Coaxial cable impedance - Page 1
Forum section: electronics / beginners

author: npelov
author_rank: Frequent Contributor
timestamp: February 08, 2014, 08:53:55 am
content: I wanted to get/make coaxial calbe(s) to use for different purposes - the one that made me think about it is testing electronics in temperature chamber. Here is my thought process and you'll tell em if I''m wrong:1. for low frequencies it could be just any shielded cable.2. for higher frequencies there should be compensation - because the cable has inductance and capacitance.3. for RF (or really high) frequencies I should switch to 50 Ohm input on the scope and use proper cable and make sure the source is 50 ohm3A. convert the source to 50 ohm using high badwidth opamp. There is AD8021ARZ in local store. I thought I can use it to probe stability of clock oscillators for MCU over temp.I could use the probe itself with a probe socket like Alan suggested:But I'm limited to temp range 0 - 50 deg. C. And I would usually reduce the range to 5-45 deg. C to not be at the limits (Rigol RP3300A probe).So here is my question: How can you say that a cable is 50 ohm (or 75Ohm). What is that impedance. It's clearly not the resistance of the cable because it's quite lower than 50 Ohm. It's not reactance of the cable because it would depend on frequency and it would have different impedance depending on the frequency. So what property of the cable makes it's impedance.

author: nihilism
author_rank: Regular Contributor
timestamp: February 08, 2014, 09:26:17 am
content: http://www.allaboutcircuits.com/vol_2/chpt_14/3.html

author: Marco
author_rank: Super Contributor
timestamp: February 08, 2014, 02:38:41 pm
content: Quote from: npelov on February 08, 2014, 08:53:55 amIs not enough, probe cables are not normal coax cables they have distributed resistance.http://www.dfad.com.au/links/THE%20SECRET%20WORLD%20OF%20PROBES%20OCt09.pdfQuoteA source doesn't have to be 50 Ohm, it just has to have a fixed impedance over the bandwidth you're interested in. It will form a divider with terminated coax. That's how the DIY high impedance coax probes work (they assume a relatively low impedance source) :http://koti.kapsi.fi/~jahonen/Electronics/DIY%201k%20probe/http://www.emcesd.com/1ghzprob.htmQuoteIdeally it doesn't act like reactance at all ... it acts like resistance and yes independently of frequency. It's counter-intuitive I know, that's why they usually make you do the math in college to hammer it in your head.http://en.wikipedia.org/wiki/Characteristic_impedancehttp://en.wikipedia.org/wiki/Telegrapher%27s_equations

author: w2aew
author_rank: Super Contributor
timestamp: February 08, 2014, 03:31:52 pm
content: Quote from: npelov on February 08, 2014, 08:53:55 amYou may want to check out my video on how to measure coax length and impedance - there's a little bit of a practical tutorial on coaxial impedance at the beginning...

author: npelov
author_rank: Frequent Contributor
timestamp: February 08, 2014, 04:51:38 pm
content: Hi, Alan. I watched this video many times. The term "sees 50 ohms" confuses me. I know that at high frequencies the speed of light begins to have effect (I mean relative to the pulse width, because it has effect in all frequencies). So I understand why you "track" the path of electricity, but sometimes it's hard to understand things from the first time. I guess I have to hear it in different ways to understand it. Also experimenting makes it easier to understand. The problem is that I don't have any 50 ohm cable and a oscillator that's capable of taking 50 Ohm load. Most digital buffers don't have that load capability. If I make a divider of 1.8k and 200 ohms will I be able to repeat the experiment on your video.All my experiments lower the voltage. I can see on your video that the voltage stays the same, except when the reflection adds with the original signal and then it makes it twice the original volgate. I guess in my case the voltage drops because the oscillator can't handle the current of 50 Ohms. Well about 100 ohms on the other end of my twisted pair cable (don't have coax yet) makes the form of the signal relatively nice.I checked all the links you folks gave me. I still find it hard to understand how exactly the cable has 50 ohms resistance. I know that I measure it it's a lot lower than 50 Ohms. A wild guess is that this resistance is only valid for the short duration before the reflection comes back. So if I try to measure the resistance of the cable that's long enough for the electricity to travel for few seconds, I 'll see 50 Ohms on my multimeter for those few seconds. Is that right?Also I read a lot of people speaking about dividers. The terminated cable and the generator form a divider. Then why the voltage on Allan's video didn't drop. It was 2V p-p all the time (except when it summed with the reflection). If the pulse generator has 50 Ohms impedance and the terminated cable is 50 Ohms and the signal from generator is 2v p-p then it would read 1v p-p on the scope.Actually I didn't hear Alan saying what voltage was the pulse generator set to.Maybe that's the source of my confusion.

