Maker Pro
Maker Pro

I need to detect a DC current of about 0.3 - 0.4 mili-amps, but uncoupled.

R

R.Wieser

Jan 1, 1970
0
Hello all,

I'm in the need to detect a phone-lines status, and the specs tell me that
I'm allowed to use the equivalent of a 80 KOhm resistor. In that case I
expect (hope) that the voltage of the line will stay at least 30 Volts,
which than results in a current of just 0.375 mili-amps.

Now the fun part : I need to detect the status while my electronics are not
DC-coupled to the phone-line (the lines should be able to float, while my
electronics should be able to connect to a computer, which will pull it to
ground-level (or close to it) ).

I thought about using an Opto-coupler. The problem is that I've only used
opto-couplers that use, in comparision with the above, massive ammounts of
current (5-30 mA). As I'm entering unknown territory I could use some
advice ....

A allso won't say "no" to a pointer to an allready existing design :)
Suggestions to do it another way are welcomed allso.

Regards,
Rudy Wieser
 
C

Chris

Jan 1, 1970
0
R.Wieser said:
Hello all,

I'm in the need to detect a phone-lines status, and the specs tell me that
I'm allowed to use the equivalent of a 80 KOhm resistor. In that case I
expect (hope) that the voltage of the line will stay at least 30 Volts,
which than results in a current of just 0.375 mili-amps.

Now the fun part : I need to detect the status while my electronics are not
DC-coupled to the phone-line (the lines should be able to float, while my
electronics should be able to connect to a computer, which will pull it to
ground-level (or close to it) ).

I thought about using an Opto-coupler. The problem is that I've only used
opto-couplers that use, in comparision with the above, massive ammounts of
current (5-30 mA). As I'm entering unknown territory I could use some
advice ....

A allso won't say "no" to a pointer to an allready existing design :)
Suggestions to do it another way are welcomed allso.

Regards,
Rudy Wieser

Hi, Rudy. Your plan to electrically isolate your computer from the
phone line is a good one.

Trying to get close to the edge of the maximum on-hook current probably
isn't a good idea though, but it's possible.

Most people who look at doing this kind of thing (phone in-use
detector) go with a small battery to provide the power for the
optocoupler. Mr. Bowden has a good circuit to drive an LED when the
phone is in use here:

http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm#inuse.gif

His circuit, "Telephone In-Use Indicator", can be modified for minimum
current to drive an optocoupler like this (view in fixed font or M$
Notepad):

|
| TIP 3V Batt +
| ___ |
| o-|___|-o------o-------o--------o
| 3.3M | | | |
| | | 47K| 270.-. VCC
| | 680K| .-. | | +
| 0.1uF | .-. | | | | |
| --- | | | | '-' o----.
| --- | | '-' | | |
| | '-' | |/ 2N3906 | .-.
| | | o------| | 1K| |
| RING | | | |< IC1 | | |
| ___ | | |/ | H11L1 _|6 '-'
| o-|___|-o------o-----| 2N3904 1| | \ |
| 3.3M |> V ~ | H )o-o-->
| | - ~ |__/ 4
| | 2| |
| '--------o |5
| | |
| 3V Batt - ===
| GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

The H11L1 is a logic-level optocoupler that will work with less than
2mA of LED current. This circuit sets LED current at 3mA or so. This
circuit will work for a long time with a 3V lithium battery if you're
space-constrained, or for over a year of moderate use with two AA
batteries. Don't forget the pullup on the H11L1 output (it's open
collector). Use an 0.1uF cap rated for at least 100V.

Good luck
Chris
 
J

John

Jan 1, 1970
0
If you have access to the phone line where it enters the building
(before any phones), then you can use what the old PBX's used - a reed
relay with the coil in *series* with one side of the phone line. This
requires zero power from the line and provides a high level of
isolation to the connected circuitry.

John
 
R

R.Wieser

Jan 1, 1970
0
John <[email protected]> schreef in berichtnieuws
[email protected]...

Hello John,
a reed relay with the coil in *series* with one side of
the phone line.

