Wireless doorbell won't work using DC wall wart

Discussion in 'General Electronics Chat' started by pkozul, Jan 29, 2013.

  1. pkozul

    Thread Starter New Member

    Mar 4, 2011
    9
    1
    Hi there,

    I've got a wireless doorbell that runs on 3 C size batteries (i.e. 3 x 1.5V). Here is the product:

    http://www.masters.com.au/product/902290351/hpm-wireless-door-chime

    I measured the DC in circuit, and it showed 4.8V.

    I had a spare 5V 2A wall wart lying around, so I thought it would be a nice way to power the doorbell unit, since it was quite fussy about batteries losing charge (would no longer make sound as soon as the batteries lost a bit of charge). I saw some 'battery eliminator' projects on the net, that showed how to do this in a nice way.

    The wall wart measured at 5.6V, so I added a diode to drop the voltage. In the doorbell circuit, the measured DC voltage was then 4.8V, exactly like it was when running on batteries.

    I tried some tests using both buzzers, but the doorbell unit did not make any sound. As a sanity check, I reinserted the batteries, and then tried the tests again. Yep, there was plenty of ringing (my wife thought someone was playing pranks outside with out doorbell!!!).

    I'm a newbie, and this has got me wondering. How can the circuit tell the difference between the batteries and the wall wart, when they both give 4.8V. I'm thinking that 2A current should be easily enough, but I'm only guessing.

    Any ideas?

    Thanks!
    Pete
     
  2. Biff383

    Member

    Jun 6, 2012
    49
    19
    Is the wall wart DC?
     
  3. t06afre

    AAC Fanatic!

    May 11, 2009
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    Do you have any old cell phone charger around. They are DC and in the range 4.5 to 5 volt. It could be that your wall wart have a AC output
     
  4. elec_mech

    Senior Member

    Nov 12, 2008
    1,513
    193
    As others have mentioned it's possible you're using an AC wallwart.
    If the unit is fairly light and the sound is definitely produced electronically then that is the most probable case.

    I agree 2A sounds like enough on the surface, but to have 3 x C batteries seems like a lot. How is the sound produced? Does it sound like something physically hitting a gong, bell, etc.? Does the unit feel heavy with the batteries removed?

    If yes, the door bell is probably using a mechanical device to produce the sound, probably a solenoid. These likely require a surge current which would explain the C batteries and perhaps why 2A isn't cutting it.

    If you have an analog meter capable of taking 10A or so current measurements, you can confirm this. I'm not sure a digital meter would catch the full value of the surge current in time. If you're unsure how to test this, let us know and we'll explain.
     
  5. gerty

    AAC Fanatic!

    Aug 30, 2007
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    Correct polarity is another thing to check for...
     
  6. pkozul

    Thread Starter New Member

    Mar 4, 2011
    9
    1
    Hi all,

    Thanks for your replies. The wall wart I'm using is definitely DC, as you can see by the photo below. I think it's from an old phone charger.

    [​IMG]

    The actual door bell is very light in weight (with batteries removed). Below is a photo of it. Definitely sounds like the sound is made electronically, rather than mechanically. There is a choice of 2 different bell tones, which I guess points towards electronic sound.

    [​IMG]

    I don't have an analogue meter, just a digital one. Can I still do some tests with that?

    BTW, one thing I forgot to mention in my original post. The very first time I connected the wall wart to the door bell, it actually worked when I first tested it (i.e. sound was produced). I didn't even have a diode connected then, so it was getting 5.6V DC from the wall wart.

    I then moved the doorbell unit from my garage into the house (where its real location should be), and then tried again, but it didn't work. Since then, I haven't been able to get the door bell working using the wall wart, and the unit is certainly still working since it runs fine on C batteries.

    I've tried plugging the wall wart into different rooms of the house, but same result.

    I took care about using the correct polarity. I connect the wall wart's red wire to the + battery terminal of the door bell, and black wire to the - one. I usually measure the DC voltage between the red and black wires, just to ensure things are correct (i.e. negative voltage means I have them the wrong way around... hope that's a good rule of thumb for checking???).

    Anything else I can try?

    Thanks!
    Pete
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Your wall wart may have too much ripple. Try adding 100uF across it. Be aware that the DC voltage will go up if this is the problem.

    EDIT: You can first check for ripple by measuring the AC voltage on the wall wart's output when connected to the doorbell circuit.
     
    Last edited: Jan 29, 2013
  8. tracecom

    AAC Fanatic!

    Apr 16, 2010
    3,871
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    It is probably just a reflection in the photo, but T2 looks damaged. Does it look ok in "person?"
     
  9. #12

    Expert

    Nov 30, 2010
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    Wild guessing here...moving from room to room sugests a wire carrying the current. Maybe the wire is too skinny (for its length) for the initial surge of current?

    If the capacitors on the board are rated for 6 volts or more, try leaving the diode out and running the bell on 5.6 volts. Diodes do not remain at .6 volts of loss when passing currents in the amp range. (Another reason to leave out the diode.)
     
  10. elec_mech

    Senior Member

    Nov 12, 2008
    1,513
    193
    Try everyone else's solutions first. If you continue to have this problem . . .

    So here's a silly question: can you repeat this - garage, no diode, test - and does it work or no?

