skin effect and 'bad' plating/coating - Page 1
Forum section: electronics / rf-microwave

author: coppercone2
author_rank: Super Contributor
timestamp: October 02, 2019, 01:42:44 am
content: What happens to the skin effect if you coat a highly conductive material with a significant layer of poorly conductive material?like if you had a bare copper coil and electroplated it with some kind of 'garbage' metal that has very poor resistivity? or a semiconductor deposition on something (lets just use pure silver as the core so everything seems extreme)?starting with say.. copper on top of silver and moving towards dielectrics on a spectrum with various thicknesses

author: radiolistener
author_rank: Super Contributor
timestamp: October 02, 2019, 04:32:37 am
content: the resistance will be higher and it leads to higher loss. The coil Q-factor will be reduced.There is even no need for contamination, you can just scrape wire surface and it will have more loss on high frequency due to rough surface.If you place dielectric over wire surface it will leads to slower wave propagation speed in the wire and additional loss in the dielectric material. Velocity factor and dielectric loss will depends on dielectric material properties.Change velocity factor due to dielectric coating leads to change coil inductance. Dielectric loss leads to higher resistance and Q-factor reduction.

author: radiolistener
author_rank: Super Contributor
timestamp: October 02, 2019, 04:50:06 am
content: Quote from: blueskull on October 02, 2019, 04:44:20 amthey are plated with good conductance material, for example silver. Use silver plated wire is more cheaper than use silver wire. Since the main part of HF current flows through skin surface, it flows through thin layer of good conductance metal and it reduces losses.If it is plated with low conductance material, it will leads to high loss. Some types of cables are designed to have high loss. For example, such cables are used for high impedance probe for oscilloscope. It needs high loss in order to attenuate standing waves in the cable.

author: coppercone2
author_rank: Super Contributor
timestamp: October 02, 2019, 12:37:35 pm
content: I thought they bent oscilloscope probes into a zig-zag

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 02, 2019, 01:55:27 pm
content: Skin depth rises with resistivity, so at some point, a plating of poor conductivity is just nonexistent, effectively, shorted out by the base metal.This is convenient as it allows us to insulate conductors, and carry EM fields in the spaces between them.Relevant examples include tarnish: Cu2O and Ag2S are common compounds, and both are semiconductors. People repeat various supposed properties of these, but no one actually measures the Q of a resonator so affected, so what good are those statements really?Tim

author: radiolistener
author_rank: Super Contributor
timestamp: October 02, 2019, 04:58:12 pm
content: nowadays electronics needs wide bandwidth. High Q is bad for wide bandwidth, because LC with high Q has very sharp resonance.Q = 1 / BW.So, there is needs some losses and low Q in order to achieve wide bandwidth. We're needs to sacrifice Q-factor to get a wide bandwidth.The problem here is to keep acceptable VSWR for required bandwidth:VSWR = (Q^2 * BW^2 + SQRT( Q^2 * BW^2 * (Q^2 * BW^2 + 4) ) + 2) / 2As we can see, the only way to keep low VSWR is to reduce bandwidth or Q-factor, there is no other way...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 02, 2019, 09:27:25 pm
content: Well, that's confusing different ways to measure Q. There is the component Q, which relates to efficiency and maximum system Q; and there is system Q, the component relative to the circuit impedance it is a part of.Higher Q components are always preferable. System Q is driven by filter design.There is good reason why cell transmitters have such high Q filters: they need high efficiency to save on transmitter cost, they need low losses to save on power dissipation and electricity cost, and they need modest pass bandwidth with sharp cutoff.Tim

