Photovoltaic Systems Engineering,
5th Edition by Roger A. Messenger
Complete Chapter Solutions Manual
are included (Ch 1 to 11)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
Chapter 1 Background
Chapter 2 The Sun
Chapter 3 Introduction to PV Systems
Chapter 4 Grid-Connected Utility Interactive PV Systems
Chapter 5 Structural Considerations
Chapter 6 Energy Storage Systems
Chapter 7 Grid-Connected PV Systems with Energy Storage (ESS)
Chapter 8 Stand-Alone PV Systems
Chapter 9 Economic and Environmental Considerations
Chapter 10 The Physics of Photovoltaic Cells
Chapter 11 Evolution of Photovoltaic Cells and Systems
,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch11-1)
Chapter 11 Problem Solutions
11.1. Assume a PV module is to have dimensions of 39.4 in x 79.2 in. Also assume that 8-in round cells are
available.
a. Calculate the percentage of the module area that can be covered with circular cells.
b. If the cells operate with 21% efficiency, and if there is no mismatch loss among cells, calculate the
overall module efficiency if the round cells are used.
c. Assume the cells are “squared up by sawing off the edges to obtain cells that measure 6 in between the
straight sides. Calculate the percentage fill of these “squared up” cells and repeat the module efficiency
calculation.
d. What is the percentage loss of material in the “squaring up” process?
a. It is first necessary to determine how many cells will fit into the module frame. Clearly, only 4 cells will
fit into the smaller dimension (39.4 in) and 9 cells will fit into the larger dimension (79.2 in). The area of
the module is 39.4×79.2 = 3120 in2. The area of 36 cells is 1810 in2. So the cells can cover 58% of the
module area.
b. The overall module efficiency will be simply 58% of 21% = 12.2%.
c. One way to determine the area of the “squared up” cell is to determine the angles θ and φ as shown in the
cell diagram. Then note that the cell will be composed of 8 triangles and 4 wedges. So if θ and φ are
known, then the areas of the segments of the cell can be computed.
For θ, note that θ = cos−1(3/4) = 41.41°. Thus φ = 90 − 82.82° =
7.18° and x = 3tanθ = 2.646 in.
Thus, the area of the triangle with sides, 3, 4 and x is X
AT = 0.5×3×2.646 = 3.97 in 2
3”
θ 4”
and the area of the wedge is
AW = (7.18÷360)×π ×42 = 1.00 in2. So the total area of the cell is
AC = 8×3.97 + 4×1.00 = 35.76 in2, compared to a 6 in square with an
area of 36 in2.
φ
Next, within the module frame dimensions, it is possible to fit 78 of the
new cells in a 6 x 13 array. So the cell area of the module is 2789 in2.
Since the module area is 3120 in2, 89.4% of the module is filled with
cells. So the module efficiency is now 18.8%.
d. In the squaring up process, the cell area is reduced from 50.27 in2 to 35.76 in2, so 14.51 in2 are lost in
processing, or 28.9%. As a final note, by reducing the spacing between cells and then reducing the length and
width of the module slightly, the overall efficiency can approach 20%.
11.2 Determine the expression for the depth of the junction after the drive-in diffusion step.
The impurity density after the drive-in diffusion is given by (11.6) The junction depth occurs when N(x,t) = 0.
So, to find the junction depth, solve for x after setting N(x,t) = 0.
58
, 1
x2
Q − Q 2
e 4 Dt = NA so x = 2 Dt ln .
4πDt N A πDt
11.3 Sketch the impurity distribution profile between the junction and the back contact of a single crystal
silicon cell to show how the annealed aluminum of the back contact creates an accelerating E-field.
The majority carrier holes from photon-generated EHPs are accelerated toward the back contact by the E-field
from the Aluminum gradient in front of the back contact. Furthermore, electrons from EHPs experience a
force away from the back contact due to the direction of the built-in E-field. This may enable some of them to
diffuse back across the junction to the n-side. So the impurity gradient has a two-fold effect on increasing
photocurrent.
