Why does the capacitance of open transmission lines vary near the end?

[dnyberg2]

I had an occasion to hook up a piece of open ended coax to a network analyzer. I know the wire has some pF per foot but I expected it to be linear over inches but it wasn't. In other words instead of 5 inches = 5X the pF per foot, the capacitance rises sharply near the end of the coax. Any idea why?

 

[yungman]

You cannot use a network analyzer to measure capacitance like this. I think you miss the concept of transmission line in length approach the wavelength of the frequency. It is not just a capacitor, you have to use solution of wave equation in phasor form to analyze the behavior.

For open end tx line, it started out as capacitance. The value increase to infinite ( become short circuit) as length approach λ/4. Then it will flip and become inductive until length approach λ/2, then it will flip back to capacitance... and on and on for every λ/4 interval.

from your example, the wave length of the frequency when you see the jump to very high value is 4X5" which is 20"

 

[dnyberg2]

This makes sense except, the coax under inspection came at a length of 42". I set the VNA to 49 MHz, Freq of OP, and made measurements as I cut 1" pieces off the end. The coax does indeed flip from capacitance to inductive at ~36" BUT, 36" is NOT a λ/4 of 49 MHz so what else am I missing? Thanks!

 

[f95toli]

The propagation speed in e.g. RG58 coax is about 0.67c (assuming PE dielectric).
This means that the wavelength at 49 MHz is roughly 4 meters.
And lambda/4 equal to 1m which is approximately 36"...

(this is a just-before-bedtime calculation, so I might have made a misstake somewhere...)

 

[yungman]

This makes sense except, the coax under inspection came at a length of 42". I set the VNA to 49 MHz, Freq of OP, and made measurements as I cut 1" pieces off the end. The coax does indeed flip from capacitance to inductive at ~36" BUT, 36" is NOT a λ/4 of 49 MHz so what else am I missing? Thanks!

If you have 42", according to f95toli, it is over λ/4, so it should be inductance.

f95toli use RG58 as an example, you have to use your coax to calculate, the dielectric might not be the same and the speed is different.

The equation is $$U=\frac 1 {\sqrt{\mu_0 \epsilon_0 \epsilon_r}}$$

Where U is the velocity of propagation.

 

[vk6kro]

The analyser would be an ideal instrument for finding the velocity factor of your coax.

Just adjust the frequency until the input impedance drops to a minimum when the opposite end is open circuited. This is usually quite a sharp dip in impedance.

You then work out the ratio of this length to a quarter wavelength at the same frequency in air.

To find capacitance, you could set the frequency as low as it would go and use a length of line that is a trivial portion of a quarter wave at that frequency.

Or, you could use a multimeter that measures capacitance at 1000 Hz. In this case, the length of the coax is not likely to matter.

 

[dnyberg2]

The analyser would be an ideal instrument for finding the velocity factor of your coax.

Just adjust the frequency until the input impedance drops to a minimum when the opposite end is open circuited. This is usually quite a sharp dip in impedance.

You then work out the ratio of this length to a quarter wavelength at the same frequency in air.

To find capacitance, you could set the frequency as low as it would go and use a length of line that is a trivial portion of a quarter wave at that frequency.

Or, you could use a multimeter that measures capacitance at 1000 Hz. In this case, the length of the coax is not likely to matter.

This is great! The coax I am using is a custom made ~50 Ohm invention that is VERY small in DIA. A wire house made it for me, NOT a coax company so, I don't have any idea what the darn dielectric speed is (also known as velocity factor right?) With the post from the one person about how to figure out the velocity factor, I can much better figure out the capacitance. Thanks VERY VERY MUCH!

 

[dnyberg2]

F95toil, What is the formula to calculate the resonant length using the velocity factor? Thanks.

 

[yungman]

F95toil, What is the formula to calculate the resonant length using the velocity factor? Thanks.

If you want to find the velocity in the coax, get a fix length coax with open end. Run it on a VNA to find the LOWEST frequency that the impedance drop to the lowest. The is the frequency where the λ/4 equal to the length of the coax. Then you times the length by 4 to get the λ. Then times the λ by the frequency to get the velocity.

With this, you can find the $\epsilon_r$ using the formula I gave you using the velocity. $\mu_0$ is the same for non magnetic material which is $4\pi \times 10^{-7}$H/m.

 

[f95toli]

F95toil, What is the formula to calculate the resonant length using the velocity factor? Thanks.

Just calculate the propagation speed cp (=speed of light in the cable) using, in this case, 0.67*c.

Then calculate the wavelength using the usual lambda=cp/f relation

 

[vk6kro]

Wavelength in air = speed of light / frequency

= 300 000 000 meters/second / frequency in Hz

= 300 / frequency in MHz... usually more convenient.

example: the wavelength of a signal of frequency 50 MHz in air is (300 / 50) or 6 meters.

A quarter wave length is 0.25 times the above value or you could modify the formula to
75 / frequency in MHz

example a quarter wave at 39 MHz is 75 / 39 or 1.92 meters.

In a transmission line (eg coax) the formula becomes (velocity factor * 75 / frequency in MHz)

example a quarter wave at 39 MHz in coax with a velocity factor of 0.72 is (0.72 * 75 / 39) or 1.38 meters.


Velocity factor is a number between 0 and 1 which is the ratio of the speed of radio waves in the transmission line divided by the speed of radio waves in air.