent
circuit causes the voltage at any given current
to shift to the left by ΔV=IRS.

17. • Finally, PV equivalent circuit including both
series and parallel resistance.
• The equation for current and voltage is:







 













 


p
S
S
SC
R
IR
V
kT
IR
V
q
I
I
I 1
)
(
exp
0
  C
at
R
IR
V
e
I
I
I
p
S
IR
V
SC
S 0
(
9
.
38
0 25
1







 



 

18. • Unfortunately, the equation is a complex equation for
which there is no explicit solution for either V or I.
• Apply KCL to the node in figure above, we can write:
• Rearrange and substituting Shockley diode equation at
250C gives:
• Voltage across an individual cell can be found from:
P
d
SC I
I
I
I 


P
d
V
SC
R
V
e
I
I
I d



 )
1
( 9
.
38
0
S
d IR
V
V 


19. • Series and parallel resistance in the PV
equivalent circuit decrease both voltage and
current delivered. To improve cell performance,
high RP and low RS are needed.

20. From cells to a module
• An individual cell produces only about 0.5-0.6V, so it is a
rare application of using just a single cell in any case.
• So, the basic building block for PV applications used is a
module consists of a number of pre-wired cells in series,
all encased in tough, weather-resistant packages.
• A typical module has 36 cells in series and often
designated as a ’12-V modules’. 72-cell modules referred
as 24-V modules.
• Multiple modules can be wired in series to increase
voltage and in parallel to increase current, the
product of which is power.
• Important element in PV system design is deciding how
many modules should be connected in series and how
many in parallel to deliver whatever energy is needed.
• Such combinations of modules are referred to as an array.

21. • When PV are wired in series, they carry the same
current and their voltages add as shown in figure below.
• To find overall module voltage, Vmodule is:
)
(
mod S
d
ule IR
V
n
V 


22. • For cell wired in series, their voltages at any
given current add. A typical module will have
36 cells.

23. Example 2
• A PV module is made up to 36 identical cells, all
wired in series. With 1-sun insolation (1kW/m2),
each cell has ISC=3.4A and at 250C its I0=6x10-10
A. Rp=6.6Ω and Rs=0.005Ω.
a. Find the voltage, current and power delivered
when the junction voltage of each cell is 0.50V.
b. Set up a spreadsheet for I and V and present a
few lines of output to show how it works.

24. Solution example 2
a. Using Vd=0.5V and the other data gives current:
A
e
R
V
e
I
I
I
P
d
V
SC
d
16
.
3
6
.
6
5
.
0
)
1
(
10
6
4
.
3
)
1
(
5
.
0
9
.
38
10
9
.
38
0












V
IR
V
n
V S
d
ule 43
.
17
)
005
.
0
16
.
3
5
.
0
(
36
)
(
mod 





W
I
V
P ule 55
16
.
3
43
.
17
mod 




25. A spreadsheet look something like the following:
Notice that: the maximum power for this module is at
I=3.16A, Vmodule=17.43V and P=55.08W.

26. From modules to arrays
• Arrays are made up of some combination of series and
parallel modules to increase power.
• For modules in series, I-V curves are added along the
voltage axis. So, at any given current (which flows
through each of modules), the total voltage is the sum of
individual module voltages.

27. • For modules in parallel, the same voltage is across each
module and the total current is the sum of the currents.
At any given voltage, I-V curve of parallel combination
is the sum of individual module current at that voltage.

28. • 2 ways to wire an array with 3 modules in series and 2
modules in parallel. Although I-V curves for arrays are
the same, 2 strings of each 3 modules (a) is preferred.
The total I-V curve of the array is shown in figure below.

29. PV I-V curve under standard test
conditions (STC)
• A single PV module is connected to load as shown in
figure below. The load may be a dc motor driving a
pump or a battery.
• Before load connected, module is in the sun will produce
VOC but no current will flow.
• If the terminals are shorted, ISC will flow but the output
voltage is zero.
• In both cases, no power is delivered by the module and
no power received by the load.
• When the load is connected, current and voltage will
result and power is delivered.
• To figure how much power, we have to consider I-V
curve of the module as well as I-V curve of the load.

31. • Figure below shows a generic I-V curve for PV module.
• At the two ends of I-V curve, the output power is zero
since either current or voltage is zero at that point.
• The maximum power point (MPP) is spotted near the
knee of I-V curve at which the product of current and
voltage reaches its maximum.
• The voltage and current at MPP is designated as VR and
IR.

