I am designing and building circuits which use 100Mb/s on a Low Voltage Differential Signaling (LVDS) bus. Some of these signals need to travel between PCBs on hand made cables. The problem is that I have no way of verifying the quality of the cables and termination.

If I was a millionaire, I'd get an expensive 'scope or a vector network analyser. But failing that, is there some way I can measure the reflected signals, or the impedance of the cable?

(I have a 150MHz bandwidth, 500MSPS 'scope available).

Added: Information about the data on the wire, taken from the ET1200 datasheet.

Ebus waveform

Added: 21 hours to go. Last chance for the bounty. Can anyone suggest even a quick and dirty way to measure impedance? Perhaps some kind of bridge where I could compare the cable against a known good cable?

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    \$\begingroup\$ "100Mb" might be a spec for a memory, but not for a communications channel since the units are clearly wrong. What then do you really mean by "LVDS" (which of course should have been spelled out anyway)? Getting basic easy stuff like units wrong means we have to assume lots of other stuff is wrong and there is no way to know what you are actually asking. \$\endgroup\$ Commented Feb 9, 2012 at 15:01
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    \$\begingroup\$ 100 MB - with a capital B is the units for memory. Lowercase b means bits, so 100Mb means 100 megabits. Yes, it should include time so 100Mb/s is correct. LVDS stands for Low Voltage Differential Signaling. \$\endgroup\$ Commented Feb 9, 2012 at 15:47
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    \$\begingroup\$ @Olin: Are you sure? Then what do people mean when they refer to 100Mb Ethernet? (Google it) Do they mean it can 'remember' 2^20 bits? No, people often use 100Mb as (very) short for 100 mega bits of data transferred per second. Agreed that the correct units are Mb/s. LVDS is always called LVDS. People practically never write it out in full: Low Voltage Differential Signalling. \$\endgroup\$ Commented Feb 9, 2012 at 15:53
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    \$\begingroup\$ @OlinLathrop, Do we now have to spell out "Transistor-transistor logic" or "complimentary metal-oxide-semiconductor logic" every time we want to talk about TTL or CMOS? \$\endgroup\$
    – The Photon
    Commented Feb 9, 2012 at 16:28
  • \$\begingroup\$ @Rocketmagnet - When people refer to "100Mb Ethernet" they are referring to it wrong. It's "100Mb/s Ethernet" (or "100Mbps Ethernet"). The fact that a lot of people incorrectly call it that does not make it correct (it just makes the people who call it that look silly). Is not typing the additional /s or ps that much of an affront? \$\endgroup\$ Commented Apr 8, 2012 at 6:30

3 Answers 3


Here's the cheapest I can come up with.

Directional coupler diagram

First, you need an rf synthesizer. If you don't have that, get your digital signal to output a pure square wave (either use the clock signal or send 1010... from your data line), and then use a lowpass or bandpass filter to transform that into a pure-ish sine wave.

Between the source and your circuit under test, connect a directional coupler, in the direction so that the coupled output is getting the reflected signal, not the source signal.

Now connect an rf power detector to the directional coupler's coupled port. Now you can use a multimeter to measure the power in the reflected signal.

If you use Minicircuits, you can get the directional coupler and power detector for something like $150, or you can probably find these parts for even less on EBay.

There will be all kinds of errors in this approach, because you don't have the equipment to calibrate it. The directivity of the directional coupler will limit the minimum reflection coefficient you can measure. But if you adjust your termination to minimize the voltage at the power detector output, you're probably close to optimizing the match.


Should add, since you're talking about LVDS, you're presumably talking about a differential line and a differential termination. Which means for this scheme you'll need a balun between the test instrument and your DUT. Which is yet another potential source of errors.

