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Telephone Ringing Circuit

Discussion in 'General Electronics Discussion' started by jiblankman, May 26, 2013.

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  1. jiblankman

    jiblankman

    3
    0
    May 26, 2013
    I am trying to make a circuit to detect a telephone ringing signal and output about a 3V signal. I am new to electronics and have been teaching myself from whatever sources I can find. I have hit a situation that I cannot figure out. Hopefully someone can help me understand what is happening here. Thanks in advance.

    I am using the attached circuit which I found at epanorama.net/circuits/telephone_ringer.html"http://www.epanorama.net/circuits/telephone_ringer.html. All parts are as listed, the zener diode is 15V 500 mW. (I tried the other circuit on that page as well.)

    The circuit seems to work, but I am confused by some things. The circuit diagram shows that I should be getting an output voltage across the collector/emitter. However, I get no voltage on my meter. When I connect the leads to the base and collector of U1 I get 0.4 volts with no current when the phone rings. Here is the optoisolator that I am using http://www.mouser.com/ds/2/427/4n25-240323.pdf.

    I thought I understood how a npn transistor operates, but was confused by this. Since it worked consistently, I was going to go with it anyway.

    Next, I tried to amplify the signal to about 3V (2.5-3.3V) is the input range, but cannot get that to work correctly either. I tried using another npn transistor with the positive output of the optoisolator connected to the base, 3V connected to the collector. I tried including a 1K and 10K resistor between the optoisolator and transistor as well. All I get is a constant 2.7V output whether the phone is ringing or not. I tried connecting the optoisolator to the collector instead, but nothing. Any help would be greatly appreciated.
     

    Attached Files:

  2. davenn

    davenn Moderator

    13,802
    1,941
    Sep 5, 2009
    Hi
    and welcome to the forums :)


    no you wont get any voltage across the collector/emitter cuz you havent supplied a voltage to the collector.
    In an opto coupler, all the light from the LED does is turn the transistor on. You need more components, not shown, around the opto's transistor to finished the circuit.

    You use this transistor to turn on a LED, drive a relay or drive other circuitry to achieve you goal :)

    cheers
    Dae
     
  3. poor mystic

    poor mystic

    1,068
    31
    Apr 8, 2011
    Hi :) jiblankman, and Dave
    I had a look at Mouser Electronics and copied this link: http://au.mouser.com/Electromechanical/Relays/Low-Signal-Relays-PCB/_/N-5g38?P=1z0x33l which will take you to a suitable 3V relay at low cost.
    The diagram shows how to use the relay in the circuit to "switch 3V".

    It is very important to completely isolate the telephone cables from the hobby project you envisage. There is room for the kind of mistakes that kill people (telephone linemen trying to resolve a fault) if you were to use a mains-powered supply, so please use only batteries!

    Telephone lines are surprisingly tricky to work with. The circuit as drawn "should" work, but almost any modification will cause subtle (or very unsubtle!) faults with the telephone exchange.

    Good luck! :)

    PS The switch marked R1/1 is a representative relay contact, which closes when the relay coil R1 is energised.
     

    Attached Files:

  4. jiblankman

    jiblankman

    3
    0
    May 26, 2013
    Thanks for the help. I was already working on Dave's suggestion when P.M. responded as well.

    I had some success. I connected a 3V power supply to the collector of the optoisolator and a second npn. The optoisolator emitter is connected to the base of a second npn and the emitter of the second npn goes to ground. A circuit diagram is attached. I measured the voltage between the second npn collector and ground and got 1.8V. When the phone rings, the signal drops to 0.2V. I can definitely work with this but am confused by the direction of the signal change. I thought it would go low to high. Do I have something wrong or is this correct?

    Additionally, I intend to connect this to a Raspberry Pi, powering the non-phone side from the 3V3 pin and monitoring the output from within an input pin. The high-low signals requirements are low <0.5 V and high >2.3 V. The low volt signal is at a good level but the high is not high enough. I have tried using different value resistors and power (12V), but there is no significant change. Again, am I missing something?

    Thanks again for your help.
     

    Attached Files:

  5. poor mystic

    poor mystic

    1,068
    31
    Apr 8, 2011
    :)
    Your circuit is fine, and does what it should.
    If you were to replace that LED with a 100k resistor you'd get a better voltage swing. There might be a bit of fiddling with the resistor value depending on what you're interfacing to.
    You could also try putting a ? resistor across the LED, which would allow you both a bigger swing (under a small load) and a LED indication.
     
    Last edited: May 27, 2013
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    All good advice on this thread, and you're doing everything right.

    The conventional way to interface an optocoupler to a micro is to ground the emitter, connect the collector through a resistor (e.g. 10k) to the positive supply rail, and connect the microcontroller input to the collector. You don't need any components other than the pullup resistor. In fact if your microcontroller has weak pullups (the AVRs do), you can avoid the resistor as well by enabling the weak pullup on that input.

    When the ringing voltage is present on the line, the light within the optocoupler makes the optocoupler's transistor conduct. It's like a switch turning on. It pulls its collector down to 0V and you will read a low state on the Arduino input. When there is no ring voltage on the line, the transistor turns off and the pullup resistor (or weak pullup inside the AVR) pulls the input line high.

    During ringing you will not see a continuous low signal. The ringing voltage is an AC signal at typically 20 Hz and it's full-wave rectified by the bridge rectifier in the circuit in post #1 so your software will see a 40 Hz signal on the input while the line is ringing, as well as longer gaps between the individual rings.
     
  7. jiblankman

    jiblankman

    3
    0
    May 26, 2013
    Thanks everyone. The 100k resistor did the trick for me. I appreciate all of your help and explanations. I definitely have a better understanding of what is happening in this circuit now and how to work with it.
     
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