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Again on ADSL line

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

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  1. Epsilon Rho

    Epsilon Rho Guest

    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. BobW

    BobW Guest

    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. ehsjr

    ehsjr Guest

    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 Rho

    Epsilon Rho Guest

    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 Rho

    Epsilon Rho Guest

    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 Herbert

    Ross Herbert Guest

    :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
    :eek: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. JosephKK

    JosephKK Guest

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

    sycochkn Guest

    What would you need? a time domain reflectometer?

    Bob
     
  9. BobW

    BobW Guest

    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 Herbert

    Ross Herbert Guest

    :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
    :eek: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. BobW

    BobW Guest

    [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
     
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