Microwave TL coax impedance - Page 1
Forum section: electronics / rf-microwave

author: nix85
author_rank: Regular Contributor
timestamp: June 09, 2019, 03:58:44 pm
content: I been researching how to impedance match microwave antenna and i am not sure about (characteristic) impedance of coax TL. Namely, we all know CI is not length dependent as all units determining it are per unit length and thus cancel out, but what about frequency?Yesterday i stumbled upon this article from IetLabs and they clearly state:Quotehttps://www.ietlabs.com/pdf/application_notes/5-Characteristic%20Cable%20Impedance-Digibridge.pdfWhat does this mean, that coax at 100KHz+ behaves purely like a resistor? If so is this because reactance of the coax is mainly due to capacitive reactance which drops with frequency?In practical example, if we use coax TL to transmit let's say 1.2GHz signal to the antenna, what impedance will the signal source see at the coax input?Few general reactance/resonance formulas.XL= 2πfLXC= 1/2πfCZ = sqrt(R² + (Xc - Xl)²)F = 1/6.28(LC)F = 1/2π√LC

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 09, 2019, 04:12:26 pm
content: Zo is the terminal impedance of an infinite transmission line, or of an infinite equivalent, i.e., terminated into the conjugate impedance. It would indeed look like a resistor in this case, or at least, even if it doesn't turn your applied power into heat, it's power you're never getting back out (of the infinite TL).For other cases, Zo is the impedance used to transform the load and source impedances according to the relevant relations. You can use a 2-port transmission (ABCD) matrix and trig identities (for the AC steady-state characteristics), or the equivalent simplified formula,https://en.wikipedia.org/wiki/Transmission_line#Input_impedance_of_transmission_lineor if you require time domain, you can use a ladder diagram to analyze the interference of wavefronts, or...Tim

author: nix85
author_rank: Regular Contributor
timestamp: June 09, 2019, 04:20:24 pm
content: Ok thnx, but can you or someone say from experience, what would be the APROXIMATE impedance of coax at 1.2GHz.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 09, 2019, 04:33:38 pm
content: "Of coax"?You'll have to fill in all the missing variables, mentioned above.ImpedanceLengthAnything else particular (losses, dispersion, leakage)Source impedanceLoad impedanceTim

author: nix85
author_rank: Regular Contributor
timestamp: June 09, 2019, 04:41:45 pm
content: Impedance - let's say standard 75OhmLength - let's say 1.5m (multiple of 25 cm wavelength)Anything else particular (losses, dispersion, leakage) - unknownSource impedance - let's say standard 50OhmLoad impedance - helical antenna so let's say standard 14OHmI am sorry but i have no more detail, i dont expect an exact answer anyway, i just want to hear from someone with experience of matching impedance at particular frequency to share their experience.

author: vk6zgo
author_rank: Super Contributor
timestamp: June 09, 2019, 04:51:08 pm
content: Quote from: nix85 on June 09, 2019, 03:58:44 pmCoaxial, or any other transmission line, may be analysed as a network of series inductance & parallel capacitance.wikipedia has a fairly good explanation:-https://en.m.wikipedia.org/wiki/Characteristic_impedanceIn the real world, however, characteric impedance is calculated using physical dimensions of the cable such as conductor radius, spacing, etc.If the transmission line was infinitely long it would always appear as purely resistive at its characteristic impedance.For instance, an infinitely long 50 Ohm CI coax cable could be replaced by a perfect resistor of that value.Unfortunately, we can't make infinitely long cables, nor would we have any use for them, so we must use real lengths of cables.if we now place our perfect 50 Ohm resistor scross the far end of our practical cable, it "looks like" an infinitely long cable at the input end.We can't make perfect 50 Ohm resistors, be we can come pretty damn close!If the 50 Ohm cable is now terminated in its Characteristic Impedance, the input will look like 50 Ohms resistive to any source, regardless of frequency, up to the design limits of the cable & the termination.Terminate the cable in any other impdance, or a short or open circuit, the input impedance to that cable will no longer be independent of the applied frequency

