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From: henry@utzoo.UUCP
Newsgroups: comp.dcom.telecom
Subject: Re: Telephone Ring Detection
Message-ID: <8704260816.AA00898@ucbvax.Berkeley.EDU>
Date: Sun, 26-Apr-87 04:03:50 EDT
Article-I.D.: ucbvax.8704260816.AA00898
Posted: Sun Apr 26 04:03:50 1987
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Funny you should ask, I posted this to sci.electronics not long ago:

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From: henry@utzoo.UUCP (Henry Spencer)
Newsgroups: sci.electronics
Subject: Re: I need a telephone ring detector chip
Message-ID: <7940@utzoo.UUCP>

The following circuit is from an HP app note; I have tested it and found
it to work well.  Sorry for the inability to just display the diagram, but
I lack the patience to try to approximate it with ASCII characters...

You have two wires coming in from the phone line.  First, each goes through
a 100k resistor.  Then there is a 10M resistor between them.  Then one of
them goes through a 0.02 uF capacitor, which should be rated for something
like 200V.  Then there are two diodes in parallel between the lines, with
opposite polarities (one with anode to top line, the other with cathode
to top line):  one a 1N4148 and one the input LED of a 6N139 opto-isolator.

The 6N139 has a Darlington output with four pins available.  The one going
to the photodiode and the collector of the inner transistor goes to +5.
The one going to the emitter of the outer transistor goes to ground.  The
one going to the collector of the outer transistor is the output.  The one
going to the internal transistor-transistor connection is connected to the
output through a 0.1 uF capacitor.

The output line then goes through a 10k resistor.  Then there is a 56k
resistor from it to +5.  Then it goes into the base of a 2N3906 transistor.
The emitter of this transistor is connected to +5.  The collector is the
final output, with a 1k pulldown resistor between it and ground.

Explanation:

The 100k resistors supply plenty of impedance between the whole 
circuit and the lines no matter what fouls up inside the circuit.  The
10M resistor bleeds off static charges that might show up when the thing
is not connected (HP actually made it 22M, but that's hard to find and
10M works fine).  The 0.02 capacitor blocks DC completely and also gives
quite a bit of AC impedance; it needs to be rated for a nice high voltage
because the ringing AC is close to 100V and you want a good safety margin.
The 1N4148 conducts when the AC voltage is the wrong polarity for the LED
to conduct; leave this out and the opto will try to block the full ringing
voltage at those times, which will fry it.

The 0.1 capacitor between the internal node and the output acts as an
integrator, so that the AC ringing shows up as one pulse rather than a
sequence of short ones.  It also gets rid of smaller and less frequent
activity like dialing pulses.  The 10k resistor limits current through
the opto and transistor when the opto is on.  The 56k resistor pulls
the transistor's input up high when the opto is off.  And the 2N3906, in
conjunction with the pulldown resistor, amplifies the output signal.

This circuit isn't particularly fussy about power voltage, although you
might want to increase some of the resistor values on the output side
for higher voltages.  The output is certainly CMOS compatible and is
probably TTL compatible, although I haven't done the numbers to be really
sure about the latter.  The 6N139 is a premium opto and will cost a few
bucks, but overall it's not what you'd call an expensive circuit, given
the logic-level output and the high degree of isolation from the phone
line.
-- 
"If you want PL/I, you know       Henry Spencer @ U of Toronto Zoology
where to find it." -- DMR         {allegra,ihnp4,decvax,pyramid}!utzoo!henry


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