NA(x)
n-side p-side
holes diffuse
−
−
Aluminum
holes from EHP −
− E
Negative ions
left behind
x
Junction
11.4 What volume and weight of tellurium is needed to produce a square meter of CdTe thin film with a
thickness of 1.0 μm? Assume the Cd and Te occupy equal volumes within the film.
If Te occupies half the volume of a sheet of CdTe that is 1.5 µm thick, it occupies a volume of 0.5×1.5×10−6
m3 = 0.75 cm3. Since the density of Te is 6.24 g/cm3, the weight of the Te will be 4.68 g, or 0.165 oz.
11.5 What volume and weight of indium is required to fabricate a 1 MW PV CIS array if the CIS layer
thickness is 1.0 µm, assuming that 18% of the layer volume is due to the In. Assume an array efficiency
of 19% and standard test conditions.
If the efficiency of the aray is 19%, then 1 m2 will produce approximately 190 W. Hence, a 1MW array will
require 5263 m2. Hence, the In volume will be (0.18×5263 m2)×(1.5×10−6 m) = 1.42×10−3 m3 = 1420 cm3.
The density of In is 7.31 g/cm3, so the weight will be 10,380 g, or 10.38 kg, or 22.83 lb.
11.6 The energy payback time (EPBT) for a PV technology is defined as the time needed for the completed cell
to generate an amount of electrical energy equal to the energy required for producing the cell. Refer to
current literature and tabulate the EPBT for the most efficient crystalline Si cells.
The answer to this problem will likely continue to decrease. A good reference for EPBT would be publications
by Fthenakis or NREL. In 2024, the average payback time is about 2 years, but it depends upon manufacturing
conditions. Since a significant amount of electricity is used in the production, the fossil fuel contribution to the
electrical generation affects the EPBT. A cell manufactured in Europe in 2024 will have a shorter EPBT than
the same cell manufactured in the USA since the European grid has a larger renewable contribution than the U.S
grid, at least at the time of publication of this book. However, shipping also contributes, so this must also be
considered for the most accurate analysis.
59
5th Edition by Roger A. Messenger
Complete Chapter Solutions Manual
are included (Ch 1 to 11)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
Chapter 1 Background
Chapter 2 The Sun
Chapter 3 Introduction to PV Systems
Chapter 4 Grid-Connected Utility Interactive PV Systems
Chapter 5 Structural Considerations
Chapter 6 Energy Storage Systems
Chapter 7 Grid-Connected PV Systems with Energy Storage (ESS)
Chapter 8 Stand-Alone PV Systems
Chapter 9 Economic and Environmental Considerations
Chapter 10 The Physics of Photovoltaic Cells
Chapter 11 Evolution of Photovoltaic Cells and Systems
,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch11-1)
Chapter 11 Problem Solutions
11.1. Assume a PV module is to have dimensions of 39.4 in x 79.2 in. Also assume that 8-in round cells are
available.
a. Calculate the percentage of the module area that can be covered with circular cells.
b. If the cells operate with 21% efficiency, and if there is no mismatch loss among cells, calculate the
overall module efficiency if the round cells are used.
c. Assume the cells are “squared up by sawing off the edges to obtain cells that measure 6 in between the
straight sides. Calculate the percentage fill of these “squared up” cells and repeat the module efficiency
calculation.
d. What is the percentage loss of material in the “squaring up” process?
a. It is first necessary to determine how many cells will fit into the module frame. Clearly, only 4 cells will
fit into the smaller dimension (39.4 in) and 9 cells will fit into the larger dimension (79.2 in). The area of
the module is 39.4×79.2 = 3120 in2. The area of 36 cells is 1810 in2. So the cells can cover 58% of the
module area.
b. The overall module efficiency will be simply 58% of 21% = 12.2%.
c. One way to determine the area of the “squared up” cell is to determine the angles θ and φ as shown in the
cell diagram. Then note that the cell will be composed of 8 triangles and 4 wedges. So if θ and φ are
known, then the areas of the segments of the cell can be computed.