author: nix85
author_rank: Regular Contributor
timestamp: June 09, 2019, 05:56:40 pm
content: vk6zgo, thank you but none of that is new to me and does not answer my question regarding FREQUENCY. I'm already subbed to Stan's channel and own his book and have read the wiki article on CI.I am asking how does CI vary relative to frequency, to quote IetLabs again, they don't mention an imaginary infinite line, it seems they are talking about real finite length TL."Although it can be represented in terms of inductors, capacitors and resistors, characteristic impedance is a complex number that is highly dependent on the frequency of the applied signal. Zo is not a function of the cable length. At high frequencies (> 100kHz), the characteristic impedance is almost purely resistive. At mid-range frequencies (1kHz), Zo is affected by capacitance (ωC) and at low frequencies (DC – 100Hz), Zo is influenced by conductance (G). Refer to Figure 2."

author: rfeecs
author_rank: Frequent Contributor
timestamp: June 09, 2019, 06:25:54 pm
content: The IET labs article is maybe badly worded at best. Figure 2 seems like it might be complete bullshit:I believe the article is explaining how they recommend to extract the line impedance by making measurements at very low frequencies with an LCR meter. They are saying that the impedance goes way off when you go to low frequencies. I don't think this is really the case. It is just based on the crude lumped model they are using for this measurement. I would disregard this article for your question of impedance matching at high frequencies.

author: Kalvin
author_rank: Super Contributor
timestamp: June 09, 2019, 06:54:41 pm
content: If you have a 50 ohm coax which is specified for 1.2GHz operation and which is terminated by 50 ohm resistor, the signal generator (transmitter) will see 50 ohm resistive impedance. If you want to double check, just measure the impedance using a VNA or spectrum analyzer+tracking generator+rf bridge.

author: nix85
author_rank: Regular Contributor
timestamp: June 09, 2019, 06:58:21 pm
content: That was my first reaction when seeing those formulas, but don't rush to conclusions. This is a professional company specialized in such matters, i would NOT call BS on them.https://www.ietlabs.com/

author: nix85
author_rank: Regular Contributor
timestamp: June 09, 2019, 07:10:01 pm
content: Quote from: Kalvin on June 09, 2019, 06:54:41 pmYea, and to be safe i'll make cable multiple of half wavelength long (taking into account velocity factor) so CI DOES NOT REALLY MATTER, since load impedance repeats at half wavelengths down the TL.

author: nix85
author_rank: Regular Contributor
timestamp: June 09, 2019, 09:37:08 pm
content: RoGeorge Double check your assumptions because i know what CI is, how it's calculated, why 50 Ohm compromise between 36 and 73 ohm, why impedance match is important, reflections, standing wave ratios, what happens when you got mismatch, both cases, i know what reactance is, what's complex impedance, real + imaginary (-capacitive, +inductive)...characterstic AKA surge impedance is closely related to complex impedance, in fact they are the same if there are no reflections. For long i understood this as a general principle comparable to water flowing through a tube, it is of course desirable that diameter of the tube remains constant cause otherwise ratio of pressure (voltage) and volume (amps) changes, and when they fall out of phase we get reflections that is energy losses...etc etc.As for IET labs article, don't dimiss their article just because you are not sure what it's about. I am not 100% sure, but looking at the Complex Equation for CI, it seems frequency does not cancel out.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 10, 2019, 01:32:06 am
content: The formulas (corrected for formatting, it would seem?) are approximations that only hold in certain regimes, and not on the transitions between them.For example, for very long lines (many km, as a telephone operator or power distributor might be concerned with), resistance and capacitance tend to dominate, and the inductance isn't so much of a problem (the electrical length may be less than a wavelength). It turns out, for normal twisted pair at telephone frequencies (500-3000Hz), 600 ohms is a closer match; preamps and frequency and phase compensation networks are needed.This is not relevant at 1.2GHz where 1.5m of cable has many wavelengths of electrical length, and normal high frequency losses apply.Tim

author: vk6zgo
author_rank: Super Contributor
timestamp: June 10, 2019, 01:34:34 am
content: Quote from: rfeecs on June 09, 2019, 06:25:54 pmI would agree.At the frequencies they used, the short will look like zero Ohms, & the O/C like infinity.Both capacitance & inductance are going to be so small at low frequencies for practical lengths of cable that they can be ignored.It seems that IET make LCR bridges---period!Such devices are not the instrument of choice for measuring the Zo of transmission lines.More effective methods are a simple RF signal generator used with an Oscilloscope & a variable resistance termination, a Time domain reflectometer, used with the same termination, or to be a bit more upmarket, a VNA.Perhaps it may be a good idea to look for application notes from HP, Rohde & Schwarz, or even venerable old General Radio.In an ideal world, Zo does not vary with frequency, but real cables will differ from ideal as they approach their maximum useable frequency, as losses will accumulate, & affect the measurable Zo.( those series resistances & parallel conductances we ignored at lower frequencies come back into the picture).