author: npelov
author_rank: Frequent Contributor
timestamp: February 08, 2014, 07:36:07 pm
content: Ok. After some more reading and testing I understood (correct me if I make a mistake) that before the electricity hit the other end and come back the cable acts almost like a resistor. And this resistor together with the internal resistance of the generator make a divider. That's why I don't see the full voltage at the start of the pulse. So if I put a series resistance after the generator to increase it's total equivalent resistance to be equal to the Characteristic impedance of the cable it'll divide the voltage exactly to half. So if I knew the internal resistance of the generator I could calculate the resistance of the cable. Or if I know the impedance of the cable I can calculate the internal resistance of the generator.So I tried to make 1:10 divider to minimize the effect of internal resistance of the generator. When I have 804 ohms the voltage is about 1/10 of the max voltage. So Zo/(R+Zo) = 1/10 and Zo = R/9. So the impedance (or the characteristic impedance) of the twisted pair LAN cable I have is about 90 ohms. Am I close? And because about 30 ohms bring it to 50% then possibly the generator has internal resistance about 60 ohms. So we can make the previous calculation again where R=804+60 = 864 ohms. Divide that to 9 and Zo = 96 ohms. That's why the signal is almost perfect when I terminate it with 100 ohms.

author: ivaylo
author_rank: Frequent Contributor
timestamp: February 09, 2014, 08:46:11 am
content: It's nice to see another Bulgarian in the forumAlthough mechanical, this explanation worked for me the best -Think how wavelenght and the lenght of your transmission line affect all this. Then see how the terminating impedance kills (or not) the reflections.

author: Rerouter
author_rank: Super Contributor
timestamp: February 09, 2014, 09:45:00 am
content: Thank you very much for the video ivaylo

author: Jon86
author_rank: Frequent Contributor
timestamp: February 09, 2014, 10:44:26 am
content: Quote from: ivaylo on February 09, 2014, 08:46:11 amI learnt more from that video than I do in an hour and a half physics lecture, thanks ivaylo!

author: VK5RC
author_rank: Unknown
timestamp: February 09, 2014, 10:53:02 am
content: What a corker of a video, you can think about quarter wave stubs, matching impedance etc for hours.

author: hgg
author_rank: Frequent Contributor
timestamp: February 09, 2014, 01:52:30 pm
content: Exceptional video!

author: ivaylo
author_rank: Frequent Contributor
timestamp: February 09, 2014, 06:40:32 pm
content: @npelov, I read your last post again and those voltage dividers you use won't let you measure characteristic impedance of your transmission line. Z0 is *not* Z=R+jX. But this is what is common between them - both are called impedances because they are defined as V/I (or cause vs. effect as the AT&T guy from the video beautifully calls them). And the most intuitive (if there is anything intuitive about this) practical way I've seen is described by c4757p here -https://www.eevblog.com/forum/beginners/cable-impedance/msg280108/#msg280108(also see his clarifications a few posts under, but see he uses exactly V/I). Now, if you want to match different transmission lines impedances, you again don't use resistive voltage dividers - see the methods described here -http://en.wikipedia.org/wiki/Impedance_matching#Impedance-matching_devices.Say for your CAT5 twisted pair case you may need to use something like this -http://www.computercablestore.com/Monoline_Coax_Balun_Male__PID7446.aspx(providing you are matching to a 50 ohm device/line). Hope this helps, took me a while to wrap my head around it as well....Edit: forgot to mention - to measure any of this accurately you need a proper VNA -http://en.wikipedia.org/wiki/Network_analyzer_(electrical)

