A friend of mine’s daughter is deaf and I wanted to build her something that would alert her when the doorbell rings. I chose to adapt an off the shelf RF doorbell from Amazon. What I thought would be an afternoon project has turned into a real headscratcher.
The idea is simple. Trigger a relay with the flashing LED which can then power an extension cord. This lets the user plug in whatever alert device they choose like a standing lamp or light etc.
I chose this doorbell because it doesn’t require batteries for the button and there are two alert devices for two sesperate rooms. I’ve already managed to blow one of them while getting meter readings and can’t get the led to trigger the relay? I have attached a jpeg to show how I’ve connected the relay and I have also tried to connect it via a transistor but I’m not that confident that I connected that correctly. If a transistor is the way to go I will add details of how I have connected that but I don’t think it is necessary. The relay is powered by the boards VCC/GND which gives me 4.39V and when the doorbell is triggered I get a reading of 2.4V across the LED points which should really be enough power to trigger the relay, but it’s not triggering? I have also tried other types of relays with no luck.
No, you don't have it wired up correctly a BC548 is an NPN, You should wire it up as such and power it from a different source (with common GND) what kind of relay is it ? 5v ? 12v ?
Hacking that device is a possibility, but with its risks as you already found out.
Didn't it occur to you that there is a way to do this in a much less intrusive way ?
Why don't you watch the LED flash, and act on that ?
You'd need a bit more electronics (but hey, this is the Arduino forum), but it would allow you to do more with the registered rings at that door.
And if you want to stick to this idea, there's also relay modules which have opto isolation and transistors incorporated.
But you'd have to search a bit and wait for its delivery.
It is not completely clear from your pictures which show only one side of the pcb, but that device appears as if it may have a transformerless power supply, that is it is not isolated from the mains supply and great care is required especially if you are bringing wires to the outside. Also the current in that case will be limited to a few mA. The opto-isolater as suggested is a good approach.
Deva_Rishi
Your diagram seems to show how to wire up a naked SRD-05VDC relay, Im not using a naked relay with no diodes, resistors etc. Its one of those easy to use 5V relays on a board. But really my lack of general electronics quickly becomes apparent when you put a diagram infront of me.... sorry
I have built up a relay before but it’s pretty complicated and requires loads of soldering etc. I really want to keep it as simple as possible if there’s a ready made relay I can use then I’d rather use that. If that’s not possible in this instance I have attached a diagram of the relay I put together before, do you think this will work?
MAS3
I am using this relay board I got off Amazon,
I have also tried it with one of these....
I've actually just ordered one of these, fingers crossed it works.
What would you suggest when you say “a less intrusive way”? I am very much up for suggestions of how to make this easier and/or better. I am using the flash of the LED at the moment, is there another way to use the LED other than wiring it to a relay?
6v6gt
I have attached a picture of the other side of the PCB.
Indeed, it is a non-isolated transformerless power supply. The value of the bright blue capacitor marked RC1 on the board will determine the power you can draw from the circuit. It appears as if it could be an X1 rated ceramic capacitor does not look very big.
I can't make out what it says on the blue capacitor, attached is a close up pic.
CVR
05043IK
It would be easier if I had a magnifying glass...
Is it critical in this instance to have this calculation?
I have managed to get the readings with my multi meter but just like the article warns there is high voltage on the board and this is how I shorted this one, while taking readings, accidentally touching the resistor positioned next to the blue capacitor. Little crackle and now it spark when plugged in. Luckily I have another to work on.
I have been working with it plugged into a switched extension lead and only powering on to test and take readings, even without much technical knowledge I could see the mains cables feeding the board so was instinctively carefull.
There is enough room in the enclosure for a relay so the only thing that will protrude is the sealed cable of the extension cord that will be running from the relay. Should be fine. If it doesn't all fit because I end up needed more stuff I plan to pack it all in a separate box. At this stage I just want to get the relay to trigger and then I can worry about making the enclosure safe...
lessmann:
What would you suggest when you say “a less intrusive way”? I am very much up for suggestions of how to make this easier and/or better. I am using the flash of the LED at the moment, is there another way to use the LED other than wiring it to a relay?
LED means Light Emitting Diode.
So that is what it does.
Less intrusive means you do not need to de-solder that LED to see it is lit or not.
A photosensitive device would suffice, like a LDR or a photosensitive diode.
Next, you could use any Arduino variant to register the lit LED, and do something with it.
Like switching relays.
But if you wanted to you could also register how often and at what times the doorbell knob was pushed.
Or even create a portable device which would vibrate when someone is at the door.
There's a lot more possibilities (smart phone ? wouldn't surprise me if she'd be using one).
But that would take a bit more than tinkering around for an afternoon.
Making a portable vibrating device or sending an email with and esp8266 was my first idea but it requires an external service provider or hub of some sort for the email and apparently she doesn't always have her phone on her (I know, teenager not keeping her phone on her sounds weird to me too!)Also she will receive emails when she is not home and if she doesn't keep her phone on her all the time she is unlikely to remember a vibrating device. After talking to the family about options we decided that a visual alert system would be ideal. I've spent a bit of time thinking of how to do this and looking at devices that I could hack instead of starting from scratch and this does feel like the simplest way. No coding and very little that can go wrong. The photosensitive sensor is a good idea but introducing an Arduino and coding sounds unecessary and adds complication and adds to things that can go wrong. (I know shuning an arduino on an arduino forum sounds like blasphemy!)
