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Derivative of I-V curve

g1lai
1-Visitor

Derivative of I-V curve

It could be a simple question, but my brain stops running ...

I have I-V test data, and try to derive the 1st derivative (dV/dI) and/or followed by smoothing. Is there any build-in function to do this?

Thanks a lot.
92 REPLIES 92
ptc-1214470
2-Explorer
(To:g1lai)

This has become a pretty long thread; I might have missed something in all the posts. Attached is my take on a more appropriate approach to working your problem.
Cheers,
Jim S.

Once you have a model you can use minerr to optimize the parameters. Eyeballing the curve is a good way to get initial estimates, but that usually needs refinement.

With a model you can usually do symbolic derivatives. No need for numeric approximations. Even where numeric derivatives are needed, with a model you can use Mathcad's built in derivative operator. No need to approximate the derivative with a somewhat arbitrary secant.
__________________
� � � � Tom Gutman

Great work, Tom - it would have taken me a week to do what you appear to have done in a few minutes.





>Why the assumption of an error in the last measurement? The current is expected to reverse at high enough voltages.





I guess that in my experience I haven't seen a voltage reversal at open circuit conditions.





>No need to reverse the sign convention for current. The sign is just in the short circuit current.





True - I just reversed it to conform with my view of the solar cell modeling.





Thanks for guiding me through the process of analytically finding the cell equivalent resistances. My original model did have max power point calculation; I just deleted it as not relevant to the immediate problem.


Thanks again,


Jim S.

>>I guess that in my experience I haven't seen a voltage reversal at open circuit conditions.<<

I don't think this is an open circuit condition. The regularity of the voltage indicates that the cell is connected to a voltage generator, and the current is measured with the voltage held at selected values. Thus voltages higher than the open circuit voltage can be generated, with the resulting reversal in the current direction.
__________________
� � � � Tom Gutman


That makes sense; in fact, looking back at some of my original modeling, I find that even the model shows a slight negative value. Looks like I should have kept on working; 8 years of retirement is taking its toll, making me go back to my archives to re-acquaint me with what I used to do.
My original model included a couple of parallel diodes with distinct characteristics, along with series and parallel resistances.
Jim S.

A long but interesting visit:

http://en.wikipedia.org/wiki/Solar_cell

On 7/11/2009 2:56:14 PM, Offroc wrote:
>This has become a pretty long
>thread; I might have missed
>something in all the posts.
>Attached is my take on a more
>appropriate approach to
>working your problem.
>Cheers,
>Jim S.
______________________________

Your input helps a lot.
Added, the Monomolecular growth model, an equivalent as good model on the account of all combined uncertainties.

jmG



Thanks Jim S. If get insight of physics, we will see that A0 (or n in eq 1 of the attached paper) is a variable that can be determined from I-V data analysis. The exponential relationship b/w I and V is nice for MathCAD analysis. However, we may have to deal with the more complicated equation, i.e. eq(1).
TomGutman
1-Visitor
(To:g1lai)

The eqation (1) doesn't look all that complicated. I don't see why it should not be easily fitted with normal tools. Good guess values may be needed, and the paper's methodology may provide a good way to get those.

The only problem is that that equation does not give I as a function of V. It gives I as a function of I and V, so getting one as a function of the other would require a solve block. Not something very esoteric, it's been done numerous times in this collaboratory. Or, somethint that has also been done, to get an entire I-V curve one can derive a differential equation from the given equation and then solve that. That is generally faster than solving for inividual I-V values.
__________________
� � � � Tom Gutman
g1lai
1-Visitor
(To:TomGutman)

>It gives I as
>a function of I and V, so
>getting one as a function of
>the other would require a
>solve block. Not something
>very esoteric, it's been done
>numerous times in this
>collaboratory.

I will very interested to know the solution in MathCAD. As far as I know, many approaches have been developed to solve this equation. None of them seems to be straightforward.

