Finding a minor Votage drop on a 50V source... It is possible?

One *cannot* depend on an un-loaded phone line to have 50V or any other value to the 10uV precision and accuracy you seem to "need".

Translation: such "modern" taps are not detectable.

Thanx John,

Nope.... Here is really dead :( maybe I have to reboot, as I'm connected more than 24 days.... I'll try to reboot later as I'm d/ling a large graphics s/w ISO, so maybe this is the problem...

Regards, Nina

time

From her numbers, it's actually somewhere close to 100uV, but we'll go with 1mV for arguments sake. At any rate, you're just as incorrect as the OP, please face facts Phil.

Wow, where did that come from? Did I make a mathematical mistake somewhere?

I didn't see anyone specify the source impedance of the CO, well except you anyway. Earlier you specified it as 2000 ohms, but you were a whole order of magnitude off on the voltage drop so, correcting for your guestimafabrications, the source impedance should really work out to be somewhere around 200 ohms. Since the OP is not reading a full 1mV drop from the load, that indicates that the actual impedance would be even less than 200 ohms, much less. Something isn't right. Maybe the OP isn't waiting long enough for the charge stored in the line capacitance to bleed thru the 10Mohm load. She should see a larger voltage drop, something like 4.8mV for a 10Mohm load [assuming a nominal 1K telephone CO source impedance and 48V], right?

@martin

A good 500V shock from a megger may "fix" it too ?

Nope, I'm looking for a more professional way :) Regards, Nina

I read in sci.electronics.design that nina.p20 wrote (in ) about 'Finding a minor Votage drop on a 50V source... It is possible?', on Sun, 6 Nov 2005:

In principle, you can sample the line voltage every 100 ms, say. You store the voltage on a low-leakage capacitor. You can't directly connect this capacitor to the line, of course. You will need to feed it from a high-input impedance buffer which has a very stable offset voltage. You can then compare the sampled line voltage with what it was 100 ms ago.

This is very difficult indeed to do if you want to see 10 uV. But it's true that the d.c. source impedance of the line is of the order of 1000 ohms. 48 V across 10 Mohms gives a current of 4.8 uA, which must cause a voltage drop of 4.8 MILLIvolts across the 1000 ohms source resistance. So I don't think you need to detect 10 uV.

I read in sci.electronics.design that nina.p20 wrote (in ) about 'Finding a minor Votage drop on a 50V source... It is possible?', on Sun, 6 Nov 2005:

You can easily do it. Just connect the 'ground' of your comparator circuit to the -48 V supply. Now you are not looking for 4.8 mV on top of 48 V but 4.8 mv on top of 0 v. Much simple.

However, there is always a snag. It's probably reasonably safe to do this at a switch/central/exchange, but if you do it at a subscriber terminal you have to take into account what happens if you get a power cross or an indirect lighting strike. Your detector box needs protection for its own circuits and protection against you getting a large electric shock from it. The latter is probably easy - a substantial plastic enclosure with NO exposed metal parts. The protection the circuit needs depends on the detailed design.

I read in sci.electronics.design that Phil Allison wrote (in ) about 'Finding a minor Votage drop on a 50V source... It is possible?', on Sun, 6 Nov 2005:

Maybe my explanation is easier to understand.

Yes there is. Those batteries' no load voltage will vary more than 10 uV due to temperature variations.

That works IF you have before and after traces of the line in question. There is enough garbage (splices, load coils, etc.) on a typical phone line so that a single snapshot will tell you little. You'd have to open every pedestal and handhole to make sure that an impedance mismatch wasn't due to the typical hairball of splices.

I read in sci.electronics.design that nina.p20 wrote (in ) about 'Finding a minor Votage drop on a 50V source... It is possible?', on Sun, 6 Nov 2005:

Phil Allison is notorious for bad language. He is only tolerated because he does know about electronics.

Plus the majority of those who would assume such authority are largely unemployable, indolent, losers of no detectable worth on any scale whatsoever- probably why they "hang" on usenet all day- and night- and weekends- and holidays...

I read in sci.electronics.design that nina.p20 wrote (in ) about 'Finding a minor Votage drop on a 50V source... It is possible?', on Sun, 6 Nov 2005:

Defeated? No, I have no idea. I can think of a number of ways that are strictly illegal, because they would damage the telephone system itself, but nothing that can be done legally (beyond locating them and removing them with the aid of regulatory officials).

I read in sci.electronics.design that Phil Allison wrote (in ) about 'Finding a minor Votage drop on a 50V source... It is possible?', on Mon, 7 Nov 2005:

Please do. I will reply in 24 hours time.

It is 1mV and you are incredibly thick.

as

Who's posturing Phil? The OP was misinterpreting the flickering last digit (the millivolt digit) as a 10uV change when, as you and I and everyone else here knows, in reality it was somewhere between 100uV and

I wasn't really "applying" any numbers, just partaking of the chit-chat.

does

I'm sorry, how bout we call it an "erroneous conclusion" then?

I was just trying to be nice since you were actually off by two orders of magnitude.

I'd say that qualifies as mean spirited. She gave you the readings and then you misread it too. Now really Phil, who's the colossal idiot?

sense.

Right. And just in case she's having trouble figuring it out, if she is using OE she can just click on "Message" and then "Block Sender" whilst highlighting your reply to this.

HAND :-)

you should measure the current in the line; possibly with a clamp on meter.

Well a lead acid battery on float (as I believe these battery banks are) has a voltage approaching 53V so maybe there is some sort of regulator on the system already? Although why the exchange would bother regulating that closely I have no idea.

Robert

"nina.p20"

Maybe the OP really meant the Fluke 179, 6000 count, hand held meter.

On its 60 volt DC range the resolution is 10 mV.

Not 10 uV.

10mV divided by 5 uA = 2000 ohms.

Seems about the right source R for a phone line.

........ Phil

have a look here

martin