I'm way over spending an afternoon on this, I've been tinkering for many days on this problem and will keep tinkering until I get it done. Failure is not an option.
The capacitor markings are impossible to see from your picture. Sometimes it helps to light the object from the side as well if the markings are recessed or etched in.
If it is a capacitor based transformerless power supply, and looking at the physical dimensions of the capacitor and considering the voltage/rating, let's say it is 100nF (but even that seems high for the size) then it would pass only around 8mA.
However, I see there is also what appears to be an inductor (near the coil antenna and next to D2) so it could well use some switching buck technology instead to step down the mains voltage. That would give significantly more power. I can't read the markings on the IC U3 but that could be a switching circuit. Again this would not be isolated from the mains supply so the same care is required even on the "low" voltage side.
I actually made a flashing light simply with the wires connecting to the doorbell once for friends who were doing a lot of woodwork in their workshop, i managed to keep it really simple with a button triggering a 555 timer that drives an analog relay providing power to it, but that did require some drilling through the wall.
6v6gt:
The capacitor markings are impossible to see from your picture. Sometimes it helps to light the object from the side as well if the markings are recessed or etched in.
If it is a capacitor based transformerless power supply, and looking at the physical dimensions of the capacitor and considering the voltage/rating, let's say it is 100nF (but even that seems high for the size) then it would pass only around 8mA.
However, I see there is also what appears to be an inductor (near the coil antenna and next to D2) so it could well use some switching buck technology instead to step down the mains voltage. That would give significantly more power. I can't read the markings on the IC U3 but that could be a switching circuit. Again this would not be isolated from the mains supply so the same care is required even on the "low" voltage side.
The capacitor markings are printed and whatl I can make out is the following:
CVR
050431K
Please find a photo attached which shows how I wired up the relay to the board. When connected it triggers so I don't think there is a load issue I'm just not connecting it correctly because it isn;t behaving as it should...
I've attached the relay which is triggered when connected, there is no change when LED is flashing.
ie. When I connect the Boards LED + terminal to the IN terminal on the relay it triggers and goes off when I disconnect it. It does not change if the doorbell is pressed or the LED is not flashing.
There is a constant current of 0.92V on the LED when off and 2.4V when flashing so I'm very confused as to why it triggers when connected and the LED is not on?
The relay (if you are using the type illustrated) has a jumper which can be used to change its behaviour from "trigger on high" to "trigger on low". You can try putting the blue trigger wire directly on the relay's DC+ and DC- terminals to see how it reacts.
If it is still unreliable when connecting to the led, follow the led tracks back and work out which side is the GND side and which side is the 'high' side and where the series resistor is.
Looking at LED 2 and LED 4 on the board, it appears that one side is directly connected to VCC . Therefore the other side must be switched to GND (almost certainly via a series resistor of say 220 ohms) to light the led.
Deva_Rishi:
No, you don't have it wired up correctly a BC548 is an NPN, You should wire it up as such and power it from a different source (with common GND) what kind of relay is it ? 5v ? 12v ?
Is a 1N4148 a good choice for a "flyback diode" for relays, motors, and other inductive loads? I have a homemade electromagnet I'm using on a project of mine where I don't know the inductance of the coil, and am not sure what the specs on a protection diode should be. (Right now I'm using a relay module that has optical isolation, a transistor, and a protection diode built into it, but I'm thinking of getting rid of the relay and powering my electromagnet directly with a transistor and flyback diode.)
And LED 3 is connected to the same VCC through 2 via's.
But it looks like those 4 LEDs are all tied together, because the cathodes seem to be all connected trough smaller via's too.
Is that correct, do all 4 LEDs light at the same time ?
The yellow wires (to a buzzer or speaker ?) and the sticker obscure my view so i can't see all i'd like to see from your pictures.
You are talking about current, but show values measured in Volts.
Voltages seem to be correct, but please try to use the correct term.
DuncanC:
Is a 1N4148 a good choice for a "flyback diode" for relays, motors, and other inductive loads?
The specs of the diode must exceed those of the coil - that is, the current must be rated greater than the coil current (because at the moment of switch-off all of that coil current will be flowing through the diode, though not for too long) and the rated voltage greater than your supply voltage (because that is the voltage which will appear across the diode when the coil is energised).
So a 1N914/ 1N4148 - a signal diode - is generally the wrong choice except perhaps for a relay operating well under 100 mA. Generally, a power diode with ratings as above will be more appropriate. On the other hand, if (but only if) you are going to use PWM, then a fast recovery or Schottky diode will be needed.
Paul__B:
choice except perhaps for a relay operating well under 100 mA.
The 12v Relays i tend to use take up around 60mA, hence the choice (actually i copied the spec from a Velleman windscreen wiper interval kit that uses similar relays and that diode)
Deva_Rishi:
The 12v Relays I tend to use take up around 60mA, hence the choice
And research on the diodes whose datasheets I quoted indicate that they are rated for over 100 mA and about 100 V so they are more rugged than I habitually assumed. Their specification also indicates they are quite fast recovery, though that specification is at quite low currents - 10 mA.
Deva_Rishi:
actually I copied the spec from a Velleman windscreen wiper interval kit that uses similar relays and that diode
One tends to trust Velleman though a recent thread concerned me. Cannot find it now.
and power it from a different source (with common GND) what kind of relay is it ? 5v ? 12v ?
