Purposely lossy coax: How to determine resistance?

George Herold wrote:
> Yup that's it... just a google find so no idea of the quality.
> (NPI)

IMO QEX magazine is often surprisingly good -- it's a great resource 
that many academics are blinded to, unfortunately!

Heck, I'd even suggest that these days, while IEEE journals certainly do 
have good articles in them, the SNR is no better or even worse than with 
QEX. :-)

On Feb 23, 6:36=A0pm, Joel Koltner <zapwire-use...@yahoo.com> wrote:
> Hi guys,
>
> As most peple here are likely aware, if you examine the coax cable for a
> passive oscilloscope probe you find that the inner conductor is made of
> a highly resistive wire -- something like nichrome, and around
> 50-200ohms per meter. =A0(For some background on this, check out this
> article:http://www.dfad.com.au/links/THE%20SECRET%20WORLD%20OF%20PROBES%2=
0OCt...
> ).
>
> What's not clear to me, though, is how one analytically determines the
> desired resistance of their coax in such a situation, given knowledge of
> the source and termination impedances of the cable. =A0The magazine
> article above mentions that, at least at Tektronix, John Kobbe came up
> with the idea... and if you then dig into some of the Tek archives, you
> can find his reminisces about doing so, where he says something along
> the lines of, "it occurred to me that lossy coax would work well here
> [to greatly extend the bandwidth of a passive probe], so I just
> calculated what the appropriate resistance would be, ran down to the
> stock room, got some and tried it out... it worked great, and Howard
> Vollum himself took me to dinner that night as a reward!" =A0(Just kiddin=
g
> on that last part...)
>
> Anyway. =A0OK, it worked... cool! =A0But... ummm... does anyone happen to
> have some pointers on how he might have gone about performing that
> little calculation? =A0I'm pretty well-versed in transmission line theory
> and imaginary characteristic impedances don't scare me. :-) =A0I'd prefer
> to read up on a little theory here than just perform the "cut and try"
> approach in a simulator that the article above uses.


Are you asking the "distortionless line" question?

          / R + jwL \
Zo =3D sqrt( --------- )
          \ G + jwC /

          / 1 + jwL/R    R \
   =3D sqrt( ---------  * --- )
          \ 1 + jwC/G    G /

Condition of distortionless:

 L     C
--- =3D ---
 R     G


propConstant =3D attnConstant + j*phaseConstant

propConstant =3D sqrt((R+jwL)*(RC/L+jwC))

attnConstant  =3D R*sqrt(C/L)
phaseConstant =3D w*sqrt(C*L)


Is the attnConstant the one you're looking for?
It seems like R & G would be built in to create the loss.

G =3D RC/L

Funny how you can put the differential quantities in Q form:

          / 1 + jQseries      R \
Zo =3D sqrt( --------------  * --- )
          \ 1 + jQshunt       G /

Since Qseries =3D Qshunt for distortionless, and G =3D RC/L, then the same-
old familiar form drops out:

Zo =3D sqrt(L/C)

Hi Simon,

Simon S Aysdie wrote:
> Are you asking the "distortionless line" question?

I might be without knowing it. :-)

> Condition of distortionless:
>
>   L     C
> --- = ---
>   R     G

...or R = L*G/C = G * Zo^2.  That would certainly work, although I get a 
feeling that passive high-frequency probes (that use this lossy coax) 
aren't distortionless: With 50ohm coax, from the article posted I know 
that R is roughly 100ohms/meter, so to satisfy the above equation it'd 
require ... G = R/Zo^2 = 100/50^2 = 40mS/meter.  I'm pretty sure that's 
orders of magnitude higher than any real coax is?

---Joel

On Feb 28, 3:15=A0pm, Joel Koltner <zapwire-use...@yahoo.com> wrote:
> Hi Simon,
>
> Simon S Aysdie wrote:
> > Are you asking the "distortionless line" question?
>
> I might be without knowing it. :-)
>
> > Condition of distortionless:
>
> > =A0 L =A0 =A0 C
> > --- =3D ---
> > =A0 R =A0 =A0 G
>
> ...or R =3D L*G/C =3D G * Zo^2. =A0That would certainly work, although I =
get a
> feeling that passive high-frequency probes (that use this lossy coax)
> aren't distortionless: With 50ohm coax, from the article posted I know
> that R is roughly 100ohms/meter, so to satisfy the above equation it'd
> require ... G =3D R/Zo^2 =3D 100/50^2 =3D 40mS/meter. =A0I'm pretty sure =
that's
> orders of magnitude higher than any real coax is?

yeah.  A distortionless line is a neat idea, but it isn't what they do
on further examination.