author: npelov
author_rank: Frequent Contributor
timestamp: February 10, 2014, 03:28:18 pm
content: What a great video!@Ivaylo What I did isn't that far from c4757p's experiment. The difference is that I did it in 2 measurements - one with and one without the cable under test. Also I didn't have a 50 ohm cable from generator because I soldered the LAN cable directly on the generator output. I guess the measurement is more accurate if I make it with two probes at the same time. But my calculation wasn't too far from the actual impedance - 100 ohms. Another difference is that I didn't short the other end of the cable. I left it open like Alan did in his video and it actually doesn't matter for the first part. The difference will be that when the reflection comes back it'll rise the voltage instead of dropping to zero.However what I added the generator impedance in the equation and I'm not sure that's correct thing to do.

author: npelov
author_rank: Frequent Contributor
timestamp: February 10, 2014, 04:42:51 pm
content: I made another test with the same cable and I attached the screenshots.Yellow is before the resistor (240 ohms), blue is after the resistorPulse amplitude = 5.9VUr = 4.18VIr = 17.416667 mAZ0= Ucable / Ir = (5.9-4.18) / 0.017416667 = 98.76 OhmsThe generator isSN65176B(RS485 transceiver) sending 0x55 at 115.2 kbps. it has a load of 120 ohms. So If I want perfect square wave I should terminate the cable with 100 ohms on both sides. Well 120 ohms did pretty nice job - almost perfect square wave over 7 meters. It was powered from battery so I could measure difference at A and B directly with normal probe.I had so much fun playing with cables that I forgot I had work to do

author: KD0CAC John
author_rank: Frequent Contributor
timestamp: February 10, 2014, 04:56:07 pm
content: I approach this from the perspective of a ham radio operator and lack a lot of training in electronics .Then suffer from having a bad experience in grade school with algebra , so laking the math skills .But on the other hand , a lot of self-taught / or finding sources like Allen's - w2aew youtube vid's and others willing to put info out there .Spent many yrs. in auto repair and was in it before the emissions & computer control came into being , but again a couple class's and yrs. of on the job learning made me one of the best techs in most shops and then my own shop and many failed repairs from dealers made it to my shop , never sent out an unrepaired car [ if the owner was willing to pay for estimate ] in about 25+ yrs. .Just saying I some knowledge / experience , and that what I say is just my perspective and open to correction / calibrationI understand impedance in feedline , in this case coax , as resistance over a frequency range - static vs. dynamic .The different impedances of feedlines is not the issue , except to match generators & loads - so as to not create common mode current , reflections ect. , as an example modern transceivers are built to work with 50 ohm feedline to at min. an antenna that is resonant to a specific frequency - which when tuned correctly , would have a 50 ohm imp.In the case of ham radio where we have VFOs and can tune to a wide selection of frequencies and bands , we need a matching system , the most common of those is a manual tuner with some combination of variable cap.s & inductors - so that the generator will see 50 ohms .It is good practice to use one of the types of baluns [ other opt.s like stub , gama etc. matching ] to transform changing impedance's between differenet feedlines like ladder-line 450 ohm etc. & coax 50 ohm , or antennas have a wide range of impedances .When there is an uncorrected mis-match this can be a source of reflected power / common-mode-current in coax and the paths in coax are - out the center conductor & return the inside of the shield of the coax , then what we do not want , at the point of mismatch we get common mode current on the outside of the coax shield , causing all kinds of issues with equipment , wether it be a transceiver or test equipment .One of the issues I am currently working on is a spectrum analyzer that has a 75 ohm input and the variable attenuatorcalibrated in dbmv , and for my uses , ham radio , I want 50 ohm input and attenuator calibrated in dbm .Any corrections or other perspectives to help my learning process ?