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# Transmission of surges through right-angle termination4

## Transmission of surges through right-angle termination

(OP)
This would be again considering 0.1 microsecond rise time pulses in a power system.  There are two different types of concern:
1 - surges from power system propogating toward the motor.. we'd like to lessen them.
2 - surges generated by partial discharge inside the motor which propogate outward into terminal box, make a right angle turn to get to coupling capacitor filter network for sending partial discharge.

In both cases, there is occasions where the surge may encounter a 90 degree turn in the copper buswork within the terminal box.  I'm guessing the buswork is 1/4" thick.

What effect does a 90 degree turn have on pulses that I assume are initially propogating in transverse electromagnetic mode.

What about a 90 degree tap.  i.e. surge has the choice to go straight or make a 90 degree turn... will any of it go down the 90 degree tap>?

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

2
"...will any of it go down the 90 degree tap?"

Of course yes.

The split depends on the total picture, including what's further downstream in each path.

90° is just 25% of a single turn coil.

### RE: Transmission of surges through right-angle termination

(OP)
I should mention, at the point where these right angles occur, it is not shielded cables with well-defined ground planes (not like a coax).  It is three separate phase conductors.

If it were shielded conductor or a coil with adjacent turns that provide a well-defined tranmission line, then I'm inclined to believe the pulse has easier time following turns.

In this case, it it is three separate buses about 1' apart from each other and 1' away from ground, undergoing 90 turn with relatiely small radius.   It is not obvious to me why that the pulse is going to make the turn since the transmission line path is not well defined by virtue of distance to the ground and other phases.  And seems even more of a challenge for wave to make a 90 degree turn at a Tee configuration.  I am picturing (right or wrong) that there is something resembling plane wave propogation within the conductor and I wouldn't think a plane wave would change its direction like that.

Do you have any suggested reference on this or further ideas? I'd be interested to hear.

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

"...not obvious to me why that *the pulse* is going to make the turn..."

A pulse is perfectly happy to split into two (or N) parts in accordance with Kirchhoff's circuit laws (adapted to account for impedance and frequency spectrum and so on and so on and so on...). Thinking in terms of "a pulse" may lead one to thinking that a pulse is an entity that is somehow going to stick together and follow only one path or the other. That's obviously not true.

Given a T-junction, it's almost a certainty that some significant fraction of 'the pulse' would be reflected back towards the original source (a perfect Z match being very unlikely). So the pulse split would be 3-way.

### RE: Transmission of surges through right-angle termination

(OP)
Thanks.  Considering the 90 degree turn:  Let's say the busbar is copper 2" wide, 1/4" thick.  There are two options for a 90, the one I'm talking about occurs the same way that it is easiest to bend the busbar....so the short thickness dimension creases, not the long width dimension  (no they're not formed by bending, but that's the easiest way to describe it).  The groundplane and other phases are 1 foot away.  Then I think there would be a significant portion of  the incoming pulse reflected back toward the source when it passes that 90 degree angle, wouldn't there be?

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

The larger the radius of the bend, the smaller the impedance bump. But it's probably not worth worrying about too much. Even 0.1us is still more-or-less HF frequency range; so an inch here or there isn't much compared to a wavelength.

We have TDR instruments that we sometimes use in our work, but we've never even contemplated using them to map out power wiring.

### RE: Transmission of surges through right-angle termination

(OP)

What about a branch connection shaped like a Tee.  It has 3 ports: two on the sides of the top of the Tee, and one on the bottom.

Will the behavior be the same in terms of splitting regardless of whether the pulse come in the bottom or come in the side of the top of the Tee?

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

(OP)
... last question assumes all three connections have the same surge impedance.  And I'm just talking order of magnitude.

I would suspect the answer will be yes based on your previous answers which suggest the behavior is approximated from transmission line theory using surge impedance and not depending on physical geometry features that are much smaller than wavelength (as long as they don't significantly affect surge impedance).

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

(OP)

#### Quote:

The larger the radius of the bend, the smaller the impedance bump.
Another way to state that would be the smaller the radius, the larger the change in impedance.
That makes intuitive sense, but it raises a question:
small compared to what?
Compared to the wavelength?  In that case we would judge that the lower-frequency-content pulses (longer wavelength) would see more impedance change for a given bend than a higher-frequency-content pulse....does not sound right.

Compared to the conductor spacing?  The example I provided has bend very small compared to spacing to ground and others?

