RF question about coax - Page 1
Forum section: electronics / beginners

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 07, 2011, 02:08:37 am
content: I've got a project where I need to send a 2.3 MHz analog signal to an LED via some cable from the main board. I'd like to know if I can use a micro-coax cable for the connection. Let me explain:My current circuit generates the signal and feeds it to an op amp with a transistor to drive the LED. The LED anode connects to Vcc (+5V). The LED cathode connects to the collector of the NPN transistor. The base is driven by the op amp through a 47 ohm resistor. The emitter goes to a 10 ohm resistor, which then goes to ground. The op amp's negative input is fed back from the emitter via a 1K2 resistor.With everything on one board, this all works great. But now I need to move the LED to a separate board about one meter away and connect it to the main board via a cable of some sorts.I know that if I just use a twisted pair, it's going to radiate like an antenna. I don't want broadcast the EM, just the light! (Probably wouldn't pass FCC testing either, which it must do eventually.)I was thinking about using a micro-coax (0.81mm dia) cable. My theory is that I can put Vcc on the shield (which would connect to the LED's anode) and drive the current through the stinger (which would connect to the LED's cathode).Q. Is this a stupid idea?Q. Would using a micro-coax in this way isolate the signal, or would it still radiate like an antenna?Q. Should I, instead, use two micro-coax cables, feeding Vcc through the stinger of one, and the drive current through the stinger of the other, grounding the shield for both, in order to prevent radiating the signal like an antenna?Any help would be appreciated!

author: IanB
author_rank: Super Contributor
timestamp: October 07, 2011, 02:27:22 am
content: I don't know anything about RF design, but you didn't mention shielded twisted pair? Network cables use this quite successfully and they surely operate at high frequencies. Are you overlooking a simple solution?

author: amspire
author_rank: Super Contributor
timestamp: October 07, 2011, 03:33:07 am
content: You can use twisted pair or coax.All your computer networking cable is just twisted pair.If you think about it, the two wires are in proximity, and each is transmitting exactly the opposite RF, so they add up to zero or close enough.Gigabit Ethenet cable uses unshielded twisted pairs transmitting a 125MHz carrier.ADSL2+ sends up to 1.1MHz frequencies over kilometers over a twisted pair.Now at only 2MHz, you will have no problem using either over a distance of a meter, as long as the output can drive the capacitance of the cable.If you use either the coax or twisted pair as a properly terminated transmission line, you will have no capacitance problems (ie if you use a 50 ohm coax, make sure the impedance for AC terminating the cable is 50ohms) but this may not suit your current circuit.Unterminated, if you are feeding the cable with non-sine wave signals, you can get some ringing in the cable around about 100MHz, so maybe a ferrite bead or two to soak up any HF problems.Richard

author: vk6zgo
author_rank: Super Contributor
timestamp: October 07, 2011, 03:44:17 am
content: What are you actually doing?Are you modulating the LED with the 2.3MHz?Or is the LED just to indicate presence of RF?If it the former,yes you can use coax or twisted pair.If it is the latter,just detect the presence of RF at the near end of the cable & use any cable to drive the LED,as it will only be DC.'VK6ZGO

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 07, 2011, 11:57:29 am
content: Quote from: IanB on October 07, 2011, 02:27:22 amI did think about shielded TP, and I might end up doing just that. The techy engineer in me loves to explore all possibilities.(Besides, how often do you get to play around with micro-coax cables? They're so tiny and cool!)Thanks IanB.Quote from: amspire on October 07, 2011, 03:33:07 amYes, that's what I was thinking too. Perhaps I'm just being overly-cautious, but this device will need FCC approval eventually, so I really want to make sure whatever wire/connection I use doesn't get flagged as a EM-radiating antenna.Quote from: amspire on October 07, 2011, 03:33:07 amRight. I hadn't planned on putting anything else besides the LED at the end of the cable. I'll have to think about impedance matching. :/Quote from: amspire on October 07, 2011, 03:33:07 amOh, I should have mentioned that the signal is an FM sine wave. But still, ferrite bead(s) might be a good idea. I'll have to R&D that.Thanks Richard.Quote from: vk6zgo on October 07, 2011, 03:44:17 amI'm sending an FM sine wave via light.Thanks VK6ZGO.