For θ, note that θ = cos−1(3/4) = 41.41°. Thus φ = 90 − 82.82° =
7.18° and x = 3tanθ = 2.646 in.
Thus, the area of the triangle with sides, 3, 4 and x is X
AT = 0.5×3×2.646 = 3.97 in 2
3”
θ 4”
and the area of the wedge is
AW = (7.18÷360)×π ×42 = 1.00 in2. So the total area of the cell is
AC = 8×3.97 + 4×1.00 = 35.76 in2, compared to a 6 in square with an
area of 36 in2.
φ
Next, within the module frame dimensions, it is possible to fit 78 of the
new cells in a 6 x 13 array. So the cell area of the module is 2789 in2.
Since the module area is 3120 in2, 89.4% of the module is filled with
cells. So the module efficiency is now 18.8%.
d. In the squaring up process, the cell area is reduced from 50.27 in2 to 35.76 in2, so 14.51 in2 are lost in
processing, or 28.9%. As a final note, by reducing the spacing between cells and then reducing the length and
width of the module slightly, the overall efficiency can approach 20%.
11.2 Determine the expression for the depth of the junction after the drive-in diffusion step.
The impurity density after the drive-in diffusion is given by (11.6) The junction depth occurs when N(x,t) = 0.
So, to find the junction depth, solve for x after setting N(x,t) = 0.
58
, 1
x2
Q − Q 2
e 4 Dt = NA so x = 2 Dt ln .
4πDt N A πDt
11.3 Sketch the impurity distribution profile between the junction and the back contact of a single crystal
silicon cell to show how the annealed aluminum of the back contact creates an accelerating E-field.
The majority carrier holes from photon-generated EHPs are accelerated toward the back contact by the E-field
from the Aluminum gradient in front of the back contact. Furthermore, electrons from EHPs experience a
force away from the back contact due to the direction of the built-in E-field. This may enable some of them to
diffuse back across the junction to the n-side. So the impurity gradient has a two-fold effect on increasing
photocurrent.
NA(x)
n-side p-side
holes diffuse
−
−
Aluminum
holes from EHP −
− E
Negative ions
left behind
x
Junction
11.4 What volume and weight of tellurium is needed to produce a square meter of CdTe thin film with a
thickness of 1.0 μm? Assume the Cd and Te occupy equal volumes within the film.
If Te occupies half the volume of a sheet of CdTe that is 1.5 µm thick, it occupies a volume of 0.5×1.5×10−6
m3 = 0.75 cm3. Since the density of Te is 6.24 g/cm3, the weight of the Te will be 4.68 g, or 0.165 oz.
11.5 What volume and weight of indium is required to fabricate a 1 MW PV CIS array if the CIS layer
thickness is 1.0 µm, assuming that 18% of the layer volume is due to the In. Assume an array efficiency
of 19% and standard test conditions.
If the efficiency of the aray is 19%, then 1 m2 will produce approximately 190 W. Hence, a 1MW array will
require 5263 m2. Hence, the In volume will be (0.18×5263 m2)×(1.5×10−6 m) = 1.42×10−3 m3 = 1420 cm3.
The density of In is 7.31 g/cm3, so the weight will be 10,380 g, or 10.38 kg, or 22.83 lb.
11.6 The energy payback time (EPBT) for a PV technology is defined as the time needed for the completed cell
to generate an amount of electrical energy equal to the energy required for producing the cell. Refer to
current literature and tabulate the EPBT for the most efficient crystalline Si cells.
The answer to this problem will likely continue to decrease. A good reference for EPBT would be publications
by Fthenakis or NREL. In 2024, the average payback time is about 2 years, but it depends upon manufacturing
conditions. Since a significant amount of electricity is used in the production, the fossil fuel contribution to the
electrical generation affects the EPBT. A cell manufactured in Europe in 2024 will have a shorter EPBT than
the same cell manufactured in the USA since the European grid has a larger renewable contribution than the U.S
grid, at least at the time of publication of this book. However, shipping also contributes, so this must also be
considered for the most accurate analysis.
59