author: vk6zgo
author_rank: Super Contributor
timestamp: June 10, 2019, 02:01:41 am
content: Quote from: nix85 on June 09, 2019, 07:10:01 pmSorry, but that's not quite so!All is rosy at the input end of the cable, except that, using your figures, you still have a matching problem between 50 Ohms & 14 Ohms.At every other point of the cable, apart from the halfwave points, the impedance looking towards the load is different than 14 Ohms.For instance, at 1/4 wavelength back from the termination, the impedance is 401 Ohms, again using your figures.You are still transferring the same power, so at that point, the line current will be low, and the voltage across the line, high.This can, at high power levels cause dielectric breakdown, damaging the cable, as well as being a source of losses.That is for a point where the impedance is purely resistive.At other points along the cable the impedance may be highly reactive, causing large variations in both voltage & current.High currents introduce greater I^2R losses, & as above, high voltages can cause losses or perhaps, break down.For these reasons, the simple use of half wavelength transmission line is not widely recommended, especially using normal flexible coaxial cable.Hams often "get away with it" by using Open Wire Parallel line, as its air dielectric, wide spacing, & low series resistance, allow the user to ignore to a large extent, the problems referred to above.

author: nix85
author_rank: Regular Contributor
timestamp: June 10, 2019, 03:41:02 am
content: Quote from: vk6zgo on June 10, 2019, 02:01:41 amYou mean 140 Ohm. And so what? I mean that's the point to terminate it exactly at halfwave point to get that load impedance repeated. Are you saying other factors will make those points not exactly where one would expect them to be so this kind of matching is not practical or?

author: nix85
author_rank: Regular Contributor
timestamp: June 10, 2019, 04:44:06 am
content: https://postimg.cc/PC5ByKLTIs this even correct, i never seen CI calculated this way. How do you even measure impedance of open circuit?

author: rfeecs
author_rank: Frequent Contributor
timestamp: June 10, 2019, 05:32:49 am
content: This seems to come out of Figure 2:Start with the "Complex Equation":This comes right out of the expression for Z0 from the telegrapher's equations:https://en.wikipedia.org/wiki/Characteristic_impedance#Transmission_line_modelThey are saying this is equivalent to the "Resistive Measurement" from Fig 2:So for a short length of transmission line (no distributed effects, much shorter than a wavelength)ZSC≈ R+jωLZOC≈ 1 / (G+jωC)It does make some sense that this BS looking equation would be approximately true for low frequencies.

author: vk6zgo
author_rank: Super Contributor
timestamp: June 10, 2019, 08:35:02 am
content: Quote from: nix85 on June 10, 2019, 03:41:02 amYou, in fact, wrote 14OHm, which it would be reasonable to read as 14The half wavelength points will be as you assumed, & the load impedance will appear at the cable input.If you read the rest of my posting, you would have realised that, at various points, along the transmission line, the relationships between current through, & voltage across the cable will not be the same as at its input or output.Your antenna & Transmitter may be happy, although you still need to match 50to 140, but remember, the transmission line exists at all these other lengths in between.At some point, the current will be high, causing I^2R losses, & at others, the voltage will be high, with dielectric losses & the risk of breakdown.,Losses in a transmission line are specified when correctly matched.In other situations, all bets are off!You seem to be ready to argue with every point brought up in response to your OP.If you feel you already "know it all", why post the query in the first place?

