when capacitance depends on voltage, how to calculate impedance 3

[mapi]

For a simple circuit with only one capacitor C, when C is constant, the impedance is Z(w)=1/jWC. But when C is dependent of the voltage, C(V), then what is the impedance?

Thanks,

Mapi

 

[IRstuff]

Not always true. The FT is transforming a time domain into frequency, but your capacitor is actually a voltage domain device.

If your capacitance is a function of DC bias, then the FT will only be valid for that specific bias.

TTFN

FAQ731-376

 

[electricpete]

Yes. To further IRStuff's point I'll return to my previous point. If the device adjusts it's capacitance on a time scale less than a cycle, then if you apply a sinusoidal voltage you will draw a non-sinusoidal current.

This type of linearity cannot be modeled by a single transfer fucntion - it should be obvious. The input has only one frequency call it w0... the output has frequency components many call them w0, w1, w2, w3 (even though the fundamental frequency will remain w0). This can't be practically addressed in the frequency domain.

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[IRstuff]

I would suggest looking at various MOS CV discussions to get some better insight, like:

http://www.ecse.rpi.edu/~schubert/Course-ECSE-2210-Microelectronics-Technology-2003/MT-26-Ch16-3.ppt

TTFN

FAQ731-376

 

[mapi]

Thanks. we may say that the impedance I calculated in my last post is the impedance for that specific input, the pressure V(t).

 

[electricpete]

I think you really really need to tell us more about what you are working on or else the thread will go around in circles forever.

What type of device do you have?

Is your system a mechanical system as might be inferred from your discussion of "For my problem, it is just a simplification of a tube with pressure inside, and so the compliance (capacitance) depends on pressure (voltage)".

Is it a mechanical system? (yes or no).

Assuming electrical...
When you say voltage dependent... you need to tell us what you mean.

Is the applied voltage purely sinusoidal? (yes or no?). Is the capacitance is dependent on the magnitude of the sinusoidal voltage? (yes or no).

Is the applied voltage a sinusoidal plus dc offset bias? (yes or no). Is the capacitance a function of the bias voltage ?(yes or no).

Please if possible address at least each of the yes/no questions directly.

Thanks.

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[mapi]

Thanks for you questions.

Yes, it is a mechanical system and I try to use circuit to simply the problem. So capacitance for example is C(V)=a*exp(-b*V)--a and b are constants. So C depends on the magnitude actually.

voltage V(t) is not purely sinusoidal, but is periodic and has complex shape, but V(t)>0 all the time.

Thanks.

 

[MacGyverS2000]

the only time I've ever needed to map a mechanical system as an electrical one (or vice versa) was for homework... what advantage is it getting you to look at this system as an electrical one?

Dan - Owner

http://www.Hi-TecDesigns.com

 

[IRstuff]

The fact that you're dealing with a supposed nonlinear capacitance is the precise reason for not using the electrical analog, since resorting to the electrical analog presumes constant coefficients for simplification.



TTFN

FAQ731-376

 

[electricpete]

Thanks for the responses.

I think analogies can in general be useful. I agree with IRstuff that an R/C analogy for a non-linear circuit loses the intuitive benefits. More importantlyl one needs to analyse/understand the system before you can properly formulate any type of non-linear R/C analogy.

I think it is most productive if you can to describe either
1 - the physical system/problem
or
2 - the differential equations.

If I take a guess at the differential equations from your commments so far.
I(t) = C(V(t)) * dV(t)/dt
C(V(t))=a*exp(-b*V(t))

Combing the two equations
I(t) = a*exp(-b*V(t)) * dV(t)/dt

Does that resemble your system? Note as I described, the instantaneous value of C depends soely on the instantaneous value of V.... is that correct?

If the problem is as described directly above, than the non-linearity precludes frequency domain analysis. You cannot characterize this as an impedance that will perform a predictable transformation on input frequency (that type of transfer function is a linear concept). On the other hand the equation as written above should be very easy to solve in the time domain. Input V(t). Calculate dV/dt. Plug and chug either analytically or numerically.

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[electricpete]

I forgot to finish my thought. Finish the sentence as shown in bold:

"Plug and chug either analytically or numerically to find I(t)"

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[electricpete]

Another typo correction in bold:

You cannot characterize this as an impedance that will perform a predictable phase and magnitudetransformation based soley on input frequency (that type of transfer function is a linear concept).

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[mapi]

Thank you a lot.

I(t)=dq/dt=d[C*V]/dt, I think we should put C(v(t)) into the derivative sign.

For non-linear capacitance (not LTI system), I didn't know we can't transform it and get the impedance. My problem is this impedance, maybe it is not correct to obtain it just like LTI system: in series or in parallel.

Yes, solving the second-order differential equation numerically is not difficult.

Thanks.

 

[electricpete]

You are right that C should have gone inside the derivative.

In general, a non-linear system cannot be characterized by a transfer function. Transfer functions like Impedance
Z(w) = I(w)/V(w) are an LTI concept.

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[Skogsgurra]

mapi

You say that "For my problem, it is just a simplification of a tube with pressure inside, and so the compliance (capacitance) depends on pressure (voltage)"

Does that mean that the volume of the tube changes with pressure? If it doesn't, I would say that the "capacitance" is constant.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[mapi]

Skogsgurra,

Yes, the tube is soft, not infinitely stiff. So the tube will expand as pressure increases.

 

[Skogsgurra]

OK, will it expand linearly with pressure? That would make the use of complex numbers meaningless (Z(w)=1/jWC not valid since it needs a constant C) and means that you have to apply a differential equation to describe the relation between applied pressure and "current" i.e. volume flow. For a capacitor, that would look like:

C = k*u
i = C*du/dt

Which combines to

i = k*u*du/dt

Which is the expression to use if there are no resonanses. And that is probably a valid assumption if your du/dt is small in comparison to the tube's "eigenfrequency" (if such an expression is allowed).

Does this really have any practical value? Or is it a mental excersise?





Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[mapi]

It is nonlinear, but we may approximate it with a linear curve, though not very good.

Thanks.

 

[Skogsgurra]

If you know the relation between pressure and volume, then use that instead of C=k*u.

It is probable that C=k*u is a very bad relation, since it means that C=0 at zero pressure. Not very likely.

C = C0 + k*u is probably better. C0 is capacitance at zero pressure.

Many containers that infalte (a balloon, for instance) are highly non-linear and are best described by

C = C0 +k*u^m, where m depends on the container's shape. The reason is that, as pressure increases, so does the surface that pressure works on and also that the walls are getting thinner as the container expands.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[VEBill]

(This thread's been going for two weeks...)

Couldn't one just enter the basic relationship equations into a spreadsheet, structure them so they can be time-stepped, copy the cells in this first row down X thousands of rows in the time domain, and then plot the results.

Repeat the exercise with and without the C(V) relationship to highlight the difference.

An easy afternoon's work.

 

[Skogsgurra]

Yes, one could. Or use Matlab or just about any system.

It was only yesterday that we really got to know what it is about (pressure in a soft tube) so we could do some real work. I was, for a long time, under the impression that we were talking about a capacitance diode or a similar device and didn't think it was worthwhile to add anything to the thread.

Have you got that afternoon? My afternoon is over already

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...