# Simple transmission line question

• posted

If I have a source terminated transmission line (think of a length of coax with a 50 ohm resistor to ground at the source.) and it goes into an open circuit. (say into an unterminated 'scope input.) Then is there a reflection from the unterminated end? And if so, the reason that there is no ringing observed is that the source termination soaks up the return signal. Or does the source termination make the transmission line 'happy' regardless of what's on the other end?

Thanks, George H.

• posted

Are you driving the coax with a voltage source with 50 Ohms "to ground", or is the source impedance 50 Ohms?

An open-terminated (50 Ohm impedance) line driven by 50 Ohms _source_ would reflect NOT-inverted.

Thus observed at the source end of the line, you'd see half your pulse height initially then popping up to full pulse height upon reflection return. (If I remember my freshman-level transmission line material correctly :-)

Likewise a shorted line will cancel the input pulse after full out-and-back transit time... think short pulse generation techniques. ...Jim Thompson

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| James E.Thompson                                 |    mens     |
• posted

El 22-04-13 18:37, George Herold escribió:

Hello George,

You get reflection at the cable to scope transition (RC=+1 theoretically). This results in twice the traveling wave voltage at the input of the scope (source EMF).

You are right with the absorption when the source is a current source (or at least with Z>>50 Ohms). The reflection travels back to the source and is absorbed by the 50 Ohms source resistance.

Though the load (scope) sees a clean signal, in the cable itself there is reflection. So if you would measure for example halfway the cable (with High-Z probe), you will notice the reflection.

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Wim
PA3DJS ```
• posted

Normally, source termination uses a SERIES resistor at the source. So, if the driving source has a 5 Ohm impedance, you would use a

45 Ohm resistor in series between source and the cable. You WOULD get a reflection from the open-circuit end, but it would be fully absorbed at the source end, so the output end would see no reflection.

For a parallel resistor to properly terminate the source end, the source would need an infinite output resistance, which prevents it from creating a signal.

Jon

• posted

One issue with this is that the output impedance of most logic devices is not very controlled, process tolerance dependent, and often also uneven between high and low side. Or as Forrest Gump would have put it, it's like a box of chocolates, you never know what you're gonna get :-)

This is one of the reasons why I prefer AC-termination.

A current source can do this.

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Regards, Joerg

http://www.analogconsultants.com/```
• posted

Not really - you can use a current source. For a proof, use Messrs Thevenin and Norton.

A different story is that most practical sources resemble more a pure voltage generator, which nees series source terminator.

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Tauno Voipio```
• posted

So I have a digital pulse and I want to send a monitor signal out to a 'scope. But the digital logic doesn't have enough poop to drive 50 ohms, so I stick in a series combo of 450 ohms and 50ohms to ground, and take the monitor output from the 'top' of the 50 ohms. (and eat the factor of ten in signal level.)

Yeah thanks..

George H.

...Jim Thompson
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• posted

Lose the 50 Ohms to ground and then do....

Signal -> 450 Ohms ----- Line ----- 50 Ohms (to GND) -- 'scope input

...Jim Thompson

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| James E.Thompson                                 |    mens     |
• posted

Thanks Jon, (and Wimpie) In this case the source doesn't have enough current to drive the 50 ohms.

George H.

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In theory, yes. Depends somewhat on what's driving and what you're measuring.

I'm a big fan of terminating the coax at the load end. That way, it provides a time-invariant resistive load to what's driving it. Can reduce driver non-linearity and crosstalk problems.

• posted

The All-American answer: Then you need a bigger source :-)

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Regards, Joerg

http://www.analogconsultants.com/```
• posted

That's an excellent solution when the input capacitance of the scope is zero...or inconsequential at the frequencies of interest.

• posted

Maybe the 'scope has a switch (or is permanently) 50 Ohms input, then no resistor needed? I haven't done this since the early days of ECL/PECL ('60's and '70's), but the 'scopes back then were 50 Ohms Zin.

...Jim Thompson

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| James E.Thompson                                 |    mens     |
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Just get it from Obama's stash >:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     |
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...Jim Thompson
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ide quoted text -

Yeah but then you have to tell people to terminate the line. This way it's 'mindless'. (Instruments for students.)

George H.

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[snip]

Omigawd, You have Larkin-level students in your class ?>:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     |
• posted

Not unless you put in a signal somewhere!

If the source is a 5 volt step generator with a 50 ohm source impedance, the initial wave will be 2.5 volts. When it hits the open end, it will double to be a clean 5 volt step. A 5 volt wave will then zip back to the generator end and die there when it hits, and it's all over. Clean 5 volt step at the far end, and the generator voltage sees

100 ohms (50 internal+50 line) as the transient load, 50 mA peak current.

LT Spice has transmission lines. If you string a bunch together, you can see what happens at intermediate points.

Source termination is cool.

And if so, the reason

The source termination will absorb any reflection, like from a mismatch or a short. The far end of the line looks like a 50 ohm resistive source to *any* load out there; it's just the 50 ohm generator terminals, extended out in time.

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John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com ```
• posted

There is always a reflection when a signal travelling on a tline hits an impedance change. For a source-terminated line, the signal gets divided, Z0/(Z0+Rt), when it enters the line, but at the other (open) end, the reflection adds to the incident signal to restore the original level. The reflection then travels back to the source, of course, and there it finally gets absorbed, since it's source-terminated. (Which requires Z0=Rt.)

'Source termination' means that looking back from the tline towards the source, one sees Z0 ohms. If your source is a low-Z voltage source, it needs a Z0 ohm series resistor. If it's a high-Z current source, it must have a Z0 ohm resistor to ground. If your source is a Z0 ohm signal generator, it needs nothing.

Jeroen Belleman

• posted

That's fine, just leave it at that. The reason the line voltage doubles at the Hi-Z termination at the monitor is because there can be no net current into the Hi-Z, so the superposition of the forward and backward wave curren ts must sum to zero there, meaning the backward wave current must equal the forward wave current thereby developing the same voltage across the line a s the forward wave (V=IxZ), making the superposition of the two twice the voltage of the forward wave. So assuming the logic is 5V your circuit laun ches a 5 x 25/475= 263mV voltage down the line, which transforms to 2 x 2

63= 526mV at the Hi-Z with a backward wave amplitude of 263mV. All but 5% of the 263mV is absorbed by your 50R to GND so nothing significant is re-r eflected back down the line again to the Hi-Z.
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Hi-Z termination at the monitor is because there can be no net current into the Hi-Z, so the superposition of the forward and backward wave currents must sum to zero there, meaning the backward wave current must equal the forward wave current thereby developing the same voltage across the line as the forward wave (V=IxZ), making the superposition of the two twice the voltage of the forward wave. So assuming the logic is 5V your circuit launches a 5 x 25/475= 263mV voltage down the line, which transforms to 2 x 263= 526mV at the Hi-Z with a backward wave amplitude of 263mV. All but 5% of the 263mV is absorbed by your

50R to GND so nothing significant is re-reflected back down the line again to the Hi-Z.

Yep. Everyone forgets their fundamentals...

Take an open-ended piece of transmission line and apply a pulse to one end of it thru a source resistance equivalent to Zo.

At the "time" of the application of the pulse, the transmission line does not "know" that the far end is unterminated. Thus its input end shows as Zo.

Thus the initial voltage at the input to the line is ONE-HALF of the applied pulse.

When the reflection comes back it adds to the initial ONE-HALF, thus your full applied height appears.

I had lots of fun in MIT's transmission line lab... all kinds of purposefully-made slow lines so that reflection studies were easily to do with cheap 'scopes.

Try a diode termination some time >:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     |