Path: utzoo!attcan!uunet!lll-winken!netsys!vector!nobody From: kitty!larry@cs.buffalo.edu Newsgroups: comp.dcom.telecom Subject: Submission for comp.dcom.telecom Message-ID: Date: 9 Oct 88 16:45:04 GMT Sender: chip@vector.UUCP Lines: 48 Approved: telecom-request@vector.uucp (USENET Telecom Moderator) X-TELECOM-Digest: volume 8, issue 159, message 3 X-Submissions-To: telecom@bu-cs.bu.edu (TELECOM Digest Coordinator) X-Administrivia-To: telecom-request@vector.uucp (USENET Telecom Moderator) In article mgrant@cos.com (Michael Grant) writes: > A long while ago someone posted a question about how to build a simple > circuit which would light a LED (or a simple lamp) when an extension > was lifted elsewhere in the house. The requirements are: > > - The LED lights when another extension is lifted, showing that it > is in use. > - The LED must be powered off the phone line. > - No extra wires must be run to the telephone. > - The circuit should be reasonable small so that it can fit inside a > desktop style phone. > > No one was able to come up with a simple schematic to do this that > worked. The upshot was that one could buy such a device for about $30 > off the shelf. Having let some time pass, anyone want to give a stab > at this? Bear in mind, that we appear to be discussing a means which bridges across tip and ring, and which operates by sensing loop voltage, and NOT loop current in series with an off-hook telephone set. There is NO WAY to properly design a circuit as you describe which will not run the risk of interfering with the proper operation of the telephone line. Proper telephone circuit design practice dictates that a bridging impedance across a telephone loop must never be less than 100,000 ohms. This means that the MOST current that one can draw is 50/100000 = 0.0005 A. One can't light an LED on 1/2 mA! The situation is really worse than the above, since we want to light the LED in an off-hook condition, where the loop voltage at the telephone set is between 6 and 25 volts. Even taking the "best" value of 25 volts, 25/100000 = 0.00025 A, which means that only 1/4 mA could be available off-hook to light an LED! Under many conditions, one can fudge the 100,000 ohms bridging impedance to 50,000 ohms without loop interference. However, for the above purposes, there is still an insufficient amount of current which can power the LED (1/2 mA off-hook at best). Trying to bridge a lower resistance across the loop to create more energy to light the LED is an invitation to trouble in the form of: (1) interferring with rotary dial pulsing by causing pulse distortion; (2) false tripping of incoming ringing signals; and (3) generation of trouble reports by automated central office equipment (both ALITS and ESS test procedures). I'm afraid that we have to hang up on this idea. Voltage sensing across tip and ring is okay - but the indicator power is going to have to come from a source other than the telephone line. <> Larry Lippman @ Recognition Research Corp., Clarence, New York <> UUCP: {allegra|ames|boulder|decvax|rutgers|watmath}!sunybcs!kitty!larry <> VOICE: 716/688-1231 {att|hplabs|mtune|utzoo|uunet}!/ <> FAX: 716/741-9635 {G1,G2,G3 modes} "Have you hugged your cat today?"