How long does a Transm Line have to be for TL theory to apply 2

[electricpete]

How long does a Transm Line have to be for TL theory to apply

As an example, see slide 1 of attached. It is an attempt to model a ferrite sleeve added onto a cable for surge suppresion as discussed here:
thread238-292631


For simplicity, everything is assumed lossless and the source and receiving ends are assumed long with characteristic impedance Z0 and matched impedances on each end. In the middle is inserted higher impedance K*ZL representing that ferrite sleeve.

Is there some minimum length (of ferrite sleeve) where transmission line theory won't apply any more? For example Lambda/4?

I am aware of a parallel restriction that we have to be careful apply lumped elements when the dimension in one direction is any significant fraction of a wavelength. But I'm not sure if that says we can't apply TL theory for short items. It seems to me that the Transmission Line Matrix Method (TLMM) is based on appling TL theory to any size elements.

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[VEBill]

Q: "How long does a Transmission Line have to be for TL theory to apply?"

A: The Smith chart is marked off in very tiny fractions of a wavelength.

One problem that would arise is that the boundaries become a bit blurred when the scale of the discontinuities is a significant fraction of the problem space. Death by details.

 

[electricpete]

Thanks. Perhaps I'll post some example dimensions to assist the discussion later.

But first I realized I have an error regarding the reflection coefficients. As shown in slide 3 I used
ReB1 := (K-1)/(K+1) (Reflection of backward wave at interface 1)
ReF2 := (1-K)/(K+1) (Reflection of forward wave at interface 2)

They are calculated just as shown here:

http://en.wikipedia.org/wiki/Reflection_coefficient

ReflectionCoefficient = (ZL - ZS)/(ZL + ZS).

These two coefficients should have been the same (not negative of each other). i.e. I should have used:
ReB1 = ReF2 = (1-K)/(1+K)

Let me correct that in my graphs and post back when I have thought about it some more.

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[electricpete]

Attached is corrected version.

Now it matches intuition better:
1 - All waveforms eventually reach the final value of 1 after awhile. (Previous were unrealistic since they steadied out at lower values like 0.02, but I rationalized that was attributable to artificial matched-impedance at the final terminations).
2 - Longer Ferrite Sleeve, Higher K give best results.
3 - For high K, shorter sleeve gives diminishing results, ending with rise time very close to initial rise time for miniscule length. That is much more intuitively satisfying.
4 - For low K, the device doesn't do well regardless of length.

Rough swag dimensions for 13.2kv motor might be 5/8" Diameter stranded conductor with 1.25" diameter insulation. Far upstream the conductor has a grounded shield at the same diameter. Somewhere around a foot or two before the ferrite sleeve, the shield is terminated. After that point the the ground plane (terminal box side) may be a foot away and other insulated conductors (from other phases) about a foot away. The sleeve might be 6" long or less and installed on individual conductors (one option... depending on saturation). Downstream without about a foot the unshielded conductors, the conductors all come together to see phase to phase voltage. Then 3 or 4' after that they see impedance change entering the motor winding.

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[electricpete]

Just a swag dimensions of ferrite sleeve a little bigger than the ones Gunnar posted. Maybe 6" diameter 0.75" thick, 1-2" long.

All of this is not exact, just trying to paint a mental picture that might help narrow down the deviations from transmission line theory that need to be explored.

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[electricpete]

I forgot to say the rise time of surge of interest is around 0.1 microsecond.

One other note - I am not trying to invent anything... more just trying to increase my understanding of propagation of surges through the system and using this as an example.

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[VEBill]

Sometimes it's cheaper to just install a $400 filter than spend 20 paid (*) hours thinking about $5 ferrite beads.

* Then again, today is Sunday... ;-)

 

[electricpete]

It is a 13.2kv system, so I'm not holding my breath for $400 filters, much less ones that will not introduce failure modes. I'm not sure where you came up with 20 hours, probably the same place as the $400. As I said in the post just above, my purpose is learning more than anything else.

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[VEBill]

In that case, it's sounds like it's well worth the time.