Telephone Bell circuit

By Robert Dvoracek

Info:

Phone calls are easy to miss while you're outside and not everyone will leave a message. Caller ID service is another $5/mo or more and doesn't work for some unlisted numbers. You could use a cordless phone but I'm always afraid I'll end up leaving it outside to get rained on. If you're working with something noisy, it's easy to not even hear a cordless phone ring at all. This is one possible solution.

Fig.1. Telephone Bell circuit

R1, R2 100k
R3 ... 22M
R4 ... 56k
R5 ... 10k
R6 ... 1k

C1 ... .02uF
C2 ... .1uF

D1 ... 1N914 (interchangable with 1N4148)
Q1 ... 2N3906 PNP General Purpose
Q2 ... IRFBC40 Power MOSFET

U1 ... 6N139 Split Darlington Optocoupler IC

Fig.2. The board is an IC prototyping board such as the Radio Shack 276-159

Description:

During idle, a phone line voltage of 48VDC exists across the ring and tip wires. This constant flow is rejected by C1 thus allowing only the AC ring voltage past this point. The 88VAC 20Hz ringing signal is allowed to pass by C1, lighting the internal infrared LED of optocoupler U1 for one half cycle and conducted through D1 during the other half. R1 and R2 limit this current to a very low value while R3 helps discharge C1. The pulsing light from the LED in U1 shines on an internal photodiode which lowers its resistance and allows a tiny current to flow through the base of the first internal transistor of the two. The transistors constitute a Darlington pair. The current gain of the first transistor is enough to saturate the second. Moving right to the outside of the IC once again, C1 smoothes this pulsing into a constant signal which is then fed to the base of Q1 through R5. R4 insures that the transistor turns off when the signal is removed. The base of Q2 is fed in a similar way by R6 and the emitter-collector junction of Q1. Q2 has sufficient power handling capacity to drive the target load.

Notes:

The circuit is flexible and doesn't necessarily have to power a bell. You could substitute a buzzer, a loud resonant piezo alarm, a strobe, or even a car horn. Use your imagination. If the MOSFET is replaced with a TRIAC, various AC and line powered devices can be used. If you do use it to switch the AC line, make sure that you don't use an autotransformer for your low voltage supply to run the IC; and that the transformer you do use isolates well.

+Vcc can be anything from 5 to 15 volts depending on the requirements of your bell or other device. You may be able to use voltages of up to 20v. The IC and Q1 did not become warm at 15v but I have not tested it. Just make sure that you don't exceed the 21v that the gate of Q2 is rated for.

R1 and R2 may be combined into a single 200k resistor. R3 may be omitted. D1 may also be omitted, however it can help protect the LED in U1 from high reverse voltage transients which may appear on the line.

You may want to experiment with different values of C2 in order to have the circuit turn off sooner after the ring signal stops. I noticed that at .1uF it would tend to stay on a few hundred milliseconds after the ring signal stopped, which is noticeable if you have your finger on the "trigger" applying a test ring signal yourself. It should be pointed out that most electronic telephones act like this anyway.

You could also put the MOSFET Q2 in the place of Q1 to eliminate a component. I may have done this myself but I didn't have any P-channel MOSFETs on hand and had already soldered most of the components in. The circuit depicted here is the exact circuit I have built so you know it has been tested and works.

You can use full 120VAC line voltage as a test ring signal or a transformer if it is AC. If you have a variac you can get the exact 88 volts, but it will still be at 60Hz. The only true ring signal will come from the phone line itself. You can call a friend and tell them to call you back or, if you're like me and don't have any friends, just dial the operator and ask for "ringback". Otherwise you'll have to wait for someone to call which is no fun.

One last thing about Q1: Make sure you test it. It may have a different pinout. If so, you will have to modify the circuit layout accordingly.

kickme.to/lightningstalker 2884 08 December 2007

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