Delay in 3 meters of wire

Hi Guys, I have question about transit time in a wire and phase shift caused by inductance. I'm going to use 3 meters of wire and 300,000,000m/s (discuss speed later)

300,000,000m/3m =10 nanoseconds, What travels? Will I measure both current and voltage as having a 10 ns delay at the other end of the wire?

Now, I wind the 3 meters of wire as a solenoid coil and I have a 30 uh coil. Now I will have a 90* phase shift between the current and voltage.

(I think I may have stumbled on my answer, but, let me continue.)

How does the 10ns add to the 90* phase shift.

Ok, to try and answer my own question:

The current and voltage both have a delay from one end of the wire to the other. If I'm looking at the voltage across and current through the inductor, the 10ns delay will not be seen. Both voltage and current will be delayed the same amount. However the current will have a 90* phase shift compared to the voltage. And both voltage and current will have delay of 10ns.

Did I get it right? Can anyone help? Thanks, Mikek

Discuss speed later, I didn't get to where I needed to discuss it. But, I understand RG58/U has .66 velocity factor, so 300,000,000m/s would reduce to 198,000,000m/s. Ladderline .91 velocity factor.

Reply to
amdx
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Yeesh, this gets MESSY! The delay from one end of a wire to the other is kind of a transmission line problem, and you need to know the impedance of the transmission line to know the propagation velocity. Just wire in open air is uncontrolled impedance, so I have no idea how that works. Now, when you wrap it into a coil, part of the coil acts as the ground plane for every segment of conductor that acts as a transmission line. Messier, still!

Jon

Reply to
Jon Elson

No need to get messy, just a yes or no. :-)

I don't want numbers, just the just the explanation of how phase shift caused by inductance and time delay are related when measuring the inductance of a long wire. ( I think it can be ignored but... )

Ok, let me try again. If I have a series R L circuit and it has a 62 degree phase shift. Does the transit time have anything to add to or subtract from the the phase shift?

Or another try. Say a straight wire has 15nh per inch and 1800nH per 10ft. If I make a series R L circuit with 10ft of wire and a 50 ohm resistor, Will this have a different phase shift than a series circuit with physically small 1800nH inductor and a 50 ohm resistor?

Mikek

Reply to
amdx

"Just wire in open air" has a reasonably controlled impedance; if you do not believe me, just look at what the Bell System did to pairs of wire for the pone system, as well as the (then) counter-intuitive practice of INDUCTIVE loading.

Reply to
Robert Baer

I think phase shift is a property of an established sinusoidal signal (at some specified, established frequency, too), while propagation delay is a transient phenomenon, so they're not really comparable.

--
John
Reply to
quiasmox

Whether or not physical length has influence, depends on the frequency. Rough guideline: when the physical length of the wire is less then 0.1*lambda for the highest frequency, you can ignore transmission line effects.

So you can calculate with lumped inductance and lumped capacitance.

For your long wire, calculate the capacitance and inductance and you can convert that to (for example) a CLC network for simulation. you need to take the return path into account as this affects the inductance and capacitance. So the 15 nH/inch you mentioned may not apply.

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Wim 
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Reply to
Wimpie

In the end, I think this transit time is a wash in calculations. But I can't seem to get a solid feel for it. I keep going back and forth.

This goes back to a couple of threads on sci.electronics.basics, Question on R L math" and "More math for "SETUP" I don't care to rehash those now, but I'll give an explanation of the "setup" and to show why I ask the question.

My drive voltage is a sinewave. I'm looking at the Voltage at the INPUT (beginning) to an inductor. I'm also looking at the current at the OUTPUT (end) of the inductor. (I'm looking at current by monitoring the voltage across a sense resistor) To repeat, I'm looking at the beginning of the inductor (wire) on channel one and the end of the inductor (wire) with channel two. Is there a delay for transit time along with a phase shift caused by the inductance?

I don't think there is, because both V and I have the same transit delay in the measurement. But... Mikek

Reply to
amdx

WOW! you have a way of asking the most mind boggling 'simple' questions. Bet like most people, I've spent a lifetime 'ignoring' such simple questions, rounding off the answers, so to speak, so is not an easy task to answer your question.

