Question

3. For a solar cell, n-Si is given to have a refractive index of 3.435. The average power incident (Pin) is 92.5 mW/cm2.

(a). Evaluate the surface reflection loss of the silicon solar cell.

(b). Design an antireflection coating to eliminate the loss for light with 0.64 μm wavelength (provide index and thickness of this coating).

Answer 1

Let's look for the design 1st so,

b) The thickness of the anti-reflection coating is chosen so that the wavelength in the dielectric material is one quarter the wavelength of the incoming wave. For a quarter wavelength anti-reflection coating of a transparent material with a refractive index n1 and light incident on the coating with a free-space wavelength λ0, the thickness d1 which causes minimum reflection is calculated by:

d1=λ0 / 4η1

where η1 is the refractive index of anti - reflecting coating,

n0 = 1.5 (encapsulated cell) for unencapsulated cell n0 = 1 (So check in the qs wether its encapsulated cell or unencapsulated)

n2 = 3.435

n1 = 2.2699

therefore, the thickness of Anti reflecting coating will be

d1=λ0 / 4η1

d1= 0.64 (micrometer) / 4*2.2699

solving we get,

d1= 0.08 micrometers

Hence, the thickness of ARC is 0.08 micrometers and refractive index of thin film or ARC is 2.2699.

Now part a)

The reflection of a silicon surface is over 30% due to its high refractive index. The reflectivity, R, between two materials of different refractive indices is determined by:

R=(n0−nsi / n0+nsi)^2

substituting n0 = 1.5 and nsi = 3.435 we get,

R = 0.153

which is around 15%

this is the reflection coefficient (If this is not what you need then please do comment I will be sure with the needful).

Thankyou..!