OK, you won't be doing this with paralleled #26 ribbon cable, either! A wide strip of copper foil can take a LOT of current, especially when it is short pulses. I have no idea how thick you can get foil on flex PCB, but I'd guess an Oz. would be available. Making your own by gluing thicker copper foil to the Kapton, Mylar or whatever should be doable. You can get Cu foil pretty thick, it is used for crushable gasket material in vacuum systems, and for making Cu gutters and such.
Jon
Oh, keeping the loop area to a minimum reduces inductance. Strap the source and return wires together tightly for as long a distance as you can, and it should help a lot.
Jon
Yes, that did.
I'm doing the calculations as a transmission line problem because that's what I'm familiar with, and I can find numbers for.
-- 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
And that's ok, because it is a transmission line. Problem is, if you drive a TX line with a couple of ohms and the load is in the same ballpark but the line impedance is many times higher, then it'll radiate and not work very well. Ferrites help to some extent.
-- Regards, Joerg http://www.analogconsultants.com/
Art and craft shops are a good source because they sell the thicker copper tape. The stuff that stained glass lamp makers use. But be careful when applying it, copper tape cuts are very messy and heal rather slowly.
-- Regards, Joerg http://www.analogconsultants.com/
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OK thanks, too much copper and not enough flex. I was remembering this old 60's video from the old Francis Bitter magnet laboratory.
George H.
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ttdesign.com
Wow, using femm 4.2 on this opened my eyes!
Stay away from foils, or ANY non-circular cross section [nature hates bumpiness, loves roundness] With a foil the current simply shoots out to the edges and most of the inner area has stopped conducting, so you end up with essentially two wires at the edge conducting and that's about it. Reason that TV antennas are hollow.
Assume a single 16 inch long 10 Awg wire is sufficient for carrying the current. Impedance calculates to approx 1.4 milliohms so no biggie dropping 700A However,...skin depth at 1MHz is around 2.6mils, which means that at 1MHz the 10 Awg wire is mainly conducting only along the outside surface and the resistance has gone up to something like 14 milliohms, still no biggie, BUT the linear inductance inf free space is around 35mH/in !! and if near a GND plane 12 inches away only drops to around 26+nH/in
Now replace that 10 Awg with interdigitated wire EQUAL in crossection area, PLUS make th diameter of the wire equal to the skin depth, that's over 400 36 Awg wires, if use 100kHz instead of 1MHz, you could go to larger 32 Awg wires, but would take 10 of them! Plus you still need to interdigitate them.
Software
Which leads to the conclusion that microstrip line impedance doesn't depend on trace width. But it does.
Or maybe to make them cheap?
-- John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation
Actually, if that were the reason, TV antennas would be, I think, hyperbolic sections. The rabbit ears anyway I'm thinking of.
Question: what conductor profile is ideal for distributing current evenly?
Consider a flat rectangular, parallel, symmetrical stripline: current concentrates at the edges, so some curvature is desired. A parallel round line construction is also not ideal, because the back sides of each conductor carry essentially no current; clearly, a circular shape is also not ideal.
Two possibilities: one, is there a single family of curves (e.g., hyperbolic, maybe a Bessel function?) which satisfies the constraint of uniform current density? Or is there a general purpose approach one can use (approximately or analytically) to, say, curl the edges of an otherwise flat profile (i.e., piecewise) to minimize current density?
I expect there is a family of curves, regardless of what shapes apply. Intuitively, an infinite parallel plate has uniform current density: it has no edges upon which current can concentrate. Such a surface has infinnitessimal curvature, a special case of whatever curves are the solution. Probably depending upon the parameter of center-to-inner-peak distance (or if a parabola or hyperbola, the focus), the curve will tend to be flatter or more pronounced.
Whatever the profile, I'm curious what the edges should look like. Presumably, straight rectangular cuts won't be a very good solution, but intuitively, it's also not necessary to taper the edges off to infinity -- once more than a good distance away from the opposite conductor, current density will simply drop off. For all cases, there must be some finite cutoff point, which means a well-defined geometry for a finite shape.
Tim
-- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
Stripline is a flat conductor between two ground planes. Is that the geometry we're talking about?