Maybe there is a compound condition implied:  "small radius of curvature compared to conductor spacing, but both large compared to wavelength?"

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

"...compared to wavelength?"

Exactly, that's why I mentioned "...HF frequency range; so an inch here or there isn't much compared to a wavelength..."

### RE: Transmission of surges through right-angle termination

(OP)

#### Quote (electricpete):

the smaller the radius, the larger the change in impedance.
That makes intuitive sense, but it raises a question: small compared to what?
Compared to the wavelength? In that case we would judge that the lower-frequency-content pulses (longer wavelength) would see more impedance change for a given bend than a higher-frequency-content pulse....does not sound right.

#### Quote (VE1BLL):

...compared to wavelength?"Exactly, that's why I mentioned "...HF frequency range; so an inch here or there isn't much compared to a wavelength..."

So, what about the bold portion of my quote above?   I don't think lower-frequency content pulses will see a larger change in impedance for a given bend radius than higher-frequency-content pulses (do you?).  Therefore I think there must be more to the story than comparing bend radius to wavelength.

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

Amend: The larger the radius of the bend, the smaller the impedance bump (for a given frequency, assuming it matters).

Many years ago there was an article in one of the radio hobbiest magazines proposing to use the medium-wave 160m (1.8 - 2.0 MHz) radio band for 'Moon bounce' (EME) because it was "fewer wavelengths to the Moon".

### RE: Transmission of surges through right-angle termination

(OP)
Thanks. That sounds similar to the compound condition I described above ("small radius of curvature compared to conductor spacing, but both large compared to wavelength"), right?

What about the Tee configuration.... does it make any difference whether the pulse comes into the bottom of the Tee or the side/top of the Tee (in terms of the fractions transmitted to each of the outputs and reflected to the input)?   (Assuming for simplicity same surge impedance in all branches... terminated with matched impedances... 2"x1/4" copper bus, 0.1 microsecond rise time.)

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

0.1uS pulse is similar to 1.6 Mhz signal which has a wavelength of 188 meters so the dimensions of the switchgear features are probably very small relative to a wavelength.

This implies the pulse will propagate without much affect from features much smaller than the wavelength.

So the tee won't matter and the pulse will split into any attached conductor.

### RE: Transmission of surges through right-angle termination

"...from features much smaller than the wavelength."

The important keyword is "much". Perhaps better expressed as "much much".

Quarter wavelength is often maximum effect.

### RE: Transmission of surges through right-angle termination

(OP)
I had been struggling to understand why my intuition doesn't accept that fact.... I think I figured it out.   I was confused by the fact that the differential equations governing voltage and current in a transmission line are analogous to the differential equations governing E and H in a plane wave.   That led me to imagine some kind of plane wave "inside" the conductor.  If that were the case, then the wave would propagate in a straight line and bounce backwards at a sharp right angle in the conductor.  But it's not the case, the fields of interest are not internal to the conductors... they're external to the conductors (duh).   If I have conductor travelling parallel to ground plane and it takes a right angle still at the same distance from the ground plane, there is no change in impedance of course.

Thanks for helping to straighten me out on that.

=====================================
(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

"...there is no change in impedance of course."

Your analysis should reveal an "impedance bump" directly *at* the right angle turn. I presume due to an increase in the series inductance part of the transmission line model at that exact point.

### RE: Transmission of surges through right-angle termination

(OP)
I think my comments about where the energy is were a little bit missing the point.  As you guys say it is the lateral dimensions of the conductor and geometry, compared to the wavelength or rise time.

I have done a very crude 2-D  "transmission line matrix method" simulation of a wave launched into a right-angle turn... for two different scenario's:

First: Short Rise time:  the rise time is half of the cross section of the conductor.
Second: Long Rise time:  the rise time is four times the width of the conductor.

There is a lot more reflection for the short rise time than the long rise time, as expected.  I was halfway expecting that the simulation would give some intuition as to why that occurs, but I'm not sure it does.

Attached is the short rise time animation.

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

"I have done a very crude 2-D  'transmission line matrix method' simulation...", electricpete, 2011.

"Please excuse the crudity of my model...", Dr Emmett Brown ('Back to the Future')

-----------   !!!    -------------

"...the rise time is [X] the cross section of the conductor..." where X = 0.5 or 4

Assuming a sensible rise-time (us scale), that's a fairly wide conductor, isn't it? (Well worth stealing for the copper value; bring a crane and a truck.) Or (for a 2-inch wide bus bar) an extremely fast rise-time (GHz band).

