Trigger both Video Doorbell and Standard Doorbell Circuit

Hi,

I just bought a Dahua Video Doorbell connected to my network via PoE (Power Over Ethernet). The thing is that some people still use my standard doorbell (I can't rewire/remove it without leaving a big hole in at my front door) so I would like to trigger the Video Doorbell when the standard doorbell is pushed (and vice-versa if the video doorbell is pushed trigger the standard doorbell so if I don't have my phone I can hear the strandard house doorbell). At first I tought maybe a simple reed relay would work but I now realized the standard doorbell is using AC (I probed the transformer and I read 21V but on the transformer it's shows 18V-10VA Load am I missing something here?), so the reed relay would need DC and therefore a bridge rectifier?

Some guys did this circuit for the same doorbell but he's using the same power source to power the video doorbell and the standard doorbell (not a possibility in my case I must use the PoE power from the network switch for the video doorbell). Source


I won't lie I'm not a expert in electronic, I do have soldering, pcb etching experience but as for circuit designing this project is a bit over my confort zone so I would like your experts advices and recommandation to do it the "right way". The main goal is to trigger the Video Doorbell when the Standard doorbell is pushed. If the other way around is not possible due that's not a big deal but would be great to get it to work both ways.

Thanks guys!

You could use an open-collector/drain opto-isolator to trigger the Video d-b. The LED inside the opto would be powered when the Standard d-b switch closed. This would require soldering the opto's output terminals to the Video's bell-switch terminals, and connection of at least a diode and resistor to the Standard bell circuit. If the Video d-b objected to repeated triggering at mains frequency then a monostable circuit or other means would be needed to control the triggering.

alec_t said:

You could use an open-collector/drain opto-isolator to trigger the Video

Click to expand...

I found this type of optoisolator compatible with AC. https://www.onsemi.com/pub/Collateral/FOD814-D.PDF



I just need to find the correct specs based on the voltage supplied on both doorbell. I know that the classic doorbell is 21VAC when probed ( 18V-10VA Load written on the transformer still confuse me tho) so I need to open my video doorbell to probe the switch's contact to find out the VDC of this circuit.

To my understanding, the optoisolator will act as a switch, is there any load that could do something with the video doorbell like overdrawning power ?

Am I missing something?

exomic said:

Am I missing something?

Click to expand...

The opto shown would conveniently eliminate the need to rectify the AC with a diode.
Don't forget a resistor in series with the opto-isolator's input diode(s), to limit the current to something a good bit less than the maximum rated current specified in the opto's datasheet.

alec_t said:

Don't forget a resistor in series with the opto-isolator's input diode(s), to limit the current to something a good bit less than the maximum rated current specified in the opto's datasheet.

Click to expand...

I’m trying to find the specs limitations in the datasheet but I’m not sure exactly were it says the max voltage/current. What is the name of the variable I’m looking for in the datasheet if I want to know the max voltage/current for the diode side and the max for the other side?

Look under Absolute Maximum Ratings. For the opto shown, for both the LED (emitter) and the detector the max current is +_50mA, so driving the LEDs at, say, 30mA should be fine. The opto (not the A version) has a minimum current transfer ratio (CTR) of 20%, so the collector current could be only 6mA. Hopefully the Video bell switch will not need more than that to trigger it. The A version of the opto has a minum CTR of 50%, so could provide a higher trigger current.

I just realized something, the doorbell circuit is not only a simple switch that ring the bell as there's a light that's always on (except when button is pressed). I believe the doorbell switch has the light plugged in parallel to the switch so when the button is pressed it short and ring the bell (and the light turn off)? I suspect the light having a high resistance so the bell doesn't actuate?

Now that I know that, the optocoupler would be getting power because of the light in parallel all the time and therefore would always be in the closed position and will ring the video doorbell. How can I detect if the button is pushed or if the button is not pressed and only the light is on the circuit?

Just connect a shunt resistor in parallel with the input side if the opto coupler. Select a value such that the current drawn by the light in the bell push causes a voltage drop less than the forward voltage of the LEDs in the opto coupler. start by measuring the current through the bell push both when pressed and when not pressed.

Les.

Here's my understanding of the circuit. So the Shunt will let the power pass and short the optocoupler. How can the shunt stop shorting the optocoupler when the current increase due to the DBSW closing and shorting the DBLight? Will the Bell be affected by all those added resistance?

Your schematic is correct in principal but requires exrea components to limit the current through the opto coupler to less than its maximum current rating.. As you have given the current through the lamp. (Button not pressed. and the current the bell takes (Button pressed then I will have to make up some numbers to answer your question. I will assume the opto coupler uses an IR LED. These have a forward voltage of about 1.2 volts. If we guess that the bulb in the bell push takes 50 mA and the bell takes 500 mA then we want LESS than 1.2 volts across the resistor. So 1.2/0.05 = 24 ohms. So the resistor needs to be less than 24 ohms. So I would pick a value of less than 15 ohms. When the bell push is pressed we need a bit more than 1.2 volts across the resistor so we can add acurrent limiting resistor directly in series with the LED in the opto coupler. so I would aim for about 1.8 volts. So to limit the current to say 10 mA through the LED in the coupler we have 0.6 volts across the current limiting resistor so this resistor would need to be 0.6/0.01 = 60 ohms. Back to the shunt resistor calculations. To get 1.8 volts across the shunt resistor with 500 mA through it we require a value of 1.8/0.5 = 3.6 ohms. This value would work as it is also less than 24 ohms. In practice I would use some diodes in parallel with the shunt resistor to clamp the voltage across it. Silicon diodes have a forward voltage rating of between about 0.6 to 0.7 volts so three in series would clamp the voltage to between 1.8 and 2.1 volts which is about what wee need. As your circuit is working with AC you woould need 6 diodes. (Two chains of three in series connected in parallel with the diodes in each chain in the opposite polarity.

