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

 

[electricpete]

The op did state he didn't think numerical solution would be a problem. I had the sense (rightly or wrongly) he was still looking to apply impedance

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.

One exception I can think of would be so-called "small signal analsysis" that electrical types are very familiar with.

If your applied pressure is something like:
V(t) = V0 + V1*sin(w*t) where V0>>V1, the you may be able to "linearize your system" as follows:
i(t) = d/dt( C(V(t)) * v(t))
recgonizing V0>>V1, then C(V)~C(V0) (it is an approximatin).
Then
i(t) ~ C(V0) * d/dt(V0 + V1*sin(w*t))
i(t) ~ C(V0) * d/dt(V1*sin(w*t))

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[electricpete]

The benefit of such approach might be that the approximate solution is more intuitive. I have to point out again (lest I be corrected) that it is an approximate solution and so of course the numerical solution using the complete function C(V(t)) is more exact.

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[VEBill]

Time stepping all the irreducible relationships is no more an approximation than Numerical Electromagnetic Code (NEC) is an approximation of Maxwell's Equations in their raw form.

Given a non-ideal (real world) vessel, the relationship of a pressure to volume should probably include some consideration that this relationship will almost certainly roll-off at high enough frequencies. If the OP will be working with high frequency pressure waves, then this might be critical.

 

[electricpete]

Time stepping all the irreducible relationships is no more an approximation than...

Sounds like you are disagreeing with someone, but who it is I have no idea.

I had suggested numerical solution 3 Mar 09 22:27 and original poster agreed it would not be difficult 3 Mar 09 22:55. The only lingering question seemed to be how the concept of impedance might be applied. So I have mentioned the idea of linearization which might apply under certain conditions, which would be an approximate solution, as opposed to numerical solution of the complete (not linearized) problem which would be more exact.

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[VEBill]

I was replying to something that actually isn't there.

Sorry. 'My bad.'

 

[yakbey]

hi,
let me solve all of your problem. step by step.
1. first of all Mr mapi the thing you are interested in was discovered many years ago. so all of us can find about this issue even with the little research
2. as far as i understand to create pressure using voltage lets say input voltage does not equal to output voltage which changes by changing capacitor. so input V(t)not = output V(t). therefore C(v(t))V(t)/dt=i(t) nonsence.
3. like i said there are many applications of your resarch. so are books about it. but since you do not talk about your project exactly, for me it is just dream, i am not ganna give you sources of that project. but i will make my point
4. you must focus on C instead of Voltage. and you must focus on the what sort of gas used in the tube istead of pressure which will be solved easily.
5. in these applications you will find capacity changes with in your example pressure. so voltage sensivity (g) of the capacitor changes. g=d/E. d is the piezo electric constant of the gas. E is dielectric constant of the gas which you must know again. so the only variable is the presssure. with the pressure the output voltage will change also. there is nothing to do with the differencial in this example even if you change the pressure with the input voltage source. V=g*t*p. P is the pressure which will change with Heat or ennarrowing or voltage. t is the distance between two edges of tube which also constant. g is the voltage sensivity of the gas. it is defined by the E and d i indicated above.

to summarize you can convert output voltage to any bigness such as power, soudnd, light. lets say output bigness is voltage V=g*t*p t=d/E. in this application you have to bulit your system according to gas so define this gas's characteristics first.
this application i mean variable capacite have very large use in very large areas. to give example i am the avionics so i find many pratics of this in aircrafts.

 

[ryanguyen]

Let put this thread once and for all... Impedance DOES NOT NOT NOT NOT depend on Voltage.. The steady state time domain impedance equation (Z(w)=1/jWC) said that Impedance depends on Frequency(W) and Capacitance (C) for a Capacitor. So if we have 0hz frequency (DC current) then Impdeance is Infinite.. because 1/j*0*C = infinity and if we have a high frequency signal such a 1Ghz, then the impdence is LOWER. Thus, this equation verify the fact that if we connect a Capacitor to a voltage source, its Steady state current flow will be ZERO, however, initially, there will be current flow until the Capacitor is charged (or reach its steady state) then there will be NO current flow...then from this point on... the Impedance of the capacitor is INFINITE for a DC voltate.. just like what the equation predicted.

If you connect this capacitor across a HIGH frequency sources, then there will be current flow at steady state, and as you increase the frequency of this sources, the impdedance will keep dropping.. So if the same capacitor is now connect to a 1Ghz voltage source, it's impdedance will be lower and thus current is flowing across the capacitor at steady state.

So again..Impedance does NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT NOT depend on Voltage... it depends only on FREQUENCY and CAPACITANCE of the device under analysis.

 

[zappedagain]

I thought this thread was put to rest 9 days ago...

You can have a voltage dependent capacitor (C vs V) - it is called a reverse biased diode (varactor).

The RF guys use them all the time to tune circuits. This effect is really useful when you want it to happen, and a real pain when you don't want to happen (like a DC offset signal having different delays depending upon the amplitude).

Good night,

Z

 

[hacksaw]

lest we for get the original poster is modelling a pneumatic circuit.

try finding a radio or modern TV that does not use varactors.

so I guess capacitance can depend on voltage after all

good morning

 

[WhiteyWhitey]

Hiya,

I couldnt help post on this long dead forum.

If C is related to V
I=C V`(t) no longer holds

Because V = q(t) / C

So V`(t) = q`(t) * 1/C`(t) - C`(t)/(C(t)^2) (Quotient Rule)

Substituting C = ae^-bV(t) gives

I = -bV`(t) * ae^-bV(t) + C`(V(t)*V`(t))/C(V(t))

I = V`(t) ( ae^-bV(t) - bae^-b(V(t) / (ae-bV(t) )

I = V'(t) * (ae^-bV(t) - b)

you can plot this parametrically to get a hysterisis type loop.

Punch this into wolframalpha:

parametric plot (Sin(t) , (cos(t)*(e^-0.5*sin(t))-0.5)),t=-100...100

Just my thoughts.

Thanks,
Andrew

 

[WhiteyWhitey]

Sorry got the plot wrong

parametric plot (Sin(t) , (cos(t)*((e^-0.5*sin(t))-0.5))),t=-100...100

Thanks,
Andrew