# Simple transmission line question

Discussion in 'Electronic Design' started by George Herold, Apr 22, 2013.

1. ### George HeroldGuest

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.

2. ### WimpieGuest

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.

3. ### JoergGuest

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.

4. ### Tauno VoipioGuest

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.

5. ### George HeroldGuest

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.

6. ### George HeroldGuest

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

George H.

7. ### mikeGuest

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.

8. ### JoergGuest

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

9. ### mikeGuest

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

10. ### George HeroldGuest

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

George H.

11. ### JeroenGuest

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

12. ### Guest

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 currents must sum to zero there, meaning the backward wave current must equal theforward wave current thereby developing the same voltage across the line as the forward wave (V=IxZ), making the superposition of the two twice thevoltage 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.

13. ### George HeroldGuest

I wish! (physics students) Some can hardly drive a 'scope. I've
spent some time on the phone explaining triggering....

George H.

14. ### George HeroldGuest

I'll have to try it. Does LT spice have directional couplers? It'd
be nice to see what's coming and going....

George H.

15. ### George HeroldGuest

Thanks Jeroen, Let's just call it 'brain dead Monday' on my part.

George H.

16. ### George HeroldGuest

Thanks... Is the 5% re-reflection bit because I 'should' have the
parallel combo of the two source resistors equal to 50 ohms... so 450
and (tap tap tap) 56 ohms?

George H.

17. ### JeroenGuest

Directional couplers in spice are trivial! Basically, it's nothing
more than a Wheatstone bridge. Replace one of the sides by the tline
and apply twice the signal to the top of the bridge. The reflection
appears across the bridge. Below an LTspice example to play with.

Jeroen Belleman
==================================================
Version 4
SHEET 1 880 680
WIRE 32 112 -320 112
WIRE 224 112 32 112
WIRE 32 144 32 112
WIRE 224 144 224 112
WIRE -320 224 -320 112
WIRE 32 240 32 224
WIRE 128 240 32 240
WIRE 160 240 128 240
WIRE 224 240 224 224
WIRE 336 240 224 240
WIRE 576 240 432 240
WIRE 32 272 32 240
WIRE 336 272 272 272
WIRE 496 272 432 272
WIRE 496 288 496 272
WIRE 576 288 576 240
WIRE 272 304 272 272
WIRE -320 336 -320 304
WIRE 336 352 272 352
WIRE 400 352 336 352
WIRE 32 368 32 352
WIRE 272 368 272 352
WIRE 400 368 400 352
WIRE 224 384 224 240
WIRE 576 384 576 368
WIRE 160 432 160 240
WIRE 224 432 160 432
WIRE 272 464 272 448
WIRE 400 464 400 448
FLAG 272 304 0
FLAG 496 288 0
FLAG 32 368 0
FLAG 576 384 0
FLAG -320 336 0
FLAG 272 464 0
FLAG 400 464 0
FLAG 336 352 reflected
FLAG 128 240 incident
SYMBOL res 16 128 R0
SYMATTR InstName R1
SYMATTR Value 50
SYMBOL res 16 256 R0
SYMATTR InstName R2
SYMATTR Value 50
SYMBOL res 208 128 R0
SYMATTR InstName R3
SYMATTR Value 50
SYMBOL tline 384 256 R0
SYMATTR InstName T1
SYMBOL res 560 272 R0
SYMATTR InstName R4
SYMATTR Value 1meg
SYMBOL voltage -320 208 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(0 2 10n 1n 1n 10n)
SYMBOL e 272 352 R0
SYMATTR InstName E1
SYMATTR Value 1
SYMBOL res 384 352 R0
SYMATTR InstName R5
SYMATTR Value 1
TEXT 80 40 Left 2 !.tran 200n

18. ### WimpieGuest

El 22-04-13 21:25, George Herold escribió:
Measure both current and voltage through the line at the same point
(that means between two connected transmission line sections.

Forward voltage = 0.5*(50*I+U)

When you add a minus sign (or flip the current measurement) you get
the reverse voltage.

The formula you can put in arbitrary source to get the output as a
voltage or current.

19. ### Guest

something like this?

Version 4
SHEET 1 880 680
WIRE -192 64 -304 64
WIRE 176 64 -112 64
WIRE 352 64 256 64
WIRE 480 64 352 64
WIRE 720 64 576 64
WIRE 720 80 720 64
WIRE 480 128 480 96
WIRE 576 144 576 96
WIRE 176 160 96 160
WIRE 320 160 256 160
WIRE 720 176 720 160
WIRE 96 224 96 160
WIRE 96 288 96 224
WIRE 176 288 96 288
WIRE 352 288 352 64
WIRE 352 288 256 288
WIRE 0 384 -32 384
WIRE 176 384 0 384
WIRE 320 384 320 160
WIRE 320 384 256 384
WIRE 416 384 320 384
WIRE 464 384 416 384
FLAG 0 464 0
FLAG 416 464 0
FLAG 96 224 0
FLAG 720 176 0
FLAG -304 144 0
FLAG 480 128 0
FLAG 576 144 0
FLAG -32 384 forward
FLAG 464 384 reverse
SYMBOL ind2 160 176 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 5 56 VBottom 2
SYMATTR InstName L1
SYMATTR Value 100µ
SYMATTR Type ind
SYMBOL ind2 272 48 R90
WINDOW 0 5 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName L2
SYMATTR Value 1µ
SYMATTR Type ind
SYMBOL ind2 272 272 R90
WINDOW 0 5 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName L3
SYMATTR Value 100µ
SYMATTR Type ind
SYMBOL ind2 160 400 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 5 56 VBottom 2
SYMATTR InstName L4
SYMATTR Value 1µ
SYMATTR Type ind
SYMBOL res -16 368 R0
SYMATTR InstName R1
SYMATTR Value 50
SYMBOL res 400 368 R0
SYMATTR InstName R2
SYMATTR Value 50
SYMBOL res 704 64 R0
SYMATTR InstName R3
SYMATTR Value 1000
SYMBOL voltage -304 48 R0
WINDOW 0 -56 -5 Left 2
WINDOW 123 0 0 Left 2
WINDOW 39 24 124 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 1 10e6 100n 0 0 1)
SYMBOL tline 528 80 R0
WINDOW 3 -5 -93 Top 2
SYMATTR InstName T1
SYMATTR Value Td=500n Z0=50
SYMBOL res -96 48 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R4
SYMATTR Value 50
TEXT 64 112 Left 2 !k1 L1 L2 1
TEXT 56 336 Left 2 !K2 L3 L4 1
TEXT -242 488 Left 2 !.tran 1.5u

-Lasse

20. ### Guest

Exactly, 56R is a better match, but even your resistor precision gets a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it's just monitor front end parasitics ....