Transformerless power supply issue

I'm trying to design a Transformer Less power supply as the circuit below,
by measuring the voltage at the capacitor with AVO in ac mode i got 220V,but by measuring the input of the regulator with AVO in dc mode i got 0V all measuring at no load, and
i don't know why i got 0V at the i/p of the regulator ?
although i connected the same circuit with load resistance 2.2K Ohm and without the regulator i got 1.3V as load voltage

also i'm confused in another thing,
the output of the diode is not a pure DC since it's half sin wave changed in value and not a direction
should i measure it with an AVO in DC mode or AC mode ?

thanks

deleted...

It's called an offline Dc regulator,,for some odd reason, because it only works when ONLINE.

Line Cap must be plastic Y safety rated . Then load after diode is a bigger Cap at low voltage.
To make a voltage divider.


But you'll be sorry about going this way when you see the cost of the caps to drive a TO-220 regulator at load with 12 W because caps must draw 20x this much in VAR due to large VA drop.

So do yourself a favour and buy a 10 or 12W offline SMPS regulator from Wallyville or PC store for about <=$1/W

Are you aware of the shock hazards of a transformerless supply?

Capacitor divider offline power supplies are often limited by cost to about 1 Watt, since the Y cap has most of the high voltage and same current across it.

We once had a wireless power meter powered offline using a full bridge using 22V zeners and 1 watt SMD ceramic coated 22K resistors to provide DC for the 2-way radio and rotating disk sensor with storage in a large low voltage electrolytic, also limited to low power loads.

Beyond this , once must use very low phase angle triac control charge pumps. which need large filters and then use SMPS for efficient conversion beyond a few Watts.
Naturally Neutral is subject to line loss from household currents and up to 10% V rise above outside transformer ground, while the V drop Y-rated Cap is "live" and exposed to line surges and lightning spikes to up 6kV after the meter, which can be reduced with a suitable line filter.

4 every youngs wrote:

i got 0V all measuring at no load, and
i don't know why i got 0V at the i/p of the regulator ?

Click to expand...

With no load on the output of the 7812 there will be very little current flowing through the 2.2uF capacitor, and so there will be very little volt drop across that capacitor.
As a result of that there will be 220v at the input of the regulator.
After about 10mS the regulator will die, probably creating a short circuit between its input and common terminals, which is why you are measuring 0 volts at the input.
This is a very bad circuit.

As others have hinted, there is no isolation between the mains and the low voltage output, which makes this a dangerous circuit, unless the circuit and its load are in an insulated enclosure.

This is a bad, dangerous circuit which can kill you.

This is probably not what you want to hear, but it is the truth.

JimB

Thanks for all of you who's modify the circuit and warn me about the dangerous of it,but really i want to build a sunlight detector and fixed at my house door which use this concept for providing the power to the circuit and i found it on the web,but when i execute it the output voltage was drop to low value (at sunlight) due to the design of the power cct so i decide to modify it to be suitable for the circuit

Hi ronsimpson
pls with respect to the capacitor 100u ,i think the max voltage to applied on it in the range of 10v to 25V or less,how could i make sure that the voltage across it will not exceed the rated values

also what's the function of D2

4EverYoungs.71 said:

also what's the function of D2

Click to expand...

Provides a path for the negative current to the capacitor.

Too much rambling. So deleted.

Thanks crutschow

4EverYoungs.71 said:

pls with respect to the capacitor 100u ,i think the max voltage to applied on it in the range of 10v to 25V or less,how could i make sure that the voltage across it will not exceed the rated values

Click to expand...

pls is any one has an answer,i could use a zener diode but is there any solution better than this

Why D2: You can not pass DC through a capacitor (C2). Current must flow both directions. For 1/2 the time current flows through D1 and to the load. For the other 1/2 the time the same amount of current must flow through D2 and back through C2. This way AC current is flowing through C2.

To limit the voltage on C3 you should use a 15V or 20V zener diode for D2. This will limit the voltage in C3.

If you want twice the current replace D1,2 with a full wave bridge. (4 diodes) The positive current and the negative current will go to the load this way. You can put a zener across C3 to keep the voltage down.

There are many stories on this forum about this project. Try searching.

4EverYoungs.71 said:

Thanks crutschow

pls is any one has an answer,i could use a zener diode but is there any solution better than this

Click to expand...

As others have mentioned, transformerless power supplies seem simple at first but end up needing many components to protect other components. The power consumption of a tranformerless supply is the same whether or not there is a load, so your zener diodes will get hot. You need to allow 2 W of heat in the zener.
Tranformerless supplies are not isolated so all points could be at mains voltage.

The heat issue is something that I have seen commercial designs fail to deal with properly, and circuit boards damaged by years of running hot are quite common sights.

If you use a transformerless supply, you have to consider:-
Capacitor inrush. The 50 Ohm resistor is for that, but ronsimpson's circuit has D2 placed where there will be a huge current through D2 and C2 if the circuit is turned on near the middle of the negative cycle of the mains.
Reverse current. Now handled by D2
Capacitor discharge. R1 is there to discharge the capacitor when the circuit is disconnected.
Excess current. This is handled by the zener voltage of D2, but it will get hot.

And you have to remember that any fault in design, manufacture or within a component could cause the whole thing to fail.

Fault-finding is dangerous, because the whole circuit can be at dangerous voltages.

Also the alternative is simple, safe and cheap.

https://www.farnell.com/datasheets/1725290.pdf

£3 for one off. To be safe, all you have to do is make sure that you connect the mains input to the correct 2 pins, and then don't touch them. The rest of the circuit will be safe.
To get 12 V DC, you still need to rectify, smooth and regulate the output.

The transformer will not generate a huge voltage if there is little or no load.

If you short it out, it will get warm but will not come to any harm. (That only applies to really small transformers like the one in the link. Larger ones will overheat if shorted. )

There are still lots of mistake you can make using a transformer like that, but none of the mistakes will endanger you, and few will cause components to be damaged.

I also include D2 and use 5 Watt Zeners (like a 1N5349B for 12 volts) in these circuits and No "R2" In-rush Resistor.
I have never had any problems with overheating or capacitor failures and I have made Many of these.

chemelec said:

I also include D2 and use 5 Watt Zeners (like a 1N5349B for 12 volts) in these circuits and No "R2" In-rush Resistor.
I have never had any problems with overheating or capacitor failures and I have made Many of these.

Click to expand...

You're in Cananda, where the mains voltage is 120 V. The OP's circuit is designed for 220 V. That will increase the inrush energy quite a lot. I don't know if that will be enough to cause problems.

chemelec said:

and No "R2" In-rush Resistor.

Click to expand...

I like the idea of a inrush resistor.
C2 has a job of keeping the current down. At 60 or 50hz the capacitor has high resistance.
There can be high frequency on the power line. Noise from switchers, motors, or distortion from loads. (also lightning) At high frequencies C2 will let through much more power. I think R2 and the filter cap makes a low pass filter that helps remove energy from high frequency sources. I think C2 and R2 should be placed so all incoming energy must pass through them. This way even a lightning strike must pass through R2 and thus limit the current in the diodes.

Yes I am in Canada, But I have built these circuit for both our 110, 60 Hz, But also for other Countries, with 220 and 240 VAC at 50 Hz.
No Problem with either ones.
But If I were to Add an In-Rush resistor, I would put it On the Line Side, Before the Capacitor.

This way it would also act as a fusing resistor, if either C1 or D2 were to short out.
Here is the Actual Circuit I use and I include a 200 mA Fuse.

Chemelec,
Please post a schematic. We have used too much words and not enough pictures.