Solar cell design by using Photonic Simulation software

 

Basic solar cell design includes P-N junction with thin N layer being at top and bulk P layer at bottom. It is designed without the ARC coating. But since bare silicon is oxidised due to reaction of oxygen so top and bottom side of solar cell is thin silica oxide layer.

Dimensions of solar cell- 

Overall dimension of cell=14 x 10 x 6 um (considered for simulation)

N – type layer at top of cell diffused doping surface concentration = 10^19 /cm3 and reference concentration = 10^10/cm3, With thickness of 1 um and junction width of 0.4 um.

Remaining material is p-type doped with density 10^16/cm3.

At upper and lower surfaces 1 um thick silicon oxide layer.

At top emitter side the silver current collector wire of  2 um breadth and 0.5um thickness.

At bottom base side aluminium current collector of 1 um thickness.

The picture of lumerical simulation model is given below-

To do the simulation in device first of all we need generation curve(wavelength range 0.3 – 1.1 um) which is obtained from the FDTD. Generation curve gives the variotion of electon hole pair generation with the solar cell depth.

After that we imported the generation rate data in the device applied  biasing across the solar cell to get the following I-V curve.

This curve shows the variation of current density of solar cell with respect to the voltage.

The short circuit current is maximum current of solar cell obtained at zero voltage, and open circuit voltage is maximum voltage obtained at open circuit i.e. zero current. It is showing that the current is nearly constant up to certain voltage after that it falls very fast.

The above curve shows the power density variation with respect to voltage. We can get the maximum value of power from the above curve so that we can find the efficiency.

 Efficiency η % =( Pmax / solar irradiance at 1.5  AM) * 100.

Results.=

Voc = 562.222 mV

Jsc = 12.37 mA/cm^2

Pmax = 5.76976 mW/cm^2

η % = 5.76976%

Voltage and current waveform of various type of loads- Simulations

In this post, I have plotted volatage and current wavefrom under various loads such as resistive, capacitive, inductive and thier combination under sinusoidal input through simulation softwares.

1. With Resistive Load- RESISTOR 10 ohm

2. With Inductive load- 10mH

3. Capacitive load- 10 mF (10 mili Farad)

4. With Resistive and Capacitive ( RC ) Load R- 10ohm & C- 10 mF 

5. With Resistive and Capacitive ( RC ) Load R- 10ohm & C- 1mF 

Comparison with C= 10mF & 1 mF

6. With Resistive and Inductive ( RL ) Load R- 10ohm & L- 10mH

7. With Resistive and Inductive ( RL ) Load R- 10ohm & L- 1mH

Comparison between with 10 mH & 1 mH

8. With Resistive, Capacitive and Inductive ( RL ) Load R- 10ohm, C- 10mF & L- 1mH

There are many simulation software such as Multisim, Matlab, proteus available for simulation trial version may be downloaded for beginning. It will help you in understand the theory in good way.

Calculation of the theoretical heat generated in the Proton Exchange Membrane fuel cell & Electrical potential Difference

Calculation of the theoretical heat generated in the fuel cell.

Basic reaction at fuel cell electrodes-

                 At the anode:       H₂ →  2H⁺ + 2e^-                                                                                                   

                 At the cathode:    ½ O₂  + 2H+ + 2e- →  H₂O                                            

                 Overall:               H₂ + ½ O₂ →  H₂O                                                                      

The overall reaction is same as the hydrogen combustion. Since combustion is exothermic process so heat will be released -

                                  H₂ + ½ O₂ →  H₂O + heat

Standard enthalpy of reaction can be calculated by difference of heat of formation of products and reactant.

       ΔH= (hf)_H2O(l)- (hf)_H2- (hf)_0.5O2

Heat of formation of liquid water is -

(hf)_H2O = -286 kJmol^-1 (at 25°C)

Heat of formation of reactant gases 

(hf)H₂₌ (hf)0.5O₂ =0 kJ/mol

So,                                                  ΔH= -286-0-0= -286 kJ/mol.

-ve sign indicates that reaction is exothermic.

But if products H2O is not cooled enough to become liquid, and it remains as vapour then the enthalpy of reaction reduced by enthalpy of vaporisation of water ie. 44 kJ/mol

     ΔH= (hf)_H2O(g)- (hf)_H2- (hf)_0.5O2

              = -286+44 -0-0= -242 kJ/mol.

All the generated energy cannot be converted into electricity due the reaction entropy. The portion of the reaction enthalpy (or hydrogen's higher heating value) that can be converted to electricity in a fuel cell corresponds to Gibbs free energy and is given by the following equation:

                                                       DG = DH- TDS    .

So Gibbs free energy in the liquid water conversion at 25⁰ C

                                                      DG= -286-(25+273)*(-0.1663)= -237 kJ/mol.

Heat produced theoretically in the reaction is = -237-(-286)= 49 kJ/mol

And Gibbs free energy in the vapour water conversion at 25⁰ C

                                                     DG= -242- (25+273)*(-0.0444) = -229 kJ/mol.

Heat produced theoretically in the reaction is = -229- (-242)= 13kJ/mol.

Electrical work:                           W _el= nFE,       

 n= no of electrons per mole. E= electric potential

Maximum amount of electrical energy generated in a fuel cell corresponds to Gibbs free energy, DG:                                              

W_el = - DG =nFE

So theoretical electrical potential   E= = - DG/ nF

                                                            =- (-237*1000)/ (2*96500) = 1.23 V 

Fore more details on fuel cell, please commment, I can share more details of fuel cell.