32. • Manufacturer provide performance data under standard
operating condition. Rated power denote as PDC,STC is the
dc power measured under STC.

33. Impacts of temperature and insolation on
I-V curves
• Manufacturers often provide I-V curves that show how
curves shift as insolation and cell temperature changes.
• Figure below shows examples for Kyocera 120-W
multicrystal-silicon module. As insolation drops, ISC drops
proportionally. Decrease insolation also reduces VOC.

34. • As cell temperature increases, VOC decreases drastically
eventhough ISC increases slightly.
• Cell vary in temperature not only because ambient
temperature change, but also due to insolation on the
cells change.
• For crystalline silicon cells, VOC drop about 0.37%, ISC
increases by 0.05% and maximum power decrease by
0.5% for each degree Celsius increase in temperature.
• Only a small fraction of insolation hitting a module is
converted to electricity and carried away, most of incident
energy is absorbed and converted to heat.
• Manufacturer often provide an indicator called NOCT
(nominal operating cell temperature) to account for
changes in cell performance with temperature.
• NOCT is cell temperature in a module when ambient is
200C,solar irradiation is 0.8kW/m2 and windspeed is 1m/s.
• To calculate for other ambient conditions, use:
S
NOCT
T
T amb
cell 







 


8
.
0
200

35. Example 3
• Estimate cell temperature, VOC and maximum power
output for 150W,BP2150S module under condition of 1-
sun insolation and Tamb=300C. The module has a
NOCT=470C and S=1kW/m2.
• Solution:
• From Table given, for this module at standard
temperature, VOC=42.8V. Since VOC drops by 0.37%/0C,
the new VOC will be:
• With maximum power expected to drop about 0.5%/0C,
this module will deliver:
– Which is a significant drop of 19% from its rated power (150W).
C
S
NOCT
T
T amb
cell
0
0
64
1
8
.
0
20
47
30
8
.
0
20













 


V
VOC 7
.
36
)]
25
64
(
0037
.
0
1
[
8
.
42 



W
P 121
)]
25
64
(
005
.
0
1
[
150
max 





36. Shading impacts on I-V curves
• The output of PV module can be reduced
dramatically even when a small portion of it is
shaded.
• A single shaded cell can easily cut output power
by more than half.
• External diode is added by the system designer
or manufacturer to preserve the performance of
PV modules.
• The main purpose of diode is to mitigate the
impact of shading on PV I-V curves. Diode is
connected in parallel with modules or blocks of
cells within a module.

37. • In figure (a), all the cells are in the sun and since they are
in series, the same current flow through each of them.
• In figure (b), the top cell is shaded and its ISC is reduced to
zero. The voltage drop across Rp as current flow through it
causes the diode to be reverse-biased, Id=0. So, the entire
current flowing through the module must travel through
both Rp and Rs in the shaded cell on its way to the load. The
top cell, instead of adding to the output voltage actually
reduces it.

38. • The output voltage of entire module, VSH with one cell
shaded will drop to:
• VC is I(RS + RP). With all n cells in the sun and carrying I,
the output voltage is V, so the voltage of the bottom n-1
cells will be:
• Combine both above equations:
• The drop in voltage ΔV at any given I, caused by shaded
cell, is given by:
)
(
1 S
p
n
SH R
R
I
V
V 

 
V
n
n
Vn 




 


1
1
)
(
1
S
P
SH R
R
I
V
n
n
V 






 

)
(
)
(
1
1
S
P
S
P
SH
R
R
I
n
V
R
R
I
V
n
V
V
V
V


















39. • Since the parallel resistance, Rp is much greater than
series resistance, Rs simplifies to:
• At any given current, I-V curve for the module with one
shaded cell drops by ΔV. The huge impact can be seen by
figure below.
P
IR
n
V
V 



40. Example 4
• The 36-cell PV module described in Example 3 had a Rp
per cell=6.6Ω. In full sun and at current, I=2.14A the
output voltage, V=19.41V. If one cell is shaded and this
current stays the same, so:
a) What would be the new module output voltage and
power?
b) What would be the voltage drop across the shaded
cell?
c) How much power would be dissipated in the shaded
cell?