On 7/12/2009 12:02:11 AM, g1lai wrote:
>>It gives I as
>>a function of I and V, so
>>getting one as a function of
>>the other would require a
>>solve block. Not something
>>very esoteric, it's been done
>>numerous times in this
>>collaboratory.
>
>I will very interested to know
>the solution in MathCAD. As
>far as I know, many approaches
>have been developed to solve
>this equation. None of them
>seems to be straightforward.
_____________________________

Which equation because I'm getting either lost or tired.
Isn't solvable by simple InverseFunction ?
Show that equation to give it a try.
There is an awful lot of expertise about Inversefunction in this collab.

jmG



>Which equation because I'm getting
>either lost or tired.

The equation in the attached paper of my reply. It is given by

I = Iph−I0*{exp[q(V+Rs*I)/nkBT]−1}−(V+Rs*I)/Rsh

where Iph, I0, Rs, Rsh, q, n, kB, and T are the photocurrent, the saturation current of the diode, the series resistance, the shunt resistance, the electron charge, the ideality factor.

On 7/12/2009 1:07:56 AM, g1lai wrote:
>>Which equation because I'm getting
>>either lost or tired.
>
>The equation in the attached
>paper of my reply. It is given
>by
>
>I =
>Iph−I0*{exp[q(V+Rs*I)/nk
>BT]−1}−(V+Rs*I)/Rs
>h
>
>where Iph, I0, Rs, Rsh, q, n,
>kB, and T are the
>photocurrent, the saturation
>current of the diode, the
>series resistance, the shunt
>resistance, the electron
>charge, the ideality factor.
_____________________________

Just posted a minute ago for both equations, yours Wiki & Jim.

Don't turn too superficial, your script is incorrect,
You better had it in real maths !

jmG



On 7/12/2009 12:02:11 AM, g1lai wrote:
>>It gives I as
>>a function of I and V, so
>>getting one as a function of
>>the other would require a
>>solve block. Not something
>>very esoteric, it's been done
>>numerous times in this
>>collaboratory.
>
>I will very interested to know
>the solution in MathCAD. As
>far as I know, many approaches
>have been developed to solve
>this equation. None of them
>seems to be straightforward.
_____________________________

Don't know who wants what in there,
for both equations, piece of cake !

Answer to:

>I will very interested to know
>the solution in MathCAD. As
>far as I know, many approaches
>have been developed to solve
>this equation. None of them
>seems to be straightforward.

jmG



TomGutman
1-Visitor
(To:g1lai)

As it turns out the equation in the paper is solvable in terms of the Lambert W function. A function for which I have an implementation in Mathcad (for reasons unknown it's not built into Mathcad).

There is a question as to the sign conventions for currents and voltages in your system.

Rsh is essentially infinite. The fit is happy with any value large enough, positive or negative, that that term doesn't significantly affect the value. This is in keeping with the previous observation that your data do not define a dV/dI at V=0.

But the fit is quite good, very little (if any) structure in the residuals.
__________________
� � � � Tom Gutman
g1lai
1-Visitor
(To:TomGutman)

>Rsh is essentially infinite.

Rsh is usually less than 1000 ohm in those solar cells.


>The fit is happy with any
>value large enough, positive
>or negative, that that term
>doesn't significantly affect
>the value. This is in keeping
>with the previous observation
>that your data do not define a
>dV/dI at V=0.

Yes, the program is great. I assume that Rsh can be derived independently (as suggested in the paper) and can also get reasonable fitting.
TomGutman
1-Visitor
(To:g1lai)

You can try applying the paper's methodology. I haven't worked it through. It seems to do a fair bit of eyeball estimates based on the plots.

But I'm still not happy with the sign conventions. The paper treats the current from the cell as positive, you consider it negative. Does your cell have different polarity from theirs, or are you just using a different sign convention?

If I change the sign of the current to match the form shown in the paper, I can get a very good fit, with a finite Rsh. But ... it insists on a negative Rs. I can force Rs positive, but then it comes out essentially zero, and the fit is decidedly poorer both in terms of the size of the error and the structure of the residuals.

