If a solar cell is subjected to a light of intensity $1000$ $\text{W/m^2}$, the output at the cell is much less than ideal case scenario. What are all the factors causing energy losses in a solar cell?
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1What do you think they might be? – Kyle Kanos Dec 30 '16 at 17:51
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if we consider the one diode model of the cell, the output current is reduced due to current flowing through the diode and also the parallel resistive branch. There are resistive losses occurring and also losses due to electron hole pair recombination, – user140592 Dec 30 '16 at 18:07
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1 Answers
The current obtained from a photovoltaic cell is $ I $ and is expressed as :
$$ I = I_g - I_d $$
where $ I_g $ is the generated current and $ I_d $ is the direct diode current:
$$ I_g = A q n K_0(1 - K_1)K_2K_3$$
$$ I_d = A q n_i^2 \left[\frac{D_h}{N_{D_n} w_{n,eff}} + \frac{D_e}{N_{A_p} w_{p,eff}} \right] \left(e^{\frac{q v_{AB}}{k_B T}} - 1\right) $$
with $ n $ being the density of incident photons measured in $ m^{-2}s^{-1} $.
The main reasons of "energy" losses in photovoltaic cells are :
(1) Because our cell is actually a huge diode it must operate below
direct conduction level, but still current will flow from $ P $ junction to $ N $ junction (this current is called direct current $ I_d $. Because of this there is a theoretical max efficiency limit because we must engineer $ N_{D} $ and $ N_{A} $ such that for our working $ v_{AB} $ we will get the smallest $ I_d $. But this also implies that there's also going to be an increase in resistivity which will produce energy dissipation in the form of thermal energy due to $ P = I^2 R $.
And so we need to balance this factors into the right proportions in order to get maximum efficiency for our operating conditions.
(2) Due to the fact that $ I_d $ increases with temperature there's going to be more loss after prolonged sun exposure. So high temperature is a factor too.
(3) Cell's surface will reflect back light, so not all light will get in the cell. This is called reflection coefficient - $ K_1 $.
(4) Photons need to arrive near space charge region (the "place" in between $PN$ regions - the junction) and this depends on the coefficient of transparency of the quasi-neutral region that is faced to the sun.
(5) Current must be collected so we need "wires" with high conduction but which are are transparent as possible or which are placed in the best manner. Bad arrangement can lower the quantity of light that get's into the cell. This effects are a part of $ K_0 $, which is called surface utilization coefficient, and of $ K_2 $, which is called transparency coefficient.
(5) Our cell needs to be resistant to shocks and UV rays and other stuff so we will have trade-of some efficiency over durability.
(6) Light ray angle. The sun moves all day so we need to follow him such that the cell surface is illuminated with the most quantity of light that is possible.
(7) The choice of semiconductor element. Not all frequencies of the light will be converted because we need the light energy to have higher energy than the band gap $ E_g \leqslant hf $; this factor is $ K_3 $, which is called generation coefficient. Too high photon energies can also affect the physical structure of the thin conductive layer, of the protective layer or of the structure of the semiconductor, leading to cracks or to donor/acceptor ion migration. For this reason advanced photvoltaic cells use multiple layer of different semiconductors (like the ones used in concentrated PV with different layers of GaAs and other semiconductors each with different band gap $E_g$ - but this is not quite the case for our classical Silicon based polymorphic $PN$ junction cells.
(8) Other various factors but which are not as important as the ones mentioned above.
PS: to understand what is which for the direct current $ I_d $ components please see the legend here : This spinning solar cell advertises a 20x efficiency improvement. How?
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So if I want to write an equation of the cell incorporating all the losses occurring would it be I = Ig - Id ? – user140592 Dec 31 '16 at 01:47
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No, I = Ig - Id is the current formulation and expresses most of the constraints. What you need is expressed in either Jules(aka Wh) or Watts - energy or power - from which you can obtain the overall efficiency by dividing produced energy/power per meter square to light energy/power per meter square coming from sun (or from other sources of light). I've updated my answer with the details needed for Ig – Mihai B. Dec 31 '16 at 13:18