Les.

I have just read back through the thread to try to understand why you want to trigger the opto isolator by sensing the current through the bell rather than the voltage across the bell. Unless you have a good reason for this then I suggest sensing the voltage across the bell. The circuit is simpler as all you need is a single resistor in series with the opto isolator input to limit the current to a suitable value. This would save you from having to measure the lamp current and bell current. It also means that you will not be reducing the voltage to the bell.

Les.

Could you draw or explain how to do that? What components would you use to trigger the optocoupler based on the « voltage change »?

Assuming you are using an AC input type opto coupler. If we call the terminals on the bell "A" and "B" then connect one of the input pins on the opto coupler directly to terminal "A" on the bell. Connect the other terminal on the opto coupler to terminal "B" on the bell via a resistor. Use ohms law to calculate the value of the resistor. As the supply to the bell is 12 volts and the voltage drop across the input of the opto coupler is about 1.2 volts we have 10.8 volts across the resistor. So for example if you want 10 mA through the opto coupler input then R = 10.8 / 0.01 = 1008 ohms (So you would use a 1K resistor.) One thing that you may not have realised is that when the bell push is pressed the output of the opto coupler will not be a steady signal. There will be short pulses at twice the frequency of the AC waveform at close the the zero crossing points of the waveform. You may have to filter this out with a capacitor before feeding it to the input of your video door bell.

Les.

Les Jones said:

Assuming you are using an AC input type opto coupler. If we call the terminals on the bell "A" and "B" then connect one of the input pins on the opto coupler directly to terminal "A" on the bell. Connect the other terminal on the opto coupler to terminal "B" on the bell via a resistor. Use ohms law to calculate the value of the resistor. As the supply to the bell is 12 volts and the voltage drop across the input of the opto coupler is about 1.2 volts we have 10.8 volts across the resistor. So for example if you want 10 mA through the opto coupler input then R = 10.8 / 0.01 = 1008 ohms (So you would use a 1K resistor.) One thing that you may not have realised is that when the bell push is pressed the output of the opto coupler will not be a steady signal. There will be short pulses at twice the frequency of the AC waveform at close the the zero crossing points of the waveform. You may have to filter this out with a capacitor before feeding it to the input of your video door bell.

Les.

Click to expand...

Thanks for your help but one thing still is unclear. How does the optocouple know when to let the video doorbell signal through? I mean the bell is always getting current since the switch has a light in parallel to it so when the button is not pressed the current still pass through the light bulb all the way to the bell and therefore how do we know how to detect if the button is pushed or if the light bulb is on (button not pressed)?

exomic said:

. I believe the doorbell switch has the light plugged in parallel to the switch so when the button is pressed it short and ring the bell (and the light turn off)? I suspect the light having a high resistance so the bell doesn't actuate?

Click to expand...

This is generally true. The doorbell voltage is typically 16 VAC which is an "ODD Value". Incadesent bulbs are a current dependent resistor, so that adds a complication.

RING has some circuits to to iterface to their doorbell system. All electronic doorbells create even more problems. HAL has adoorbell interface and do does Legrand. Legrand hasn't a clue how their's operate though,

RING, "power steals" to charge a battery inside the doorbell.

KeepItSimpleStupid said:

This is generally true. The doorbell voltage is typically 16 VAC which is an "ODD Value". Incadesent bulbs are a current dependent resistor, so that adds a complication.

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So basically how can we make the optocoupler to closes when the doorbell button is pressed and stays open when only the bulb is on (button unpressed)?

This https://www.discounthomeautomation.com/ELK-Doorbell-and-Telephone-Ring-Detector-ELK930 guy detects the current to the doorbell.

KeepItSimpleStupid said:

This https://www.discounthomeautomation.com/ELK-Doorbell-and-Telephone-Ring-Detector-ELK930 guy detects the current to the doorbell.

Click to expand...

Would it be that hard to design my own circuit?

Most lighted door bell buttons use AC for the light and DC for the button close they do this by adding a diode.
The diode is only a part of the circuit when the button is pressed

I ran into this a bit back I installed A nice new sys the owner wanted to keep his basement bell and button it didn't work with the new sys.

I did fix the problem by adding a diode to the basement bell it blocked the AC for the light that was buzzing the basement bell and then passed the switched DC that the new bell was seeing when the press was made.

When the bell push is not pressed there will only be a small current through the bell. (The current taken by the lamp.) As the current through the bell is small the voltage across the bell will be small. When the bell push is pressed there will be the full 12 volts (Or so.) across the bell and also the resistor and opto coupler input in series. You will need to select a resistor so the opto isolator is not triggered by the small voltage across the bell when the button is not pressed but is triggered when there is 12 volts across the bell. Measure the voltage across the bell with and without the bell push pressed and report the readings.

Les.