Trigger both Video Doorbell and Standard Doorbell Circuit

[KeepItSimpleStupid]

I looked at your initial post, actually for the first time.

For the Video bell to real bell, what I think you can do is:

a) Locate the 12V power within the doorbell
b) find out what's going on with the button. let's say it uses 3.3V or something. and the button is grounded when pushed.

Aside: An optomos relay such as https://www.digikey.com/product-det...cuits-division/CPC1014NTR/CLA314CT-ND/1973289 only takes 2 mA to operate.

The LT6700 https://www.analog.com/media/en/technical-documentation/data-sheets/6700123fh.pdf comparitor is a dual comparitor that comes in all sorts of varieties. There is a single version.

Put the contacts of the OPTOMOS part across the normal doorbell and use the comparitor to detect that the button is prssed.

That gets you the ring when you push the video doorbell.

To interface to the regular doorbell, you might be able to use something like: https://docs.broadcom.com/doc/AV02-2153EN

You have AC and it would drop to close to zero when the button is pushed.

I don;t know if you get a "double-feedback" sort of thing. Again, use an optomos relay on the video bell side.

There are my thoughts.

 

[exomic]

I just probed the doorbell and I got 21.0vac. I also probed the transformer and I got 21.4vac. When I press the doorbell the voltage across the bell goes to 0vac. So could we understand that the doorbell is drawing 0.4vac ? I just getting more confused...

 

[Les Jones]

I think you are measuring the voltage across the bell push NOT the actual bell.

Les.

 

[exomic]

I think you are measuring the voltage across the bell push NOT the actual bell.

Les.

I probed the doorbell (the switch).

 

[KeepItSimpleStupid]

So could we understand that the doorbell is drawing 0.4vac ?

No, you can;t do that. AC, inductive loads complicate things, so lets pretend:

One switch
One light bulb
One 1.5V battery

Complications:
The battery has an internal resistance of say 1 ohm
the wiring has a resistance of 1 ohm,
The light bulb is say 10 ohms cold and 100 ohms hot.
All fake numbers.

All in series.

Voltage across the switch.

It will read 1.5 V, The meter acts as a 10 M resistor. 10 M + 100 +2 is still about 10 M ohms.​

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So, monitor the voltage accoss the switch. If you don't take too much current away, the bell won't ring.​

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At low currents or low DC currents, the bell acts as a piece of wire.​

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Monitoring the voltage across the doorbell is an effective way of determining if the button is pressed, A lighted doorbell MIGHT add complications.​

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There might be a possibility that with dual monitoring, you could end up with both buttons indefinatey pressed. The circuit may have to intentionally prevent that "somehow".​

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[exomic]

Ok so if I understand I need a to find a optocoupler that will close the video doorbel side with a voltage that is close enough the doorbell switch voltage when pressed. How do I know what is the voltage on the line when the doorbell button is pushed ?

 

[KeepItSimpleStupid]

I gave you some ideas. You need to steal some power from the video doorbell which is likely 12V DC

Activating the real doorbell is likely easier. You may have to pulse the real doorbell making it "harder:.

The OPTOMOS series of optocouplers, some will handle AC, but the cool partis that it only requires a few m, like 1 or 2 mA. So the series R might be less than (12-2.1)/1e-3. 12V is the operating voltage, 2.1 is the drop across the LED and 1e-3 is 1 mA.

the voltage, current, series resistance all have to be in range too.

--

You can likley use a similar device to activate the video doorbell. I'd suggests finding what's the highest resistance that will work. I'm suggesting that a DC version with an on resistance of 500 ohms might work on that side.


If you have a LED that lights up on the video dorbell that might be useful too.

==

Doing it one way may easily work. Doing it both ways may be catch-22 and the buttons latch.
unless logic like.

Pulse real doorbell for 1s when video doorbel is pressed unless real doorbel was pressed < 10s ago

or some such nonsense.

 

[exomic]

I gave you some ideas. You need to steal some power from the video doorbell which is likely 12V DC

Activating the real doorbell is likely easier. You may have to pulse the real doorbell making it "harder:.

I'm currently only planning to trigger the video doorbell switch when the real doorbell switch is actuated. On the video doorbell I only need to short 2 cable together to make it ring so no big issue there (I like the optocoupler idea to avoir mixing 2 different current circuit together AC/DC).

So to do that if I understood I need to know what is the running voltage when the Doorbell switch's light is on (switch open) so that I can find a optocoupler that won't trigger the video doorbell unless the current is higher than that (so when the switch is closed it will trigger the video doorbell and not when the light is on).

I'm new to electronic so how can I find the running voltage when the light is on using my multimeter? Perhaps I should turn off the doorbell switch and probe it to find the amperage drawn? For the AC source I probed the transformer and I read 21.4VAC so is that normal or should I unplug the doorbell circuit from the transformer before reading the value?

Thanks for your help!

 

[KeepItSimpleStupid]

I think you have like 3 ways:

1) detect current
2) "relay at the bell" sort of thing.
3) "Sense" voltage across the doorbell switch.

All are difficult for one reason or another.

 

[KeepItSimpleStupid]

Like I said, this https://docs.broadcom.com/doc/AV02-2153EN may work, but if you have a lighted doorbell with LED lights, it might cause a problem. The equations are messy.