In defense, my justification for not thinking about what you just asked is this: the delay of the transmission line can be EXACTLY duplicated with a series of distributed very tiny inductors with capacitance to GND. Therefore as the overall inductance increases when you bend the whole wire into a loop it becomes IMPOSSIBLE to tell exactly where the contribution to the inductance is coming from. You can't tell if it comes from the self-inductance and coupling because you bent the wire into a loop, or if it comes from the 'length' of the wire. It just is. And, that original straight wire inductance, being small contribution, just gets scooped into the overall inductance and buried in the error budget.

From memory working with HP's Network Analyzer, the length of a single turn inductor is the max you'll ever get [air core] you can loop it 2, or

3 times, but the diameter gets smaller so you're left with approximately the SAME inductance no matter how you bend up your wire. Anybody confirm that memory? Or, verify using femm?

To show you how much I've been ignoring simple questions, I'm not even going to get into the 'coupling' to free space and launching energy, which obfuscates everything even more.

Reply to
RobertMacy

Measure how? Your single wire in free space is going to act like a lot of things, but the one thing it won't act like is a clean transmission line.

The most likely thing it'll act like is an antenna, which means that it'll radiate away the energy of the sharp edges of your pulses, or delay it in hard to predict ways.

Again, you're so far from a transmission line that the answers will be nearly meaningless.

Are you trying to do something real, or just pondering?

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Tim Wescott 
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Reply to
Tim Wescott

I think that means you understood the question. However useless the question may be. Again, in the end I think it washes out. But I can't get my head around how.

Oh, my first thought, that's not right. I can wind a high Q, 240 uh inductor with 55 ft of #18 wire, (turns spaced 1 wire width apart). A straight wire, 55ft long of #18 wire has 35uh of inductance.

Mikek

Reply to
amdx

yes.

no.

you haven't described a test circuit. I am assuming current tests with high impedance source and low impedance sensor, and voltage tests with the opposite.

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Reply to
Jasen Betts

I driving with a sinewave.

Mostly pondering, And again, I'm driving everything with a sinewave, (up to 10MHz) The question arose when I attempted to measure an 11ft piece of RG58/U with a bunch of ferrite around it, (big beads). I'm measuring just the shield from one end of to the other. Because some questioned my measurement indicating over 3.6k of loss resistance, it was suggested I measure just the cable. So, while making this measurement and seeing the the phase shift, in this case a

61* phase shift*, I wondered, with this 11 ft of wire to travel where does the travel time delay show up in that 61* phase shift.

I think it's a wash because both current and voltage have the same transit delay. But then I think again, and my measurement at both ends is a voltage measurement (E across an R) there should be a transit delay to go the 11 ft.

I'm so very confused and I'm not even doing math. :-)

Mikek

*partly from Rloss and partly because of my 47.5 ohm sense resistor.
Reply to
amdx

When you put many ferrite beads arount a long piece of wire, you can't ignore the capacitance from the wire to ground (via the ferrite beads). When you use the ferrite in its lossy region, your circuit may act as an almost distributed RC line. That may introduce lots of error in your induction calculation based on Vin and Vout across a resitor.

Reason: current that goes in IS NOT current that goes out (due to the capacitance to ground).

If you use the ferrite in its inductive region, your wire with many ferrites behaves like a transmission line with a propagation speed below c0.

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Wim 
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Reply to
Wimpie

Oh, sure, for voice frequencies, and with miles of wire stretched about 15' up over the earth, that could be quite controlled. Just like microstrip! Only, instead of mils, the units are feet.

Jon

Reply to
Jon Elson

The usual concern is, the group velocity (i.e. the energy packets from the driven end, to the receiving end); so, what travels is energy.

But, that doesn't have anything to do with the wire! If you have a resistive load on the end, its current and voltage are in phase, so the current and voltage have the same phase delay. Other loads will have other answers. The group delay goes with neither voltage nor current, but rather the product of the two.

That's true if the load on the end is zero ohms (i.e. you short the second terminal of the coil to ground).

