Discussion in 'Electronic Design' started by Epsilon Rho, Feb 29, 2008.

1. ### Epsilon RhoGuest

A few weeks ago I posted a message regarding some problems with my ADSL
line. Following the intervention of the AT&T technician, the line is now OK.
In an effort to understand the original problem (a mismatch between the Ring
and Tip impedance) I measured again such impedances. Now that everything
seems to be operating properly, I measured approx. 2,200 ohm (DC) on the Tip
and on the Ring line. I know that at the central office (CO) there is a 200
ohm resistance, so each wire resistance should be 2,000 ohm. AT&T told me
that they cannot upgrade the speed to my location because I am too far from
the CO. I learned that this distance is 18,000 feet.

Armed with these numbers, I tried to calculate how far I am from the CO. And
this is what is puzzling. Assuming that I am at 18,000 feet away, the AWG
that would give a reading of 2,000 ohm at 18,000 feet is pegged at around
30-31 AWG and that is impossible, because this kind of gauge is far too thin
for a phone line. As a matter of fact, I read that the smallest AWG used in
the phone line is 24 AWG.

Has anybody an idea of what is going on?

TIA

Gene

2. ### BobWGuest

Gene,

First, you are using the word "impedance" to describe the loop resistance.
The impedance of a phone line is an a.c. characteristic and the magnitude
and phase of the impedance varies greatly with frequency.

The telcos use 24awg and 26awg copper. You haven't said what method you used
to calculate the loop resistance, but unless you can get someone to short
the pair at the central office or the remote terminal, you will never know
what its true loop resistance is. This is because not all line circuits feed
the loop the same way.

There are some good books out there that describe telephony in detail. One
of them is:

"Subscriber Loop Signaling and Transmission" by Whitham Reeve.

http://www.amazon.com/Subscriber-Si...=sr_1_2?ie=UTF8&s=books&qid=1204321176&sr=1-2

This will get you much closer to a true understanding of what is going on.
Otherwise, you're just guessing.

Bob

3. ### ehsjrGuest

I get a significant difference in calculation using
http://www.bnoack.com/data/wire-resistance.html

I get AWG 27, figuring 18000 feet (2 wires).

If we go the other way, figuring AWG 24 at 27.3 ohms
per kfoot and 2000 ohms, it comes to ~73,000 feet of
wire. Figuring 2 wires, that's ~36,000 feet distance.

Probably time to draw it up with some precision. At this
point, every number is in question - impedance, distance,
wire gauge, even the 200 ohms in the telco.

Ed

4. ### Epsilon RhoGuest

I get a significant difference in calculation using
http://www.bnoack.com/data/wire-resistance.html

I get AWG 27, figuring 18000 feet (2 wires).

If we go the other way, figuring AWG 24 at 27.3 ohms
per kfoot and 2000 ohms, it comes to ~73,000 feet of
wire. Figuring 2 wires, that's ~36,000 feet distance.

Probably time to draw it up with some precision. At this
point, every number is in question - impedance, distance,
wire gauge, even the 200 ohms in the telco.

Ed

5. ### Epsilon RhoGuest

Thank you all for the replies. As I pointed out, I measured the impedance
(in this particular case is reduced to resistance) using a DC method, i. e.
by loading the line with a known resistor of 15Kohm connected to RING & TIP
and measuring the voltage drop at TIP & RING. This is straightforward Ohm's
law. Nothing to do with AC and associated reactive impedances. Each leg
shows 2,200 ohm to ground. I used the phone ground and just to be sure it is
a reliable ground, I checked also against the power ground of the house,
without noting any discrepancy. I am absolutely certain that the figure of
2,200 is correct. I did the same measurement at the phone line of my
friend's house a few miles away and I measured 450 ohm for each leg.

AWG 30 shows 113 ohm per 1000 feet, therefore 18 x 113 = 2,034 ohm as I have
calculated for each leg.
In a nutshell by loading the line with 15Kohms I get a voltage drop of 11.41
V. The open loop voltage is 50.80 V; when the line is loaded, the voltage
across the 15Kohm is 39.39 V, the current is 39.39 V/15Kohm = 2.63 mA. The
line resistance is therefore (50.80-39.39)/2.63 mA = 4,338 ohm (both wires)
and when divided by 2 we get 2,169 ohm for each wire!
TIA
Gene

I get a significant difference in calculation using
http://www.bnoack.com/data/wire-resistance.html

I get AWG 27, figuring 18000 feet (2 wires).

If we go the other way, figuring AWG 24 at 27.3 ohms
per kfoot and 2000 ohms, it comes to ~73,000 feet of
wire. Figuring 2 wires, that's ~36,000 feet distance.

Probably time to draw it up with some precision. At this
point, every number is in question - impedance, distance,
wire gauge, even the 200 ohms in the telco.

