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External circuit elements

The above equations will simply solve the diode equation for an ideal diode. They neglect, the parasitic components of the diode, namely the shunt resistance, the series resistance and the parasitic capacitance. For simulating a JV curve, it is a simple matter to convert from the ideal diode voltage/current curve to the non ideal curve. One simply performs a JV curve sweep for an ideal diode, then uses the following equations to generate the non-ideal JV curve, including parasitic components.

$\displaystyle j_{external}=i_{ideal}+\frac{V_{applied}}{R_{shunt}}$ (58)

where $ i(V_{applied})$ , is calculated from the current at the contacts $ (j_{left}+j_{right})/2$ .

$\displaystyle V_{external}=V_{ideal}+R_{series}j_{external}$ (59)

This is an efficient procedure. For some measurements however, such as transient photovoltage, the situation becomes more complex, as one has to find $ V_{applied}$ , to suite the desired conditions. For example when finding $ V_{oc}$ (or indeed performing a TPV simulation), one is looking for the value of $ V_{applied}$ , which will give $ j_{external}=0$ . Clearly one way to do this would be to search a reasonable range of $ V_{applied}$ until the condition $ j_{external}=0$ is met. This would however be slow, especially if one is having to repeat the procedure multiple times, such as in a time domain simulation. A better method is to first guess the value of $ V_{applied}$ which could cause the condition $ j_{external}=0$ to be met (a good guess may be the value of the built in bias), then to iterate using an external newton solver, to update $ V_{applied}$ until the condition $ j_{external}=0$ , is met to the desired degree of accuracy. This is more efficient than a linear search for $ V_{oc}$ , however it is still inefficient, as it will require two Newton solvers, one for the diode equations (described above) and one to enforce $ j_{external}=0$ . When one couples together two Newton solvers, convergence is not guaranteed, in fact they can oscillate and diverge. Also, convergence will probably proceed in a linear fashion, rather than a super-linear fashion. The equations to external circuits, in an external newton solver is included in gpvdm as an option, mainly for debugging. Another difficulty with simulating TPV, is the precision one can solve the diode current to is of the order of $ 1\e -6$ -$ 1\e -7$ Amps, and a typical value of shunt resistance is around $ 1\e 6$ Ohms, this the numerical noise in the solver can very easily be amplified by the shunt resistance to a level where the whole system of equations becomes unstable. Solving, the system of equations in a coupled form expatiates this problem.

A better way to solve for the presence of external circuit elements is to, include another solution variable in the main Newton solver, namely $ V_{applied}$ . This avoids the need of two coupled solvers. This is a single value of potential, which defines the applied voltage to the device. The updated solution matrix will look like this

$\displaystyle \begin{bmatrix}\frac{\partial W_{\phi}}{\partial \phi} & \frac{\p...
...W_{nt}  [0.3em] W_{pt}  [0.3em] W_{V_{applied}}  [0.3em] \end{bmatrix}$ (60)

Note, the derivatives of $ W_{\phi}$ ,$ W_{n}$ ,$ W_{p}$ ,$ W_{nt}$ ,$ W_{pt}$ , which are functions of $ V_{applied}$ , now also have to be calculated. In practice, only the far right hand (or left) of the finite difference grid next to the contacts will be non zero. Now, to solve any external circuit under any desired condition, one simply has to couch the equation describing the external circuit in the form of a residual equation,

$\displaystyle W_{V_{applied}}(\phi,\zeta_{n},\zeta_{p},\zeta_{nt},\zeta_{pt},V_{applied})=0 .$ (61)

For example, in the most simple case, if one wants to solve for a given internal diode voltage, one can write

$\displaystyle W_{V_{applied}}(\phi,\zeta_{n},\zeta_{p},\zeta_{nt},\zeta_{pt},V_{applied})=V_{applied}-V_{desired}$ (62)

in this case,

$\displaystyle \frac{\partial W_{V_{applied}}}{\partial V_{applied}}=1.0$ (63)

where $ V_{desired}$ , is the desired output voltage. Or for another example, if one wants to solve for $ V_{oc}$ one can write

$\displaystyle W_{V_{applied}}(\phi,\zeta_{n},\zeta_{p},\zeta_{nt},\zeta_{pt},V_{applied})=V_{applied}-i(V_{applied})*R_{shunt}$ (64)

which forces the current going through the shunt resistance to equal that the diode is generating. Appropriate, derivatives of $ W_{V_{applied}}(\phi,\zeta_{n},\zeta_{p},\zeta_{nt},\zeta_{pt},V_{applied})$ , would then have to be calculated.


next up previous
Next: Average free carrier mobility Up: Solving the electrical equations Previous: Numerical clamping
rod 2017-12-08