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Where is a good location to connect a 3-Volt relay?

Discussion in 'General Electronics Discussion' started by LucidSymmetry606, Mar 20, 2015.

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  1. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    With my wireless doorbell project, I originally had my relay connected to the speaker outs on the receiver's PCB. I assumed that since the outs power a speaker, they'd pump out the voltage necessary to trip the relay. However, it did not work when I tried it. Curious, I hooked up a voltmeter to the speaker outs and during the "ding-dong" cycle I only got up to .09 volts. The 3 AA batteries are fresh. I then connected a red LED to the speaker outs and sure enough, it lit up.

    I guess I'd achieve the proper effect I'm going for if I connect the relay to different points on the PCB. The question is where...? I included a picture of the PCB. If anyone would be so kind as to simply put a red dot or a mark of some sort where my relay should be connected to, I'd be greatly appreciative. I'd rather ask for help then go my usual route of trial/error - I don't want to short anything and cause problems.

    2015-03-20 02.jpg
     
  2. davenn

    davenn Moderator

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    Sep 5, 2009
    without a circuit of the receiver, it would be almost impossible to tell where :(
     
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  3. hevans1944

    hevans1944 Hop - AC8NS

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    Well, there you go! Just substitute an optical isolator for your red LED and use the transistor side of it it to pull in your relay.. Who knows why the "speaker" wires would light up an LED, but if it works for you, go with it.
     
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  4. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    I'm on Wikipedia reading about optical isolators now. I've never heard of 'em until now. Why aren't the 'speaker out' wires putting out 3+ volts yet somehow lighting up an LED? That seems bizarre...

    Could you explain to me how an optical isolator will pump out the voltage necessary to trip the relay?
    Thank you in advance for your time in helping me w/ this.
     
  5. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    After further reading, I think I'm beginning to understand the implementation of the opto-isolator. Would you say that's the easiest method to get the wireless doorbell receiver to trip a 3-Volt relay when pressing the button on the corresponding transmitter? It seems that the solution to my problem is under my nose somewhere. If I connected one input on the relay to one side of the receiver's power source would I then be able to connect the relay's other input somewhere to the PCB in which it would push out three volts upon activating?
     
  6. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    I'm not sure I follow. The picture I included is the receiver's circuit. Are you talking about a detailed circuit diagram?
    Isn't there a relatively uncomplicated way to get a doorbell receiver to trip a relay? Any idea why it isn't working when I simply connect the relay to the speaker outs?
     
  7. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    Just to confirm my understanding of your suggestion... Is the drawing I included what you were thinking of? 2015-03-20 08.jpg
     
    hevans1944 likes this.
  8. davenn

    davenn Moderator

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    looks good :)
     
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  9. davenn

    davenn Moderator

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    because there isn't enough power out of the speaker line to operate a relay
     
  10. KTW

    KTW

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    Feb 22, 2015
    What is the purpose for the relay?
     
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  11. duke37

    duke37

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    Jan 9, 2011
    You might have been measuring DC when the output is AC with an average voltage of 0.09V. A DC relay will be reluctant to respond to AC.

    I would try a resistor feeding the base of a transistor with the relay between the collector and the + supply. A diode must be connected across the relay as a flywheel diode.
     
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  12. hevans1944

    hevans1944 Hop - AC8NS

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    Yes, although I do not understand why the LED in the optical isolator would "light up" when connected to the "speaker" wires.

    Well, we actually don't know that it will "light up," but you said an LED you tried did "light up" so I am assuming the LED (which you can't see) in the optical isolator will do the same. The transistor part of the optical isolator should be capable of providing enough current to operate the coil of a small relay providing the "speaker wires" can "light up" the LED part. Who knows what sort of waveform is present when the doorbell rings? It may be an AC audio signal (ding-dong) that happens to be strong enough to visibly "light up" an LED when the AC peaks are the correct polarity. In which case, those peaks may be pulses of insufficient duration to "pull in" a relay. If that be the case, you need to go to "Plan B". But try the circuit first, and if it doesn't work, we can work together to figure out a Plan B.

    BTW, this is a good example of "trial and error" experimentation. I used to do this a lot when I was a kid because (1) I didn't know anything and (2) I didn't have the right measurement tools to use to learn anything. So, I did a lot of reading (real paper books because there was no Internet then) and eventually saved up enough money to buy an RCA vacuum tube voltmeter (VTVM) kit, which I soldered up using a Weller soldering gun. It was many years later that I bought my first "pencil" soldering iron, but that's another story. I am not suggesting you need a VTVM (these are obsolete now) or any other instrument to get your relay to actuate, just that if you did have something more sophisticated than a multimeter, you might be able to figure out what is actually going on at the speaker wires when the doorbell rings. Still, without a schematic diagram showing the actual circuit, you are working pretty much in the dark here.

    We can all learn by trial and error if we are observant and take notes about what does and doesn't work. Sometimes that is the only practical way of "getting something to work" if we can't discover the actual "how" something works. Please let us know what progress you are making.
     
