Transformerless power supply with only Live wire

[Ibob]

It says less tha 6mA standby, plus operating the relay

The ones you gave the link to are momentary, which means the light only stays on when you hold the button down, I would have thought latching would have suited better, unless you only want the light on for a short time like walking along a short path.

Yes, you're right they are momentary. Therefore I've developed an 555 circuit to go to constant. I will put button like manual switch on the wall, so the light will be remotely and/or manually controlled thus way.

 

[Ibob]

Hi,

Looks like 6ma. Should be ok.

I've seen ones that screw into the bulb socket, but you dont have to add a separate power supply. Couldnt you get one of those kinds?

If I use one connected to the bulb socket, then when it is off, no one will be able to manually turn the lights on from the wall switch, right?
Being momentary will help me control the lights by pressing a button (on/off) on the wall switch as well. This way one can switch it from the room, and I will be able to control it from the internet. Because the remote control I will connect to a Ethernet controller

 

[Ibob]

Hi,

1. The jumper is just drawn to show how the two power supplies (upper and lower) connect their positive outputs together. The grounds (GND) also connect together.

2. The logic gets powered from the output of the two supplies which are tied together.

3. The bulb goes between the triac and the line as usual, but also in series with that is the primary of the current transformer.

4. You can get pretty small current transformers as long as the current on the output is light. I use one of these daily that is about 1.2 inches in diameter and about 0.3 inches thick. So it's a small donut shape.

How much current does the 'radio' switch require?

Hi MrAl,
In order to confirm I got your circuit right is that what you are suggesting?
Apologies for the lame graphic but I need to get the basic idea for myself.



It seems that when light is OFF, both supply exits are connected together providing 12V. When light is ON, only the serial power supply provides 12V. Is it correct what I've understood?

Actually, on a second thought, both of these supplies are in serial.

Also, where I can put a fuse? On the Live wire, before the parallel power supply, between the bulb and the serial power supply?

 

[MrAl]

Hi,

When the switch is off the serial supply puts out very very little power, mainly the parallel supply does it then.
When the switch is on obviously the parallel supply gets no input, but the current goes up so the serial supply kicks in with higher output.
One note is that the output cap should be big enough to keep the output voltage high enough for the circuit during the switchover. Probably at least 100uf but maybe as high as 1000uf.

Yes put a fuse in series with the Live wire before the switch and parallel supply.

It looks like you have it connected properly.

If you want to use a regulator put it before the 12v circuit, using the output from both supplies as you have drawn as the input to the regulator.

 

[Ibob]

Hi MrAl, Hi All,

I wish I could avoid that current transformer as it will consume a lot of space I don't have and also, I see (but don't understand yet) other solution is possible (see below):

I've opened one of these apart:
http://www.lexing.com.cn/en/proview.asp?P_ID=51

And here what's inside:








I will analyse the circuit, if I can.... but you're welcome with comments if you want...

What I've figured out so far is that it provides 6V to the sensor PCB which controls back one transistor - which most probably controls the triac.
On the other side, when I measure the big cap, it gives me 19V. I can see 2 zener diodes - most probably one is on 19V the other on 6V. The sensor gives some buzz noise when the triac is active. I guess that's because of the bad rectifying. As soon as I get the basics, my idea is to use improved elements: bridge rectifier, maybe a regulator, optron to control the triac, etc. (with someone's else help i guess)...

The three cables connected to the sensor PCB are power 6V, and one that should control back the transistor switch.

 

[MrAl]

Hello again,


Well the other solution myself and MrRB were talking about is to control the triac so that it does not turn on until maybe 10v or so, and during that time the power supply takes in some power. So when we tell the triac to turn on we really are telling it to turn on later in the cycle, not enough to dim the light, but enough to tap off some power before we turn it on all the way.
That requires a bit more of a circuit however im not sure you want to get into or not.

If you can build up a complete schematic for that unit we may find a way to deal with it better, but it would have to be very complete and VERY accurate or we could really mess something up. We have to know exactly what we are dealing with first before we can modify anything. There are subtle reasons for doing things sometimes so we'd have to know everything about the circuit including component values and ratings. We'd also have to do some complete testing of the unit once it is modified to make sure it is safe and works properly.

The small current transformers are pretty small, but yes it's another thing to have to buy too.

How about a photo cell or two to keep the box running when the light is on Just kidding there.

 

[Ibob]

MrAl, thanks for the details and for the support so far!

I berеly understood what you've said above in the first paragraph, mostly because of my lack of knowledge, but I did get the point that it will be too complex.

So then I have 2 options:

1) to rip off the power supply the way it is now from such a sensors... expensive option.

2) to go with the current transformer and the circuit you've proposed above.

