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I came across the following article on impedance matching for pulses which have very sharp rising and falling edges: https://www.avtechpulse.com/appnote/techbrief12/

  • They claim that the output impedance of the generator (50 ohm) and the load impedance should be equal

  • They assume the pulsers have \$Z_{out}\$ = 50 Ohm Output Impedance

Therefore in the case of a load which is less than 50 ohm, a series resistor needed as shown below:

enter image description here

Similarly if the load is greater than 50 ohm, a shunt resistor can help as shown below:

enter image description here

I have three questions regarding this issue:

  1. When they make matching why they only take into account the load resistor and series or shunt resistor but not the 50 ohm resistance from the coaxial cable? For example in the first example they set: \$Z_{out} = R_s + R_{load}\$ = 50 ohm. Why do they omit the coaxial cable resistance in this matching?

  2. Why would unmatched output impedance and the equivalent load resistance cause ringing or reflections? Is there a way to describe this reason illustratively?

  3. Is there a specific reason historically to standardize output impedance 50 ohm in general but not 10 ohm for instance?

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  • \$\begingroup\$ Well, for your 1), I would say they match it correctly. You pass from coaxial cable which have a 50 ohm signal impedance to a load so therefore you want your load to have a 50 ohm too. 2) I am certain you can find the mathematical proof for that. 3) I think it is because of radio or RF but this is just a hypothesis. \$\endgroup\$ – MathieuL May 23 '16 at 14:16
  • \$\begingroup\$ I dont understand isnt that Zout is output impedance? Coaxial is like a series resistor. Im confused \$\endgroup\$ – user16307 May 23 '16 at 14:20
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    \$\begingroup\$ It isn't a 50 ohm from ohmic resistane. It is a 50 ohm from line impedance. If you take a ohmmeter and measure the resistance, mostly it will be around a very small value because coaxial cable are made of copper. You should check out the lossless line section en.wikipedia.org/wiki/Characteristic_impedance. The impedance of 50 ohms represent the inductive & capacitive parasite in your lines that will result in delay and reflections. \$\endgroup\$ – MathieuL May 23 '16 at 14:25
  • \$\begingroup\$ So do u mean what ever measurement I make with a coax i can neglect its resistance? Even for a pullse at too high freq.? \$\endgroup\$ – user16307 May 23 '16 at 14:43
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    \$\begingroup\$ @user16307, from a theoretical perspective, the amplitude is not diminished by the characteristic impedance. But as a practical matter, at 2 GHz, all real cables have significant attenuation. You are asking a LOT of questions about a BIG topic. You really need to read about transmission line theory. Start with lossless transmission lines. After you sort of understand that, you can read about skin effect, and dielectric losses in cables. \$\endgroup\$ – mkeith May 23 '16 at 15:28
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The characteristic impedance of any cable at high frequencies is determined by the inductance per unit length and the capacitance per unit length. It should not to be regarded as a conventional lossy resistor - characteristic impedance is simply the impedance that the cable should ideally be terminated with to prevent reflections.

So, reflections happen when there is a mismatch between the termination and the cable's characteristic impedance.

Consider a longish piece of coax fed at one end with an instantaneous voltage of 5V. That 5V will take some finite time to travel down to the load (lets say the load is 1 kohm) so it cannot know how much current the load needs. However, the cable "informs" the source how much current to flow - if it's 50 ohm cable then 100 mA will flow. So you have 5 V and 100 mA rapidly travelling down the cable and they reach the load to find that it's 1 kohm.

In other words, too much current is flowing for a 1kohm load with 5V applied. So a reflection occurs to combat the excessive current. After a few cycles of "there and back" things settle down.

Here's a nice picture of a transient wave passing through a mismatch (vertical black line) - note the energy reflected back to the source: -

enter image description here

Because I'm old and sometimes wise I can tell you that the left half of the cable has a higher characteristic impedance than the right half. There are two clues that tell me this. First clue; the width of the pulse shortens in the right half implying the velocity has dropped as the pulse entered this right hand area, Clue 2; there is a negative voltage reflection.

Now, think about that vertical black line - can you imagine that the black line is a solid wall and you're holding a rope attached to that wall. You wiggle the rope to induce a transient pulse and you'll get a reflection coming back just like the picture above. Same phenomena, same maths.

