[sci.electronics] YAIC

edhall@rand.org (Ed Hall) (06/06/91)

Other people have taken a shot at the problem of a line-driven in-use
indicator.  Here is my attempt:

					  2.2K
		 +--------------+-----+---v^v^v-----+
		 |              |     |            \-/
		 |              >     > 2.4M       _V_ LED
		 | +      2.4M  <     <             |
	     --------           >     >     | ------+
	 O---|200PIV|           |     |_____|/      |
 Phone       |Bridge|           |     |     |\      |
	 O---| Rect.|           |  | /      | V  | /
	     --------           |__|/          \_|/
		 | -            |  |\            |\
		 |              >  | V           | V
		 |         100K <     \             \
		 |              >     |             |
		 |              |     |             |
		 +--------------+-----+-------------+


All transistors are NPN and should have a BVce of 150V or so, and an
Hfe of about 100.  The LED should run on 2 or 3 mA.  The 100K
resistor may need to be adjusted if the on-hook voltage is less
than 48V, or if the phone doesn't hang up properly.  It should
be set to allow the LED to come on when the line is down to 12V
or so.

Of course, conecting anything that isn't type-approved to your
phone line is a violation of FCC rules, so this circuit is for
educational purposes only.  :-)

The theory of operation is simple.  The bridge rectifier guards
against polarity reversals.  When the voltage divided between the
first 2.4M and 100K resistors goes above 0.6V or so, the first
transistor will begin to conduct; this happens at a line voltage of
about 12V.  This conduction shunts the current passing through the
second 2.4M resistor so that the bias current through the other two
transistors (connected in a gain-multiplying Darlington configuration)
is no longer able to make them conduct, extinguishing the LED.

When the phone line is in use, the voltage reduces to 9V or less.
This prevents the first transistor from conducting, and thus allows
the other two to conduct, lighting the LED.  The 2.2K resistor is
chosen so that the LED doesn't draw enough current to keep the line
below 12V (assuming the line will source a minimum of 5mA or so);
thus when the line is no longer in use, the circuit can shut the
LED off.  Increasing the current through the LED by reducing the
value of this resistor might increase the current enough to keep
the circuit from allowing the line to go back "on-hook".

I'm afraid I can't give part-numbers for the transistors.  The
BVce is likely to be the limiting factor in choosing them; you
can choose a smaller breakdown voltage, but you'll risk frying
things the first time the phone rings.  (The junkbox transistors
I tested the circuit with were hand-picked and didn't have
intelligible part numbers.)

Putting a 150V MOV at the input might not be a bad idea, come to
think of it.

		-Ed Hall
		edhall@rand.org