After making my post I downloaded that Tek paper someone referenced.
It explains what they are doing in there.

They basically say G approaches 0, so

          / jwL      R  \
Zo =3D sqrt( ----- + ----- )
          \ jwC     jwC /

As w gets big, it is back to same the same old formula.  It doesn't
matter that Zo is bigger at low frequencies.

It is an very interesting topic for me.  I had known that the probe
lines were lossy for a long time, but only had a vague feel about how
they operated.  I knew it was crap without them though, based on
experience.  lol!

Yet you know, I'm sure, that the highest performance today does not
use that technique, and instead amounts to active probes and a 50 Ohm
doubly termed setup, which perhaps is not surprising.

Thanks to your question, I now have a new set of interview questions.
Then, if I don't like their "tatoos," I can say in the wrap-up "the
dude doesn't even know how a simple scope probe works!"

Simon S Aysdie wrote:
> They basically say G approaches 0, so
>
>            / jwL      R  \
> Zo = sqrt( ----- + ----- )
>            \ jwC     jwC /
>
> As w gets big, it is back to same the same old formula.  It doesn't
> matter that Zo is bigger at low frequencies.

Good point, thanks.

> Yet you know, I'm sure, that the highest performance today does not
> use that technique, and instead amounts to active probes and a 50 Ohm
> doubly termed setup, which perhaps is not surprising.

I generally agree, although a passive probe wins on sales price and -- 
in many cases -- performs well from millivolts to many tens of volts, 
whereas active probes usually aren't designed (from a noise perspective) 
to cover that many decades.  How many users actually care about these 
factors is likely a significant function of whether you ask the Tek 
salesguy vs. the LeCroy or Agilent salesguy...

(Tek's presently the only company offering a 1GHz passive probe; they 
even include one per channel with their 1GHz scopes.  If you want the 
*fastest* scopes and probes out there, LeCroy is where you'd look. 
60GHz real-time via 50ohm coax and something like 25GHz via high-Z 
probes... amazing!)

> Thanks to your question, I now have a new set of interview questions.
> Then, if I don't like their "tatoos," I can say in the wrap-up "the
> dude doesn't even know how a simple scope probe works!"

I didn't know that the "magic" of passive probes was the use of lossy 
coax until I was some years out of college, although I *do* remember 
thinking the first time I saw the simplified schematic of a probe that 
no way could you just blithely ignore the meter-long coax with the 
then-already-400MHz passive probes that were available!

---Joel

On Tue, 28 Feb 2012 17:58:29 -0800, Joel Koltner
<zapwire-usenet@yahoo.com> wrote:

>
>> Thanks to your question, I now have a new set of interview questions.
>> Then, if I don't like their "tatoos," I can say in the wrap-up "the
>> dude doesn't even know how a simple scope probe works!"
>
>I didn't know that the "magic" of passive probes was the use of lossy=20
>coax until I was some years out of college, although I *do* remember=20
>thinking the first time I saw the simplified schematic of a probe that=20
>no way could you just blithely ignore the meter-long coax with the=20
>then-already-400MHz passive probes that were available!

I join you in having learned stuff from this thread.