author: deephaven
author_rank: Frequent Contributor
timestamp: October 07, 2011, 12:28:28 pm
content: Twisted pairs only really work properly with true differential signals, otherwise you'll get a lot of radiated emmision. You could put the signal through a transformer to make it differential, but that sounds like an overkill, plus you would still need to factor in the power. The wavelength of a 2.3 MHz signal is pretty long, so a 1 metre long coax cable is going to be a fraction of that wavelength, so impedances don't really come into play. The coax will have some capacitance which will load the signal. It might be better to try a low(er) resistance drive to the coax to overcome this. Connecting the coax outer to V+ is fine, just make sure V+ is decoupled to ground close this connection.

author: amspire
author_rank: Super Contributor
timestamp: October 07, 2011, 01:14:31 pm
content: Deephaven, you are absolutely correct. The signal won't be differential, so the twisted pair is not the right solutionl.Go for coax.

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 07, 2011, 02:24:44 pm
content: maybe you can simulate differential signal using 2 mcu pin, 1 is the signal and another is inverse of it.

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 07, 2011, 11:23:10 pm
content: Quote from: deephaven on October 07, 2011, 12:28:28 pmYes, that's what I am worried about. I don't want the line to become an antenna, no matter how poor of an antenna it may make.Quote from: deephaven on October 07, 2011, 12:28:28 pmBut since that's not a true differential signal, will the coax still isolate the EM?Quote from: Mechatrommer on October 07, 2011, 02:24:44 pmHmm. That gives me an idea. Maybe I can substitute the op amp in the circuit for an op amp with a differential output. The only drawback to that idea is that I would need to add another op amp next to the LED in order to receive the differential signal and drive the LED. So instead of 2 wires, I would need 4 (Vcc, GND, +s, -s) :/

author: gregariz
author_rank: Frequent Contributor
timestamp: October 08, 2011, 12:32:49 am
content: Coax is only an EM transmission line when properly terminated ie 50 ohms for most coax.When it is not terminated with 50 ohms it is no longer a transmission line and currents will flow on the outside of the cable. ie you will get radiation from the cable.When driven by digital logic with a high internal impedance this usually just means you need to place a 50 ohm resistor across the source end of your cables. If you are driving an LED at the other end with a reasonably high impedance drop resistor you should also place a 50 ohm across the terminating end. Under these conditions the cable will behave itself.

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 08, 2011, 12:45:03 am
content: you'll use standard ethernet cable then. i think there's still way if you want only 1 TP. set both line at Vdd/2, the +ve will swing Vdd/2-Vdd, the -ve will swing Vdd/2-0, so the led will still see a swing of Vdd-0. never tried, but i think its worth it.

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 08, 2011, 02:02:45 am
content: But if I use two micro-coax cables, grounding the shield of both, and sending Vcc through one and the drive current through the other, that should prevent EM leakage, right?

author: Kiriakos-GR
author_rank: Super Contributor
timestamp: October 08, 2011, 04:43:03 am
content: I believe that the micro-coax will work, the distance looks sort and probably there will not be any issues.There is only once chance to get in to problems, for example the micro-coax to cause losses to the point that the signal is not strong enough so to activate the transistor.In this case you will have to use something more appropriate, like the RG58 cable.

author: vk6zgo
author_rank: Super Contributor
timestamp: October 08, 2011, 05:15:15 am
content: If the coax isn't terminated,it will still look like a shielded cable.Unterminated shielded cables were used in quite a lot of equipment back in the day.As far as radiation from the shield is concerned,this is more of a problem when a cable of substantial length terminates in an antenna.The cable length at this frequency is 0.0077 wavelength,so,no,I don't envisage a lot of radiation.Of course,as you are feeding a diode with RF,you should check for harmonics on 4.6MHz,6.9MHz,8.4MHz,etc,as well as intermodulation products between 2.1MHz & the local AM radio stations!Seriously,though,I doubt if you will have any problems at all!VK6ZGO

author: jahonen
author_rank: Super Contributor
timestamp: October 08, 2011, 06:00:58 am
content: It might be worthwhile to note that radiated emissions are measured only from 30 MHz up per FCC or CE. Unless you have much harmonics (like high edge rate square wave carrier), don't worry about it.And regarding the coax termination, one can terminate it only at the driving end. From the load perspective, it works just fine and is the most convenient technique in PCB level point-to-point terminations. Another advantage is that it does not cut the signal amplitude into half. The trick is that while the load end is then mismatched, but back-reflection from the driving end is suppressed, so the load does not see them.Regards,Janne

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 09, 2011, 01:35:31 am
content: Quote from: jahonen on October 08, 2011, 06:00:58 amI didn't know that. Thanks!Quote from: jahonen on October 08, 2011, 06:00:58 amLike this? (See pic)

author: amspire
author_rank: Super Contributor
timestamp: October 09, 2011, 01:06:02 pm
content: The new circuit does not have a 50 ohm output as the collector of the transistor has a high impedance output - much greater then 50 ohms.But you are running at 2.3MHz, and reflections will cause ringing at around 100MHz.. That is why I suggested putting ferrite beads on the lines. Hopefully they will not absorb too much of your 2.3MHz, but they will damping any 100MHz ringing.I think that to make a true 50 ohm terminated connection is more effort then your application needs. Your first circuit looks fine.Richard.