It would be interesting if you could get the raw data for the examples in the paper, and then compare results.
__________________
� � � � Tom Gutman
g1lai
1-Visitor
(To:TomGutman)

>But I'm still not happy with
>the sign conventions. The
>paper treats the current from
>the cell as positive, you
>consider it negative. Does
>your cell have different
>polarity from theirs, or are
>you just using a different
>sign convention?

The photocurrent (Iph) is always negative because its direction is reverse to the direction of internal electrical field. The light has the effect of shifting the IV curve down into the fourth quadrant where power can be extracted from the diode. Since solar cell is generating power, the convention is to invert the current axis.
TomGutman
1-Visitor
(To:g1lai)

Current direction is a matter of convention. That it is opposite the electric field does not make it negative. And clearly conventions differ. The paper you posted shows currents as positive, not negative.

The usual convention for positive current is current flowing from the more positive terminal to the more negative terminal. Alternatively, it is opposite the electron flow (for normal electron carrier currents).

As near as I can tell the conventions used in the paper are that the output end (upper in the diagram) of the cell is positive (open circuit), the positive direction for the generated current (Iph) is upwards, the positive direction for current through the diode and Rsh is downward and the positive direction for current through Rs is from left to right. This seems to be consistent with the output voltage being positive and the voltage at the top of the diode and Rsh being more positive than the output voltage V. All seems consistent ... except that the fit requires a negative Rs. This makes no sense, unless the current sign convention is wrong (or the model is simply incorrect).
__________________
� � � � Tom Gutman
g1lai
1-Visitor
(To:TomGutman)

On 7/13/2009 1:36:56 AM, Tom_Gutman wrote:
>Current direction is a matter
>of convention. That it is
>opposite the electric field
>does not make it negative.
>And clearly conventions
>differ. The paper you posted
>shows currents as positive,
>not negative.
>
>The usual convention for
>positive current is current
>flowing from the more positive
>terminal to the more negative
>terminal. Alternatively, it
>is opposite the electron flow
>(for normal electron carrier
>currents).
>
>As near as I can tell the
>conventions used in the paper
>are that the output end (upper
>in the diagram) of the cell is
>positive (open circuit), the
>positive direction for the
>generated current (Iph) is
>upwards, the positive
>direction for current through
>the diode and Rsh is downward
>and the positive direction for
>current through Rs is from
>left to right. This seems to
>be consistent with the output
>voltage being positive and the
>voltage at the top of the
>diode and Rsh being more
>positive than the output
>voltage V. All seems
>consistent ... except that the
>fit requires a negative Rs.
>This makes no sense, unless
>the current sign convention is
>wrong (or the model is simply
>incorrect).
>__________________
>� � � � Tom Gutman

Please take a look of Ch 4.6 in the following link which addresses the details of I-V characteristic (much better than wiki).
http://pvcdrom.pveducation.org/

>Please take a look of Ch 4.6 in the following link which addresses the details of I-V characteristic (much better than wiki).<<br> ___________________________

Same model equation than Wiki, just copied from Wiki !

Fromt the basic solar cell model, any component can be solved for via either
1. Lambert W(x)............................... [Mathcad]
2. solved directly via symbolic.......... [Mathcad]
3. or via symbolic RootOf................. [Mathcad]

Fill-in the blanks your project, come back for help.

jmG
TomGutman
1-Visitor
(To:g1lai)

Clear as mud. You still need to know your conventions for the sign of the current. The reference shows the equation for the current twice, each with a different sign. And compared with the previously cited paper it covers only the diode portion of the circuit (the two leftmost components) ignoring the currents and voltages involving Rs and Rsh, which affect the measured I-V curve.
__________________
� � � � Tom Gutman
g1lai
1-Visitor
(To:TomGutman)

On 7/13/2009 5:26:27 PM, Tom_Gutman wrote:
>Clear as mud. You still need
>to know your conventions for
>the sign of the current. The
>reference shows the equation
>for the current twice, each
>with a different sign. And
>compared with the previously
>cited paper it covers only the
>diode portion of the circuit
>(the two leftmost components)
>ignoring the currents and
>voltages involving Rs and Rsh,
>which affect the measured I-V
>curve.
>__________________
>� � � � Tom Gutman

Good point! Thanks Tom.