There are plenty of great pictures on the web that demo this. Here's one that shows how a transmitted square wave becomes misshaped: -

enter image description here

Note that the "ringing" does look like traditional LC type ringing but, if you inspect closely you'll see that they are slightly rounded-off square wave reflections adding and subtracting from what would be the perfect received signal.

Here are two animations of a pulse hitting an open circuit (top) and a pulse hitting a short circuit (bottom). Note the polarity of the reflection heading back to the left: -

enter image description here

enter image description here

Regarding the value of 50 ohms, 50 ohms is a compromise between power (power prefers lower impedances) and attenuation characteristics (75 ohms is much preferred to reduce high frequency attenuation).

And finally, the Shive wave machine: -

enter image description here

You can watch a 30 minute video on youtube that is really great I reckon. It deals with all sorts of reflections and loading effects but uses a mechanical analogy of the transmission line. Video HERE

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  • \$\begingroup\$ Great explanations of the 50 ohms value. \$\endgroup\$ – MathieuL May 23 '16 at 14:29
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    \$\begingroup\$ Reflections happen at all frequencies but are more obvious to see at higher frequencies. As the cable becomes longer than about one-tenth the wavelength of the signal, "things" start to happen. Clearly at 20 kHz, one wavelength is 15 km so it's pointless trying to see things happening at this low a frequency. At exactly one quarter of one wavelength a short becomes an open circuit and vice versa - there are some strange things happening! \$\endgroup\$ – Andy aka May 23 '16 at 14:40
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    \$\begingroup\$ @user16307, when voltage is first applied to the 50 Ohm cable, it looks like a 50 Ohm resistor UNTIL the reflection comes back from the far end. If the cable is infinitely long, or if it is terminated in 50 Ohms, then no reflection comes back, and it is a true 50 Ohm load. The signal, and the reflection, travel at the "speed of light," however, they travel in the dielectric of the cable, so the speed of light will be less than speed of light in vacuum. \$\endgroup\$ – mkeith May 23 '16 at 15:23
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    \$\begingroup\$ Just one last question: Imagine we have a very sharp pulse train but only 50Hz freq. and we have 50cm short coax cable. 50Hz has huge wave length we can conclude we don't need matching because our cable is 50cm much less than the wavelength where no reflection occurs. But if we think the high freq. components of the pulse we might have reflection. Is that right? So in case same signal is not a pulse but a pure sine, we don't need matching at all but in case of sharp pulse we need because it has ultra high freq. components which might reflect due to their short wavelength? Did I understand it? \$\endgroup\$ – user16307 May 23 '16 at 17:50
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    \$\begingroup\$ You did understand it and if the 50 Hz pulse train was actual data there could still be corruptions around the points where the data changes state. Of course if the top data rate was only 50 Hz then simple filters would remove any trace of anything that would corrupt data receivers but generally a terminator wouldn't be used. RS232 transmits quite happily at several kbaud without a terminator for instance. We also need terminators with sinewaves - you can get standing waves without a properly terminated line and these can be problematic. Power reflections into a transmitter can destoy it. \$\endgroup\$ – Andy aka May 23 '16 at 17:59
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Here's a link explaining the "50 ohm" history: http://www.microwaves101.com/encyclopedias/why-fifty-ohms

That, as well as 75 ohm, most likely were settled on by early coaxial cable manufacturers dictated by manufacturing issues.

For what it's worth, there's nothing "magic" about either. But as the article will tell you, 50 ohm is a very good match on a coaxial cable of a certain construction. On a PCB you could use whatever as long as it a) does allow for your rise/fall times to happen and b) does not overwhelm your driver.

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  • \$\begingroup\$ I found this might be a good introduction to the topic: allaboutcircuits.com/textbook/alternating-current/chpt-14/… eventhough in their switch lamp example they don't take into account the special theory of relativity when it comes to cause effect observation. \$\endgroup\$ – user16307 May 23 '16 at 16:05
  • \$\begingroup\$ youtube.com/watch?v=I9m2w4DgeVk \$\endgroup\$ – user16307 May 23 '16 at 16:37
  • \$\begingroup\$ @user16307 there are a lot of resources for signal integrity but that microwave article explains the origin of those values. PTFE was invented only after 50 ohm cable was settled on.. \$\endgroup\$ – Barleyman May 24 '16 at 1:51

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