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I used a silicon coax cable (characteristic impedance 45 ohms) and sent a square wave with frequency 1 Mhz and amplitude 5V through it. The cable was teminated openly with no end resistance. I could find a reflection in-phase with the input signal.

Now i tried to hit the cable in between with a force around 80N so that the cable is deformed. But i dont find any reflection in the area that i tried to deform in the oscilloscope. The reflections are still like before for the open end and every time i bend or squeeyze or deform the cable, i dont find any reflection for that area. Whats going wrong here and how can i have reflections on the cable during impact with an object ?

Thanks

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  • \$\begingroup\$ I might be wrong but I always thought that you measure these reflections with the best termination you can get... \$\endgroup\$ – PlasmaHH Jan 19 '15 at 11:54
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    \$\begingroup\$ I believe you need a fast rise time to make a reflectometer. whats the rise time for your 1MHz signal ? \$\endgroup\$ – efox29 Jan 19 '15 at 12:27
  • \$\begingroup\$ @PlasmaHH A properly terminated line will minimize the reflection. If impedance control is maintained from source to driver, you wouldn't get any reflections. \$\endgroup\$ – efox29 Jan 19 '15 at 12:29
  • \$\begingroup\$ How long is the cable, how sharp is the edge of the 1 MHz square wave? I might try a short pulse rather than a square wave. \$\endgroup\$ – George Herold Jan 19 '15 at 14:48
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I think that the massive reflection from the unterminated far end is masking anything you might expect to see. I also think you might benefit from putting the measurement circuit inside a Wheatstone bridge arrangement. The bridge stimulus will be the driving signal - feed it via (say) 2x 47 ohm resistors to (a) the terminated cable and (b) to a reference impedance to ground.

The reference impedance to ground will be 45 ohms (possibly adjustable to compensate for the basic cable being a few ohms different). Use a differential amplifier to measure the difference voltage between what is applied to the cable (via one 47R) and what is applied to the reference resistor (45R) by the other 47R.

schematic

simulate this circuit – Schematic created using CircuitLab

With a perfect far-end termination and perfectly homogeneous cable, there should be zero signal from the difference amplifier. Deforming the cable should produce a small reflection and this should be detectable.

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  • \$\begingroup\$ Could you explain me whats the role of the newly introduced resitors in the circuit ? kindly provide me with some theoretical background. \$\endgroup\$ – user3396084 Jan 19 '15 at 14:34
  • \$\begingroup\$ The above circuit is not going to show the point of damage on the cable. Measuring the voltage difference between the two points even with a scope may not produce a visible difference. Note that this arrangement is comparing a (DC)resistance to a potentially changing (mixed DC/AC) impedance. To see the change you would need to sync the scope to the original V1 source then look for a phase change in the differential signal. A more sensitive circuit could be made by substituting R1 with an identical cable/resistor combination. \$\endgroup\$ – Nedd Jan 19 '15 at 14:45
  • \$\begingroup\$ @user3396084 see wheatstone bridg en.wikipedia.org/wiki/Wheatstone_bridgee \$\endgroup\$ – Andy aka Jan 19 '15 at 14:49
  • \$\begingroup\$ @Nedd I'm trying to demonstrate to the OP that a sensitive measurement circuit is needed. I'm not trying to go down the path of explaining all the technical details. Of course the circuit won't show the point of damage - synching a scope is needed at the very least - but the circuit is needed to allow decent amplification of anything that can be seen due to cable damage. \$\endgroup\$ – Andy aka Jan 19 '15 at 14:52
  • \$\begingroup\$ Actually make that "substituting R3" instead of R1. \$\endgroup\$ – Nedd Jan 19 '15 at 14:54
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Most coax cable is purposely made to withstand a significant amount of abuse. Most bending and denting will hardly at all result in a measurable impedance change at the abused point.

In terms of an impedance change at the "deformed" area, when you deform a round cross-area it becomes oval like, the oval shape thins at two points but also widens at two other points. So the averaged capacitance from center to shield at the deformed point is hardly changed. For the center conductor's inductance, note that you are unlikely to change this as it is very much protected at the center of the material layers. Even when coiled the center conductor does not see (magnetically) the other close conductor loops due to the shielding.

If you start to abuse the cable to the point where the conductive shielding becomes significantly damaged or cut then that should begin to produce a measurable reflection. You might test this by running the reflection test while slowly crushing the cable in a strong metal twist clamp or vise. You will likely find that the cable needs to be significantly crushed before seeing a reflection.

If the need is to create a cable that detects a mechanically altered point try using a simple piece of flat cable. On it's own a conductor in this type of cable is only partially shielded by the nearby conductors. A sharp bend or other damage to this type of cable may be much more detectable. Using two or more flat cables laying over each other might make an even more sensitive detector of a point that is being crushed or abused.

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You can't see anything because the signal you're using is not suitable for reflection measurements. Generally, for manual measurements you don't want more than one stimulus pulse present inside the length of your conductor and time between pulses must be sufficient for all reflections to die off. A squarewave would be a poor choice here. Imagine yourself shouting in a canyon and hearing the echo. Now imagine another thousand people doing the same while standing next to you - will you be able to recognize your own echo?

To better understand what is happening take a look at pics below - I took them while fixing a sampling head in my TDR. This instrument generates its own stimulus signal with ~25 ps rise time ~300 ns wide with rep.rate of 1 KHz. Note the time scale.

  1. This is the DUT. It is a sampling head of the TDR - the device that "sees" the signal. Upper right connector has bad contact inside the head. All connectors SMA.

  2. That's how the stimulus looks like when 50 ohm terminator is placed on the connector. You can see the step traveling a short distance inside the head ( horizontal section at the third division ) before being reflected.

  3. That's the same signal but now the terminator is removed. You can now see more reflections coming from the open end of the head. They are travelling back and get reflected again. Now, this head was specifically designed to dampen those reflections and when you use some general purpose input like an oscilloscope you'll get many more of those.

  4. The signal after the head has been fixed. No large reflection can be seen anymore. You can now see the small reflections from mating surfaces of the connectors getting smaller and smaller as time passes. That is what you will see when the cable you're studying has some non-uniformity caused by mechanical stress. The signal will be quite small and in order to see it you want it to be clean.

Before continuing with reflectometry build a source of fast risetime pulses. Any general purpose transistor driven to avalanche will give you better than 1 ns.

Sampling head Bad output connector - terminated Bad output connector - unterminated No Reflection

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  • \$\begingroup\$ What is the stimulus signal pattern that is generated by the device you are using ? Is it sine wave ? I have used a normal function generator and oscilloscope. I dont know much about TDR device. I could not understand what is a DUT and a sampling head connector. \$\endgroup\$ – user3396084 Jan 20 '15 at 9:33

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