Sorry, my device goes in parrallel to an allready existing telephone.

Allso : I'm attempting to monitor the line while the attached telephone is
*on*-hook, so there will be no current flowing thru that relay, one way or
the other (no pun intended :) )

Thanks for the response, suggestion though.

Regards,
Rudy Wieser
 
R

R.Wieser

Jan 1, 1970
0
Chris <[email protected]> schreef in berichtnieuws
[email protected]...

Hello Chris,
Hi, Rudy. Your plan to electrically isolate your computer from the
phone line is a good one.

I thought so too :)
Trying to get close to the edge of the maximum on-hook current probably
isn't a good idea though, but it's possible.

Well, its the specs, so it must be usable. Even better : at certain times
it's even adviced to draw that current (around the time a DTMF CLID is put
on the line) :)

Thanks for the suggestions and pointers.

Regards,
Rudy Wieser
 
C

Chris

Jan 1, 1970
0
R.Wieser said:
Chris <[email protected]> schreef in berichtnieuws
[email protected]...

Hello Chris,


I thought so too :)


Well, its the specs, so it must be usable. Even better : at certain times
it's even adviced to draw that current (around the time a DTMF CLID is put
on the line) :)

Thanks for the suggestions and pointers.

Regards,
Rudy Wieser

De nada, Rudy. By the way, I just noticed I got the 2N3906 (PNP)
upside down, & reversed e and c. It should look like this:

|
| TIP 3V Batt +
| ___ |
| o-|___|-o------o-------o--------o
| 3.3M | | | |
| | | 47K| 270.-. VCC
| | 680K| .-. | | +
| 0.1uF | .-. | | | | |
| --- | | | | '-' o----.
| --- | | '-' | | |
| | '-' | |< 2N3906 | .-.
| | | o------| | 1K| |
| RING | | | |\ IC1 | | |
| ___ | | |/ | H11L1 _|6 '-'
| o-|___|-o------o-----| 2N3904 1| | \ |
| 3.3M |> V ~ | H )o-o-->
| | - ~ |__/ 4
| | 2| |
| '--------o |5
| | |
| 3V Batt - ===
| GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

As I said, a 3V lithium will last for months in this circuit if you're
constrained for space, and a couple of AA alkalines should last a year
at least.

Good luck with your project
Chris
 
D

Don Bowey

Jan 1, 1970
0
Hello all,

I'm in the need to detect a phone-lines status, and the specs tell me that
I'm allowed to use the equivalent of a 80 KOhm resistor. In that case I
expect (hope) that the voltage of the line will stay at least 30 Volts,
which than results in a current of just 0.375 mili-amps.

What is your country?

What is the spec that allows 80k Ohms and where in the telephone current
path is it located? Where are you measuring .375 mA?
Now the fun part : I need to detect the status while my electronics are not
DC-coupled to the phone-line (the lines should be able to float, while my
electronics should be able to connect to a computer, which will pull it to
ground-level (or close to it) ).

If your equipment is not DC coupled, what phone line parameter are you going
to measure for status?
I thought about using an Opto-coupler. The problem is that I've only used
opto-couplers that use, in comparision with the above, massive ammounts of
current (5-30 mA). As I'm entering unknown territory I could use some
advice ....

A allso won't say "no" to a pointer to an allready existing design :)
Suggestions to do it another way are welcomed allso.

Regards,
Rudy Wieser

What ate you attempting to do?

Don
 
P

Phil Allison

Jan 1, 1970
0
"R.Wieser"
I'm in the need to detect a phone-lines status, and the specs tell me that
I'm allowed to use the equivalent of a 80 KOhm resistor. In that case I
expect (hope) that the voltage of the line will stay at least 30 Volts,
which than results in a current of just 0.375 mili-amps.

Now the fun part : I need to detect the status while my electronics are
not
DC-coupled to the phone-line


** I doubt that is true.

You must not connect the line to ground - but there is no need for galvanic
isolation.

(the lines should be able to float, while my
electronics should be able to connect to a computer, which will pull it to
ground-level (or close to it) ).