author: radiolistener
author_rank: Super Contributor
timestamp: June 10, 2019, 10:54:33 am
content: You can measure your coax cable capacitance (with open second cable end) and it's inductance (with short circuit on the second cable end), then you can use this formula to get the cable impedance :Z = sqrt( L / C )also you can calculate wave propagation speed of your cable by using this formula [m/s]:NOTE: for this formula you're need to divide measured L and C values by cable length in order to get distributed L and distributed C.v = 1 / sqrt( L * C )and then calculate velocity factor of your cable by using this formula:vf = v / 299792458L should be in [H/m] and C should be in [F/m].You can transform these formulas for L in [uH/m] and C in [pF/m]:Z = 1000 * sqrt( L / C )v = 1000000000 / sqrt( L * C )But note, you're needs to measure L and C for your cable at the working frequency, because coax cable impedance depends on the frequency. Coax cable has too high impedance deviation at very low frequency (smaller than 500 kHz). So, you're needs to use LC meter with working frequency for about 1 MHz or above that. Usually coax cable has stable impedance from specification at 5-10 MHz and more. At low frequency coax cable has too high impedance deviation which depends on the exact frequency value.Also it will be hard to measure too short piece of coax cable, because it has too small capacitance (just some pF) and too small inductance (just some nH) and your your probes will affect your measurements. For long cables (several meters and more) this measurement method works very well.PS: in the same way you can calculate impedance and wave propagation speed of any environment by measuring distributed impedance and distributed capacitance of that environment.For example you can calculate impedance and wave propagation speed in the free space vacuum by using it's distributed inductance (also known as magnetic constant or vacuum permeability μ0) and distributed capacitance (also known as electric constant or vacuum permittivity ε0):L = μ0 = 4 * pi * 1e-7 [H/m]C = ε0 = 8.854187812813e-12 [F/m]Z = sqrt( 4 * pi * 1e-7 / 8.854187812813e-12 ) = 376.7303136 v = 1 / ( 4 * pi * 1e-7 * 8.854187812813e-12 ) = 299792458 [m/s]these values known as free space impedance (377 Ohm) and speed of light (299792458 m/s)

author: nix85
author_rank: Regular Contributor
timestamp: June 10, 2019, 04:59:26 pm
content: Quote from: vk6zgo on June 10, 2019, 08:35:02 amFirst of all, i have no intention to argue nor i feel "know it all", i came here FOR ANSWER. It's just that people often answer without thinking through or really knowing.Again, helical antenna radiation resistance at resonance is ~140 Ohm, i don't know where you saw 14."If you read the rest of my posting, you would have realised that, at various points, along the transmission line, the relationships between current through, & voltage across the cable will not be the same as at its input or output."You keep bringing that irrelevant and obvious fact up although it is assumed cable is rated for given current/voltage.I did not say i will just make cable multiple of halfwavelength, i said i will do it as additional measure to impedance matching.

author: nix85
author_rank: Regular Contributor
timestamp: June 10, 2019, 05:16:17 pm
content: Quote from: radiolistener on June 10, 2019, 10:54:33 amYea, i was considering that but i also thought these values should be known by the cable manufacturer. I'm planning to buy this LCR meterhttps://www.ebay.com/sch/i.html?_from=R40&_nkw=Peak+LCR45&_sacat=0&LH_TitleDesc=0&_sop=15(for some reason link above starts listing with $156.66 unit altho original link displays cheaper ones starting at 61$ + 17$ shipping)but i think it can only measure up to 200KHz, far from 5-10MHz. And LCRs that can measure that high are quite expensive, will consider those too tho.And tnx for additional explanations of free space impedance and speed of light based on LC values of free space.

author: rfeecs
author_rank: Frequent Contributor
timestamp: June 10, 2019, 05:42:54 pm
content: Quote from: rfeecs on June 09, 2019, 06:25:54 pmUh, yeah I was wrong about that. But you do have to typically go down below about 1 MHz before you see this effect:"Transmission Lines at Audio Frequencies, and a Bit of History"http://audiosystemsgroup.com/TransLines-LowFreq.pdfNot a concern for matching an antenna at 1.2GHz.Not really a concern unless you are dealing with miles long cables at audio frequencies.Section 5.9 of Keysight Impedance Measurement Handbook shows the exact same method as EIT Labs for measuring cable impedance with an impedance analyzer:https://literature.cdn.keysight.com/litweb/pdf/5950-3000.pdf