 

[KeepItSimpleStupid]

The LT999 https://www.analog.com/en/products/lt1999.html#product-overview might be aplace to start if you were monitoring current. But you need a 5V supply.

 

[KeepItSimpleStupid]

Your going to have to do some sort of reverse engineering. probably +12, +5 or even 3.3 VDS is available in the video doorbell.

Your going to have to find power and ground.

Write down the codes on the IC's and look for "SMD marking code" to try to identify the IC. This can help identify power. You might need a little bit of 5V.

 

[ChrisP58]

Here is how I would do it.

The problem that the lamp across the doorbell switch causes there to always some voltage across the bell.

The two zener diodes subtract about 7 Volts from the peak voltage that shows up on the doorbell. These values assume that the voltage across the doorbell is less that about 5 volts AC when the button is not pressed.

The values may need to change. The voltage of the zeners needs to be about 1.5 to 2 times the AC voltage measured across the bell when the button is not pressed.

The value of R1 may need to change depending on the final zener voltage is chosen.

The opto coupler can either be the AC input unit you found, or 2 generic opto couplers wired as shown.

Also, be aware that the output of the opto coupler will be pulsing at twice the AC line frequency. C1 should flatten those pulses out, but it's difficult to choose a value without knowing what the front end circuitry of the video doorbell lokks like.

 

[exomic]

Also, be aware that the output of the opto coupler will be pulsing at twice the AC line frequency. C1 should flatten those pulses out, but it's difficult to choose a value without knowing what the front end circuitry of the video doorbell lokks like.

What would be the result if I don't put a capacitor and let the video doorbell switch receive the pulsing?

Never heard of those zener diodes but seems to my understanding that they have a minimum current limit that when reached will all current to pass so I simply need to find out what is the minimum current. Do you know how to find out using a mulitmeter?

 

[ChrisP58]

What would be the result if I don't put a capacitor and let the video doorbell switch receive the pulsing?

I don't know. You'll just have to try it and see.

Never heard of those zener diodes but seems to my understanding that they have a minimum current limit that when reached will all current to pass so I simply need to find out what is the minimum current. Do you know how to find out using a mulitmeter?

Zener diodes are voltage and polarity dependent devices. When they are forward biased, they act like a regular silicone diode with a forward voltage of about 0.6V. When reverse biased, and the voltage across the diode is less than the zener voltage, they are nominally open circuit. When the reverse voltage exceeds the zener voltage, they start to conduct. You do need limit the current through the zener to make sure that the total power dissipation is less than the power rating of the zener.

The reason that two end to end zeners are needed in this circuit, is because the applied voltage is AC. In my circuit, the effective AC zener voltage is the forward voltage (~0.6V) plus the reverse voltage (6.2V).

Please measure the AC voltage at the bell when the button is NOT pressed.

 

[Les Jones]

If you take the voltage measurements I requested in post #20 it will help us to give you a circuit that is more likely to work. We seem to be using different terminology. When I say "bell push" I mean the the push button switch by the door that the visitor presses. When I say "bell" I mean the thing that actually makes the noise. It is the voltage across the part that makes the noise that I wanted you to measure both with the button the visitor presses pressed AND not pressed. I did something similar many years ago to trigger the wireless push button for a remote door bell from the original hard wired door bell. I had a look at this and I has used a single diode (Half wave rectification), a capacitor, a DIL (Dual inline.) relay and a resistor as the bell circuit was 12 volts and the DIL relay coil was 5 volts. This has been working for about 20 years without any problems. The circuit that Chris has provided is basically the same as was first suggested in post #2 but Chris has used two DC opto isolators in place of the AC one and has used zener diodes to prevent the voltage across the bell due to the lamp current from triggering the opto isolator. This is a better solution to using a shunt resistor as I suggested. You seem to be trying to make a very simple problem difficult by not providing information requested.

Les.

 

[exomic]

The voltage across the doorbell when the doorbell switch is not pressed is 0.5vac

as for the voltage when the button is pressed I would need to wait for one of my friend to come see me home to help me as I can’t do it alone but now we at least know that the voltage to ignore is 0.5vac?

 

[ChrisP58]

The voltage across the doorbell when the doorbell switch is not pressed is 0.5vac

as for the voltage when the button is pressed I would need to wait for one of my friend to come see me home to help me as I can’t do it alone but now we at least know that the voltage to ignore is 0.5vac?

Then you can use lower voltage zeners. There is a slight advantage to using a lower voltage zener. The width of the pulse on the output of the opto will be a bit wider with a lower zener voltage. That may, or may not, make any difference to your video doorbell input.

The 1N5221B is a 2.4 volt part, and is the lowest value in that family. Here is a possible source:
https://www.digikey.com/product-detail/en/on-semiconductor/1N5223B/1N5223B-ND/977556

The voltage at the bell when the button is pressed will be a bit lower than the transformer output. (lower due to wire resistance)

At 21VAC the LED current with 1K for R2 will be about 17mA. The average dissipation, when on, is about 0.3 watts, so use at least a 1/2 watt part there.

 

[exomic]

Is this optocoupler suitable? https://abra-electronics.com/opto-i...817-dip-4pin-general-purposephotocoupler.html

 

[ChrisP58]

Is this optocoupler suitable? https://abra-electronics.com/opto-i...817-dip-4pin-general-purposephotocoupler.html

Yes, that'll be ok. But you will need to use two of them wired as shown in my schematic.