I choose the 2nd one. I will start looking for such transformer and will initiate some tests on the breadboard.
I will get back with questions I might face or hopefully with the mock-up up and running!

Thanks a lot and looking forward to collaborate again.
Bobby

p.s. I think there is a DIAC element in this sensor power supply as well. (DB3 - in the middle of pic 3). I see some circuits with also a DIAC controlling the TRIAC. Now I wonder why is it for?

 

[Ibob]

OK, I can't stand it so I post it... look how simple it looks like:


Can I draw power from that one somehow...
I don't want to dim the bulb... just to turn on/off..

 

[MrAl]

Hi,

Where did you get that schematic? If that is it, then you might be able to tap some power from right across the triac.

A diac is a device like a zener diode, except that it works for current in either direction and voltage of either polarity, and when the threshold voltage level is reached the device turn off voltage threshold goes low. So it's like two zener diodes back to back but they both have a lot of hysteresis.

What this means is that if the conduction angle is high enough, you might be able to use a diode and capacitor to tap off power. The only catch is that you have to make sure that one side of the triac is already connected to one lead of the 12v power input to the module. If it's not, then it may be more difficult to implement...we'd have to have the entire schematic.

It could be that they are using the diac so that THEY can tap power off of the triac leads too. That's if it is the same schematic.

 

[Ibob]

I took it from some web page.

Regarding the sensor above, I tried to follow and it looks like this:


These 2 caps creating an isolated point maybe is the key here?

Hope you can finish it somehow with bridge rectifier and everything till 12v output ..

 

[MrAl]

Hi,


Sorry i dont understand your diagram because it shows the fuse right across the triac for one thing. That would not be possible.

 

[Ibob]

Hi,


Sorry i dont understand your diagram because it shows the fuse right across the triac for one thing. That would not be possible.

Big apologies here. My fault.
Here is the correct drawing:

 

[MrAl]

Hi,

Ok here is the first trick we can try. Use a diode that is rated for the full line voltage times two. Use a cap with a nice high rating like 500v. Measure the voltage across the cap as shown in the diagram with the triac turned ON (bulb lit brightly). That voltage will be a DC voltage hopefully high enough to use for something useful.

By all means be careful as this is a little dangerous because of the shock hazard as well as the explosive hazard of electronic parts subject to extreme operating conditions due to mistaken wiring or accidental temporary shorts as well as other mistakes. Obviously wear eye protection and stand back when doing tests whenever possible.

Also, check and recheck your schematics to make sure they are accurate because one little thing wrong could make whatever we do useless

 

[Ibob]

Hi MrAl,

This high voltage cap you are talking about is your blue cap on the diagram, right? Or not - I guess not. The blue one I suppose should be a DC cap? The caps in series are 400V or something.

 

[Ibob]

OK,
This is what I am going to test. The idea is to see if I will be able to switch the light ON/OFF with a switch, and the same time to have DC 12V no matter the switch position.

Before I do it, can you give the "green light" that I didn't miss something. I hope I am not going in wrong direction here. I think this floating point between the two caps in series is THE KEY.

 

[alec_t]

RED light! Be aware that both the + and - of your DC output can be at peak mains voltage (~330VAC). Also, if C4 has no load across it the voltage across the cap will also rise to ~330VDC.

 

[Ibob]

I know it is dangerous, but it will be inside the wall junction box.
The first point you mention is because it is the way the transformerless supply works, not because of a mistake, right?
There will be a load always - the rf receiver and the triac control.
But I guess you're right I need a load for the test and to measure the voltage across it.
What type of load I could use for the test?
Eventually do you see errors in my circuit or you just warn me?
Thanks a lot!

 

[alec_t]

The RF Rx and the triac control will present a variable load. That means the DC voltage will also vary. You would need to regulate it, with a zener at the very least. Once you've decided on the voltage (12V?), then you can determine what load could be provided via the capacitors (or conversely what cap values you would need for a given load). Don't mess with mains voltage unless you take all due safety precautions. You should also ensure you won't invalidate any house insurance policy you may have, or infringe any local wiring regulations. This circuit would leave the lamp bulb socket in a permanently live state and I have concerns over its safety.

 

[Ibob]

Thanks alec_t,
Then I will put a zenner before and in parallel with C4 at the end.
I agree with your remarks for the safety.
I have all arround these sensors I was talking about above, several dimmers working the same wsy, i.e. requiring incandescent bulb to operate, meaning the socket is under voltage in a way. The same time all of those are not prohibited here and you can buy freely.
Thank you again.

 

[Ibob]

OK, based on the alec_t considerations, I've modified the circuit by adding a zenner diode and R1 (fat one).
Instead of the zenner diode I could use regulator, but I guess for the RF receiver driving a relay and the 555 IC + optocoupler elements driving the triac I don't need such, right?