My instinct says that the waveform in the .avi sims is showing way too much 'internal' reflections for the scales to be correctly matched. Almost as if you could shave the corner at 45° to 'bounce' the waveform edge around the corner like a beam of light.

Double-check that the scale of the waveform and the scale of the conductor actually match reality.

### RE: Transmission of surges through right-angle termination

(OP)
The purpose was to give a qualitative demonstration showing more reflection for shorter rise time and less for slower rise time so we could see the mechanism by which the relationship between rise time and conductor width affected the result.  For that I chose one rise time less than conductor width and one more, but neither dramatically more.   What I think I learned  (and kind of already suspected) is that for fast rise time relative to width, there is very little time for diffusion (per TLM or Huygen's principle) of the wave toward/into the 90-degree direction, so that most of the wave hits the far wall and almost doubles/bounces before it has had a chance to diffuse.  In contrast the longer wave has more opportunity to diffuse and never hits the wall head on.

To make it easy and general purpose, I tried to model all my boundaries using and impedance mesh. Low impedance in conductor area and high impedance everywhere else including a 1-cell-wide strip around the edges.  Then I blocked out the values in the center because they were distracting from the picture. There are some features of the boundaries that are quite obviously not exactly correct like the ridge on the left side of the wave, but I didn't spend a lot of time correcting them... maybe later.   Another is that simple 2-D mesh creates anisotropy in the medium (wave propagates sqrt2 faster along the mesh than 45 degrees from the mesh). There is a correction available, but for simplicity I ignored it.  There are the reasons I called it crude.

I am well aware of the relationship between rise time and dimension, but the dimensions I selected suited my purposes better (demonstration, not calculation of a magnitude).  I don't think that picture of doubling and reflection would have been as clear comparing two rise times which were both many orders of magnitude higher than the width.  I learned the speed of light in 3rd grade, and this speed will not be too much lower.  But thanks for your assumption that I did not have that 3rd grade knowledge... we all know what they say about assumptions.

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

(OP)

#### Quote:

My instinct says that the waveform in the .avi sims is showing way too much 'internal' reflections for the scales to be correctly matched. Almost as if you could shave the corner at 45° to 'bounce' the waveform edge around the corner like a beam of light.
That is what the model uses... high impedance in the non-conducting areas, so wave will bounce off of those.  Perhaps the real world has a blurrier boundaries due to fringing field effects, but as a first approximation that is correct, isn't it?

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

(All of this is within the context of this and your other thread on this forum. I'm assuming that both threads are related, and bother are on the subject of surges within bent bus-bars.)

If the purpose of this exercise is to refine your instincts, then it's best to start in the middle before working out to the extreme cases.

Assuming that the rise time of surge of interest is still around 0.1 microsecond, then your model bent conductor is the size of a building. That's pretty big. It's not that you didn't know that (reading minds is not my specialty), it's that the implications of this extreme scale ratio choice are significant.

The mental construct of an internal "wall", where the waveform bounces around internally, does not exist within correctly (more real-world) scaled conductors. Most conductors are essentially one-dimensional structures, often bent into other dimensions.

If I may re-purpose a famous phrase - Within most conductors as typically applied, "God does not play..." billiards with charge carriers.

### RE: Transmission of surges through right-angle termination

(OP)
For 2" conductor, 3E8 m/sec(round numbers...  should be lower) speed, the wave travels one foot per nanosecond and 1000 feet per microsecond. 1000 feet would be 6000 times the conductor width.  That poses some challenges in my programming environment, but I tried attached a rise time 40x the conductor width.

To keep the avi file size managable, snapshots are taken every time the wave advances 20 grid points (before it was 5).  If you look carefully you can see the leading edge jumping along the path for the first 14 snapshots (1st 5 seconds of avi) and you can see the trailing edge (where it turns constant) passing through between 20 and 25 seconds.

There is no discernible (from the graphic) reflection by this 2-D model even for rise time 1/40 of conductor cross width.  How well that related to real life scenarios I'm not sure.

=====================================
(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

(OP)
Correction:

#### Quote:

There is no discernible (from the graphic) reflection by this 2-D model even for rise time 1/40 of conductor cross width.  How well that related to real life scenarios I'm not sure.
should've obviously been

#### Quote:

There is no discernible (from the graphic) reflection by this 2-D model even for rise time 40 times the conductor cross width.  How well that related to real life scenarios I'm not sure.