41. Solution example 4
a) The drop in module voltage will be:
– The new output voltage will be 19.41-14.66=4.75V.
– Power delivered by the module with one cell shaded would be:
– For comparison, in full sun the module is producing 41.5W.
b) All of 2.14A current goes through parallel plus series
resistance (0.005Ω) of shaded cell, so the drop across
the shaded cell will be:
(normally a cell in the sun will add about 0.5V to the module, this shaded
cell subtracts over 14V from module)
V
IR
n
V
V P 66
.
14
6
.
6
14
.
2
36
41
.
19







W
VI
P ule 17
.
10
14
.
2
75
.
4
mod 



V
R
R
I
V S
P
c 14
.
14
)
005
.
0
6
.
6
(
14
.
2
)
( 





42. c) The power dissipated in the shaded cell is voltage
drop times current which is:
Most of power dissipated in the shaded cell is
converted to heat, which can cause a local hot spot
that may permanently damage the plastic
laminates enclosing the cell.
W
I
V
P c 2
.
30
14
.
2
14
.
14 




43. • Module under full-sun conditions and with one cell 50% shaded, one
cell completely shaded and 2 cells completely shaded. The vertical
line at 13V is a typical operating voltage for module charging a 12V
battery. The impact of charging current is obviously severe. With just
one cell shaded out of 36 in the module, the power delivered to the
battery is decreased by about 2/3 of power delivered at full sun.

44. Bypass diodes for shade mitigation
• Shading – can shift drastically I-V curve and create hot
spots that can damage the cell.
• In figure (a) below, a solar cell in full sun operating in its
normal range contributes about 0.5V to the voltage
output of the module. In the equivalent circuit in (b), a
shaded cell experiences a drop as current is diverted
through parallel and series resistances. This drop is
considerable, it is over 14V.

46. • The voltage drop in shaded cells can be corrected by
adding a bypass diode across each cell as shown in
figure.
• When a solar cell is in the sun, there is a voltage rise
across the cell so the bypass diode is cut off and no
current flow through it.
• When the solar cell is shaded, the drop that occur if
the cell conducted current would turn on the bypass
diode, diverting the current flow through that diode.
• The bypass diode, when it conducts, drops about
0.6V.
• So, the bypass diode controls the voltage drop
across shaded cell and limit it to relatively modest
0.6V instead of large drop that may occur without it.

48. • Refer to the graph, PVs delivering charging current at
about 65V to 60V battery bank. In full sun, about 3.3A
are delivered to the batteries. When shaded cell without
bypass diode, the current drop by one-third to about
2.25A. With bypass diode across shaded module, I-V
curve is improved considerably.

49. • Figure below explain how bypass diode do their job. 5
modules wired in series, connected to battery that forces
the modules to operate at 65V battery.
• In full sun, the modules deliver 3.3A at 65V.
• When any of the cell shaded, they stop to produce
voltage and instead begin to act like resistors that cause
voltage to drop as the other modules continue to push
current through the string.
• Without a bypass diode to divert the current, the shaded
module loses voltage. The other modules try to
compensate by increasing voltage, but the net effect is
that current in the whole string drops.

50. • If bypass diodes included, current will go around
the shaded module and only a slight voltage drop
across the whole modules.
• Bypass diode help current go around shaded or
malfunctioning module within a string.
• This will improve the string performance but also
prevent hot spots from developing in individual
shaded cells.

52. Blocking diodes
• When string of modules wired in parallel, a similar
problem also arise when one of the strings is not
performing well.
• A malfunctioning or shaded string can withdraw
current from the rest of the array.
• By placing blocking diode (isolation diodes) at the top
of each string, the reverse current drawn by a shaded
string can be prevented.

53. • Blocking diodes prevent reverse current from flowing
down malfunctioning or shaded strings.

54. Crystalline silicon technologies
• Currently, about 80% of all PV are thick cells and
the remaining 20% are thin-film cells used in
calculator, watches and other consumer electronics.
• PV technologies can be categorized into:
i. Single crystal – dominant silicon technology
ii. Multicrystalline – cell made up of a number of relatively
large area of single crystal grain (1mm to 10cm).
iii. Polycrystalline – many grains having dimensions of 1μm to
1mm. Example cadmium telluride.
iv. Microcrystalline – grain sizes less than 1μm.
v. Amorphous – no single-crystal regions as in amorphous
silicon.