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Its actually a hack and a concept of reuse of scrap. I have a 4 LED night bulb and a spare not working adapter. I took out the actual components out of bulb and then replaced it with faulted PCB of adapter and I again got a 12 V adapter which I basically used to power 12V LED strips.

In below circuit diagram you can make one more change. We got two points for input P and N, you can insert a resistor of low value and High wattage like 1/2 Watt to use as current limiter with point N...I have 220 ohms in my circuit.

Note:

You can use 1M ohms if you dont have 820K ohms.

The next circuit is more safe and you can use it for 12V generation also.

Also be careful while testing the circuit, its 220-240V AC not DC, its dangerous.
The circuit you show in the first image: What exactly keeps the output voltage limited to 12v? <br> <br>I think you've missed something important here in the disassembly steps :) <br> <br>As there is no regulator, no zener diode, nothing ... the voltage on the nominally 50v max capacitor will rise, and rise, and rise until it is almost the full rectified mains voltage. Of course, it will go BANG! at some point before that. <br> <br>In the original application, the LED string served as a regulator to hold the voltage down to the forward voltage of the LED string. With this removed, there is nothing to control the voltage. <br> <br>Suggest you put a 12v zener across that electrolytic, at the very least! <br> <br>I know the capacitor and resistor limit the current flow, but all that means is that the capacitor will charge over time to bring the voltage up to way more than 50v.
<p>Those curious how the voltage is limited in the 1st circuit : the 474K/250V capacitor before the bridge rectifier drops most of the 230V due its resistance (impedance) to AC. The input to the bridge (in both circuits) is only about 15Vac due to that.</p><p>Of course the output is not regulated, and will vary a volt around 12V as the mains varies. But for lighting LEDs etc, it will do. If you want to power one or more ICs, then it needs to be regulated with a resistor/zener or better, a 7805 regulator. </p>
<p>The capacitors CX and C1 form a voltage divider, with CX having the larger resistance because it has a smaller capacitance, and C1 having smaller resistance because it has a larger capacitance.</p><p>So, the voltage across C1 will be significantly smaller than that across CX.</p>
<p>I don't believe that's true. If both capacitors were in series across the AC, I might concede that point. However they are not, and C1 can only charge, and charge, until it reaches close to the supply rail (mains) as there is no discharge path without a load (effectively acting as a shunt regulator).</p><p>Look at this way, C1 starts at 0v,and every &quot;positive&quot; mains cycle, a packet of current is passed through CX/R into C1. Every negative cycle, the same happens -- the bridge steering the current to still charge C1 despite CX reversing. </p><p>At first the voltage will be low because C1 will &quot;load&quot; CX. As C1 charges it will appear to be less and less of a load, as its voltage rises to meet the line, until Cx is passing nothing much, and C1 is charged.</p><p>End point is moot: C1 will explode long before it is fully charged.</p>
<p>Simple Test: </p><p>Is the current through CX also flowing through C1 ? Answer: YES.</p><p>Therefore CX and C1 are in series. Therefore the voltage divider effect.</p><p>Note: The current through CX cannot disappear ... it has to return back through C1 (and load, if any), the one and only path!</p>
<p>Nobody says the current through CX disappears.</p><p>You don't take into account how the bridge rectifier is affecting this &quot;divider&quot; you appear to see.</p><p>The current in CX is alternating. The current in C1 is only EVER charging it up, steered to always be DC, by the bridge.</p><p>Without a shunt regulator/zener/LED load C1 will charge until it reaches mains level. It will NOT magically produce 12V. It <strong>will </strong>magically produce smoke.</p>
<p>Just to be clear -- I was talking about picture one (no zener). I have no problem with picture 2 (with series resistor and zener), other than the general &quot;shock hazard&quot; caution :)</p>
There is no zener diode in first circuit...I checked it too but its giving me 12.3 V as an output. I posted the second circuit which I've tried long ago and it has zener diode :)
When you measured the 12.3v, was it still connected to the LEDs at this point, or just to a meter? <br> <br>If it was connected to the LEDs, I'd believe you 100% - 4 white LEDS, 3.0v forward drop each. Those LEDs regulate the voltage by holding it down. <br> <br>If you measured the voltage with the LEDs disconnected, or (heaven forbid!) one of the LEDs failed ... then there's nothing controlling the voltage.
just tested it by muiltimeter no LED connected...i disconnected the LEDs and then measured it....after that I connected the LED strip and it worked.
<p>please tell us all the parts name used in the four led night bulb circuit kindly</p>
The circuit worked for me.....but I also suggest to put zener diode to make it better :). Thanks for guiding me.
I made this circuit...but at the output when i check with the multimeter in ac i am getting 26 v ac voltage as wel as 12v dc voltage... How can i remove ac voltage
MikB is 100% right. This circuit as is is very bad.
<p>Those curious how the voltage is limited in the 1st circuit : the 474K/250V capacitor before the bridge rectifier drops most of the 230V due its resistance (impedance) to AC. The input to the bridge (in both circuits) is only about 15Vac due to that.</p><p>Of course the output is not regulated, and will vary a volt around 12V as the mains varies. But for lighting LEDs etc, it will do. If you want to power one or more ICs, then it needs to be regulated with a resistor/zener or better, a 7805 regulator. </p>