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 09, 2011, 02:03:35 pm
content: Quote from: amspire on October 09, 2011, 01:06:02 pm(Forgive my ignorance in this area. I'm still trying to understand how to terminate a coax "at the transmission end".)So you're saying that I don't need R1?Quote from: amspire on October 09, 2011, 01:06:02 pm(The current version has everything on one board. I need to move the LED to a separate board that can be moved around.)So do you think I should just go with shielded twisted pair then?

author: jahonen
author_rank: Super Contributor
timestamp: October 09, 2011, 02:40:17 pm
content: Ah, it certainly helps to see the schematic so what you describe is easier to visualize. I see now that the series resistor is not going to achieve the termination here. Series resistor will only increase this high impedance, so it will not work there. It only works if the source has low output impedance (like a logic gate). The current source has high output impedance so the resistor should be in parallel to the coax, but that will of course steer some of the intended led current through the resistor.Alternatively, you could perhaps put the 50 ohm resistor in series of the led at the end, that might work as an end termination in this case (if we can make a slightly dodgy assumption that led has essentially zero impedance). That will of course tax some voltage compliance of the transconductance amplifier.Regards,Janne

author: amspire
author_rank: Super Contributor
timestamp: October 09, 2011, 03:03:57 pm
content: Quote from: SoftwareSamurai on October 09, 2011, 02:03:35 pmTerminating either end of the cable would ultimately mean you cannot drive the LED directly - you would send voltage over the cable to some kind of receiver that would drive the LED - as I said - it may be more complication then you need.Where you might have a problem is the cable capacitance will attenuate the modulation to the LED. If, say, your cable had a capacitance of 200pF. Put a 200pF cap across the LED, and see if it still works as well.If the capacitance is a problem with your current circuit, then you may be forced to move the LED driver to the remote LED circuit, and use s properly terminated cable to drive the input of the LED driver. In the case of your current circuit, it would mean moving U1 and Q1 to the LED, and sending "SIG" on a properly terminated cable.QuoteI do prefer the coax with a unbalanced signal, but there is every chance that in practice a twisted pair would be adequate. We don't know the kind of modulation you are doing, but if it is sinusoidal type modulation, you will have no problem with higher harmonics. If the modulation is digital with sharp transitions, then you are more likely to have high harmonics radiating and you will have to be more careful with trying to damp the high resonances.But if you have an oscilloscope, do a test. Connect the LED remotely via a coax. Connect the ocsilloscope probe earth to the sheils at the start of the cable. Probe the shield at the end, and if you have a problem, you will see the 2.3MHz. Then try it with a twisted pair.

author: amspire
author_rank: Super Contributor
timestamp: October 09, 2011, 03:15:17 pm
content: Quote from: jahonen on October 08, 2011, 06:00:58 amI don't really agree with this in principle. If the end is mismatched, then the reflected wave changes the voltages at the end of the cable. Now with a short cable, we are talking about effects in the order of 10nSecs which the LED couldn't care less about, so you could get away with a 50ohm termination at the driving end.I would normally prefer to have the receiving end terminated, and if the transmission is from a 0 ohm source, then there is no signal attenuation.Richard

author: jahonen
author_rank: Super Contributor
timestamp: October 09, 2011, 03:48:57 pm
content: But it worksAnd it is recommended practice for point-to-point connections in all high-speed books I know of. This is because the reflected and incident waves are in the same phase at the load end, so the load does not see the reflection. And the reflected wave is then absorbed by the driver series termination. Here is a small simulation, to show the principle. Here we have three 50 ohm transmission lines with 10 ns propagation delay, and each source terminated by different impedances. As expected, if the driving impedance is lower than the line impedance, the first reflection will overshoot the voltage. One can also see that ringing is in fact series of steps (if the signal rise time is short enough). And respectively if it is higher than the line impedance, then it will not rise to full height until after several reflections. Perfect matching with 50 ohm source impedance does not show any reflections at the load.If we would measure the voltage at the line driving end just after the series resistor, we would see the step first rising to half the final voltage and only after the first reflection (input impedance of the low loss transmission line is initially resistive), see it rise to full voltage. I use this method to tune the termination resistors (adjust the initial step to midpoint).Regards,Janne