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  13. hevans1944

    hevans1944 Hop - AC8NS

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    It will be used to remotely ignite an "electronic match" a small-wattage, low valued resistor used to ignite the engine of a model rocket.
     
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  14. KTW

    KTW

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    Feb 22, 2015
    LucidSymmetry606 likes this.
  15. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    Perhaps. The OP is trying to "make do" with what he already has available, i.e., a wireless doorbell set. Personally, I would just use a pair of insulated wires of appropriate length to fire off a model rocket. Less chance of inadvertent ignition. But then, I am a kind of conservative guy, especially when it comes to pyrotechnics.
     
    LucidSymmetry606 likes this.
  16. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    Guys, wow. These responses are gold. Thanks for all the help.
    I've spent a bit of time hoping to get lucky looking for electrical schematics of the PCB. Google found nothing. :(
    It never even crossed my mind that the signal going to speakers would be AC. Indeed, Duke37, I was measuring in DC.
    KTW, it would be much simpler. But as Hevans1944 stated, I am re-purposing this doorbell because it happens to be what I have and it'd be a waste to see it sit in a drawer somewhere for years. If all else fails, the dedicated transm./receiver will be option D.

    Guessing by your name Hevans, you were born one year before the bombs in Japan went off. Isn't it amazing how in one lifetime you've seen the change from having to go to the local library (if open) and find a specific book on a specific topic (if available), check it out, drive home, look for the information and copy it down... ... Now you go online and within seconds, anything you're looking for is virtually available. You can go to a forum and speak to groups of like-minded people of all ranges of expertise. It makes you think what things will be like in a few more lifetimes long after I'm gone.
     
  17. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    It's so strange that this blue LED is lighting up from the speaker out. Don't LED's have a minimum voltage required to light up? Idk. Perhaps it's different for AC. Throughout the ding-dong cycle, the voltage jumps through the following values when testing it with my multimeter on the AC setting:

    0.1
    1.5
    0.7
    1.4
    0.6
    0.3
    0.2
    1.5
    0.8
    1.5
    0.6
    0.4
    0.3
    0.2

    0.1

    Values in blue are the ding-dong cycle. When the doorbell's not activated, it's constantly showing a value of 0.1 Volts. What's that about?

    I'm on a Polish version of Ebay browsing through optical isolators. I see they have different values for voltages. Is there anything specific that I should watch out for? The one I'm looking at is linked below:

    http://www.soselectronic.pl/a_info/resource/d/ltv814.pdf
     

    Attached Files:

  18. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    Those LTV8x4 devices look like they have back-to-back LEDs in them. If so, that means they will illuminate the transistor nicely with AC input to the diodes. The current transfer ratio is small... about 20% or so, meaning you may not get sufficient current through the photo-transistor to directly operate a 3V relay. If that is the case, you can connect the photo-transistor emitter to the base of an NPN transistor, put the relay coil in series with the NPN collector and an external power source, V+, and connect the emitter of the NPN to V-. Try a value for V in the range of 3 to 9 volts. The collector of the photo-transistor in the optical isolator is also connected to V+ so when the photo-transistor is conducting it provides base current from its emitter to the base of the NPN. And you might need a diode connected across the relay coil, cathode to V+ and anode to the NPN collector, to dampen the inductive "kick back" across the relay coil when the optical isolator turns off.

    I have no idea why the doorbell shows a small AC voltage when it isn't going "ding-dong." It isn't enough voltage to turn on the LEDs in the optical isolator, so I wouldn't worry about it. The voltages you measure while the "ding dong" is playing are not a reliable indication of the actual waveform at the speaker terminals because the update rate of your multimeter is too slow to follow whatever AC waveform is making the sound. Much more information would appear if only you had an oscilloscope... or as the Tin Woodsman in the Wizard of Oz said, "If only I had a heart!" Well, he didn't, but things worked out okay anyway. As an experiment, you might connect a diode in series with a polarized 10 μF electrolytic apacitor across the speaker wires. Anode of diode to one speaker wire, cathode to the plus side of the capacitor, and the other side of the capacitor to the other speaker wire. You should measure a small DC voltage across the capacitor when the "ding-dong" occurs. This voltage should more or less slowly decay as the input impedance of your multimeter discharges the capacitor between "ding-dongs". Can you do anything with this voltage? Maybe. Do the experiment and let us know what you find out.
     
  19. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    Excellent. Thanks.
     
  20. LucidSymmetry606

    LucidSymmetry606

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    Mar 15, 2015
    Oh, man. I got effin' lucky. I connected one side of my multimeter to the power supply's negative terminal and connected the other node to the (-) speaker out. When hitting the doorbell on the transmitter, I get a beautiful cycle that starts at ~4.5 volts and stays there for a little bit before trailing off back to completely zero. My relay will be very, very happy. I'd upload a quick video, but I see that I can't...

    Seriously. I'm excited. This project has cost me a lot of energy. I'm glad I have the toughest nuts cracked. It's all thanks to you guys.
     
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