Path: utzoo!telecom-request Date: 6 Jun 91 15:03:42 GMT From: Julian Macassey Newsgroups: comp.dcom.telecom Subject: Re: Busying Out a Phone With a Resistor? Reply-To: Julian Macassey Message-ID: Organization: The Hole in the Wall Hollywood California U.S.A. Sender: Telecom@eecs.nwu.edu Approved: Telecom@eecs.nwu.edu X-Submissions-To: telecom@eecs.nwu.edu X-Administrivia-To: telecom-request@eecs.nwu.edu X-Telecom-Digest: Volume 11, Issue 434, Message 5 of 9 Lines: 66 In article mju@mudos.ann-arbor.mi.us (Marc Unangst) writes: X-Telecom-Digest: Volume 11, Issue 425, Message 15 of 15 > I recently had one modem in the middle of a 15-line hunt group go out. > Not wanting to move the rest of the modems up a line each, I did some > investigation with a VOM and a telephone, and discovered that plugging > a 270 ohm resistor into the phone line (across tip and ring) should > have almost the exact same effect as an off-hook telephone does -- > thus busying out the line. Now, my question is, is there anything > wrong with doing something like this? Will a 1/2W resistor be enough? > Is this an "accepted" way of busying out a phone line? A regular telephone measured with an Ohm Meter looks pretty much like a 300 Ohm resistor. As I recall the "DC Resistance" Used for testing phone devices is considered to be 200 Ohms. But a real phone (AT&T, ITT, Comdial et al), has "Loop Compensation", this is a usually a silicon carbide varistor. The varistor resistance drops as the voltage is raised. This works as an "Automatic Gain Control (AGC)", so if you are near the CO, the audio won't clean the wax out of your ears. When the phone is on a short loop (Short loop means close to the CO and the loop of wire between your phone and the CO is short), the resistance is quite low. This means there is no absolute DC value for a phone. But read on, and you will see that it doesn't really matter. Part of the whole equation is, how far are you from the CO. Here is why that matters. You can consider that the average distance from a CO is 15 Kilo-Feet (This is another dumb US measurement, the rest of the world uses Kilometers). The wire is a pair, that is a total of 30,000 feet of wire. The wire is usually 24 AWG (0.5mm). The resitance of this wire is 26.17 Ohms per thousand feet. So you will have about 785.1 Ohms worth of resistance between you and the CO. Add to this, about 400 Ohms for the CO. So if you shorted your wires, the total current that can flow assuming the CO voltage is 48V DC is 40 Ma. So for the hell of it, clear the short and add a phone. This adds 200 Ohms more to the loop resistance for a grand total of 1385 Ohms. Your loop current is now 34.6 Ma. My point is, it doesn't really matter whether you put a paper clip across the line or a 300 Ohm resistor. But yes, a half Watt resistor is fine, as is the paper clip. The paper clip has better transient handling capabilities though. Note that after all this resistance drop, what was 48 Volts at the CO is now between 3 and 9 volts depending on your loop length. I have used 1,000 Ohm resistors in the past to busy out lines. Note that many COs will go off hook and feed dialtone supplying only 8 Ma of current. Not many phones will work with 8 Ma. The US minimum current is 20 Ma. The lowest current spec is Sweden with 12 Ma. Sweden is a big country with not many people. Like the Western US, they have some very long loops. I recall when this question last came up, Brian Kantor mentioned that he used an LED and resistor combo which lit up and reminded him that he had a bad line or modem at that location. This is an excellent idea. It is also cheap and simple to implement. Note that the above discussion has been strictly about the DC characteristics of a phone line. They do not address the complex impedances of miles of wet cable, aerial wire, bad splices and crummy quad cable. Julian Macassey, julian@bongo.info.com N6ARE@K6VE.#SOCAL.CA.USA.NA 742 1/2 North Hayworth Avenue Hollywood CA 90046-7142 voice (213) 653-4495