IV characteristic equation of triple junction solar cell

[harikasri]

Homework Statement

The given quantities are the shunt resistances across each of the 3 diode junctions,assume them to be Rsh1,Rsh2,Rsh3; Tunnel diode resistances as Rt1 and Rt2, Photocurrent for each of the junction be Ip1,Ip2,Ip3 and bandgap of each subcell be Eg1,Eg2,Eg3 are given.Let V & I be the voltage and current across the complete triple junction cell.
Now I need to find out (I-current) equation in terms of bandgap of each subcell and also it is a function of exponential terms(e^(function of V)) (diode current). I also need to find the voltage V interms of bandgap of subcells.

Homework Equations

The Attempt at a Solution

I could get the current equation with the diode current( I=Is*e^(qVd/kT)) where Vd is the voltage across the diode (say for 1 subcell). The Vd is not given and I need to represent it interms of overall voltage V across the subcell.That is e^(function of V).

 

[Redbelly98]

Welcome to Physics Forums!

Since no one else has responded, I'll give it a try. Do you have some way to relate the bandgap energy Eg to the diode voltage Vd for one subcell? (Review your class notes or textbook, if needed.)

Also, you might as well simplify by combining the three series resistors Rt1, Rt2, and Rsel.

 

[harikasri]

Sir, I require the current equation in the following generalised form: I= exp(q*function(V)/kT)+linear first order function of V+constant. The diode voltage of each subcell is to be represented in terms of overall voltage V and the final equation should be independent of diode voltages of each subcell(sayVd1,Vd2,Vd3). All these should be represented in V only.
Sir, I don't need my equation in terms of bandgap now.
Thank you for your help

 

[Redbelly98]

Sorry about the delay in responding.

Without actually solving this myself, it looks like you should apply Kirchhoff's laws to this circuit. In other words:

Total voltage V is the sum of voltages across each series branch -- express that in an equation
For each subcell, the total current I is the sum of currents through each element in the subcell -- express that in 3 equations, one equation for each subcell.

However . . . the current-voltage relation for each diode is an exponential function, which complicates things when it is combined in parallel with a shunt resistor. Is it possible you have been given a simpler I-V relation for a standard diode, that might be used here?

 

[harikasri]

sir, I have tried solving it using KCL and KVL and I have assumed that the voltage available across each subcell is divided in the same ratio of their bandgaps. that is v1=(eg1/(eg1+eg2+eg3))*total voltage available for all the three cells given the voltage drop across the resistances are subtracted from the total voltage(V). I know that V=(bandgap/charge). Then
v1=(eg1/(eg1+eg2+eg3))*(V-IRse-IRt1-IRt2). similarly for the others v2 and v3.
substituted these values and found the characteristic equation. I=∑_(i=1)^3▒〖[I_pi+(V_i/R_shi )-〖I_si (e〗^(((qV_i)/kT) )-1)]〗 i=subcell .
Thank you sir