Transmission of surges through right-angle termination 4

[electricpete]

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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[VEBill]

"...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. ;-)

 

[electricpete]

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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[VEBill]

"...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.

 

[electricpete]

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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[VEBill]

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.

 

[electricpete]

ok, I can buy that.

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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[electricpete]

... 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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[electricpete]

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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[VEBill]

"...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..."

 

[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.

...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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[VEBill]

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".

 

[electricpete]

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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[2dye4]

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.

 

[VEBill]

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

The important keyword is "much". Perhaps better expressed as "much much". ;-)

Quarter wavelength is often maximum effect.

 

[electricpete]

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.

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[VEBill]

"...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.

 

[electricpete]

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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[electricpete]

And attached is the long rise-time simulation

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[VEBill]

"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')

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"...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.