** Not if the detecting circuitry is high impedance and differential.

Think in terms of a differential op-amp circuit with high value input
resistors ( say 1 Mohms each) and lower value feed back and divider
resistors, say 100 kohms each. Such a circuit will have a gain of 0.1
times.

So a +/- 30 volt input will generate a +/- 3 volt output.

Impedance to ground is very high.

Common mode voltages or signals are rejected

Op-amp input can be protected from spikes etc by reverse parallel and input
to rail diodes, as usual.

A bridge rectifier prior to the 1 Mohm resistors will give single polarity
operation.




......... Phil
 
R

R.Wieser

Jan 1, 1970
0
Don Bowey <[email protected]> schreef in berichtnieuws
C08F77CA.329B9%[email protected]...

Hello Don,
What is your country?

The Netherlands.
What is the spec that allows 80k Ohms and where in the telephone current
path is it located?

When I am understanding the second part of your question correctly its
nowhere. The resistor will be placed over the line, parrallel to allready
existing equipment.
Where are you measuring .375 mA?

Nowhere. I calculated that current from the (assumed) minimum line-voltage
and the required minimum resistance.

If your equipment is not DC coupled, what phone line parameter are you going
to measure for status?

Its phase. What I ment is that there is no DC connection between the
phone-line and my electronics.

What ate you attempting to do?

Detecting the phase of the line : reversed or normal. (+ and - are reversed
or normal)

Regards,
Rudy Wieser
 
R

R.Wieser

Jan 1, 1970
0
Phil Allison <[email protected]> schreef in berichtnieuws
[email protected]...

Hello Phil,
** I doubt that is true.

You must not connect the line to ground - but there is no need for
galvanic isolation.

Lets put it this way : I had a DTMF-decoder connected to the line in a
differencial way (as described in its specs), and it refused to work. It
turned out that the line had/has an AC component of 70 volts top-top. And I
realize that I did not even bother to check the DC component.

I would need to have a gain of at most 0.05, but maybe less. It allso means
that any DTMF signal would be just 1/20 of its normal strength, and I'm not
sure that the DTMF-decoder (an MT8870DE) would be able to coope with it.
Although I must say that I did not check that in any way :-( :)

Maybe I should take some time re-think (and check) my options ...

Thanks for the help.

Regards,
Rudy Wieser
 
P

Phil Allison

Jan 1, 1970
0
"R.Wieser"
Lets put it this way :


** Lets not.

You are now moving the goal posts to another ballpark.

I had a DTMF-decoder connected to the line in a
differencial way (as described in its specs), and it refused to work. It
turned out that the line had/has an AC component of 70 volts top-top. And
I
realize that I did not even bother to check the DC component.


** Ever hear of capacitors ?

A wonderful invention that stops DC and allows AC to pass.

I would need to have a gain of at most 0.05, but maybe less. It allso
means
that any DTMF signal would be just 1/20 of its normal strength,


** You only ASKED for line status detection originally.

SO:

In my example - if you bypass each 1 Mohm resistor with a 100Kohms and
0.022uF in series, the overall gain will return to unity at audio
frequencies while remaining at 0.1 for DC.



......... Phil
 
R

R.Wieser

Jan 1, 1970
0
Phil Allison <[email protected]> schreef in berichtnieuws
[email protected]...

Hello Phil,
** Lets not.

You are now moving the goal posts to another ballpark.

Are you sure ?

What you do not know/realize is that the input stage of the mentioned
component is actually an op-amp, thereby comparable to your suggestion.
** Ever hear of capacitors ?

Yes. Those are placed in the input-lines to the DTMF-decoder (which only
needs the AC-component).

But what about measuring the (difference in) actual voltage of those
phone-lines (my origional question) ? I don't think that putting
capacitors in those lines would be a good idea. :) And those lines *will*
need to be able to cope with the AC as well as the DC components.

** You only ASKED for line status detection originally.

Correct. But I hope you agree that changing the defined requirements *by
you* allso means I have to re-evaluate the rest of the current apparatus
involved.