<p><em>The capacitors CX and C1 form a voltage divider</em>, with CX having the larger AC resistance because it has a smaller capacitance, and C1 having smaller AC resistance because it has a larger capacitance.</p><p>So, the voltage across C1 will be significantly smaller than that across CX.</p>
<p>assist me getting 12v 2 amp. power supply for LED strip (circuit diagram)</p>
<p>You have mentioned R3 to be 220 Ohm 1 watt. However when circuit draws 330 mA, then power dissipated by 220 Ohm (when circuit is drawing full power) is I*I*R = 0.3A*0.3A*220Ohm= 19.8 watt. Hence 1 watt rated resistor will burn. Should it be 1 watt rated?</p>
<p>I have done this circuit and it works! As long as you put it in an enclosure then you will be fine. or in my case, I put in a paper tube and sealed one end, then filled it with epoxy and peeled the paper tube off of it when it hardened. It doesn't get hot or anything so it doesn't burn the epoxy, for all I care I could have put it in hot glue!</p>
<p>I have done this circuit and it works! As long as you put it in an enclosure then you will be fine. or in my case, I put in a paper tube and sealed one end, then filled it with epoxy and peeled the paper tube off of it when it hardened. It doesn't get hot or anything so it doesn't burn the epoxy, for all I care I could have put it in hot glue!</p>
<p>I have done this circuit and it works! As long as you put it in an enclosure then you will be fine. or in my case, I put in a paper tube and sealed one end, then filled it with epoxy and peeled the paper tube off of it when it hardened. It doesn't get hot or anything so it doesn't burn the epoxy, for all I care I could have put it in hot glue!</p>
<p>I have done this circuit and it works! As long as you put it in an enclosure then you will be fine. or in my case, I put in a paper tube and sealed one end, then filled it with epoxy and peeled the paper tube off of it when it hardened. It doesn't get hot or anything so it doesn't burn the epoxy, for all I care I could have put it in hot glue!</p>
<p>I have done this circuit and it works! As long as you put it in an enclosure then you will be fine. or in my case, I put in a paper tube and sealed one end, then filled it with epoxy and peeled the paper tube off of it when it hardened. It doesn't get hot or anything so it doesn't burn the epoxy, for all I care I could have put it in hot glue!</p>
<p>I have done this circuit and it works! As long as you put it in an enclosure then you will be fine. or in my case, I put in a paper tube and sealed one end, then filled it with epoxy and peeled the paper tube off of it when it hardened. It doesn't get hot or anything so it doesn't burn the epoxy, for all I care I could have put it in hot glue!</p>
<p>I have done this circuit and it works! As long as you put it in an enclosure then you will be fine. or in my case, I put in a paper tube and sealed one end, then filled it with epoxy and peeled the paper tube off of it when it hardened. It doesn't get hot or anything so it doesn't burn the epoxy, for all I care I could have put it in hot glue!</p>
<p>I have done this circuit and it works! As long as you put it in an enclosure then you will be fine. or in my case, I put in a paper tube and sealed one end, then filled it with epoxy and peeled the paper tube off of it when it hardened. It doesn't get hot or anything so it doesn't burn the epoxy, for all I care I could have put it in hot glue!</p>
<p>The use of transformerless power supply dated back at least 15-20 <br>years ago if not earlier. It was a historic innovation at that time when <br> transformer power supply that is heavy, bulky and cost-ineffective is <br>the mainstream power supply in small electrical and electronic gadgets.</p><p>Transformerless power supply are normally for loads that consume a <br>small current ranging from a few mA to a few tenths mA. In recent years, <br> they are also designed to supply larger current like those inside a LED <br> light bulb. Every transformerless power supply is in fact custom design <br> specific to the current load of the application, and is put inside an <br>insulated enclosure. The risk of electric shock is minimal. In <br>comparison to its transformer counterpart, it is more cost-effective in <br>manufacturing and more compact. They are not supposed to be used as a <br>general power supply.</p><p>For anyone who are interested in this kind of circuits, treat them <br>for educational purposes, learn how and why the circuits work. One can <br>surely try to build one to suit a specific application, but certainly <br>not using it as a general-purpose power supply.</p><p>As the contributor mentioned, the first circuit is for powering up a night light with 4 LEDs. If this circuit is used for the same purpose or any load that consumes around 34mA, and put inside an insulated enclosure, it will be as safe as any types of power supply. Just remember transformerless power supply is not built to be used as a general-purpose power supply.</p><p>All component ratings in circuit 1 look fine except Cx. Since AC is 230V, so ACpeak = 230 * 1.414V = 325.22V, so the rating for Cx should be 350V or higher.</p><p>Circuit 2 is not a good circuit, ratings for a few components are under-rated, R3 is redundant. The circuit is supposed to supply roughly 150mA. Don't try this circuit.</p>
MikB is correct. There is nothing to regulate the voltage here and you will likely end up with an exploding electrolytic capacitor. It would probably work out better if you tried to design a 120/220V DC power supply and stepped it down with a voltage divider and regulator network. <br> <br>The other problem that I wanted to point out is that in the drawing, you have a warning printed about a 230 V shock hazard shortly after the Diode bridge. After AC passes through the bridge it is DC with basically a large ripple voltage. <br> <br>Also your warning about 220-240V implies that AC is more dangerous than the same voltage in DC. Having worked with both, I would caution that at these levels it's all dangerous, but DC volts have the capability to be more damaging.