author: amspire
author_rank: Super Contributor
timestamp: October 09, 2011, 03:55:16 pm
content: Janne,Sound like you have more experience in this then me so thanks for the correction.Richard

author: Mechatrommer
author_rank: Super Contributor
timestamp: October 09, 2011, 06:38:06 pm
content: you dont really like to argue with a person named Janne Ahonen

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 09, 2011, 09:27:13 pm
content: Quote from: amspire on October 09, 2011, 03:03:57 pmIt's an FM sine. All analog. No digital.And as far as harmonics is concerned, the LED I'm using has a 40ns rise time, which effectively acts as a LPF at 12.5 MHz (If I'm not mistaken in my calculations.)So frequencies higher than that just won't matter to the LED, right?Quote from: amspire on October 09, 2011, 03:03:57 pmYeah, I suppose some R&D testing is in order here. But if terminating the coax causes the circuit to have to dump more current into the signal in order to drive the LED through the coax, it might not be worth the effort (as suggested before).My whole goal in using the coax was to provide a way to transport the signal just a meter away without any additional load/components and to provide EM shielding. I know that if I bite the bullet and supply power, ground, and signal, I can use the op amp on the board as a line driver, and put another op amp + transistor next to the LED as an LED driver circuit. I was just trying to get away with needing only 2 wires instead of 3 or 4.

author: vk6zgo
author_rank: Super Contributor
timestamp: October 10, 2011, 02:20:55 am
content: Although it is a sine wave that is fed to the LED,the LED is a diode,so it will create harmonics.It is doubtful that any of these will radiate,but bear in mind,they do exist.As I said before,the coax is extremely short,compared to a wavelength at 2.3MHz,& even at those harmonics which might be generated.PAL video cables carry frequency components at up to 5MHz &,if incorrectly terminated do not radiate.(OK,before somebody says it,they are terminated correctly at the source end).I feel you are over thinking this thing,& I would suggest,just go with your original idea,& see if you have any significant radiation.Buy a cheap all band portable radio,& wave the thing around near the coax.Even though these things are junk radios,they are more than sensitive enough to show up any problems.I know it doesn't sound very "EE like",but then,I'm just a Tech!VK6ZGO

author: jahonen
author_rank: Super Contributor
timestamp: October 10, 2011, 03:33:26 pm
content: Tried out the series resistor with the LED in LTSpice. Seems to work quite nicely, certainly much better than with no termination at all. Slight undershoot is visible (mismatch) but I think that it is in acceptable limits here.Regards,Janne

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 10, 2011, 03:45:09 pm
content: Quote from: jahonen on October 10, 2011, 03:33:26 pmVery interesting! I'll have to try that!So is that 47 ohms + LED resistance = 50 ohms?Thanks Janne!

author: jahonen
author_rank: Super Contributor
timestamp: October 10, 2011, 05:14:31 pm
content: Quote from: SoftwareSamurai on October 10, 2011, 03:45:09 pmYes, that's basically it. Of course, LED "resistance" is not constant so termination is never perfect. But good enough, certainly. I assume that you have relatively high bias current and small frequency modulated sine on top of that?Regards,Janne

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 10, 2011, 06:20:19 pm
content: Quote from: jahonen on October 10, 2011, 05:14:31 pmMakes sense. Got it.Quote from: jahonen on October 10, 2011, 05:14:31 pmThe "current version" circuit that I posted earlier is the one I need to modify for moving the LED to a "movable" platform. The op amp together with the transistor provide the driving current for the LED, which seems to work well.The signal is an FM sine, 2.3 MHz with a +/-75 kHz modulation.The signal itself comes out of an IC, to a 22n cap, then a 2k4 resistor, then to the op amp positive. The op amp positive input is pulled up by a two-resistor voltage divider:+Vcc -- 24k -+- 2k7 -- GND.

author: elCap
author_rank: Regular Contributor
timestamp: October 11, 2011, 06:41:30 am
content: Very interesting thread!As understand it, it's only the LED that will be outside of the main board, would an optical fibre work?