It doesn't. Your 'coil' inductance results from mutual inductance phenomena, your distances are now related to the shorter length that results from magnetic induction distance (like, the length of the solenoid, not the wire length that was wound to make it). Or, if you wind both the ground and the signal wire to make a double-helix solenoid, your terminal point ISN"T connected to ground, it's connected through a secondary transformer-winding to ground.

Reply to
whit3rd

A few things that might help.

Propagation velocity in coax varies quite a bit. from as low as 0.65c to as high as 0.92c. Typical coaxes are mostly in the 0.80 to 0.85 range.

Sometimes i have found that my confusion stems from not setting up the math (often i end up having to hit the books to handle once i have). Once the math is clear the rest of my confusion usually clears up with it.

YMMV

?-)

Reply to
josephkk

I meant comparing a single LOOP of wire to two loops of the same wire, not a straight wire compared to a loop of wire. How does a single loop of 55ft compare to multiple loops of that 55ft wire?

Back to your original question, been bugging me for the longest time, how about...a straight wire in free space has 'distributed' inductance and 'distributed' capacitance along its length. Thinking in terms of calculus, where you can break this up into tiny segments of inductors, which add up to the total value. And, you get to make each of those segments as small as the accuacy you need, in the limit, zero length. Now keep that image of all those tiny inductors strung together along your straight wire. Now roll the wire into a loop and suddenly the fields interact a bit more, creating mutual inductances between them. As you remember just like in a simple pair of inductors where the total inductance gets increased to L1+L2+2*M; the tiny segments interact, increasing their overall total inductance and now you have loops of wire becoming much larger inductance, but all based upon those original tiny inductors.

Uh, wait. That's inductance of the straight wire becoming a larger inductance, NOT much to do with the speed of light delay, hmmm...nevermind.

However your point is important when trying to understand high frequency inductors. At low frequency where the current is completely inphase, everywhere around the loop the fields arrive at the same time, so the contributions of mutual inductance add up, but with high frequency, and the inherent delay, the current, and therefore its field, at one location along the wire is NOT in phase with another location a little further down the wire, so the mutual inductance doesn't add as much [better to say, the same]. Thus, at really high frequencies the coil is NOT going to have the same inductance it has at lower frequency because the fields are actually canceling and destroying what little inductance you were accumulating. To envision, guess you have to think very small and mentally 'walk' along the wire, while looking at nearby other segments of the wire to get a feel for what nearby wire's current is in phase and out of phase add adding or subtracting.

How important is all this? Not sure. But keep in mind that it is really, really difficult to make impedances above 300 ohms, because free space is around 377 ohms and tends to 'short' out everything.

Reply to
RobertMacy

If You are applying the signal between the screen and some external ground connection you do form a transmission line between the two lines. The transmission characteristic (ie. velocity factor) of that line is affected greatly of the high permeability of the ferrite beads around the screen conductor. The apparent electrical length of the line increases compared to no ferrite. And so does the inductance and also the transmission loss.

If you on the other hand apply the signal between the two conductors of the coax the magnetic fields from the two conductors cancel each other outside the coax screen. The transmission characteristics is therefore not affected by the ferrite beads but only of characteristics of the space between the center and screen conductor, and the conductors them self.

One very common reason for putting ferrite beads around a coax cable like what you describe is to dampen common mode (unbalanced ) currents. Witch is what you intentionally introduce when try to run a current only in one direction trough the coax.

Might not be what you was asking for, just had to try and remember some old knowledge.

You could try to google for: Baluns, transmission line transformers, or differential and common mode excitation.

Cheers, Rune

Reply to
Rune

not

55ft

I think i am remembering parts of the formula for short inductors at low frequencies. It had two clear dependencies, L is proportional to n^2 and the area enclosed by the loop. For a fixed length of wire in a circular loop these seem like they would cancel out. Diffrent shapes would have very interesting effects though.

?-)

Reply to
josephkk

Not quite.

The characteristic impedance of a (lossless) transmission line equals:

sqrt(L/C) , where L and C are the inductance and capacitance per unit length.

That's dependence on the *quotient* of L and C

The propagation delay equals:

1/sqrt(LC)

That's dependence on the *product* of L and C.

You can have lines with identical characteristic impedance and widely different propagation velocities.

--
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Reply to
Fred Abse

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