author: radiolistener
author_rank: Super Contributor
timestamp: June 10, 2019, 05:53:21 pm
content: Quote from: nix85 on June 10, 2019, 05:16:17 pmit's too expensive and has too small working frequency. And doesn't have way to connect picofarad capacitors and nanohenry inductors with no wires (to avoid influence).I recommend you to buy vector antenna anlyzer:https://www.aliexpress.com/item/Lusya-4-3-inch-LCD-Mini600-HF-VHF-UHF-Antenna-Analyzer-0-1-600MHz-SWR-Meter/32912877741.htmlit will allow you to measure L, C, SWR, complex impedance (R,X), S11 and smith chart at 0.5 - 450 MHz range.Also it will help you to check or tune your antenna.Also you can use it as TDR reflectometer to check RF cables.And you can use it as a simple RF generator for 0.5 - 450 MHz.With custom firmware you can also use it as spectrum analyzer, frequency meter, crystal Q-factor tester and otherIf you want just a cheap and simple LC meter, you can buy this one:https://www.aliexpress.com/item/Digital-LCD-Capacitance-meter-inductance-table-TESTER-LC-Meter-Frequency-1pF-100mF-1uH-100H-LC100-A/32829227933.htmlit works at about 700 kHz

author: radiolistener
author_rank: Super Contributor
timestamp: June 10, 2019, 06:10:59 pm
content: Quote from: rfeecs on June 09, 2019, 06:25:54 pmit really goes way off at low frequency. You can connect a piece of coax cable to a vector analyzer and make sure that this is a truth.here is example how 10 meters of Chinese RG316 cable behave at low frequency.Unfortunately I cannot show you frequencies below 500 kHz, but I know that it goes to even higher impedance deviation.

author: nix85
author_rank: Regular Contributor
timestamp: June 10, 2019, 08:44:11 pm
content: @radiolistener Tnx, White Croat brother.

author: nix85
author_rank: Regular Contributor
timestamp: June 10, 2019, 11:14:01 pm
content: I got another question about coax cable common mode.https://en.wikipedia.org/wiki/Coaxial_cable#Common_mode_current_and_radiationI'd like to know why..QuoteI mean if shield is the return path, in it current flows opposite to inner conductor, why does this unwanted electric field formed between shield and antenna flow in the SAME direction as inner conductor?

author: nix85
author_rank: Regular Contributor
timestamp: June 10, 2019, 11:25:20 pm
content: Also, does current alternate on the shield or not, according to wikipedia, it does and inner and outer fields cancel out just like in ladder line.From coax wiki page above:QuoteHowever this guy (and others) says there is no oscillation on the shield.https://youtu.be/GMeOMwf2DJU?t=120Obviously this is contradictory and confusing, may someone clear this up.Or does it simply depend on how coax is connected, that is, is shield grounded or is it too connected to the oscillator circuit?

author: vk6zgo
author_rank: Super Contributor
timestamp: June 10, 2019, 11:41:55 pm
content: Quote from: nix85 on June 10, 2019, 04:59:26 pmDo whatever you like, I give up!

author: nix85
author_rank: Regular Contributor
timestamp: June 10, 2019, 11:45:51 pm
content: Quote from: vk6zgo on June 10, 2019, 11:41:55 pmAgain, you try to make me look as if i am arguing with you while everyone sees i am respectful and just trying to learn.

author: ejeffrey
author_rank: Super Contributor
timestamp: June 11, 2019, 03:15:42 am
content: Quote from: nix85 on June 10, 2019, 11:25:20 pmI think you are misinterpreting that video. Or possibly the video author intends your interpretation, but in that case it is wrong.Coax absolutely has oscillating charge and current waveforms on both the center conductor and the shield. The point is that those currents and charges, combined with the currents and charges on the center conductor result in zero field outside the shield and therefore zero voltage on the shield.In the balanced line configuration the voltage on the lines is equal and opposite. While the fields fall off rapidly with distance, they are not exactly zero.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 11, 2019, 12:08:57 pm
content: The shield is, well, a shield. It supportstwo independentcurrent flows: currents inside the line (the image current of the signal line's current flow), and currents outside the shield (image current carried by surroundings, or free space).(When that current reflects a current in free space, that's also known as an antenna.)I think it makes more sense when described in this way. It is equivalent to talking about directions of current flow, but that may be confusing to the beginner who only sees alternating currents everywhere -- the sign (or more generally, the phase) is easy to get confused about.For further reading:look up the image current, displacement current, path of least impedance, stuff like that.Tim