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 11, 2011, 11:41:40 am
content: Quote from: elCap on October 11, 2011, 06:41:30 amPerhaps so.The LED I'm using has high-intensity and a fairly wide FOV. For my project, that's important because it needs to be movable (within certain limits), yet still received by the receiver board. Fiber optic cable is pretty good at transporting light, but only in a narrow FOV (AFAIK). I'd have to add a lens to it in order to regain the wide FOV.But that is a good question!

author: ejeffrey
author_rank: Super Contributor
timestamp: October 11, 2011, 12:04:41 pm
content: How big of a FOV do you need? Cheap plastic optical fibers typically have a fairly high NA / wide FOV. A 60 degree cone is pretty typical. Glass fibers and those used for high speed communication (ethernet, not TOSLINK) are typically smaller, on the order of 10-20 degrees.

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 11, 2011, 11:46:29 pm
content: Quote from: ejeffrey on October 11, 2011, 12:04:41 pmThe LED I'm planning on using has an angle of half-intensity of +/- 60 degrees.Still, your idea of using a fiber optic cable does have merit. I'll have to think about that.

author: IanB
author_rank: Super Contributor
timestamp: October 12, 2011, 01:36:30 am
content: This thread is full of discussion about 50 ohms. What exactly is the "magic" to 50 ohms? In a world full of numbers, why is 50 ohms special? Could someone explain?

author: SoftwareSamurai
author_rank: Regular Contributor
timestamp: October 12, 2011, 01:58:05 am
content: Quote from: IanB on October 12, 2011, 01:36:30 amI was considering the different ways I could relocate an LED off of a main board, but keep it connected. One idea I had was to use a micro-coax cable. (e.g. 0.81mm). Those types of coax cables are typically 50 ohm cables, as opposed to, say, CATV cable that's typically 75 ohm. So the question turned into how best attach an LED to a meter-long micro-coax cable, but still match the impedance of the cable. If you simply connect the LED across the coax, it won't work very well due to impedance mis-match. Janne suggested adding a 47 ohm resistor in series with the LED (which has a little resistance itself) in order to better balance the coax impedance.As for "why 50 ohms", there's a pretty good write up at Wikipedia (http://en.wikipedia.org/wiki/Coaxial_cable), under "Choice of impedance". In a nutshell, and typical of engineers, it's the best compromise between "power handling capability and attenuation".

author: amspire
author_rank: Super Contributor
timestamp: October 12, 2011, 04:23:41 am
content: Quote from: IanB on October 12, 2011, 01:36:30 am50 ohms is a common impedance for coax cables, but there are many other impedances around like 75 ohm coax and 100 ohm twisted pair.In a way it is pretty "magic" and "special" because what is actually happening in these cables is amazing.As you probably know, when two wires are near each other, these is a capacitance between them, so a very long cable has lots of capacitance. This is a bad thing as capacitance tends to attenuate AC.All wire also has a self inductance, so the longer the wires, the more inductive impedance which again attenuates signals.Now if you have a pair of wires that are separated with a constant spacing, such as in coax or twisted pair, then at one particular ratio of voltage to current, the self inductance exactly counters the effect of the capacitance. So no attenuation except for resistive and dielectric dissipative losses, even for very high frequencies. This explains why you are able to send a 1GHz signal down a 10 meter coax cable with little attenuation, even though from the apparent capacitance, there should be massive signal loss. This one particular ratio of voltage to current is the cable impedance and so if it is terminated at the far end by a resistor of the same impedance, a signal can travel down the cable with almost no deformation or attenuation.What is actually happening is that each microscopic slice of the cable with its inductance and capacitance is temporarily storing the energy from the previous slice, and then passes all the energy on to the next slice at somethings a bit less then the speed of light. So each microscopic slice is actually a bit like a memory, as each different slice could have a different voltage. This sequence of differing voltages get faithfully passed from one slice to the next down the cable.As an example, analogue oscilloscopes need to delay waveforms in the the channels by something like 5 - 100nS so that they can have the trigger firing before the edge you want to look at reaches the screen. The way they used to do this was to have a long coiled length of 50 ohm coax. If the sweep rate was set to 1 nSec per division, the complete waveform you see on the screen was stored at one point in the delay cable.Why it is magical is that with discrete inductors and capacitance, their impedance only cancels out exactly at one frequency. With the case of the distributed inductance and capacitance in a coax cable, it surprisingly cancels out at every frequency.This is incredibly lucky for us. Without this, the phone system wouldn't work and it would be very hard to get the signal from your TV antenna to your TV.Richard