Thanks rneff for suggesting the right concept, I'll update it once I test the things :) <br>
The whole thing is a shock hazard (not a criticism of the design or author!) because it is not isolated from the line in any way (usually a transformer). This is perfectly okay as long as everything is contained and insulated, and there is nothing AT ALL that can end up getting in contact with fingers, external electronics etc. So that limits what you can usefully power with it, and limits anything like connectors on the PSU output. <br> <br>This is where the current fuss about Chinese knockoff &quot;Apple&quot; iPhone chargers is coming from: People getting hold of a 5v USB connector to slap into their phone and being thrown across the room or KILLED because of the lack of proper insulation from the line. <br> <br>Be safe!
I know it may cause problems in charging batteries and not suitable for powering things which run on 12 V....but as long as I use it for LED lightning I find this circuit OK to me. I used 3 of this kind of LED bulbs which have just 4 LEDs and I removed those 4 LEDs network and powered 2 meters of LED strip and 4 LEDs (in series) array in parallel and everything till this time working for me :)
I have to step in here with a thought. <br>Your &quot;it works for me&quot; mentality is the same sort of thinking that people have about certain pets. <br>Unfortunately, saying that your pet never ever bit anyone or any other animal UNTIL the day it ate your baby is little comfort. <br>Saying that a circuit works perfectly until it electrocutes you or burns down your house will be of little comfort after it happens. Just like with the pets, it may very well work fine for years - and for that entire time you will be telling everyone how it works fine for you...but what about the amateurs out there that aren't quite as safety conscious - or worse yet - we have 12 year olds that come here and build these things figuring it must be OK to build BECAUSE it's here. How would you feel if you found out some 12 year old was killed because he latched onto the output which according to you only has 12V on it - and electrocutes himself and dies. How much comfort will &quot;It works for me&quot; be to you then? There simply isn't any isolation from the mains, so, yes - there is enough voltage at the output to house ground to electrocute someone. <br>Do yourself a favour - don't just measure the output...measure any output point to GROUND and see how much voltage is there. <br>Another point to make is that if your house burns down, and they find one of these things ANYWHERE NEAR the origin point, your insurance company will be quite flatly denying your claim before the smoke has even settled. <br>When a family pet kills someone, the fact that it never bit anyone BEFORE that day is little comfort. I REALLY must object to your &quot;It works for me&quot; attitude. If it works for you, then you may want to keep this one to yourself. By putting it here, you are putting others who may not be as careful as you at risk. <br>Sure you can blame the one that built it for not being careful - however if they don't have access to the circuit in the first place, they won't even try.
And please note that I am referring to family pets that through genetic design are much more dangerous than your average kitten. <br>I just wanted it to be clear that we are talking about obviously (obvious to everyone else in the universe) dangerous pets - regardless of how safe the owners claim them to be.
Even the second version can be lethal, the power supply is not isolated, so the outputs may only have 12v across them, but they will have line voltage (120 or 230) to ground. <br>Please read this: http://sound.westhost.com/articles/power-supplies2.htm#s8 <br>AKA: Cheap Death (or How Not to Design a Power Supply)
I wasn't going to say this, but I think this instructable should be removed, and Transformer less power supplies banned from instructables. <br>
talking this much and going against of something doesn't solve the problem. After taking all precautions and safety I used it and it worked for me. Thinking of hazard, that should be the concern of the person who is going to build it...and I know many wont try to directly do work on AC current :)
Yah. This has a current regulator and nothing more. The capacitor controls input current but there is nothing here determining voltage. The voltage will be based solely on the load. 240 volts ac is in rms. multiply that by 1.4 for its peak value and you get 340 volts peak. Rectify that to dc and you get 340 volts dc with massive ripple. With no load connected that capacitor will charge up to a maximum of 340 volts, causing it to fail and or blow up. 340 volts at dc potential is quite close to the break point of air which is ~350 volts depending on air humidity temp and movement. So on top of being unsafe and completely out in the open, if left unloaded you have a potential arc flash which could heat up and send molten bits of metal everywhere. This type of power supply should be handled by someone who knows EXACTLY what there doing and what the consiquence could be. This is clearly no you. Sorry to rant on but for your own safety next time, do the research first.
MikB is 100% right. This circuit as is is very bad.

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Bio: Hi, I'm actually an IT consultant but love engineering. I love to try, do DIY stuff which are cost effective and solve my purposes.
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