Multi-conductor transmission line impedance

If I lay two insulated wires side to side, or lightly twist them, then I end up with a characteristic impedance of around 100 ohms (more if the insulation is unusually thick, less if it's enamel or something).

Is there a way to start from that, and predict what happens if I add more conductors?

Specifically, start with my wires -- call them circuit A and circuit B, laid out as AB in a "ribbon". Now add one more, so that I have two wires in circuit A and one in B: ABA. Now add more yet, in a ribbon (no quotes), not twisted, so I have N/2 wires in A, and N/2 wires in B: ABAB...BAB.

Is there a way to predict -- even approximately -- what the impedance of the whole cable will be?

And no, this isn't a test question -- I have a need for an unusually low impedance transmission line (really a really low inductance way to connect point A with point B), and I'm trying to reduce the problem to the part that I really don't know. I'm also considering two parallel plates, with insulation in between -- I've found equations for that, but fabrication may be harder than "ribbon cable".

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply to
Tim Wescott
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es

ttdesign.com

How long is the distance?

What frequencies (waveshape) are you trying to pass?

How much loss can you accept?

Reply to
brent

Use a field solver, like ATLC or ATLC2.

The impedance for "standard" ribbon cable will be roughly 90 ohms divided by N/2, something like that. For a really lo-z cable, stick copper foil tape to both sides of a piece of thin mylar or kapton.

--

John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
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Reply to
John Larkin

es

well if each AB pair is 100 Ohms, if you put N AB pairs in parallel and you do a simple analysis that ignores the coupling between AB pairs, then the Z=3D 100/N no?

If you include the coupling, the Z will be even lower, so I think 100/ N is an upper bound.

Mark

Reply to
Mark

Whatever it is, it's going to get better and better. We have some high-amp cables that I'm told are nothing more than bundles of twisted pairs. One is rated for 2kA and is 30' long (yeah, it's the size of your forearm), it has < 0.3uH total inductance. Shove a screwdriver across the terminals while it's in operation and CRACK! The line is filled with insulation and water, so the dielectric constant is fairly high, and the capacitance large. It discharges very quickly with a nice pop of current. The characteristic impedance is probably less than an ohm.

Tim

-- Deep Friar: a very philosophical monk. Website:

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Reply to
Tim Williams

Distance does not affect a transmission line's characteristic impedance.

Nor do waveshapes.

That's a different problem.

But, to answer your questions: a few inches, 500ns edges at hundreds of amps, and not much (compared to hundreds of amps at hundreds of volts).

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply to
Tim Wescott

I was hoping for an analysis (or at least results) that took the coupling into account.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply to
Tim Wescott

Ick. I was hoping someone would have an answer. Preferably already in a bowl, with a spoon handy.

I think the pairs will interact in my favor though -- that's what I was hoping someone would know.

I'm trying to do a compare/contrast between the copper plates & Kapton route and the ribbon cable route. Copper & Kapton presents some interesting fabrication issues for this particular application that the ribbon cable would simply not have.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply to
Tim Wescott

I'm having a hard time wrapping my head around this. Inductance and impedance are closely related, but it's not clear that they're interchangeable in this case????

In order to make use of a wide transmission line, you need to launch a plane wave into it. That may not be a trivial task.

But, to answer your question... If it were me, I'd start with an 80-pin hard drive cable. Terminate every other wire into a buss bar. Poke it with a TDR to measure the impedance. Rip off two wires at a time and repeat. Graph the results. Math is good, but not always easier than experiment.

I'm also considering two parallel

Reply to
mike

Well, no, but we're dealing with frequencies and a length of wire where a shorted transmission line acts like an inductor. I'm really thinking "transmission line" here because I know how to go from a transmission line of known characteristic impedance and length to an inductance -- I don't know how to calculate apparent inductance of coupled wires otherwise.

Eh, true. But I'm more interested in making it look like a really low inductance, not in sending a pulse down it and getting anything intelligible back.

That's not the "Egghead Way". (Good suggestion, though).

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply to
Tim Wescott

What do I look like, Ben & Jerry?

I doubt it.

One thing I have done, for driving nmr/mri gradient coils, is make my own lo-z coax. Start with that MIL-type fine-strand wire with the very thin teflon insulation, #8 or 10 maybe. Slip that inside a braid, snug that down, and finish with shrink tubing. That makes for very low inductance.

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
Reply to
John Larkin

Maybe a 30-something pair SCSI cable, with all the pairs paralleled. It's available from stock, as well as PCB-mount connectors for both ends.

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
Reply to
John Larkin

2 100 ohm transmission lines in parallel will have a 50 ohm impedance same as 2 100 ohm resistors in parallel

If you start with 100 ohms and add N conductors the resistance should approach 100/n as a ribbon, and possibly less if laid out in a checkerboard grid instead of a row

You seem to be describing a 2 sided flex PCB

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Reply to
Jasen Betts

es

Something like this:

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latcablezo.pdf

-Lasse

Reply to
langwadt

,
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Distance does not, but distance and waveform characteristics will determine whether transmission lines are needed or not. Your description of a few inches and 500nsec edges does not strike me as a situation where it is going to matter much whether you have a 20 ohm line or a 40 ohm line. it sounds like the key is passing the hundreds of amps, not maintaining the waveshape due to differing reflections vs frequency.

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Reply to
brent

es

ttdesign.com

without having read all the replies...

USE femm 4.2 !!!!

No matter HOW complicated you make the geometry, it will give you the right prediction, including eddy current losses, if any!

Simply use the 2D, draw the structure of wires and voila!

A slab conductor will lower the inductance, ...at DC but it can also create a weird distribution across the width of the connector as eddy current fields shove the current out to the extreme edges. You end up with effectively two conductors carrying all the current at the edgees, resistance can skyrocket on you.

However, it is true that any connection wider than long is a conntection, not much of an inductor.

If you need low impedance, interdigitate the conductors A B A B compared to AA BB Nothing lowers inductance like the way the IC manufacturers have finally started connecting their chips.

Now, to predict what you have when you do A B A B A B ?

USE femm 4.2 and you'll be spot on.

Reply to
Robert Macy

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Hear, hear.

Less than 1/2 day, done, well almost.

Reply to
Robert Macy

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If it's only a few inches, could you get a pcb house to put heavy copper layers down on each side of flex?

George H.

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Reply to
George Herold

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That was my thought too. I've never done any flex thing. But can you get more than one layer? That might really keep L low.

George H.

t.net ---

Reply to
George Herold

That's the way to do that on the cheap. The deluxe method would be flex circuit. Both can be spiraled up and put in a jacket, to make a cable that is still somewhat flexible.

It is also possible to get coax with a really fat center conductor and thus low impedance.

--
Regards, Joerg

http://www.analogconsultants.com/
Reply to
Joerg

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