author: radiolistener
author_rank: Super Contributor
timestamp: October 03, 2019, 01:03:40 am
content: Quote from: T3sl4co1l on October 02, 2019, 09:27:25 pmyes, but higher system Q-factor leads to narrow bandwidth. So, the maximum Q-factor is limited by required bandwidth. Also, there is thermal noise limitation. Because high Q-factor leads to higher thermal noise level.For the system which needs very wide bandwidth high Q is bad and if you use high Q components you will need to add resistors in order to make losses.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: October 03, 2019, 11:29:45 am
content: Quote from: radiolistener on October 03, 2019, 01:03:40 amYou really shouldn't add resistors to make your stuff wider band. You can reduce Q with lossless components.When you design a filter, you want your resonators to be as high-Q as possible - lower component Q will lead to higher insertion loss. Filter Q (which is honestly not something I've ever used outside of single-resonator-notch filters) is set by designing all of your different resonators to spread out their notches. This does not add losses - it just widens the bands.Sure, at lower frequencies, where gain is easy, and power is cheap, you often see resistors used to widen up filter bands. But once you get into the MHz or higher, you really don't want to be throwing away your power like that in many situations - it hurts NF. (not to mention that resistors are a pita to do at high frequencies - The reason we use those 'voodoo-rf-black-magic' filters, at least in my field, has little to do with them being better, and more to do with the fact that lumped components are just a pita. You can emulate inductors and capacitors with transmission-line stuff, but you can't emulate resitors).One application where you do need resistance is for oscillation: You cannot get rid of oscillation with only lossless components. So you have two options - push the resonant pole to a point where you have no gain (and then your transistors become your lossy component) or add loss in the feedback-path, if possible.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 03, 2019, 03:22:31 pm
content: Quote from: radiolistener on October 03, 2019, 01:03:40 amYou start by seemingly accepting the distinction: you specify "system Q" at first.Then you move to ambiguous terms. Presumably, "maximum Q-factor is limited by required bandwidth" also refers to system Q, give or take the particulars of a given filter design. "Maximum" is erroneous: the system-Q of each resonator will be higher or lower as needed for the filter type; thegeometric averagewill always be conserved, and equal to the bandwidth.Thermal noise, is also the fraction of resistance in the filter that contributes to loss: component Q. Component Q can be infinite (lossless elements), while system Q is finite (and freely adjustable), giving a noise figure of 0dB. (This should be correct; anyone who's a better study of noise, please correct me.)Finally, you say a high Q factor gives a high noise level, but do not specify which Q. If it were component Q, it would be the opposite relation. If it were system Q, it wouldn't be quite true -- lossless components give the same noise factor independent of system Q. Finite-Q components however give more noise factor as system Q approaches component Q (the maximum system Q).I hope this helps clarify things.Tim

author: radiolistener
author_rank: Super Contributor
timestamp: October 03, 2019, 04:06:19 pm
content: Quote from: T3sl4co1l on October 03, 2019, 03:22:31 pmFor example resonant impedance of LC circuit depends on Q:Z = 2 * pi * f * L * QThe thermal noise isEnoise = sqrt( 4 * k * T * R * df ) we can approximate it with simple:Enoise ~ sqrt( R * df) / 8 So, the thermal noise of LC circuit will be:Enoise ~ sqrt( Z * df ) / 8As we can see, Z depends on Q, and higher Q will leads to higher thermal noise.The source of thermal noise is random electrons movements in the inductor wire.

author: PartialDischarge
author_rank: Super Contributor
timestamp: October 03, 2019, 04:36:36 pm
content: Quote from: radiolistener on October 03, 2019, 04:06:19 pmYou are somewhat drowning yourself in formulas while getting confusedQuoteNice trick, but now filter thatwhite noiseby the sharper filter response of the LC (higher Q remember?) and you will get the same rms noise, simulate it and you will see.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 03, 2019, 04:39:33 pm
content: You're just substituting Z for R, without justification. This is meaningless...Tim

author: radiolistener
author_rank: Super Contributor
timestamp: October 03, 2019, 04:53:13 pm
content: Quote from: MasterTech on October 03, 2019, 04:36:36 pmHm... that's true, I forgot to take into account that increasing Q will leads to decreasing df... Needs to check

author: radiolistener
author_rank: Super Contributor
timestamp: October 03, 2019, 05:15:04 pm
content: Quote from: T3sl4co1l on October 03, 2019, 04:39:33 pmNo, that is correct approach. I found such approach in some literature.LC circuit has zero reactance at resonant frequency, so there is active only resistance and you can replace R with Z.There is another issue, since we increase Q, it will leads to decrease bandwidth. So, df will be smaller. And it seems that MasterTech are correct, the RMS noise within bandwidth will remains the same...Since Q = 1 / BW, increasing Q by 2 leads to increase Z by 2. But at the same time it leads to decrease BW by 2, as result we have the same thermal noise:Enoise ~ sqrt( (2*Z) * (df/2) ) = sqrt( Z * df )Also it means that we cannot reduce thermal noise level with LC filter. Noise reduction due to the filter narrow bandwidth will be eliminated with noise gain due to higher Q