telephone in use circuit

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johnsmith123

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Hello guys, I found this simple circuit to build with simple parts that I have here.
Led is supposed to turn on when the phone is off hook.

I have a problem in that the LED is always ON, no matter if the phone is off hook or not.
I checked the voltage at the 680K resistor, and the polarity of the Tip and Ring, ALL is good.
I found other circuits but I have all the parts for this one.
Can someone help?

Thanks
 

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When the phone line is not being used then there is about 50VDC across it. So the voltage across the 680k ohm resistor causes the base voltage of the NPN transistor to be a little negative which turns it off and then nothing turns on the PNP transistor so the LED doesn't light.
 

Thanks for your reply. I took some measurements on hook, and off hook.

Voltage across TIP & RING on hook 51.2 V, off hook, 12.96 V.

Voltage across 680K resistor on hook = 3V
off hook, same 3V.

I am using a simple 3V power supply built from a LM317 circuit so it is exactly 3 Volts.
 
I worked in the telephone industry for 40 years.

What you need is a dual colour LED - one that has 2 wires with the red and green LEDs in parallel anode to cathode so if you apply a voltage one way, the red lights and if you reverse the voltage, the green lights.

This ensures that the LEDs are protected against over reverse voltage (they protect each other) and it does not matter which way they are connected - one or the other will light.

Connect this in SERIES with the phone (or phones) it may also be advisable to connect a 2 uF capacitor in parallel with the LEDs in order to prevent audio distortion. However, try it without the cap first.

However, first measure the current through the phone when it is off hook and check the LED data sheet to ensure that the current is not greater than the LEDs can handle. I don't think it will be but it would be best to check.

If the current is too high, connect a resistor in parallel with the LEDs to reduce the LED current to a safe level.
 
I use an old VU meter which will give a 3 status indication of the phoneline.

Across the phoneline A and B terminals is usually around 50 volts dc.

Connect via a 330 khm: series resisitor a fullwave brige rectifier to drive the VU meter.
Set the R value so that the VU meter reads about 3/4 schale. This is the line voltage. ( R. values between 220 and 470 khm: )
If phone is off hook, the voltage drops to about 1/3 rd of VU meter schale ( ± 10 Volts )

If ring tone is present the VU meter goes full schale.

I got this circuit permanently connected in my house on 3 different phones. Also the beauty is that it does not need a separate power source.

I think i listed the circuit on this forum before a while ago.
 

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I suspect the NPN may have been damaged by the ring voltage.

It would be advisable to connect a diode such as a 1N4148 across the B/E of the NPN, cathode to the base.

This will protect the transistor from reverse voltages.
 
ljcox said:
I suspect the NPN may have been damaged by the ring voltage.
Click to expand...
The revese-biased emitter-base junction of a silicon transistor has avalanche breakdown at about 6V like a zener diode. The current from the peak of the ringing voltage is about only 19uA which won't damage the transistor.
 
You should all be aware that Any Resistive circuit connected across the phone lines CAN Be detected by the phone company.

And should you have a problem with your telephone line and the detect your device, it could easily result in you having to pay for the service call. Possibly also a Fine!

Years ago I had a simular device that had greater than 22 Meg-ohm across the line. "The phone company detected it".
That Cost me Money.

A Good Telephone line has Considerable Capacitance as a result of parallel phone lines, but they Exibits Infinitive Resistance. This capacitance Varies from one persons phone to all others.

And It is possible to detect a Phone-in-Service by using capacitance changes. I have a circuit Somewhere, I'll see if I can dig it up.
This Can't be detected by the phone company.
 
audioguru said:
The revese-biased emitter-base junction of a silicon transistor has avalanche breakdown at about 6V like a zener diode.
Click to expand...
That's not quite true zener breakdown is totally different to avalanche breakdown. Avalanche break down is harsh, the reverse junction suddenly changes from a good insulator to good conductor above the breakdown voltage. It posesses a large hysteresis loop, for example the reverse biased transistor might start conducting at 6V but it won't stop untill the currrent goes below 100:mu:A. Zener breakdown is soft, the zener diode lowers its resistance more slowly and regulates the voltage across its junction past the breakdown voltage.

But you probably already know all of this, I'm just giving more detail for other posters.
 
chemelec said:
I have a circuit Somewhere, I'll see if I can dig it up.
This Can't be detected by the phone company.
Click to expand...

Very interesting. I would love to see it!

What about using an opto isolator? Can this be detected?
 
Hero999 said:
That's not quite true zener breakdown is totally different to avalanche breakdown.
Click to expand...
I have used the reverse-biased emitter-base of a silicon transistor as a "zener diode" a few times. It works.
The point is that it limits the voltage so that high voltage damage does not occur.
 
audioguru said:
I have used the reverse-biased emitter-base of a silicon transistor as a "zener diode" a few times. It works.
The point is that it limits the voltage so that high voltage damage does not occur.
Click to expand...
My recollection is that the transistor's gain degrades if it is allowed to go into zener breakdown.

I don't know whether it requires a larger current (than the current in this case) to cause the damage, but I would use a diode to be sure to be sure.
 
johnsmith123 said:
Very interesting. I would love to see it!

What about using an opto isolator? Can this be detected?
Click to expand...
An opto isolator would need to draw several milliamp of current from the phone line which:-

  1. would be detectable by the company
  2. would probably cause a hum to be heard due to line unbalance.

Also, it may trip the ring, ie. when an incoming call arrived, the phone would ring briefly (about a quarter of a second) then stop.
 
I know that an avalanche diode and zener both exhibit characteristics of each other but I thought that it depended on the temperature. As far as I'm aware reverse biased transistors go though avalanche breakdown when a certain voltage is exceeded I suppose it could be used as over-voltage protection but it would possibly latch up causing destruction of the transistor.

Anyway, if a reversed biased transistor behaved more like a zener than an avalanche diode then a reverse bias oscillator wouldn't work would it?
 
My frst job was with Philips. I was told to NEVER allow a transistor to avalanche-breakdown its reverse-biased emitter-base junction.
At only 19uA I don't think much damage will be caused. At 1000 times more current then the "bumps" in the junction might be burned off and the current-gain of the transistor will disappear.
 
So you think it is a thermal issue. If so, then I agree 19 uA should not be a problem.

But I would prefer to protect the junction anyway - the cost of a diode is minimal.
 
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