Enjoy

?-)))

Joel Koltner wrote:

> Simon S Aysdie wrote:
>> They basically say G approaches 0, so
>>
>>            / jwL      R  \
>> Zo = sqrt( ----- + ----- )
>>            \ jwC     jwC /
>>
>> As w gets big, it is back to same the same old formula.  It doesn't
>> matter that Zo is bigger at low frequencies.
> 
> Good point, thanks.
> 
>> Yet you know, I'm sure, that the highest performance today does not
>> use that technique, and instead amounts to active probes and a 50 Ohm
>> doubly termed setup, which perhaps is not surprising.
> 
> I generally agree, although a passive probe wins on sales price and --
> in many cases -- performs well from millivolts to many tens of volts,
> whereas active probes usually aren't designed (from a noise perspective)
> to cover that many decades.  How many users actually care about these
> factors is likely a significant function of whether you ask the Tek
> salesguy vs. the LeCroy or Agilent salesguy...
> 
> (Tek's presently the only company offering a 1GHz passive probe; they
> even include one per channel with their 1GHz scopes.  If you want the
> *fastest* scopes and probes out there, LeCroy is where you'd look.
> 60GHz real-time via 50ohm coax and something like 25GHz via high-Z
> probes... amazing!)
> 
>> Thanks to your question, I now have a new set of interview questions.
>> Then, if I don't like their "tatoos," I can say in the wrap-up "the
>> dude doesn't even know how a simple scope probe works!"
> 
> I didn't know that the "magic" of passive probes was the use of lossy
> coax until I was some years out of college, although I *do* remember
> thinking the first time I saw the simplified schematic of a probe that
> no way could you just blithely ignore the meter-long coax with the
> then-already-400MHz passive probes that were available!
> 
> ---Joel

I'm just wondering (out loud) if the added R gives you a situation where the
real as well as the reactive impedance components of the transmission line
match.

If you view a transmission line as a bunch of lumped impedances strung
together, then you'll want to match the dielectric loss of the cable
(parallel component) with a series loss that have same ratio as the
imaginary components (L and C).
 
-- 
Paul Hovnanian     mailto:Paul@Hovnanian.com
------------------------------------------------------------------
The world is coming to an end ... SAVE YOUR BUFFERS!!!

Hi Paul,

Paul Hovnanian P.E. wrote:
> I'm just wondering (out loud) if the added R gives you a situation where the
> real as well as the reactive impedance components of the transmission line
> match.

I believe that gets you back to a distortionless line again, right? -- 
But in the case of the scope probe coax, AFAIK while the center 
conductor (R) is significantly lossy, there dielectric is still almost 
lossless (G approaches 0).

I wonder if anyone has ever mass produced distortionless coax?  I'm 
guessing the answer is "no" in then it'd be difficult to provide the 
requisite hi-Z termination much above some many MHz... whereas you can 
provide a pretty good 50ohm termination to at least some GHz... and just 
making the center conductor lossy is sufficient if your goal is to not 
have to worry about termination impedances too much.

---Joel

On Wed, 29 Feb 2012 11:53:09 -0800, Paul Hovnanian P.E. wrote:

> I'm just wondering (out loud) if the added R gives you a situation where the
> real as well as the reactive impedance components of the transmission line
> match.

There's no difference between  "unavoidable" series resistance (copper
cable", and "intentional" resistance (resistance wire). Both contribute to
to the resistance per unit length in the same way.

> 
> If you view a transmission line as a bunch of lumped impedances strung
> together, then you'll want to match the dielectric loss of the cable
> (parallel component) with a series loss that have same ratio as the
> imaginary components (L and C).

Given the transmission line equation:

Zo = sqrt(R + jX / G - jB) 
(note the sign of B, by convention, capacitive susceptance is negative)

Then for nonzero R and/or G, Zo *must* be complex, ie. any line with
finite losses has complex Zo.

Only in the case of a theoretical lossless line does this reduce to the
familiar:

Zo = sqrt(jX/-jB) = sqrt(jwL/jwC) = sqrt(L/C).


RG223 (nominally Zo=50) has Zo = 50.34 -j5.38 at 100KHz, and 174.56
-j167.25 at 1KHz ! Only above about 1MHz does it get to 50 + 0j, to one
significant figure. 
[Based on Belden figures for R and L, with G dielectric
losses assumed negligible]

The use of resistive cable in oscilloscope probes damps ringing on fast
transients (lower Q), but brings its own problems in high frequency
probes, since it degrades rise time, needing receiving end series-parallel
LCR compensation to sharpen the response up again. The probes I use (PMK
500MHz), have the usual series LF compensation adjustment at the probe
end, and two HF adjustments at the scope end. They always seem to come
from the factory slightly overpeaked at HF, needing some TLC with a tunnel
diode pulser to get really flat.


-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)