Ed

6. ### Ross HerbertGuest

:A few weeks ago I posted a message regarding some problems with my ADSL
:line. Following the intervention of the AT&T technician, the line is now OK.
:In an effort to understand the original problem (a mismatch between the Ring
:and Tip impedance) I measured again such impedances. Now that everything
:seems to be operating properly, I measured approx. 2,200 ohm (DC) on the Tip
:and on the Ring line. I know that at the central office (CO) there is a 200
hm resistance, so each wire resistance should be 2,000 ohm. AT&T told me
:that they cannot upgrade the speed to my location because I am too far from
:the CO. I learned that this distance is 18,000 feet.
:
:
:
:Armed with these numbers, I tried to calculate how far I am from the CO. And
:this is what is puzzling. Assuming that I am at 18,000 feet away, the AWG
:that would give a reading of 2,000 ohm at 18,000 feet is pegged at around
:30-31 AWG and that is impossible, because this kind of gauge is far too thin
:for a phone line. As a matter of fact, I read that the smallest AWG used in
:the phone line is 24 AWG.
:
:
:
:Has anybody an idea of what is going on?
:
:
:
:TIA
:
:Gene

As BobW has indicated you can't measure the actual dc line loop resistance
without disconnecting the line at the MDF and placing a short circuit across it
while measuring at the distant end. Looking down the line towards the exchange
you effectively have a 48V battery connected between the wires at the exchange
end plus the battery feed retard windings (200+200ohms) plus the line
resistance. Depending upon the type of exchange (electromechanical analog or
digital) the line interface arrangements may differ slightly.

In most cases a 26AWG line loop resistance of not more than 1800ohms is
allowable for reliable telephone service on a POTS line, but this can vary
depending upon the exchange type and line construction. It is usual for a POTS
line construction using 26AWG which has a dc resistance of 133.8568ohms/km
(total length = 13.4km). For a loop resistance of 1800ohms this would put the
maximum distance at about 6.7km (straight line) from the exchange. In practice,
because the cable route will follow street layout, this would limit the radial
distance to probably around 5.5km max from the exchange, give or take. ADSL1 is
generally limited to a line length of 3.5km from the exchange while ADSL2 has an
even shorter distance limit.

7. ### JosephKKGuest

Measurements to ground are not even meaningful. Measure ring to tip.

8. ### sycochknGuest

What would you need? a time domain reflectometer?

Bob

9. ### BobWGuest

TDRs have their limitations on long loops. The high frequency loss of the
loop is large enough such that you can't see a "sharp" reflection. A long
pulse TDR will get there, and back, but the "hump" that returns if very
tough to pick out-of-the-noise.

Also, If the other end of the loop happens to be terminated with 100ohms at
high frequencies (as the case if that line is served with DSL) and 600ohms
at lower frequencies (depending on the POTS line card when it's "on hook"),
then you won't get a farthest-end reflection, anyway. So, you won't be able
to deduce its length with a TDR. Either way, you need to do something at the
other end of the loop.

The phone companies spend big bucks on test equipment. I used to work for a
company that made some of this stuff. We made large amounts of money selling
to the telcos -- at least until the market got saturated with our equipment.
If they (telcos) could have figured out a way to do things like loop length
determination WITHOUT the expensive central office equipment or some
craftsperson sitting at the MDF then they would have done it.

BobW

10. ### Ross HerbertGuest

:TDRs have their limitations on long loops. The high frequency loss of the
:loop is large enough such that you can't see a "sharp" reflection. A long
ulse TDR will get there, and back, but the "hump" that returns if very
:tough to pick out-of-the-noise.
:
:Also, If the other end of the loop happens to be terminated with 100ohms at
:high frequencies (as the case if that line is served with DSL) and 600ohms
:at lower frequencies (depending on the POTS line card when it's "on hook"),
:then you won't get a farthest-end reflection, anyway. So, you won't be able
:to deduce its length with a TDR. Either way, you need to do something at the
ther end of the loop.
:
:The phone companies spend big bucks on test equipment. I used to work for a
:company that made some of this stuff. We made large amounts of money selling
:to the telcos -- at least until the market got saturated with our equipment.
:If they (telcos) could have figured out a way to do things like loop length
:determination WITHOUT the expensive central office equipment or some
:craftsperson sitting at the MDF then they would have done it.
:
:BobW
:
:

The telco's do have the equipment to determine distance to fault testing by
remote control. As an example, several months back, on a Friday evening, a
reticulation contractor cut through the 50 pair local street distributor cable
about 100M away from my house. Being an ex telco tech myself I managed
(eventually) to talk to the rostered testing officer who was located in Tasmania
about 4,000km away. He tested my line and came back with the result that the
cable was broken 961M from the exchange. In this case, the distance to fault
would have been determined by a TDR test because the broken cable produces a
discontinuity reflection.

Even the local fault techs can do line tests from their laptops. These access
the cellphone network and the telco's remote test facility. However, such tests
cannot determine line loop resistance unless the fault tech disconnects the
customer wiring and places a loop on the line.

There are devices which can be fitted to the line at the customer premises
http://www.heuer.com.au/tmsa41/r41_datasheet.pdf which will allow remote
disconnnection and line loop resistance testing to be conducted without human
intervention.

11. ### BobWGuest

[snip]

Yes they do. That's what I was saying. What they cannot do is *all* of their
diagnostics without something or someone at each end.

BobW