Simple transmission line question

[George Herold]

Aren't the carbon films spiral cut?

Oh, I don't know. Hang on I've got some in the garage...
Well all the 47, 56 ohms are gone, but 68 ohms has spirals,
maybe 2 or three turns from end to end.

More things to do to avoid real work!

Oh not on my account, I thought you might have done it already.
(Ummm, but if you're at it how about a carbon comp too, is it really
any better?)

 

[George Herold]

Aren't the carbon films spiral cut?

More things to do to avoid real work!

Surface mounts, even the cheapies, are very good.

--

John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot comhttp://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
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Say If you had a long transmission line could you send a pulse down
it, with no source termination, and then switch in the termination for
the return pulse? Waste energy only when you need to. (Assuming the
pulse width is less than the length of the T-line.)

George H.

 

[mike]

Say If you had a long transmission line could you send a pulse down
it, with no source termination, and then switch in the termination for
the return pulse? Waste energy only when you need to. (Assuming the
pulse width is less than the length of the T-line.)

Sure,
All you gotta do is make sure the voltage at the source end of the line
doesn't change when you insert the resistor.

 

[mike]

I guess you could drive the line with current, turn off the current,
then terminate.

Ok, but to what voltage do you terminate?

If you start with a discharged line and stuff some electrons into it,
the terminal condition will be V=Q/C. The voltage on the line may never
go to zero. The logical line voltage is whatever the voltage
was at the source end of the line when you decided to insert said
termination. And the at the line to which you tie the resistor and the
current
into the line at that instant all determine the voltage on the other end
of the resistor required to keep the line voltage constant.
Simple in theory, but rather difficult if you don't know what's
on the other end or the elapsed time or the width of the pulse or...or...
Doing it all dynamically likely takes WAY more energy than you'd save.

If you pick the right voltage, there's one reflection at the end that gets
absorbed at the source as the resistor current goes to zero.

 

[Phil Allison]

"dave"
"Phil Allison"

meters times megahertz equals 300 so 1m 1/4 wavelength is 75 MHz

** The " velocity factor " for typical ( RG-6U, RG58/U, RG59/U etc) co-ax is
66%.

So the wave travels at 200,000,000 m/s.



.... Phil

 

[Robert Baer]

Here's a cheap 51 ohm carbon film resistor:

https://dl.dropboxusercontent.com/u/53724080/TDR/51R_setup.JPG

and its TDR step response

https://dl.dropboxusercontent.com/u/53724080/TDR/51R_TDR.JPG

The main error seems to be from the length of the resistance element,
which is over in about 200 picoseconds. Then it rings at 9.6 GHz for a
bit.

Plenty good up to maybe 1 GHz. Carbon comps are probably better.

Try two 100 ohm resistors, "Y" from cable to ground, about 60 degrees?

 

[George Herold]

I guess you could drive the line with current, turn off the current,
then terminate.

Anyone want to write-up what waveform you'd see ?

                                        ...Jim Thompson
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I was thinking you could do it either way. 50 ohms to ground with a
current source or added in series with a voltage source. Hey maybe
just a diode in parallel with 50 ohms driven by a voltage source...
ohh time to play with Jeroen's spice file a bit!

George H.

 

[George Herold]

I'm toying around with just such a design. Imagine a 50 ohm pullup
resistor from Vcc and a fast transistor that can pull it to ground.
The transistor collector drives a coaxial cable, through a series
capacitor, out to the load.

If I turn the transistor on hard, for some 10s of picoseconds, it
launches a pulse of -Vcc down the cable. By the time any reflections
return, the transistor is off and the 50 ohm terminator is back in
place.

Fun! I don't have any use for such a scheme, but if you had an
unknown or variable load at the end of the coax it might be a nice
trick.

George H.

 

[Fred Abse]

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?

There is always a reflection from the unterminated end. The 50 ohm source
termination absorbs the reflected energy.

A hi-Z 'scope input isn't an open circuit, BTW, there's capacitance to
consider.

My Gould pulse generator's 50V-into-50 ohms output module has a source
impedance of about 7 ohms. I made a 43 ohm source terminator (Caddock 10W
precision resistors) for that. Minimum ringing from an open cable.

