Transformerless power supply issue

[tvtech]

I am a bit slaughtered now to post my original diagram of my little light..well ten of them (LEDS that is). And the PP3 and every thing else.

When things get tough now I go drinking. Only beer. No Spirits for me. Beer is good. Many minerals and nutrients are to be found in beer. Like a really healthy way to get sloshed

Loveu you all
tv

 

[chemelec]

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

See Above POST Number 19.

 

[ronsimpson]

I see post 19.

I use a resistor as a fuse and filter.
and
Put a resistor across the cap to blead down any charge. UL, CSA gets after me for this.

 

[chemelec]

In Theory it is Illegal to use any devices that are not Approved by CSA, UL, Provincial or State Certification.
Resistor, Fuse or Nothing.
Especially if they run on 110 VAC, 220 VAC or if they Create Heat, like a wood burning stove and other devices that can have safety problems.

A Resistor as a Fuse/Filter is OK, But Keep the Resistor UP in FREE AIR, to Prevent a possible Fire.

 

[Tony Stewart]

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

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

There are far better ways to implment this.

First define what output you need.
e.g. Logic 0 or 1 isolated or triac drive output non-isolated or contact closure isolated or Optoisolator saturated output?

 

[MrAl]

Hello there,

The original circuit is improved with the addition of another diode because the capacitor can not function on AC with one diode rectifying the current into DC. The capacitor charges and stays charged and therefore can no longer pass AC. So the second diode is mandatory.

In spite of that improvement, you really should use the full wave bridge version. That's because for one thing you already know now that you need two diodes, so for the price of two more diodes you get to use a capacitor value that is one half of what it would need to be if you only use the two diode solution. This is because with the two diode solution the cap only passes current that the circuit can use for one half of the cycle, so you need a cap twice larger to pass the same average current as you could with a full wave bridge. This becomes more important when you try to draw a little more current than a few milliamps.

So start with a full wave bridge and go from there.
The only time this doesnt work is when you need a circuit connection that is common to one side of the line, but when that is not needed then the bridge rectifier is the preferred method.

You also need a relatively low value series resistor to limit inrush current when the unit is plugged into the wall. That will limit the max peak current through the circuit. Only catch here is the resistor should be rated for the expected power, and the expected power could go up if the cap shorts out. A fuse is also a good idea.

By now you have seen the precautions when using a circuit like this. It is not isolated from the line so it MUST not be used in equipment that has ANY circuit node that can possibly contact a human body part. That said, these circuits are quite common.

 

[4EverYoungs.71]

Hi chemelec
with respect to your schematic U determined one terminal is Hot and the other is Neutral,
what will happen if i plugged it in reverse(cause when i put the plug i don't know where is neutral and where's the line).
Regards

 

[JimB]

what will happen if i plugged it in reverse

 

[ronsimpson]

A non isolated power supply is never to be used where it is connected to another item.
It is never to be used where a human can touch any wire. (even "ground")
It should be double isolated. (insulated)
They work for night lights, smoke detectors, etc.

What you called "ground" on the schematic is not ground. It could be neutral or line. NOT ground.

 

[atferrari]

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.

Inherently short circuit proof transformer, the data sheet says ???

How comes?

 

[Diver300]

Inherently short circuit proof transformer, the data sheet says ???

How comes?

I think that the resistances of the windings is sufficiently high that the short-circuit current is no more than a few times higher than the running current, so the transformer will not get hot enough to destroy itself if shorted.

Smaller transformers have worse regulation than large ones, so that their output voltage varies more as the load changes. A result of the poor regulation is the low short circuit current. The poor regulation comes from the high winding resistance. Small transformers have less area of core, so need more turns, not fewer, than a large transformer. I can't find the input resistance specification for a short circuit-proof transformer, but **broken link removed** has a primary resistance greater than 1 kΩ

 

[MrAl]

Hello,

Another example of where these offline power supplies are used would be in a lamp dimmer. The entire circuit is inside a box with no nodes accessible to the human operator. In this case it could be advantageous to use the two diode solution so that we can ground one side of the DC supply.

 

[Tony Stewart]

Designing a direct line AC -DC converter or low power is easy but along with the capacitive transformer step down ratio N= C2/C1 voltage comes the stepup impedance ratio where the load Impedance must be 10 times (10*) greater than the smaller input cap, C1 for 10% drop in Vdc and rise in Vac ripple or R=20*Z(c1) for 5% ripple. Where 1/20=5%.

Since the bridge acts as a frequency doubler for the input line , f and only conducts during the peak voltage rise in ripple as the narrow rectangular current pulse, pumps up the secondary Cap.

The result is ;
Rload=10*2π f*2/C1, thus C1=40πf/RL @60Hz , thus if RL =2k , the C1=40*3.14*60/2k=3.77mF or 3,770 uF @200V nonpolar film

Meanwhile 1mF costs $150 which why I said this was a BAD IDEA, in the beginning... And why either an inductive transformer or SMPS are your only choices.

Even 100uF at 250Vac costs~$40 which would mean 37.7*10k as your average load resistor. You can get much more in peak power for a short burst in a fraction or a cycle, but thats it.

Offline SMPS start at $1/W and can go down to <$0.05/W in volume off the shelf

If you want to get fancy use a precise phase controlled FET or IGBT with over V, C, T protection and run the phase to get 2x Vdc out then smaller cap ratios are required and use a choke., but will still cost more. a triac is risky with false triggering unless very well protected with a transient line filter.

 

[tvtech]

I never thought I would do this...but what the hell.

Bottom and top traces here..this is as reliable it gets.

Same layout that seems to survive anything thrown at it. SMPS go to hell and burst Main Smoothing caps...this one thrives on Crap.

Anyway, my design, my layout and my little project. Sharing it with the World here you can all learn stuff.

Regards,
tv

 

[MrAl]

Hi,

Yes the capacitor has to increase significantly as the current goes up. The current is approximately proportional to the capacitor value. The actual current available is:
Iac=(Vac*w*C)/sqrt(w^2*C^2*R^2+1)

and because C is usually very small (uf range) we get as an approximation:
IacApprox=Vac*w*C

and so with w=around 350 we have:
I=Vac*350*C

and with Vac=120 and C=1uf we have:
I=42ma

so with half that capacitance we'll see about 20ma, and with double that C we'll see about 80ma.
To get 800ma we'd need 20 times that capacitance as a rough estimate, or 20uf, and it would have to be able to handle the entire line voltage peak times 2. For 120vac this is 340 volts.
You can get some smaller values like 3 or 4uf for a decent price, but after that the price goes up.

 

[ronsimpson]

with Vac=120 and C=1uf we have: I=42ma

That is why it is good for lighting a LED or driving a small micro (smoke detector) (light detector)

Look at the price of power line rated capacitors. .22uF, .47uF 1uF.....at some point the price increases beyond reason.
Then you need to move to a cell phone USB charger. They are isolated from the power line! Much safer.

 

[ramesh0007]

This is extremely dangerous circuit to be built, you need to be extremely careful while building. I have built one prototype, not tried to use for long time, its works fine for me. I have used 1 watt Zener to regulate the voltage instead the Voltage regulator IC. I will try with voltage regulator IC. Make sure that one have used 1 watt more rated resistors and 50 v rated capacitors, other wise they will burn. X rated capacitor should be chosen according to table. I found a demo working video here : Transfomerless Power Supply
The best way to build one is to open a LED bulb and copy its circuitry and value of resistors and capacitors, like done in this article: Transfomerless Power Supply out of LED bulb