# Delay in 3 meters of wire

Started by February 4, 2014
```On 2014-02-05, amdx <nojunk@knology.net> wrote:
> 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?

yes.

> 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.

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.

--
For a good time: install ntp

--- news://freenews.netfront.net/ - complaints: news@netfront.net ---
```
```On 2/5/2014 12:25 PM, Tim Wescott wrote:
> On Tue, 04 Feb 2014 20:39:48 -0600, amdx wrote:
>
>> 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?
>
> 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.
>
I driving with a sinewave.

>> 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.
>

> 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?

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.
```
```El 05-02-14 20:23, amdx escribi&oacute;:
> On 2/5/2014 12:25 PM, Tim Wescott wrote:
>> On Tue, 04 Feb 2014 20:39:48 -0600, amdx wrote:
>>
>>> 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?
>>
>> 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.
>>
> I driving with a sinewave.
>
>
>>> 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.
>>
>
>> 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?
>
>
> 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.

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.

--
Wim
PA3DJS
www.tetech.nl
Please remove abc first in case of PM
```
```Robert Baer wrote:

>    "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
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

```
```On Tuesday, February 4, 2014 6:39:48 PM UTC-8, amdx wrote:

>   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?

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.

> Will I measure both current and voltage as having a 10 ns delay at the
>
> other end of the wire?

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.
>
>
> 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.

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

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

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.
```
```On Wed, 05 Feb 2014 13:23:44 -0600, amdx <nojunk@knology.net> wrote:

>
>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=
=20
>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.

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

?-)
```
```On Wed, 05 Feb 2014 11:29:02 -0700, amdx <nojunk@knology.net> wrote:

> On 2/5/2014 10:08 AM, RobertMacy wrote:
>> On Wed, 05 Feb 2014 08:11:09 -0700, amdx <nojunk@knology.net> wrote:
>> ...snip....
> 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.
>
>
>> ...snip...comment about the same inductance.
>
>   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.
>
>> ...snip....
>
>                   Mikek

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.
```
```On 05. feb. 2014 20:23, amdx wrote:
> On 2/5/2014 12:25 PM, Tim Wescott wrote:
>> On Tue, 04 Feb 2014 20:39:48 -0600, amdx wrote:
>>
>>> 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?
>>
>> 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.
>>
>    I driving with a sinewave.
>
>
>>> 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.
>>
>
>> 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?
>
>
> 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.

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

```
```On Thu, 06 Feb 2014 06:51:43 -0700, RobertMacy <robert.a.macy@gmail.com>
wrote:

>
>I meant comparing a single LOOP of wire to two loops of the same wire, =
not =20
>a straight wire compared to a loop of wire. How does a single loop of =
55ft =20
>compare to multiple loops of that 55ft wire?

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.

?-)
```
```On Tue, 04 Feb 2014 20:57:37 -0600, Jon Elson wrote:

> you need to know the impedance of
> the transmission line to know the propagation velocity.

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.

--
"Design is the reverse of analysis"
(R.D. Middlebrook)
```