I do not mind at all, but please do not accuse me of "changing the rules".
It would be like the pot calling the kettle black :)
In my example - if you bypass each 1 Mohm resistor with a 100Kohms and
0.022uF in series, the overall gain will return to unity at audio
frequencies while remaining at 0.1 for DC.

Thank you. A good suggestion. One I should have remembered myself.

Regards,
Rudy Wieser
 
P

Phil Allison

Jan 1, 1970
0
"R.Wiesarse Fuckhead "


** Well O-Kaye !!

This WIESARSE wog **** has just made himself into a prize target for me.

The shooting gallery commences now folks:



Are you sure ?


** I sure as hell know when some WIESARSE has moved the goal posts.

And am just about to shove the LONGEST one right up his stupid, fat arse.




** ROTFL - this vile cretyn is utterly clueless !!


But what about measuring the (difference in) actual voltage of those
phone-lines (my origional question) ?


** Exactly what * both* my circuit suggestions do.

Shame WIESARSE is a congenital mental defective or he might see that.




** Soooo, now the WIESARSE wog **** admits he IS is a goal post bandit.

On top of being a rabid, autistic arse bandit too - no doubt.


Thank you. A good suggestion.


** Go shove it where the sun never reaches - WIESARSE.


One I should have remembered myself.


** ROTFL -

WIESARSE is a * WISEARSE* to the very last !!!







........ Phil
 
R

R.Wieser

Jan 1, 1970
0
Phil Allison <[email protected]> schreef in berichtnieuws
[email protected]...

Hello Phil,
"R.Wiesarse Fuckhead "

** Well O-Kaye !!

This WIESARSE wog **** has just made himself into a prize target for me.

The shooting gallery commences now folks:

I'm sorry to see you took my reply that way.

Its sad/funny to see that you are responding with such a rage to a remark
(pot and kettle) that I hoped was put in a friendly way.

And why I put that remark ? Because you gave a COUPLED solution where I
specifically asked for a UNCOUPLED one.

Although I did/do not mind (I accept any help, as long as it *solves* my
problems), it did mean that I had to re-think a few things. And yes, that
allso ment I turned my attention to similar parts of my apparatus, parts
that have given me grief in the past. Sharing those problems and their
causes is normally considered a good response.

In your case that did not work. And I cannot say that I know why.

But that is life I suppose.

Regards,
Rudy Wieser
 
R

R.Wieser

Jan 1, 1970
0
R.Wieser <[email protected]> schreef in berichtnieuws
[email protected]...

In your case that did not work. And I cannot say that I know why.

I should have looked at some of your other posts before I wrote that. It
looks like its your normal behaviour. Being abusive against people that are
"dumb" or just do not agree with you.

Oh well, every community has them and newbies (like me) in those groups
allways get to "enjoy" their attention (most "old hands" will simply ignore
them, or allready have them in their block-list)

Which is a pitty, as you do seem to know what you are talking about in
regards to electronics. If you would be able to brush-up on your
presentation of it you could be a really helpfull member of the community.

Regards,
Rudy Wieser
 
P

Phil Allison

Jan 1, 1970
0
"R.WiesarseFuckwit "


** Lemme tell you something - asshole

When seeking free advice from experienced technical folk on a public
orum - it pays to keep your insanely bloated ego and pig fucking arrogance
under control.

Or you will get an even bigger slap next time.





......... Phil
 
R

R.Wieser

Jan 1, 1970
0
it pays to keep your insanely bloated ego and pig fucking arrogance
under control.

Funny : My previous message told you the same, only with a bit less ...
ranting in it :)
Or you will get an even bigger slap next time.

I'm sorry, did I miss something ?

Regards,
Rudy Wieser
 
J

Jonathan Kirwan

Jan 1, 1970
0
Hi, Rudy. Your plan to electrically isolate your computer from the
phone line is a good one.

Trying to get close to the edge of the maximum on-hook current probably
isn't a good idea though, but it's possible.

