Solve more I-V equations

[g1lai]

Since I don't have MC11, could you please solve more I-V equations? Thanks a lot.

[g1lai]

Sorry Eq2 2 and 3 are wrong. See the update.

[g1lai]

Sorry Eqs 2 and 3 are wrong. See the update.

[ptc-1368288]

On 7/22/2009 10:48:58 PM, g1lai wrote:
>...
________________________

Make sure of:
1. You correct your own mistakes
2. You follow the procedure previously outlined.

jmG

[g1lai]

On 7/22/2009 11:17:36 PM, jmG wrote:
>On 7/22/2009 10:48:58 PM, g1lai wrote:
>>...
>________________________
>
>Make sure of:
>1. You correct your own mistakes
>2. You follow the procedure previously
>outlined.
>
>jmG
>

Thanks jmG. There is a mistake in Eq3 probably due to format loss when I save as a MC11 file. As Tom said, the 2-diode will not solve symbolically. Nevertheless, Eq2 is good enough for me.

[TomGutman]

No, there was no loss in saving as MC11. The error occurred when jmG rewrote the equations. You can download your own file and see the equations as you posted them. Compare to what jmG did.
__________________
� � � � Tom Gutman

[TomGutman]

eq1 you already have. eq2 and eq3 are probably wrong, as they imply current in different directions through the diode and the shunt. eq2 can be gotten from eq1 by setting I_ph to zero and changing some signs (different sign conventions -- I suggest just setting I_ph to zero and keeping the same sign conventions as for the illuminated condition). eq3 is the two diode model and will not solve symbolically (as previously discussed).
__________________
� � � � Tom Gutman

[g1lai]

On 7/23/2009 4:05:05 AM, Tom_Gutman wrote:
>eq1 you already have.

What I have the current (I) as a function of (Iph, V, I0, Rs, Rsh, n). What I am looking for is the series resistance Rs as a function of (Iph, I, V, I0, Rsh, n), and also the Source Voltage V as a function of (Iph, I, I0, Rs, Rsh, n)


>eq2 and eq3 are probably wrong, as
>they imply current in
>different directions through
>the diode and the shunt.

For dark I-V, the source voltage is greater than the junction potential. It is a good idea to set Iph=0 for dark I-V, however, junction potential is higher than the source voltage for light I-V, and smaller than V for dark I-V.

[ptc-1368288]

On 7/23/2009 11:29:29 AM, g1lai wrote:
>On 7/23/2009 4:05:05 AM, Tom_Gutman
>wrote:
>>eq1 you already have.
>
>What I have the current (I) as a
>function of (Iph, V, I0, Rs, Rsh, n).
>What I am looking for is the series
>resistance Rs as a function of (Iph, I,
>V, I0, Rsh, n), and also the Source
>Voltage V as a function of (Iph, I, I0,
>Rs, Rsh, n)
>_______________

Follow the same solving process for those as solving for I(V).

[g1lai]

Unfortunately the solver does not work in my MC14.

[ptc-1368288]

On 7/23/2009 12:09:17 PM, g1lai wrote:
>Unfortunately the solver does not work in my MC14.<<br> ____________________

I gave the solution eq2, eq3 in the reply.

[TomGutman]

The physical system doesn't change with light. The exact same equation applies. The obvious sing convention may differ, as one's perspective changes from the cell as the current source to the cell as the load. But one need not make this change and one can use the same sign convention for both cases.

As to whether the junction voltage is higher or lower than the terminal voltage, that depends on the sign of I. And the sign of I depends on the relationship of V to V_oc. If V<V_oc current is flowing out of the cell (positive in my preferred sign convention, negative in your data's convention) and the junction voltage is higher then the terminal voltage. If V>V_oc then current is flowing into the cell and the junction voltage is lower than the terminal voltage. For illuminated measurements (as in your data) on generally has V<V_oc (indeed, if one uses a passive load resistor, this is forced). For dark measurements V_oc is zero, so one generally has V>V_oc.

The diode and the shunt resistance are both passive devices in parallel. The currents in them will be in the same direction (using the refence diagram and polarity, from top to bottom). These two currents should add to each other, not subtract.
__________________
� � � � Tom Gutman

[TomGutman]

Here are the solutions for R_s and V, for eq1. I'll leave the setting of I_ph and, if you really want to, the change of sign convention up to you.
__________________
� � � � Tom Gutman

[g1lai]

Thanks Tom.

It is interesting that my MC14 cannot solve I, however, it solves Rs and V!

[TomGutman]

That is odd. Just goes to show MuPad is flaky.
__________________
� � � � Tom Gutman