Transformer Less 12V Power Supply Out of LED Bulb

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Introduction: Transformer Less 12V Power Supply Out of LED Bulb

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.

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  • I made this circuit....-smanikanta

    smanikanta made it!

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user

We have a be nice policy.
Please be positive and constructive.

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42 Comments

user

The circuit you show in the first image: What exactly keeps the output voltage limited to 12v?

I think you've missed something important here in the disassembly steps :)

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.

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.

Suggest you put a 12v zener across that electrolytic, at the very least!

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.

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.

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.

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.

So, the voltage across C1 will be significantly smaller than that across CX.

user

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).

Look at this way, C1 starts at 0v,and every "positive" 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.

At first the voltage will be low because C1 will "load" 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.

End point is moot: C1 will explode long before it is fully charged.

Simple Test:

Is the current through CX also flowing through C1 ? Answer: YES.

Therefore CX and C1 are in series. Therefore the voltage divider effect.

Note: The current through CX cannot disappear ... it has to return back through C1 (and load, if any), the one and only path!

user

Nobody says the current through CX disappears.

You don't take into account how the bridge rectifier is affecting this "divider" you appear to see.

The current in CX is alternating. The current in C1 is only EVER charging it up, steered to always be DC, by the bridge.

Without a shunt regulator/zener/LED load C1 will charge until it reaches mains level. It will NOT magically produce 12V. It will magically produce smoke.

user

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 "shock hazard" caution :)

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 :)

user

When you measured the 12.3v, was it still connected to the LEDs at this point, or just to a meter?

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.

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.