 

[Tauno Voipio]

Typical coax is around 0.66 or thereabouts...

http://en.wikipedia.org/wiki/Wave_propagation_speed

...Jim Thompson

I use Times Microwave LMR series. (Or BuryFlex if I don't have to pass
code.) Velocity <sub>Propagation depends on the dielectric constant of
the spacer between the outer surface of the inner conductor and the
inner surface of the outer conductor. I did not have to look that up. I
grew up reading Reference Data for Radio Engineers (Bill Orr.et al), and
have some of it memorized.

http://www.rfparts.com/old_site/coax_specs.html[/QUOTE]


The mechanical layout of coaxial cable limits the available
propagation speed range and impedance range. The logarithmic
relation of diameter ratio to the capacitance / length unit
and inductance / length unit keeps all practical cables
pretty the same.

If you change the dielectric constant, you have to change
the diameter ratio also, to preserve the wave impedance,
and the speed factor will stay quite near 2/3.</sub>

 

[Fred Abse]

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.)

That will present less than 50 ohms to the line, do this:

T-pad:- 500 ohms in, 50 ohms out, -20dB, theoretical values:

Version 4
SHEET 1 880 680
WIRE 192 96 144 96
WIRE 240 96 192 96
WIRE 192 176 192 96
WIRE 192 288 192 256
FLAG 192 288 0
SYMBOL res 160 80 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 478
SYMBOL res 336 80 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 19
SYMBOL res 208 272 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R3
SYMATTR Value 32
TEXT -48 176 Left 2 ;500 ohms in
TEXT 376 184 Left 2 ;50 ohms out

 

[Fred Abse]

In this case the source doesn't have enough
current to drive the 50 ohms.

What will it drive? How much attenuation can you stand?

 

[Fred Abse]

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.

If the source R>50, use a parallel resistance to bring the effective
source resistance to 50 ohms, You might need to properly pad down to 50,
if the source doesn't like 50 ohm loading. For example, you can get a 1000
to 50 ohm pad, accurate in both directions, for a 20dB insertion loss.

If source R<50, use a series resistor. See my previous post regarding
Gould pulse generator's 7 ohm source resistance.

The source resistance is part of the source termination. You just need to
include it when calculating a series or shunt source termination, and
allow for its effect on the launch voltage.


A true 50 ohm source (if you are lucky enough to find one) doesn't need
any further correction. That's why TDRs give you a clean step on an open
line without ringing, they have expensively accurate 50 ohm sources.

 

[Fred Abse]

GR used to sell true precision air core 50 ohm lines;
probably someone still does.

Wouldn't surprise me. Looking through MWJ, I never cease to be amazed at
the number of companies still making niche stuff, like interseries adapters
at 400+ bucks a pop.

 

[Jeroen Belleman]

On Thu, 25 Apr 2013 07:25:23 -0700, dave <

Do I multiply 75 X 0.87 = 62.5?

Typical coax is around 0.66 or thereabouts...

http://en.wikipedia.org/wiki/Wave_propagation_speed

...Jim Thompson

I use Times Microwave LMR series. (Or BuryFlex if I don't have to pass
code.) Velocity <sub>Propagation depends on the dielectric constant of
the spacer between the outer surface of the inner conductor and the
inner surface of the outer conductor. I did not have to look that up. I
grew up reading Reference Data for Radio Engineers (Bill Orr.et al), and
have some of it memorized.

http://www.rfparts.com/old_site/coax_specs.html[/QUOTE]


The mechanical layout of coaxial cable limits the available
propagation speed range and impedance range. The logarithmic
relation of diameter ratio to the capacitance / length unit
and inductance / length unit keeps all practical cables
pretty the same.

If you change the dielectric constant, you have to change
the diameter ratio also, to preserve the wave impedance,
and the speed factor will stay quite near 2/3.
[/QUOTE]

That isn't true. The propagation speed goes as 1/sqrt(epsilon_r).
It does not depend on the ratio of diameters. Coax in which the
dielectric is mostly air will have a speed factor much closer to
unity. Look up 'air dielectric coax'.