=====================================
(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

(OP)
I have made plenty of mistakes and I don't mind being corrected/questioned whether I'm right or wrong. However when I correct/question people, I generally don't try to do it by turning it into a joke at their expense.

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

"...a joke at their expense."

Sorry.

Do you mind to point out exactly where I crossed the line?

### RE: Transmission of surges through right-angle termination

(OP)
Hmmm. First pictures and quotes comparing me to this  eccentric scientist.  That alone could be good or bad,  so we look for the context that follows.....

What follows is your discussion where you percieve my simulation to be based on complete ignorance of the scale (*).     And to up the comic value of your comments, let's throw in "Well worth stealing for the copper value; bring a crane and a truck."

I hope you can see where that might strike me negatively. At any rate I suspect now it was not intended that way, maybe I just need more sense of humour and lighten up a little.

(*) By the way, I hope you can see that comparing 40xwidth to 4000x width would show nothing at all interesting.  The comparison of 1 / 2 width to 4x width showed a lot more that we might possibly learn from the simulation.  Relevance of that "learning" to a particular scenario's is an important item to judge and an open question for me.

#### Quote:

The mental construct of an internal "wall", where the waveform bounces around internally, does not exist within correctly (more real-world) scaled conductors. Most conductors are essentially one-dimensional structures, often bent into other dimensions.
It may be the case.  What is the explanation for the impedance bump that you referred to earlier?

=====================================
(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

(OP)

#### Quote:

It may be the case
It appears to be the case from the simulation. 40x -> nothing reflected.  Much longer rise times relative to cross section -> even less.

For right angle bend, there is also the fact that the effective width gets longer as you pass through the turn.  But this is seen in the simulation and does not cause any reflection at 40x and beyond.

I wonder if there is some effect similar to self inductance... the wave at position X has different linkage to the wave at position X-deltaX when the line is curved ?  That might be mixing quasistatic/field concepts with wave concepts... I'm not sure.

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(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

(OP)
I should have waited for you to answer. My last post was trying to address the previous post question "What is the explanation for the impedance bump that you referred to earlier?"

I would be interested to hear more about what kind of impedance bumps you see... including wiring configuration and rise time of incoming wave.

=====================================
(2B)+(2B)'  ?

### RE: Transmission of surges through right-angle termination

Pete, my reference to the world-famous meme ("Please excuse the crudity of my model...") was intended as an amusing comparison of your quoted statement, certainly not of you personally. I apologize if it caused offense.

It's worth noting that your expertise is held in the highest possible regard by myself and many others on Eng-Tips. In comparison, all that I can offer is the occasional real-world 'boundary condition' that may occasionally guide and constrain the theoretical analysis.

...

What is "...the explanation for the impedance bump..." of a right angle bend?

Series inductance.

90° is obviously one-quarter of a full turn (excuse the embedded humour, it's unavoidable).

As a gedanken experiment, replace the 90° bend with a very smooth and continuous curve that continues to curl around and around. Make it as many turns as you wish.

It's a coil; an inductor.

Lots of series inductance.

Plenty of 'impedance bump' to restrict (choke) the flow of higher frequencies.

And not a single "wall" anywhere.

### RE: Transmission of surges through right-angle termination

There are situations where the impedance discontinuity of a right angle bend may cause noticable effects. One such case is the DIN 41612 connector. See http://www.erni.com/DB/PDF/DIN-PDF/ERNI-DIN-HFE160.pdf

Lots of data, geometrical and electrical, plus results from measurements on actual connectors to test different simulations against. The measurements show all aspects you can think of - including TDR.

The phenomenon does not have any practical consequences in power applications where voltage levels are orders of magnitude higher than the voltage induced in the discontinuity. Add to that that rise-times very seldom are shorter than 100 ns in power circuits while sub-ns are met in 'signal' electronics.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

### RE: Transmission of surges through right-angle termination

(OP)
VE1BLL - thanks for the kind words. They are probably not deserved, but I won't dwell on that.  I certainly appreciate your knowledge and help.  I'll have to think about that quarter turn.  Makes some sense.

Gunnar -  That is pretty much the same conclusion I was reaching. For most purposes the right angle makes no difference in power systems.  Compared to computer systems, the dimensions are larger which makes wave effects more likely, but the rise times are lower by a much larger factor which makes wave effects less pronounced.

In contrast, tees/branches should have an effect on pulses even in a power system, although I can understand now that it doesn't make much difference whether you go straight through the Tee or make a turn.

=====================================
(2B)+(2B)'  ?

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