sundance-communications.com/forum Forums General General Telephone & Systems Information loop current

hello. i have a customer who had blown line port one on 2 ksus. the loop current is at 65 miliamps and the tel co says they cant do anything about it. what is the best way to handle this situation. thanks dave

Sandman has loop current regulators. I have used them and they work real well. Sandman loop current regulators

i may have to go that route but at 40 bucks a line i hope to find an alternate solution.

Not unless you can convince the LEC that is too high, I tried for a very long time. Now I buy the regulators. If you'll search on loop current, you'll find quite a bit of discussion on this.

Why not add a resistor or potentiometer?

Okay, I'm gonna bite. Please show us how to wire that in MP.

Actually you can mess around with resistors to lower the loop current just like adding loop length to lower it, but you also lower voltage and volume. The loop current regulators at Sandman are constant and keep the current right at 25 milliamps. Even if you could get resistors to get it right the LEC (at least here) is always messing around and the current changes. How Sandman's work, don't know, don't care, they work.

according to sandman the telco can be up to 125ma and be within FCC specs so I don't think your going to get them interested in dropping it

what kind of equipment is seeing damage at 65ma ?


I carry a half dozen or so of his and install them when needed Ive never had a customer balk at the price

(his early ones had dip switch's and you had to set it up to regulate the current down to 25ma , problem was if the telco dropped then the regulator would drop it a corresponding amount. the new ones just put out 25ma no matter what is going in , like Bill said I don't care how he does it , It works)

The way you would use a resistor or potentiometer (variable resistor) is to just add it in series to the line.

Lets actually calculate what size resistor would work. Presently, since R=E/I, E=48 V and I=0.065 A, it follows that R must equal 48/0.065 or 738 Ohms. That is the resistance of the entire telephone circuit. To get the desired current of 0.025 A, that resistance should be 48/0.025 or 1920 ohms. Therefore you must add 1920-738 = 1182 ohms of resistance to the circuit.

The wattage of the resistor needs to be at least 1182/1920 x 48 V x .025 A = 0.73 W, or as a practical matter, ¾ Watt. You might want to go to 1 Watt.

You can buy a few of these resistors for a dollar or two. The color code for a 1200 ohm resistor is brown-red-red.

If anyone would like to check my calculations, I’d appreciate it. I haven’t done this for a while.

I've used the resistor method to cut down on loop current and not run into any problems.

I was taught that you should add the same amount of resistance to both tip & ring so I always follow that advice.