author: radiolistener
author_rank: Super Contributor
timestamp: June 11, 2019, 12:40:17 pm
content: coax cable has low radiation loss due to skin effect in the braid conductor. High frequency current flows mainly at skin surface. In case of coax cable it means that RF current flows on the inner side of the braid surface. So, RF current cannot go from the inner side of braid to the outer side of braid due to skin effect. This is why bad braid quality leads to higher radiation loss for coax cable.Actually RF energy flows as electromagnetic wave inside insulator (between center conductor and braid). Conductors here just keep this wave inside insulator.The same thing happens with symmetrical transmission line. The RF energy flows as electromagnetic wave inside insulator between two wires. But since there is no shield, the part of this electromagnetic wave flying away. This is why it has radiation loss.

author: nix85
author_rank: Regular Contributor
timestamp: June 11, 2019, 03:18:10 pm
content: I don't think i am misinterpreting the video, nor i think he is referring to cancelation of voltage on the shield, you can hear in the vid he makes it clear there is no current oscillating on the outer shield 'cause it is "just ground".If that is how coax is connected THEN HE IS RIGHT, except for the secondary induced currents due to transformer effect.If however both inner conductor and shield are connected to the balanced source then HE IS NOT RIGHT.So, like i assumed, it depends to which type of source coax is connected.Yea, i know about skin effect, that is, energy flowing around the conductors at higher frequencies as EM wave (poynting vector) as described by Oliver Heaviside.As for image current, i assume that's the 1:1 transformer effect between inner and outer conductor which is secondary effect and not really an answer to my question if shield carries oscillating current fed directly from the signal source.I also know what displacement current is.

author: radiolistener
author_rank: Super Contributor
timestamp: June 11, 2019, 04:21:12 pm
content: Quote from: nix85 on June 11, 2019, 03:18:10 pmShield don't carries oscillating current and center conductor don't carries it.Insulator carries it in the form of EM wavesWhen EM wave falls to a conductor surface, it makes oscillating current and this oscillating current emits back EM wave. So, conductor works like mirror for the light and keeps EM wave inside insulator of the coax cable by reflecting it back when it trying to fly away. In such way EM waves traveling through coax cable in it's insulator. Current oscillations on conductor surface is just an effect of traveling EM wave in the insulatorSo, technically, current oscillations are present on both - on center conductor surface and on inner surface of the shield. But they are carried by EM wave in the insulatorThe signal source just initiating this process by making first current oscillation, which leads to emit EM wave. And then coax cable carries this EM wave in the insulator between conductors. When you consume energy of current oscillations at the end of cable, this energy will not be emitted as EM wave again, because it was consumed by load. And the path of RF energy will ends. If you will consume just a part of energy of these current oscillations, the rest will be emitted as EM wave and will travel back to the source.The average speed of electrons in the conductor is very-very small, they cannot carry current oscillations with a speed of light. EM wave doing it.

author: ejeffrey
author_rank: Super Contributor
timestamp: June 11, 2019, 06:37:08 pm
content: Quote from: nix85 on June 11, 2019, 03:18:10 pmThat is wrong, and I still think you are misinterpreting the video. There is charge and current oscillation on the shield. At the ends of the transmission line those currents flow into the ground connection at either end. The electrons don't care whether they are "driven by the source" or "induced by transformer effect". Those are distinctions that only exist in your mind. That is like saying the neutral wire in your electrical supply doesn't carry current because it is just ground.QuoteIf you already "know" the answers and are going to ignore the answers, please don't bother asking.

author: vk6zgo
author_rank: Super Contributor
timestamp: June 11, 2019, 09:10:58 pm
content: Quote from: ejeffrey on June 11, 2019, 06:37:08 pmI rest my case!

author: nix85
author_rank: Regular Contributor
timestamp: June 11, 2019, 11:15:16 pm
content: Quote from: radiolistener on June 11, 2019, 04:21:12 pmOk, sure it travels outside of conductor due to skin effect, we got itbut you didn't address the difference if both inner conductor and shield are connected to oscillating circuit aka balanced feed, or if just inner conductor is fed oscillating signal and shield is connected only to ground.