As seen in the attached paper (Eq. 1), in my case the current is given by

I = -Iph+I0*{exp[q(V-Rs*I)/nkBT]−1}+(V-Rs*I)/Rsh. Note the sign change in the V-Rs*I term.
TomGutman
1-Visitor
(To:g1lai)

It is not at all clear that eq(1) in that paper applies to your system as measured. That paper does not even show the assumed circuit for the solar cell, much less show the sign conventions used. Presumably you are supposed to know that from other sources. You need to know the sign conventions (which can differ from paper to paper) to be able to interpret the results.

To the extent that they show data (I don't see an actual I-V curve) they seem to use conventions that result in positive values for I and V (for their solar cell). But their very first plot seems inconsistent. They show Voc varying with Rs. But under open circuit conditions I is zero, and Rs should have no effect whatsoever. It's hard to know how seriously to take anything else in this paper.

But it's easy enough to use their equation is the work sheet. And since their equation differs from the previous one only in the convention for the sign of I we get (using the uninverted current figures) exactly the same negative Rs.

I notice that the authors of the paper note that negative values for Rs do occur. While they put it down to a large error, it is possible that it is simply a attribute of this model, and that the model is just not really right.
__________________
� � � � Tom Gutman
IRstuff
12-Amethyst
(To:TomGutman)

Negative resistances for conventional PN junctions not biased into breakdown are almost always errors, though.

The Esaki diode does have negative differential resistance because of quantum mechanical tunnelling:
http://en.wikipedia.org/wiki/Esaki_diode


TTFN,
Eden

The negative resistance is, in terms of the model used, in error. The question is the source of the error. The paper seems to claim that the problem is just that the parameter estimations are inaccurate, and result in negastive resistances. I suggest the possibility that the parameter estimates are accurate enough, and that in terms of the model the resistance has to be negative. That would indicate that the problem is with the model itself.
__________________
� � � � Tom Gutman
IRstuff
12-Amethyst
(To:TomGutman)

On 7/14/2009 3:44:23 PM, Tom_Gutman wrote:
>The negative resistance is, in
>terms of the model used, in
>error. The question is the
>source of the error. The
>paper seems to claim that the
>problem is just that the
>parameter estimations are
>inaccurate, and result in
>negastive resistances. I
>suggest the possibility that
>the parameter estimates are
>accurate enough, and that in
>terms of the model the
>resistance has to be negative.
>That would indicate that the
>problem is with the model
>itself.
>__________________
>� � � � Tom Gutman

Hypothetically, given the modelling components being used, a negative Rs should be impossible, particularly since it's in series with everything else. From the text, there appears to be a bunch of assumptions that may not be completely valid at all times, and errors in those values could certainly muck up the calculation of Rs.

TTFN,
Eden

>Hypothetically, given the modelling components being used, a negative Rs should be impossible, particularly since it's in series with everything else. From the text, there appears to be a bunch of assumptions that may not be completely valid at all times, and errors in those values could certainly muck up the calculation of Rs. <<br> ________________________

As you say so well: a case of "unknown model".
Then, pretty difficult to collaborate.

"Deceptively simple or simply deceptive".

jmG

The solution in my sheet makes no assumptions beyond that the model is correct. And that is coming up with negative values. That's why I suggest that the model may be wrong. Mainly in the calculation of the current through the diode. Note that the latest reference does not accept this simple model as adequate, but uses a two diode model.
__________________
� � � � Tom Gutman

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