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Transformerless power supplies circuit, cheatsheet and questions
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soldar:
Max forward voltage for the LEDs is about 4.0 V. You could put a 5V zener in parallel with each LED and that way if one LED dies the rest will keep working.

That circuit is putting about 20~22 mA through the LEDs which is a bit too close to the absolute max of 25 mA, especially if it is enclosed and cannot dissipate heat well. I would feel more comfortable bringing the current down to about 13~15 mA by decreasing C1 to about .22~.25 uF. That is the first thing I would do. Decrease current and increase refrigeration.

As has been said, when one LED fails, there goes the capacitor (which is not really needed).

Contrary to what has been said, this type of power supply approximates an AC current source, not voltage. Short the LEDs and you get a certain current through the short. Now place the LEDs and the current is almost the same while the voltage rises. Open the circuit and the voltage keeps rising.
waste:
Thanks for all the interesting comments, I will try to adress and keep asking if you may :)


--- Quote from: Benta on February 05, 2019, 08:40:06 pm ---I've also built a lot of these capacitor-drop supplies, and a couple of points stand out.
For 470 nF X2 drop-caps I use 330 ohms wirewound for R1 These are very robust when it comes to current surges, which you'll certainly have here when switching on. I never had any problems
I also always use a 1 W Zener after the bridge to have a somewhat constant DC voltage.

Now to your failure mode: what happens is, that one of your LEDs fails (probably due to a high current spike when switching on). When this happens, you have no voltage regulation any more, and the voltage over your filter cap rises until destruction.
BTW: where's the current limiting for your LEDs? I see no resistor.

--- End quote ---

Can you elaborate in what way the wirewound is more robust? It attenuates the surge better for the rest of the elements, or it just more robust for itself. I never got a failing resistor. I think the failure mode you described is what most people agree on :)

You don't need a resistor for the LEDs, as the X2 Capacitor (C1) acts as the current limiting factor. Capacitive droppers to my understanding are current limiting devices. You can add up as many LED's as you want in series, it will still give the same amperage.




--- Quote from: not1xor1 on February 06, 2019, 08:19:01 am ---a couple of notes:
-1) it is better to use 2-3 resistors in series for discharging the input capacitor, you do not need a power resistor and low power ones have a limited range for the maximum voltage

--- End quote ---

1) about the low powered resistors (ex 1/4watt) . What do you mean they have a limited range for maximum voltage? that they fail easier (acting as fuse) or attenuating the surges better?




--- Quote from: StillTrying on February 06, 2019, 08:52:56 am ---Start the Vin Sine source with a delay of 90 Deg. to check for start up surges.
On the odd occasion when you get an intermittent switch on or switch off, I think the peaks could be over 1A, mostly going into the 33u cap.

--- End quote ---

Would be using 5-6μF ceramic caps, or higher voltage rated electrolytic caps help prevent this problem??





--- Quote from: soldar on February 06, 2019, 09:23:21 am ---Max forward voltage for the LEDs is about 4.0 V. You could put a 5V zener in parallel with each LED and that way if one LED dies the rest will keep working.
That circuit is putting about 20~22 mA through the LEDs which is a bit too close to the absolute max of 25 mA, especially if it is enclosed and cannot dissipate heat well. I would feel more comfortable bringing the current down to about 13~15 mA by decreasing C1 to about .22~.25 uF. That is the first thing I would do. Decrease current and increase refrigeration.

As has been said, when one LED fails, there goes the capacitor (which is not really needed).

--- End quote ---

puting a zener diode in parallel with each LED kind of defeats the simplicity of this project. Most of the times I use 3825LED's in 12V strips which are already connected in parallel and then you can connect them in series of 3s together (see attached pictures).
Using lower current is a solid and good idea, I usually don't go over .22nF and the smoothing capacitor decreases the max current even more apart from minimizing flickering. Maybe even that is too much though and should have gone with 150nF X2 Capacitor.



--- Quote from: soldar on February 05, 2019, 07:57:21 pm ---The only way I see for that damage to happen is that there was a spike in voltage and current and for that to happen R2 and C1 are suspect. Are they OK? How about the four diodes of the bridge? It may be that a high voltage transient caused a spike in voltage.
R2 is not strictly necessary and may be a cause of failure so you might consider removing it
C1 should be of adequate voltage. for 230 volt circuits I like to put 630 volt caps.
After the bridge you could place a zener as extra protection.
I cannot think how else it could have failed.
In any case, I like to make things easily repairable for this very reason.

--- End quote ---

adding a zener in parallel I think it's not going to work with LED's so well as the zener should be really precise to the Vf of the LEDs . 1-2 volts higher and the LED s are toast anyhow :)
Everything else apart from the smoothing capacitor is working nicely. R2 is "needed" if you dont want to get a little sting every time you take the plug out and touch the metal, I dont want to scare my friends with my present :)



Do you have any suggestions about how to probe correctly?

Thanks again a bunch for all the help and ideas
not1xor1:

--- Quote from: waste on February 06, 2019, 11:40:00 am ---
--- Quote from: not1xor1 on February 06, 2019, 08:19:01 am ---a couple of notes:
-1) it is better to use 2-3 resistors in series for discharging the input capacitor, you do not need a power resistor and low power ones have a limited range for the maximum voltage

--- End quote ---

1) about the low powered resistors (ex 1/4watt) . What do you mean they have a limited range for maximum voltage? that they fail easier (acting as fuse) or attenuating the surges better?

--- End quote ---

The datasheet specifies the maximum working voltage.
With higher voltage there is loss of insulation so you risk to burn all your leds.
Probably that is affected by environmental conditions (temperature/pressure/humidity) too.
Zero999:
A transformerless power supply is a current source if the LED forward voltage is low, compared to the mains.

My advice is to avoid them where at all possible. Use a switched mode power supply instead. The only reason to use a transformerless PSU is for cost, which perhaps makes sense at low power levels, but at more than about 50mA, a switched mode becomes more attractive, because decent, large capacitors aren't cheap.
Benta:
"Can you elaborate in what way the wirewound is more robust? It attenuates the surge better for the rest of the elements, or it just more robust for itself."

They are much more robust when it comes to surge overloads.
Think about it: if the circuit is switched on at max. AC voltage (~320 V), we're talking about almost 1 A in 330 ohms, as all caps are discharged.
This works out as a peak power surge (very short, admittedly) of ~300 W. A wirewound will withstand this, a film resistor will not in the long run.
If the resistor is 680 ohms, we're still talking ~150 W.
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