  • \$\begingroup\$ Thanks Photon. This is more like what I'm after. Since this is LVDS, do I need a pair of directional couplers? Why use a pure sin wave? Why not use my real signals? Surely this would make it a more accurate test. \$\endgroup\$ Commented Apr 7, 2012 at 21:18
  • \$\begingroup\$ Maybe I could calibrate it. I can create a situation with maximum reflection by leaving out the terminator. And I can create minimum reflection using a 100R HDMI cable and 100R terminator. \$\endgroup\$ Commented Apr 7, 2012 at 21:24
  • \$\begingroup\$ The directional coupler and rf detector won't have particularly uniform response over frequency. So you won't really know what's the response to your square-ish wave input. Also, you won't know the phase response at each frequency, so overall its a pretty limited measurement. But it could get you pretty close to a good match. \$\endgroup\$
    – The Photon
    Commented Apr 7, 2012 at 21:25
  • \$\begingroup\$ Cool. I'm reading up on directional couplers now. Is there such a thing as a differential directional coupler ? \$\endgroup\$ Commented Apr 7, 2012 at 21:39
  • \$\begingroup\$ @Rocketmagnet - Don't ground the coupler? Or use a balun to go from single-ended to differential. \$\endgroup\$ Commented Apr 8, 2012 at 6:28

You don't have to spend a million dollars do get a decent VNA. Since you have the skills to build circuits, you can build one yourself for about $400 USD. I've been building up a N2PK VNA over the past several months. You don't need any special tools, just a steady hand, and a good soldering station. There's an active Yahoo Group, in the files section there are plenty of completed projects. I've sourced most of the parts through Digikey, with a few from Mouser, and MiniCircuits. I've been writing up my progress on my website too.

  • \$\begingroup\$ Is that one suitable for differential signals? I.E. Could I use it to test some home made twisted pair cable? \$\endgroup\$ Commented Feb 9, 2012 at 16:06
  • \$\begingroup\$ @Rocketmagnet - Possibly: picosecond.com/objects/AN-21.pdf Something I'm going to try once my VNA is done. \$\endgroup\$ Commented Feb 10, 2012 at 5:48

In my opinion, the most obvious thing to do is to build an oscillator (or a multivibrator, to have square waves) with variable frequency and look at the signal at the other end of the cable if the degradation is acceptable.

But first you should define some dimensions: 100 Mb/s it's the overall bandwidth or only for the payload? You should first convert it into a signal frequency (in Hz), and then check the length of the cable to make sure that it's of the proper length.

I think that measures make sense when you have an hypothesis to check, otherwise you won't know what to do of the results.

  • \$\begingroup\$ Problem is that my oscilloscope isn't fast enough to do any meaningful measurement at this speed. A scope like that costs thousands of pounds. \$\endgroup\$ Commented Apr 1, 2012 at 14:21
  • \$\begingroup\$ @Rocketmagnet you can still make sure that a 100 MHz sinewave is passed, and that the line is of the proper length to have stationary waves \$\endgroup\$
    – clabacchio
    Commented Apr 1, 2012 at 14:38
  • \$\begingroup\$ Well, the cable is pretty short (about 300mm) compared to the wavelength. I'm not sure exactly what the wavelength would be, because apparently the propagation velocity depends hugely on the medium. \$\endgroup\$ Commented Apr 1, 2012 at 14:40
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    \$\begingroup\$ Also, critical line length depends more on the rise & fall times of your source signal than on the bit rate. So what are your rise & fall times? Problem is, you don't have the budget for the test equipment that could tell you...I thin the reason you haven't gotten many answer to your question is, there's no magic bullet. If you want to work at 100 MHz and up, you need to budget for proper test equipment at those frequencies. \$\endgroup\$
    – The Photon
    Commented Apr 1, 2012 at 16:27
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    \$\begingroup\$ If the termination R is greater than the line impedance Z0, you get a reflection with positive voltage; so if you measure voltage at the termination, it is higher. If R < Z0, you get a reflection with negative voltage, so you measure a lower voltage at the termination. But if you are doing this by varying Z0 instead of R, you also create at mismatch at the source end, and you get multiple reflections in the line. What is the end result measured in terms of peak voltage at the termination end is not trivial to predict. \$\endgroup\$
    – The Photon
    Commented Apr 1, 2012 at 17:24

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