Jeroen Belleman</sub>

 

[George Herold]

That will present less than 50 ohms to the line, do this:

T-pad:- 500 ohms in, 50 ohms out, -20dB, theoretical values:

Version 4
SHEET 1 880 680
WIRE 192 96 144 96
WIRE 240 96 192 96
WIRE 192 176 192 96
WIRE 192 288 192 256
FLAG 192 288 0
SYMBOL res 160 80 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 478
SYMBOL res 336 80 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 19
SYMBOL res 208 272 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R3
SYMATTR Value 32
TEXT -48 176 Left 2 ;500 ohms in
TEXT 376 184 Left 2 ;50 ohms out

OK, Thanks Fred. I don't really need a new design. I'm just trying
to understand what I already did :^) Is the above "tee" circuit any
better than 450 followed by 56 ohms to ground? It certainly
attenuates more.

George H.

 

[Jeroen Belleman]

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.)

That will present less than 50 ohms to the line, do this:

T-pad:- 500 ohms in, 50 ohms out, -20dB, theoretical values:
[deleted ....]

OK, Thanks Fred. I don't really need a new design. I'm just trying
to understand what I already did :^) Is the above "tee" circuit any
better than 450 followed by 56 ohms to ground? It certainly
attenuates more.

George H.

It's -30dB (20log(Uo/Ui))! That's a waste!

A minimum loss matching L-pad between 500 and 50 ohms would
have 474.3 ohms in series and 52.7 ohms to ground at the 50
Ohm side. That's about 4dB better and the best that can be
done with resistors.

Jeroen --Every dB counts-- Belleman

 

[Mark]

speaking of charging up cables...

we had an interesting failure...

we had a small GPS active antenna on the end of about 100 ft of
coax... the ant / LNA has a simple circuit powered by DC on the cable
and we had bias T to feed +5 volts into the cable at the lab end.

one day, the pig tail / adapter at the LNA end developed an
intermittent short... should be no big deal the +5 v supply was
current limited to an Amp or so..

but after the short was repaired we found the LNA was fried...

I thought about it and realized that when the cable was shorted about
1 Amp was flowing, and when the short cleared, 1 Amp continued to flow
for the duration of the cable prop delay. That was like a current
source and the voltage at the LNA end when up to whatever it takes to
keep that 1 Amp flowing for the duration...

Mark

 

[Jasen Betts]

speaking of charging up cables...

we had an interesting failure...

we had a small GPS active antenna on the end of about 100 ft of
coax... the ant / LNA has a simple circuit powered by DC on the cable
and we had bias T to feed +5 volts into the cable at the lab end.

one day, the pig tail / adapter at the LNA end developed an
intermittent short... should be no big deal the +5 v supply was
current limited to an Amp or so..

but after the short was repaired we found the LNA was fried...

I thought about it and realized that when the cable was shorted about
1 Amp was flowing, and when the short cleared, 1 Amp continued to flow
for the duration of the cable prop delay. That was like a current
source and the voltage at the LNA end when up to whatever it takes to
keep that 1 Amp flowing for the duration...

That doesn't work. Coaxial cables have basically no parasitic inductance,
there's nothing to charge up unless perhaps the cable was grounded at
both ends. or some other wierdness with the powersupply.

 

[Jeroen]

speaking of charging up cables...

we had an interesting failure...

we had a small GPS active antenna on the end of about 100 ft of
coax... the ant / LNA has a simple circuit powered by DC on the cable
and we had bias T to feed +5 volts into the cable at the lab end.

one day, the pig tail / adapter at the LNA end developed an
intermittent short... should be no big deal the +5 v supply was
current limited to an Amp or so..

but after the short was repaired we found the LNA was fried...

I thought about it and realized that when the cable was shorted about
1 Amp was flowing, and when the short cleared, 1 Amp continued to flow
for the duration of the cable prop delay. That was like a current
source and the voltage at the LNA end when up to whatever it takes to
keep that 1 Amp flowing for the duration...

Mark

Not quite. A coaxial cable isn't a pure inductor. It's true that
after the short clears, 1A continues to flow into the cable for
another cable delay. At the LNA end, neglecting the LNA current,
that would translate into a 50V step that propagates back towards
the power supply and which persists for twice the cable delay.
What happens next depends much on the power supply's behaviour.

But you were right that this can kill an LNA that isn't protected
against such events.

In this context, coaxial cables can hold nasty surprises. Just
last week I wanted to check the connectors on a piece of cable
that had been sleeping in a cupboard for years. It was just a few
meters long. I foolishly neglected to discharge it before connecting
it to my Tektronix S-6 sampler head. That was the end of that one.
What a shame to break such a fine piece of vintage hardware for a
trivial measurement attempt. (Snarl!)

Jeroen Belleman