variable capacitor Impedance

[mtnkh]

Hi,

I have a question regarding variable capacitors. Consider there is C(t) and its voltage is V(t). So Q(t) = C(t)V(t) and I = dQ/dt = C(t)*dV/dt + V(t)*dc/dt.
My question: SO for variable capacitor, Impedance(Zc = 1/cjw) cannot be defined. Is it correct? since its differential equations is different. Am I right? if not. Please explain me how we define Impedance for variable capacitors.

Thanks

 

[KlausST]

Hi,

are you talking about
* mechanically adjustable capacitors, or
* varicaps

But in either case the formula: Zc = 1/cjw is correct.
You just need to know the value of C. If you adjust it to 20pF, then use 20pF in the formula. If you adjusted it to 30pF then...

Now you may say "but what if the capacitance is not known" ....
then you can´t calculate the impedance.

But the same applies to the frequency (or "w" in your case).
If "w" is unknown you can´t calculate the impedance.

Klaus

 

[mtnkh]

Hi,

are you talking about
* mechanically adjustable capacitors, or
* varicaps

But in either case the formula: Zc = 1/cjw is correct.
You just need to know the value of C. If you adjust it to 20pF, then use 20pF in the formula. If you adjusted it to 30pF then...

Now you may say "but what if the capacitance is not known" ....
then you can´t calculate the impedance.

But the same applies to the frequency (or "w" in your case).
If "w" is unknown you can´t calculate the impedance.

Klaus

Thanks for your response. I am talking about capacitive sensors which because of the movement the value of capacitor would be variable, as distance is changing (C=ϵA/d)-----> C(t)
Actually my question is we cannot write:
I = V/Zc, which Zc = 1/cjw, (this equation is written for constant capacitors)
As the current equation would be I = dQ/dt = C(t)*dV/dt + V(t)*dc/dt instead of I = dQ/dt = C(t)*dV/dt

 

[KlausST]

Hi,

I don´t understand.
If you use C(t) in the forumula you get Zc(t) as result.

That´s how capacitive sensors work.

If C(t) did not change, then Zc(t) does not change too. But if it did not change you can not measure it. And if you can´t measure the variation in impedance ... then it´s no longer a sensor.
It´s the nature of sensors that the values change.

Klaus

 

[FvM]

As the current equation would be I = dQ/dt = C(t)*dV/dt + V(t)*dc/dt instead of I = dQ/dt = C(t)*dV/dt

Yes. Your question is however very general. If the dV/dt or dC/dt term or both actually matter depends on the application.

 

[mtnkh]

Yes. Your question is however very general. If the dV/dt or dC/dt term or both actually matter depends on the application.

Actually I am a bit confused about the differential equations and Laplace equations of the variable capacitor. maybe it is better to ask my quastion in this way. Consider the circuit below, the parameters with (t) are time-varient, the others are time-invarient.
I would appreciate if any one can write differential and also Laplace equations for this circuit.

 

[danadakk]

LaPlace just annotate the reactive components with their s plane representations/freq dependent
value, and write the loop equations.

Diff I profess I would have to open the text....

Give the problem a try, and ask the forum to review your work.


Regards, Dana.

 

[mtnkh]

LaPlace just annotate the reactive components with their s plane representations/freq dependent
value, and write the loop equations.

Diff I profess I would have to open the text....

Give the problem a try, and ask the forum to review your work.


Regards, Dana.

When we cosider C is time-invarient and i = Cdv/dt, We get Laplace from both sides we will have ----> I(s) = C*(S*V(S) - v(0)).
But if C is time-varient and I = dQ/dt = C(t)*dV/dt + V(t)*dc/dt (However I am not sure this equation is correct for time-varring capacitor or not and if not what is the correct form) then we would not have the same Laplace equation. Isn't it?
In this condition I(s) = L(C(t)*dV/dt + V(t)*dc/dt) = C(s)conv(SV(s)) + ...
Or maybe Laplace is just defined for time_invarient RLC?

 

[danadakk]

When we cosider C is time-invarient and i = Cdv/dt, We get Laplace from both sides we will have ----> I(s) = C*(S*V(S) - v(0)).
But if C is time-varient and I = dQ/dt = C(t)*dV/dt + V(t)*dc/dt (However I am not sure this equation is correct for time-varring capacitor or not and if not what is the correct form) then we would not have the same Laplace equation. Isn't it?
In this condition I(s) = L(C(t)*dV/dt + V(t)*dc/dt) = C(s)conv(SV(s)) + ...
Or maybe Laplace is just defined for time_invarient RLC?

I think you have that correct, eg. LaPlace can be applied to both time varying as well as LTI. And C could also be a f() of rotation angle or distance or.....whatever.


Regards, Dana.