Most people who look at doing this kind of thing (phone in-use
detector) go with a small battery to provide the power for the
optocoupler. Mr. Bowden has a good circuit to drive an LED when the
phone is in use here:

http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm#inuse.gif

His circuit, "Telephone In-Use Indicator", can be modified for minimum
current to drive an optocoupler like this (view in fixed font or M$
Notepad):

|
| TIP 3V Batt +
| ___ |
| o-|___|-o------o-------o--------o
| 3.3M | | | |
| | | 47K| 270.-. VCC
| | 680K| .-. | | +
| 0.1uF | .-. | | | | |
| --- | | | | '-' o----.
| --- | | '-' | | |
| | '-' | |/ 2N3906 | .-.
| | | o------| | 1K| |
| RING | | | |< IC1 | | |
| ___ | | |/ | H11L1 _|6 '-'
| o-|___|-o------o-----| 2N3904 1| | \ |
| 3.3M |> V ~ | H )o-o-->
| | - ~ |__/ 4
| | 2| |
| '--------o |5
| | |
| 3V Batt - ===
| GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

The H11L1 is a logic-level optocoupler that will work with less than
2mA of LED current. This circuit sets LED current at 3mA or so. This
circuit will work for a long time with a 3V lithium battery if you're
space-constrained, or for over a year of moderate use with two AA
batteries. Don't forget the pullup on the H11L1 output (it's open
collector). Use an 0.1uF cap rated for at least 100V.

Here is another possibility, where the opto section on the phone line
side doesn't require a power supply at all (and it loads the line
perhaps a little less than your example, even so) and where it is
possible to actually observe the line voltage by counting pulses. It's
a voltage to frequency converter. The other side of the opto will
require power, of course. Another difference is that the pulse
frequency would need to be handled:
: Voltage
: Trigger SCR Switch
: R6 R5
: TIP---/\/\----+----|>|----/\/\---+---------+-----------------+---------,
: 4.5Meg | OPTO 10 | | | |
: | | | | |
: | --- | --- |
: | \ / D2 | \ / D1 |
: | --- 1N4148 | --- 1N4148 |
: | | | | |
: | | | | |
: | | | | |
: | | | | |
: | | | | |<e Q2
: | \ | +-------| 2N3906
: | / R4 | | |\c
: | \ 4.5Meg | | |
: | / | | |
: | | | ,-----------------'
: | | | | |
: | | |<e Q1 | |
: | +-------| 2N3906 | |
: | | |\c | |
: | ,------+ | | |
: --- C2 | | | | |
: --- 22nF | | | | |
: | 200V | | | | | C1
: | | | | | | || 100pF
: | | \ | | +--------||----,
: | | / R2 | | | || |
: | | \ 4.5Meg | | | |
: | | / | | | |
: | | | | | | |
: | | | | | | |
: | | | | | |/c Q3 |
: | | | +---------+-----| 2N3904 |
: | | | | |>e |
: | | \ | | |
: | | / R3 | | |
: | | \ 4.5Meg \ | |
: | | / / R1 | |
: | | | \ 4.5Meg | |
: | | | / | Positive |
: R7 | | | | | Feedback |
: RING---/\/\----+------------------+---------+-----------------' to trigger |
: 4.5Meg | |
: '-------------------------------------------------'

I kept all the resistors of the same value, except for R5 which is
needed to help current limit at times, in case that helps any. I
think the frequency will vary roughly linearly with voltage, at about
10Hz/volt or so.

Jon
 
R

R.Wieser

Jan 1, 1970
0
Jonathan Kirwan <[email protected]> schreef in berichtnieuws
[email protected]...

Hello Jon,

Here is another possibility, where the opto section on the phone line
side doesn't require a power supply at all

That sounds good :)

I think the frequency will vary roughly linearly with voltage,
at about 10Hz/volt or so.

And that sounds good too. At 10 to 50 volts that gives me a pulse-time
short enough for me to be usefull.

It looks like it could be the solution to my problem (although I'm not to
keen at handling this little current : I'm just an amateur :) )

Regards,
Rudy Wieser
 
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