    If not, remove the wall wart connection, install the batteries, set your meter to VDC, trace the battery wires to where they are soldered on the PCB and measure the voltage there. Roughly 4.5VDC? Good. Now remove the batteries, attach the wall wart leads, plug it the wall wart and measure the same two points again with the meter.

    Do you see about 5VDC? If not, one or both of the leads between those two points and the wall wart is open. Remove the wall wart from the board and verify there is 5VDC across the wall wart leads. If yes, then there is an open between where you attach the leads and PCB.

    If you do see 5VDC, keep the meter attached and press the door bell. If the voltage suddenly drops a good bit and there is no sound, the wall wart isn't providing enough current to the circuit. Doubt this is the case with an electronic bell, but won't hurt to check.
     
  11. pkozul

    Thread Starter New Member

    Mar 4, 2011
    9
    1
    Hi again,

    Been busy lately, so only tonight did I find some spare time to continue with this. I tried some of your suggestions...

    AC ripple - with the doorbell connected to the powered wall wart, I measured 01.V AC at the battery terminals (which is where the wall wart + and - wires are connected). That would be a very minor reading?

    [​IMG]

    T2 - looks fine, as you can also see on the latest photos.

    [​IMG]

    [​IMG]

    Diode - removed the diode, and allowed the doorbell to be powered directly by the wall wart (5.6V DC). No difference - still no bell sound.

    Retest in garage - tonight I performed the same tests in the garage, and in the kitchen. No difference. The same symptoms as last time. If I connect the batteries (4.6 V DC), the doorbell works perfectly. If I immediately remove the batteries, and connect the wall wart to the battery terminals (5.6V DC), no bell sound is ever made.

    Battery reading:

    [​IMG]

    Wall wart reading:

    [​IMG]

    I couldn't easily get to the actual connectors on the PCB, so I've just been using the battery terminals are the points at which I connect my multimeter.

    With the multimeter connected to the circuit, I got my wife to press the doorbell buzzer a few times. I observed the DMM and it stayed at exactly 5.6V and did not move from that reading. It's as if nothing at all was going on.

    Got me totally baffled. Is it possible for the circuit to somehow 'know the difference' between a battery source and a wall wart? I assumed that since they are both DC that it's equivalent from the circuit's perspective.

    Another crazy thing I can't explain is how the doorbell actually managed to make the ring sound that one time (in the garage, when I performed the very first test). Since then, no such luck.

    Any more ideas?

    Thanks!
    Pete
     
  12. strantor

    AAC Fanatic!

    Oct 3, 2010
    4,302
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    I know almost nothing about RF and I'm pretty noobish all around so this is probably stupid, but...

    If the wall wart is the transformerless type, which it probably is, then your DC voltage may have some connection to earth ground.

    If your door bell was meant to operate on batteries, then the designer may not have taken any measure to isolate the antenna portion from the power (battery) circuit.

    So when you connect the wall wart which is connected to earth, to the battery terminal which is connected to the antenna, you effectively ground the antenna, which renders it inoperable.

    So there's my theory, now let the experts chime in.
     
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  13. elec_mech

    Senior Member

    Nov 12, 2008
    1,513
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    Does the voltage change when you have the batteries installed and press the doorbell? If it doesn't take much current it may not show a change in the voltage reading.

    Wow Strantor, that would've never occurred to me. Excellent observation. Would that happen with a switching power supply, an old school wallwart, or both you think? I know the classic wallwarts use transformers and wonder if the transformer is isolated and if that would make a difference?

    OP, do you happen to have a UPS handy? You plug the cord into the UPS and unplug the UPS from the wall and test your doorbell then. This should tell you if you're having an RF issue.
     
    strantor likes this.
  14. pkozul

    Thread Starter New Member

    Mar 4, 2011
    9
    1
    Interesting information. Thanks for your input.

    Unfortunately, I don't have a UPS so won't be able to try that out.

    Any other tests I can do, to determine whether it's an issue with grounding or not?
     
  15. strantor

    AAC Fanatic!

    Oct 3, 2010
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    Not without more stuff that you probably dont have. You could try an isolation transformer or an old school wall wart that has a transformer inside.
     
  16. strantor

    AAC Fanatic!

    Oct 3, 2010
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    With your wall art plugged in, test both wires voltage to ground. The the 3rd prong of the outlet should work. If you can read a voltage on one of the wires then that will confirm half my theory.
     
    elec_mech likes this.
  17. pkozul

    Thread Starter New Member

    Mar 4, 2011
    9
    1
    Hi all,

    Thanks for your help. Just wanted to let you know that it works now.

    I ended up getting this Isolated Power Supply Module Adapter DC to DC Converter:

    http://www.ebay.com.au/itm/261156224638

    I had a spare 12V wall wart lying around, and after soldering the above converter between this power supply and the doorbell, the doorbell finally started making noise.

    I guess it all works now thanks to the converter providing an 'isolated' ground, as strantor hinted...

    Thanks again!!!

    Regards,
    Pete
     
    strantor likes this.
  18. #12

    Expert

    Nov 30, 2010
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    F'n brilliant, strantor.
    Makes me wonder if a dual inductor filter attached between the battery terminals and the 5 volt cord on the grounded wall wart would exclude the RF from leaking out to the wall ground. Probably not as well as the complete isolation module.
     
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