December 17, 2012

Fig. 1: Schematic of the nanowire solar cell. |

China has become the second biggest country in energy consumption and the demand is still increasing. Developing the renewable energies should be an urgent topic in China as well as the whole world with the consideration of the energy crisis and the global environmental problems such as the Green House Effects and Climate Change. Solar energy has been the most attractive renewable energy source, and how to effectively transform it to the energies available to human beings has been discussed for many years. With the development of nano materials science, nanowire solar cells have shown its potential to be efficient and cheap. [1-4] In this paper I will explain the advantages and show some good performance of nanowire solar cells, which should be produced in a large scale in China to replace some energy sources and protect the environment.

In order to lower the cost of thin film solar cells,
people use silicon with high level of impurities and density of defects,
which results in low minority-carrier diffusion lengths. [2] According
to the equation I = I_{0}e^{ - αz}. We also want
to increase the thickness of p-n junction to absorb as much light as we
can and improve its efficiency. These are contradictory because the
carrier collection is limited by the short minority-carrier diffusion
length in thick solar cells. However, a nanowore solar cell with a p-n
junction in the radial direction can separate light absorbtion and
carrier extraction, which is shown in Fig. 1. Each individual p-n
junction nanowire in the cell could be long in the direction of incident
light, allowing for optimal light absorption, but thin in another
dimension, thereby allowing for effective carrier collection.

To simplify the analysis, the carrier transport was
taken to be purely radial. We assume the minority-carrier diffusion
length in n and p region is L_{n} and L_{p}
respectively. We also set that the depletion region of each side is
d_{n} and d_{p}. In order to take all the minorities, we
can imagine a nanowire with the radius equals to L_{n} +
L_{p} + d_{n} + d_{p} = r. The cross-section
area of the solar cell is S=πr^{2}. The length of the
nanowire is also an important parameter which influence the absorption
of incident light. We set the absorption coefficient as α. To get
95% of the light we choose the length of the nanowire to be 3/α.

We assume the doping concentration in n and p type
silicon is the same, with N_{A} = N_{D} =
10^{18} cm^{-3}. With the doping information, we can
firstly settle down all the other parameters such as L_{n},
L_{p}, and so on. From some physics classes we know that the
size of the depletion region should be:

Fig. 2: The numerical calculation of V_{max}. |

and

The left parameters are L_{n} and
L_{p}, which have the expression as L_{n} =
(τ_{n}D_{n})^{1/2} and L_{p} =
(τ_{p}D_{p})^{1/2}. From Einstein Relation
we know that

The question is how to calculate the lifetime of
minority carriers in quasineutral in each type of region,
τ_{n} and τ_{p}. Learnt from other semiconductor
materials, we get to know that with highly doped semiconductors
τ_{n} =
1/σ_{n}N_{r}υ_{th}, and
τ_{p} =
1/σ_{p}N_{r}υ_{th}, where σ
_{n} and σ_{p} are the cross sections for electron
and hole capture, respectively, N_{r} is the density of
recombination centers which can be simply assume as the density of
doping, and υ_{th} is the thermal velocity. To the first
order, σ_{n} = σ_{p}, so that
τ_{n} = τ_{p}. With all the equations described
above, we can now calculate all the parameters about this nanowire under
room temperature: [2]

and

So the radius of our nanowire is

and the area is

Scientists have calculated the number of photon flux
with energies over 1.12eV to be 2.73 ×
10^{21}m^{-2}s^{-1}. So the light current in
this nanowire is

We must also calculate the saturation current:

So

In order to get the theoretical efficiency of this
nanowire solar cell, we need to calculate the maximum power. It's
obvious that the V_{max} is given by the equation

where V_{oc} = 0.593V, V_{th} =
0.0259V. We can simplify this equation like the following:

We can calculate V_{max} numerically and
obtain the following figure. From Fig. 2 we can know that
V_{max} = 0.570V. So the I_{max} can easily be obtained:

So the fill factor for this ideal nanowire solar cell is

and importantly, the efficiency is

In this paper I just set up a very simple model and use a large amount of knowledge from what I have learnt in my physics courses during my undergraduate and the first year graduate study to estimate some key parameters of a nanowire solar cell without a detailed consideration of recombination, series resistance, shunt resistance, and so on. Although it may be not so accurate for actual cells, but I am sure these data can give us a basic concept about this kind of solar cells. Such a high efficiency should be tremendously useful in the energy area in China and will have a huge market in both China and the world.

© Haotian Wang. The author grants permission to copy, distribute and display this work in unaltered form, with attribution to the author, for noncommercial purposes only. All other rights, including commercial rights, are reserved to the author.

[1] E. C. Garnett *et al.*, "Nanowire Solar
Cells," Annu. Rev. Mater. Res. **41**, 269 (2011).

[2] B. M. Kayes, H. A. Atwater and N. S. Lewis,
"Comparison of the Device Physics Principles of Planar and Radial p-n
Junction Nanorod Solar Cells," J. Appl. Phys. **97**, 114302
(2005).

[3] E. C. Garnett and P. Yang, "Silicon Nanowire
Radial p-n Juncion Solar Cells," J. Am. Chem. Soc. **130**, 9224
(2008).

[4] L. Tsakalakos *et al.*, "Silicon Nanowire
Solar Cells," Appl. Phys. Lett. **91**, 233117 (2007).