In a message dated 01/08/2002 10:32:35 GMT Daylight Time,
apptech@PARADISE.NET.NZ writes:


> >   I am thinking of making a simple on/off hook
> > indicator. If the phone is "hung up", there are 42v on
> > the line. If it is off hook, there are about 4v on the
> > line. Would this work (in general)??
> >
> >   A bridge rectifier to a 12v reg. to a 5v reg to an
> > I/O line on a pic (or an LED, for that matter).
>
> May work but the telephone company will not be your friend.
> Any load placed across the phone line MUST have a very high effective
> resistance.
> Most regulators will draw substantial current in the above scenario. Also,
> the line will be unbalanced and this will cause hum due to noise being
> coupled into the line. The arrangement MAY also stop the line releasing
> after a call. There are also regulatory/permission issues about connecting
> things to phone
> liones.
>
> A simple idea to play with is to use a zener diode of say 10 to 20 volts in
> series with a large resistor joined to a PIC pin with another resistor
> pulling the same PIC pin to ground. When the line voltage is greater than
> the zener voltage
> the PIC pin is pulled high. When lower than zener voltage the PIC pin is
> low. This is just the basic principle - have a play and see if you can work
> out a practical circuit.
>
> Use of a simple differential amplifier is liable to more easily produce a
> more "real world" result.
>
>
>
>
>         Russell McMahon
>
>
>
>
> >
> >   My thinking is that if there is < 12v, the 12v
> > regulator wont power the 5v reg. and the LED (or IO
> > pin) will be 0v.
> >
> >   Will this work? As you can tell, I am fairly new to
> > this. I am reading everything I can get, though.
> >
> >   Also, Op-amps are used in quit a few circuits -
> > anyone know a link to a GOOD paper on theory and such?
>

Keep the impedance high so as not to upset the Telecommunication company line
tests and stop any "hang up" problems!

For experimenting -Forget trying to power regulators from the line- it will
drop the voltage!

Try using an opto-isolator via a bridge rectifier via 150R approx. resistors
to the line. Adjust the bias to suit your on hook voltage (about 50v UK) and
off hook voltage.
The transistor side of the opto can be powered by whatever without any
inteference to the Telephone line- common emitter and open collector would
not require another power source apart from the attached gadget.

If you are using a PIC then the collector of the opto could go direct to an
input with a pullup resistor to + , as soon as the line is answered the
output would move from low to high (based on npn type opto)
Remember to connect the transistor emitter to -ve of your circuit!

Hope this helps,

Let me know, I will mail you diagrams ( off list) if you get stuck!

regards

Kev Howard

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