author: nix85
author_rank: Regular Contributor
timestamp: June 11, 2019, 11:54:51 pm
content: Quote from: ejeffrey on June 11, 2019, 06:37:08 pmYou are wrong, i am not misinterpreting, listen again to what he says in the vid "all the going back n' forth is being done JUST by center conductor"."The electrons don't care whether they are "driven by the source" or "induced by transformer effect"." Hah, there you go, you didn't get my question at all cause that is exactly what i was pointing to, not if current forms on the shield as a transformer effect but if shield is DIRECTLY driven on not. And the answer is, of course, it can be, but as an unbalanced line it is usually not.QuotePlease, from now on, if you don't really know the answer, refrain from answering.

author: nix85
author_rank: Regular Contributor
timestamp: June 11, 2019, 11:55:49 pm
content: Quote from: vk6zgo on June 11, 2019, 09:10:58 pmYou had no case. Chill.

author: EEVblog
author_rank: Unknown
timestamp: June 12, 2019, 02:27:26 am
content: Quote from: nix85 on June 11, 2019, 11:54:51 pmThat's not how forum's work. You don't get to tell people to stop providing their answers or opinions.You won't make friends here with that sort of approach.

author: EEVblog
author_rank: Unknown
timestamp: June 12, 2019, 02:30:41 am
content: Quote from: nix85 on June 09, 2019, 03:58:44 pmCoax's are designed to act as transmission lines with a constant impedance across a wide bandwidth, that's their job.

author: radiolistener
author_rank: Super Contributor
timestamp: June 12, 2019, 05:51:32 am
content: Quote from: nix85 on June 09, 2019, 03:58:44 pmTechnicaly thats not correct. Coax cable is not resistor, this is transmissionline. Resistor consumes energy and transforms it into heat. Coax cable transfer energy, it doesn't consume it.From source feed point side coax line is really looks like resistor equivalent, but this is just for some limited period of time. When EM wave in coax cable runs to the end of cable and will not be consumed, it will be reflected back and when it runs back to the source the things will be changed. The source will see that moment like coax line input resistance was suddenly changed. The change will depends on the load impedance at the second end of the coax cable.If load impedance on the second end of cax line will be equal to coax line impedance, all energy will be consumed by load and reflection will not happens. So the source will continue to see the same resistor equivalent on the coax cable input. But when the load impedance will not be match with coax line impedance, a part of EM wave will be reflected back and the source will see the change of input impedance at coax cable input. It will depends on frequency and cable length.

author: nix85
author_rank: Regular Contributor
timestamp: June 12, 2019, 06:16:06 am
content: Quote from: radiolistener on June 12, 2019, 05:51:32 amTechnicaly thats not correct : ) cause there will always be losses.

author: nix85
author_rank: Regular Contributor
timestamp: March 04, 2020, 09:29:23 pm
content: Just to add, i did not misinterpret that video, guy is clearly showing no current on the shield and this is of course wrong as everyone here agrees. And i know electrons hardly move, that energy is carried by EM wave, poynting vector, but that is beyond the scope of this thread.Some food for thought... Ask yourself how does flux running through the core of a transformer induce voltage in the secondary altho flux is totally confined in the core and coil is outside of it, no secondary wire cuts a single line of flux.There is much for you to learn, to expand upon.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: March 06, 2020, 03:32:52 pm
content: Quote from: nix85 on March 04, 2020, 09:29:23 pmSome food for thought.... Ask yourself why you feel the need to phrase this as some riddle you solved and the rest of the world hasn't, even though this was all solved in the 1800's.

author: nix85
author_rank: Regular Contributor
timestamp: March 11, 2020, 04:38:44 pm
content: Quote from: TheUnnamedNewbie on March 06, 2020, 03:32:52 pmSince you think it was solved in 1800's, explain how is voltage induced in the secondary altho not a single line of flux cuts a single wire.

