TDR of Wire on the ground

May 11, 2020 3 Replies
A

I did some TDR measurements on the 260ft wire on the ground. As a dipole on the ground, it is resonant at 1070kHz, (R=185 ohms X=0 ohms) That calculates as a 59.4% VF. I did several iterations while altering the TDR output impedance*, I ended with 439 ohms output impedance. This is the scope shot, I obtained with a minimal 1/8? 2ft ground rod. (it could easily be 200 ohms in my limited experience measuring ground resistance)



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I have the total impedance (calculated from the waveform voltage) as



515 ohms. (includes Ground resistance) Vout = Vs x R2 / (R1 + R2) // 2.7 = 5 x 515 / (439 + 515). Therefore, 515 ohm Total antenna impedance.

The time to the end of the wire is 0.6us. At a VF of 59.4%, that is



107 meters, while the measured value is 126 meters.

Comments?



Mikek


I was attempting to get a 50% step. I missed, but got tired of iterating.


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A

OK, is the second step roll off from losses in the wire/earth?

Mikek

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R

ohms X=0

e*, I

d
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e) as

%, that is

g.

I was just wondering myself. You are saying the wires are laid out like a dipole, so the differential impedance of the wires would vary a great deal along the length. When the wave reaches the wire end a high impedance woul d reflect with increase in amplitude. Your signal increases slowly with th e ultimate value being higher than 5V. I can't see losses causing that. W hatever is happening in the wire, the ultimate effect should be complete by 1.2 uS when the reflected wave reaches the origin. Can the velocity facto r be a be a strong function of frequency? Is the wave edge spreading as it travels?

Nicely done image, btw.

Rick C. - Get 1,000 miles of free Supercharging - Tesla referral code - https://ts.la/richard11209
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A

Sorry to have confused you, I'm now back to just a long wire, 260ft long. I'm resting at one end.

Yes.

Yes, and I think it's because of wire and ground losses, But I don't really understand why it happens. The pulse does not go over 5v.

Huh, its does seem wrong that the first step is 0.675us and the second takes about 2.1us. I would think they would be equal periods.

Interesting question about whether VF is a function of frequency.

It came out better than expected, The trace is very dim in expanded time base. The sun was shining bright, and I could not see the image on my cellphone camera. I did fashion a cardboard box to help. I took multiple pictures and I picked the best. Some shots I missed 1/4 of the oscilloscope screen. It was a pain.

Mikek

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