Why does fringing capacitance increase with frequency?

[drkirkby]

I know the coaxial "open" standard used for calibration of Agilent/HP vector network analysers (and possibly other makes), is modelled as some offset (in ps) and a polynomial representing the capacitance as a function of frequency.

C(f)=C0 + C1 f + C2 f[SUP]2[/SUP] + C3 f[SUP]3[/SUP]

In a book recently published on VNAs by Dr. Joel Dunsmore at Agilent

**broken link removed**

there's plot of capacitance vs frequency for an open, and that shows an increasing capacitance vs frequency, but the reasons are not explained. (I've asked on the Agilent VNA forums, so I'm sure Joel will answer). I highly reccommend that book to anyone using VNAs seriously.

Someone else asked the same sort of question some time ago on the EDA forums, when his simulations with Sonnet showed an increasing capacitance with frequency, and he did not know why.

https://www.edaboard.com/threads/64338/

Someone posted a PDF for him to read



But having read that short paper, it only says the capacitance does increase with frequency, until the self-resonate frequency is reached. That models a practical capacitor as a seriel R L C circuit, where R is the ESR and L_s is the self-inductance. There's an equation there, which gives the effective capacitance C_eff at some frequency f, as a function of the capacitance at 1 MHz C_m.

C[SUB]eff[/SUB](f)=C[SUB]m[/SUB]/(1-(2 π f)[SUP]2[/SUP] L[SUB]s[/SUB] C[SUB]m[/SUB])

(I purposely changed the symbols used, so not to conflict with the Agilent one, since both use C0, but for a different purpose).

But the paper gives no physical understanding of why capacitance changes with frequency.

I can understand why a capacitor becomes self-resonate at some frequency - that would happen even if the capacitance remains fixed. But what I can't understand is why capacitance changes as a function of frequency.

(For what it is worth, the inductance of a coaxial "short" also changes with frequency, but the effects of that on VNA calibration can be ignored below about 6 GHz, whereas the change in fringing capacitance of a coaxial "open" are significant at much lower frequencies).

Any ideas why fringing capacitance should vary with frequency?

Dave

 

[volker_muehlhaus]

Someone else asked the same sort of question some time ago on the EDA forums, when his simulations with Sonnet showed an increasing capacitance with frequency, and he did not know why.

https://www.edaboard.com/threads/64338/

Someone posted a PDF for him to read



But having read that short paper, it only says the capacitance does increase with frequency, until the self-resonate frequency is reached. That models a practical capacitor as a seriel R L C circuit, where R is the ESR and L_s is the self-inductance. There's an equation there, which gives the effective capacitance C_eff at some frequency f, as a function of the capacitance at 1 MHz C_m.

C[SUB]eff[/SUB](f)=C[SUB]m[/SUB]/(1-(2 π f)[SUP]2[/SUP] L[SUB]s[/SUB] C[SUB]m[/SUB])

(I purposely changed the symbols used, so not to conflict with the Agilent one, since both use C0, but for a different purpose).

But the paper gives no physical understanding of why capacitance changes with frequency.

Dave, I think that's a different story. The question refers to capacitors with finite size, where the physical size causes series inductance. That series inductance leads to an increase of the effective series capacitance.

IMO, this is not directly related to your case (fringing capacitance).

 

[FvM]

My intuitive explanation says that the effect should be related to the radiation resistance of the open coaxial end.

 

[drkirkby]

Dave, I think that's a different story. The question refers to capacitors with finite size, where the physical size causes series inductance. That series inductance leads to an increase of the effective series capacitance.

IMO, this is not directly related to your case (fringing capacitance).

Yes, I agree. I did think that after I wrote my post.

But do you know why fringing capacitance increases with frequency? I'm particulary interested in knowing how to make a coaxial "open" standard, such that the fringing capacitance is as small as possible, and with as little change with frequency as possible. (At least I think they are probably desirable atributes). But in order to do that, I need to have some understanding of why fringing capacitance increases with frequency.

I've asked the same question here,

http://www.home.agilent.com/owc_discussions/thread.jspa?threadID=34169&tstart=0

on the Agilent VNA forum dedicated to calibration of VNAs. I might get an answer there. I suspect the Agilent staff designing calibration kits have a good idea about this.

I guess a good hunt though IEEE papers might get me an answer, since it is not obvious to me.


Dave

- - - Updated - - -

My intuitive explanation says that the effect should be related to the radiation resistance of the open coaxial end.

So is your logic something along the lines of the following?

* Open acts as an electrically short antenna
* As the frequncy is increased, the antenna becomes electrically longer, so radiation resistance rises.

If so, I still don't see how that leads to a change in capacitance.

An electrically short dipole has a capacitive input impedance, but if the frequency is increased, whilst the antenna remains of the same length, so the input reactance becomes less capacitive, and eventually resonates at around 0.48 wavelengths long. If the length is increased to 0.5 wavelengths, then the input impedance is inductive with an inductive reactance of about + j 42 Ohms. That's the dead opposite way to how the fringing capacitance changes, as that increases with frequency.

I'm glad I don't seem to be the only one who finds this a non-trivial problem!

Dave

 

[biff44]

Fringing capacitance involves the electric field intensity. Anything with sharp corners, like steps or open circuited ends of coaxial lines, is going to have a sharp mechanical transition. As the frequency goes up, the wavelength gets smaller and the "sharpness" of that mechanical transition gets more pronounced--bunching the E field in that region. Higher E field means more fringing capacitance. Just a guess, but it seems to make sense.

 

[drkirkby]

In the Agilent VNA forums, Dr_joel replied

"has to do with the bessel function response of a truncated cylinder. I read the original paper one time but can't find it now."

I can't seem to find the paper, though there are a lot of papers mentioning both Bessel functions and fringing fields. I'm not able to download them without paying from this computer, so I'll revisit this later.

Dave

Dave