author: Kalvin
author_rank: Super Contributor
timestamp: March 11, 2020, 05:29:23 pm
content: Quote from: nix85 on March 11, 2020, 04:38:44 pmThe operation principle of a transformer can be explained using Lentz's law and Faraday's law. So, you may be wrong stating "not a single line of flux cuts a single wire".Like others have already written, you can measure transmission line impedance vs frequency with a network analyzer, for example. At very low frequencies (like below 100 kHz or so), a transmission line has frequency dependent impedance, but at higher frequencies (1 MHz or so) the characteristic impedance remains pretty constant although the attenuation increase with the frequency.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 11, 2020, 07:42:39 pm
content: Lines of flux aren't a terribly great explanatory tool. They lead to confusion with the spinning magnet-disk problem (doesn't matter if the magnet spins, it's radially symmetric -- but you might be tempted to think the lines are bonded to it like a brush), and the transformer just looks weird.Consider the de-energized transformer. Where are the flux lines? There aren't any inside. But there must be flux lines somewhere, if we imagine them to be some kind of conserved physical material. The answer is: they are at infinity. In infinite supply, unreachably far away, so their presence there can't possibly matter otherwise. When the transformer is energized, lines of flux are brought in, from infinity, into the local loop -- thus cutting wires and inducing voltage, as the story goes.Far better to just stick with enclosed area and field intensity...Tim

author: nix85
author_rank: Regular Contributor
timestamp: March 13, 2020, 01:10:18 pm
content: Quote from: Kalvin on March 11, 2020, 05:29:23 pmFalse and false.

author: nix85
author_rank: Regular Contributor
timestamp: March 13, 2020, 01:31:37 pm
content: Quote from: T3sl4co1l on March 11, 2020, 07:42:39 pmThat's all nice and sweet, flux exists "in infintiy" etc, but you did not address the issue. Good that you touched upon the Faraday's disc where induction happens altho there is no relative motion between magnet and copper disc.So, from the father of "wire must cut flux" we have the negation of the same.Let's hear Joseph Henry, discoverer of self-inductance and mutual inductance and what is less known but far more important, two kinds of induction, one weaker which cannot be screened except with iron and another that can be screened by any metal. He concludes suddent breaking of contact produces both kinds of induction and that this mysterious first type is akin to induction produced by motion.This is from Scientific writings of Joseph Henry, Section 11, PDF file pages 178. - 185. >>The book >>https://ia800302.us.archive.org/27/items/scientificwriti03henrgoog/scientificwriti03henrgoog.pdfFeynman on two kinds of inductionAbout ether....read/watch these if you want to gain any deeper understanding in this life, and all these are but basics.http://freenrg.info/Physics/Scalar_Vector_Pot_And_Rick_Andersen/Rick_Andersen_Ortho.htmlhttp://villesresearch.com/ether.htmlEtheric Rainmaking with Trevor James ConstableAnother example...We can see in this videohowflux prefers to loop on itself through the air, rather than go through the core surrounded by aluminum ring. Why would flux in the core care about a ring that is OUTSIDE of a core. This is the whole point.In simple words, induction happens by etheric disturbance, magnetic field of the primary affecting the subtler energy present all around us which is then picked up at the secondary and transformed, that is, slowed down back into magnetic field.Magnetic field is just an effect, ether is the true medium of transmission.

author: ejeffrey
author_rank: Super Contributor
timestamp: March 13, 2020, 10:34:24 pm
content: Transformer induction is described by maxwells equations. The fact that you can't or won't understand the math is irrelevant. Its fine (good!) to try and find better mental pictures and explanations of why or how it works, but if you are claiming that maxwells equations -- in particular - Faraday's law of induction do not work, you are just plain wrong.Also, while maxwells equations were completed in the 19th centry, they were basically defective until the 20th centry as they cannot properly be understood without relativity -- in fact the failure of maxwells equations to maintain a nice form under newtonian coordinate transforms was one of the main motivations for the development of relativity. If you try to reason about electrodynamics in a newtonian mechanics formalism you are going to get confusing or contradictory results.Quoting Henry is also not particularly relevant. It is enlightening to read the writings of the "old masters" to understand how they thought about the world and made their discoveries, but our collective understanding of their discoveries have progressed considerably since then. Most 3 year physics undergrads understand electromagnetism better than faraday or maxwell or henry ever did.

author: KaneTW
author_rank: Frequent Contributor
timestamp: March 14, 2020, 02:52:20 am
content: Can we please not feed the troll?