I would like to confirm the soundness of a simple steady-state energy balance over a solar cell.
Assume we take a solar cell as the control volume for an energy balance where the electrical conversion efficiency is defined with respect to the incident solar power.
As shown in the attached energy balance and energy flow diagrams, are the following general equations correct for steady-state operation:
Where:
My uncertainty is rising from the fact that P_ele is defined with respect to P_inc not P_abs and how that would affect the energy balance.
Also, what is the energy form of P_abs? Clearly, P_abs is an intermediate step in the conversion of solar power into electrical and thermal powers. Does that mean P_abs is in the form of internal chemical power within the semiconductor?
Any help is highly appreciated.
There are two "etas":
Usually the "photovoltaic (PV) power conversion efficiency" is defined by the quotient of electrical power output P_ele divided by the incidencing irradiance P_inc (but not on the cell, it is accounted on the PV module surface: there is a glass sheet and a plastic encapsulation material (e.g. EVA) above the solar cell).
The efficiency accounting for the absorbed irradiance (or photons) is usually called "quantum efficiency" and is the quotient of absorbed irradiance (photon flow) P_abs and electrical current (electron flow) being "generated" P_ele.
The energy of the photons is eV.
Unit of irradiance is W/m², multiplied by the area of the module (or cell): W.
Omar,
Your energy balance is okay except there are some modifications on it. Which may explain your ambiguity. The incident solar radiation is Pin, part of it will be reflected R pin where R is the reflectance, and part of it Ptras will be transmitted The reflectanc is magnitude square of the reflection coefficient roh. The transmitted power ptrans is then= pin- R pin. The reflection coefficient roh is a a property of the surface receiving the radiation. It is known and it is function of the wavelength of the incident radiation. The transmitted power will be either absorbed in the material Pabs or get out of the material. If the thickness of the material is small and the absorption coefficient is not large an appreciable portion will get out of the material.
In solar cells however, the back side is covered with metal which prevents the radiation from coming out and reflects it again in the interior of the material. The transmitted radiation inside the material will be absorbed with an absorption coefficient alpha such that I= I0 exp - alpha x where I is is the radiation intensity at position x and Io is the intensity at x=0, the surface of the material. I0= Ptra/ area of the incidence.
The absorbed power will be controverted partly to electrical power and partly to heat. The heat is dispatched from the material convection, conduction, and radiation.
This is the overall budget of the incident solar radiation.
wish you success
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