Two wide, flat conductors separated by a dielectric is more like microstrip.
cccccccccccccccccccccccccccccccccccccccc copper dddddddddddddddddddddddddddddddddddddddd dielectric cccccccccccccccccccccccccccccccccccccccc copper
current
Why would current concentrate at the edges? All of the dielectric has got to be charged.
Transmission lines like this (copper foil, thin dielectric) are commonly used to drive low impedance laser diodes.
-- 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
It doesn't have a good unique identifier. "Parallel plane transmission line"? "Broadside differential stripline"?
It's most certainly not microstrip, which has an infinite ground plane. Sure, when w >> h, all these planar TLs reduce to the same thing, but that's not what we're interested in, we want the dirty fringe effects too.
On a related subject, coplanar stripline is handy. Take an 0.062" thick, two layer board and regular microstrip takes a whopping 120 mils width to achieve 50 ohms (and you can't put anything relatively near it if you need to guarantee impedance and minimize crosstalk). Bring ground pours up within 10 mil of that trace and its impedance just dropped to 38 ohms; the
50 ohm trace width is fully halved. Handy to know when scratching away on copper clad breadboards. Obviously, not as handy around multilayer boards, when your ground plane is so close, microstrip of nominal impedance is just barely within design rules.Dielectric doesn't matter, it's induction from the fringing magnetic field.
We use them a lot. The bigger machines have >6" wide buses made out of 1/4" copper plates and teflon slabs.
Tim
-- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
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Read that wrong! I thought that said a 10 Awg wire.unless a pair meant one down one back.
Oh well, I analyzed comparison of a single 10 Awg wire to replacing with nearly the same =3D 8 18Awg wires, BUT with the return interdigitated so here are the results, you'll have to put back in the tabs and spaces.
16 inch long 10 Awg wire carrying 700 Areplace in a 4 x 4 pattern of +/- wires 8 wires of 18 Awg
Single wire 10 Awg wire assume 100 mil diameter with 50 mil coating of insulation
FREE SPACE L(in nH) R(in milliohms) DC 289 1.38
100kHz 275 4.57 1MHz 270 13.8 10MHz 269 46.2 8 and 8 =3D 16 conductors of 18 Awg wire: 40 mil diameter with 20 mil coating of insulation 4 x 4 in- + - +
- + - +
to obtain: FREE SPACE L(in nH) R(in milliohm) DC 11.5 1.15
100kHz 10.7 1.62 1MHz 9.32 4.6 10MHz 8.84 14.8
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Did the analysis again using two 10 Awg wires, one supply, one return. The insulation is thick, but maybe the wires don't get too close anyway. You can see the effect of the conduction 'bunching' due to being near the other wire, and how the Reff is larger than that for a single wire in free space where the conduction can uniformly distribute around the surface.
should add the following entry:
Double wire 10 Awg wire - one supply, one return assume 100 mil diameter with 50 mil coating of insulation
FREE SPACE L(in nH) R(in milliohms) DC 133 1.38
100kHz 115 5.15 1MHz 109 15.9 10MHz 107 53.4700A --
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Cool, a cross between litz wire and twisted pairs. Tim may have been on the right track from the git go.
?-)
If you apply a microstrip solver, you'll get about the right impedance. Extending the ground plane won't change things much.
That's usually referred to as "coplanar waveguide." Coplanar stripline would be a trace between two ground planes, with additional grounds on its own inner layer.
Then the dielectric in the middle of the trace doesn't get charged?
-- 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
And if you use half the thickness in the microstrip solver and then double the result, you'll get even closer.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
At those frequncies you don't need more metal, and it has the advantages of being light weight. Lower shipping costs, easier to install, and if wind destroys it, the shrapnel does a hell of a lot less damage.
-- You can't have a sense of humor, if you have no sense.
One reason for using relatively large hollow tubes instead of thinner rods (with sufficient mechanical strength) is the large bandwidth required by a TV antenna, with up to 1:1,5 frequency range.
A thin rod would be good at one frequency only, dropping off quite rapidly at other frequencies. An element with a larger diameter works reasonably well at a much wider frequency range.
On HF, gage dipoles are sometimes used in order to increase the effective element diameter in order to get a large bandwidth. In this constructions, several wires are used in parallel, separated by round spaces, creating an effective element diameter of 1 m or more.
Even as low as 54MHz the skin depth in aluminum is around 0.5 mils [12 microns]
That's not a lot of cross sectional area UNLESS there's some diameter.
Any pics?
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