Capacitive dropper question

[Dannyx]

Good day folks. A friend of mine bought himself a 230v 50w LED from China really cheap and secured it to a PC heatsink which also happened to have a fan on the other side. Instead of using a separate supply for that fan, I thought I'd try putting together a capacitor dropper for him to run that fan straight off of the 230v the LED uses. Safety aside, as I'm aware of the dangers and how such a supply doesn't insulate from the mains (the contraption is gonna be inside an enclosure anyway), I'd like some advice on, you guessed it, the numbers. I did my research beforehand and have some values in mind, but decided to ask for advice anyway. Here we go: I understand the circuit looks like this, just with the fan in place of that shunt regulator (which might actually be a good idea to have). The formula (and I'm going to do this 1st-grade and use the actual words for the operations) is 1 divided by 2 times Pi times Hz times capacitance. For the capacitor, I chose 2uF, so that would give us 1/6,2831*50*0,000002. This gives us the resistance of the cap, which I calculated to be 1591,57 ohms. Now input that into ohm's law: V divided by R equals I, and we get a current of 0,144A, or...144mA ? Did I even get the math right ? Assuming I did, this brings me to my next question: I see an R1 in series with the whole circuit, its role being to limit spikes - that I understand, but I don't know where that goes into the calculations, since the current must flow through it no matter what. Also, there's no mention of the voltage at the output ! How do I calculate that ? At the end of the day, it's all about running a fan, so it doesn't need to be too precise, but the voltage still has to be within the ballpark so the thing doesn't go pop. Not too savvy with math I'm afraid. Cheers.

[stj]

Re: Capacitive dropper question

dont do it, the led needs pretty precise current regulation.
and the fan needs a pretty specific voltage because it's pulse-driven internally using a multi-pole motor.

[Dannyx]

Re: Capacitive dropper question

dont do it, the led needs pretty precise current regulation.
and the fan needs a pretty specific voltage because it's pulse-driven internally using a multi-pole motor.

The LED already runs of the mains, so no worries there, since it would just be before our circuit. What we're trying to achieve here is running the fan.

[stj]

Re: Capacitive dropper question

well it needs to be smooth DC.
other than that, the current is usually marked on them - so i suppose it's possible with a bit of safety like a zener diode.

[Dannyx]

Re: Capacitive dropper question

well it needs to be smooth DC.
other than that, the current is usually marked on them - so i suppose it's possible with a bit of safety like a zener diode.

A random fan I found in my junk bin says 0.1A on it, so assuming that's 10mA, I guess 20mA would be a reasonable to have, to ensure it starts properly, since a fan draws more current when it's stationary and first takes off.

[stj]

Re: Capacitive dropper question

.1 = 100ma

[mariushm]

Re: Capacitive dropper question

0.1 A is 100 mA
Fans need slightly more current when they turn on and then they can work with lower current.

For all the formulas you need, you can read AN954 Transformerless Power Supplies: Resistive and Capacitive : https://cdn.badcaps-static.com/pdfs/...5df14516be.pdf

If you're too lazy to read the pdf:





Keep in mind that those cheap 50w leds from China will have a forward voltage that will vary quite a bit with temperature.
You should use a current limited power supply to figure out at what voltage the led uses about 45w (or whatever the led is actually capable of outputting, minus around 5%) and aim for that voltage with your power supply.

For example, the led may have a forward voltage of 40v at 50w (1.25A) but once it gets to normal high temperature of let's say 60c, its forward voltage could drop to 38v and now you're pushing ~ 1.32A through the leds and you could damage it.

So it would make sense to set the voltage to around 38v and configure the peak current to around 1.25A and initially the led would be dimmer but within minutes as the led heats up it will reach the full brightness.

[Dannyx]

Re: Capacitive dropper question

Ok, need to clarify once again that the LED itself is not relevant here - I'm not trying to drive the LED with my capacitor dropper, since it already runs off the mains just fine. This thing only needs to drive the fan at roughly 12v or 5v.

[mariushm]

Re: Capacitive dropper question

Most 12v fans are designed for computers where the standards say they must tolerate up to 10% extra (so they should work at up to 13.2v)

Basically, aim for 12v and don't worry if you get 11 .. 12.5v whatever

12v zener diodes are cheap as hell and very common, as are 9.1v zener diodes and 10v zener diodes are also very easy to find (12v fans would still work fine with 5v..13v)

[Dannyx]

Re: Capacitive dropper question

Most 12v fans are designed for computers where the standards say they must tolerate up to 10% extra (so they should work at up to 13.2v)

Basically, aim for 12v and don't worry if you get 11 .. 12.5v whatever

12v zener diodes are cheap as hell and very common, as are 9.1v zener diodes and 10v zener diodes are also very easy to find (12v fans would still work fine with 5v..13v)

Ok but what about the actual output voltage of the dropper ? I understand there's no way it's going to be spot-on and I have to regulate it to 10-12v using a zener, but is there a way to approximate what it's gonna be ? Like 18-20v ? Struggled so damn much with those zeners before - getting the formulas for calculating the resistor they require has always been the bane of my existence...

[stj]

Re: Capacitive dropper question

aim for about 15v with the dropper, then a current limiting resistor, then the 12v zener.
that should cover any fluctuations.

[Dannyx]

Re: Capacitive dropper question

I went over this before, but by my calculations based on this formula, the value of the resistor before the zener (assuming the voltage of the dropper is 15v) should be about 13ohms, as follows: Vs-Vz is 15-12 = 3 so we get that out of the way. Iz is what always puzzles me, as I don't know what it should be: the maximum current allowed by the diode, or the minimum one required to keep the diode conducting ? What I did was go to the datasheet for the 1n4742a and choose the 20mA used for the "test current", as the datasheet calls it. So on the bottom side of the fraction we now get 0.02+0.2 (assuming the fan requires 200mA during startup so we have some headroom). That gives me 3 divided by 0.22 = 13.63, so let's just say 13ohms...damn...should open up a series like "monday math with Danny"

[stj]

Re: Capacitive dropper question

??? dont ask me, i'm a digital engineer!

[Dannyx]

Re: Capacitive dropper question

??? dont ask me, i'm a digital engineer!

Well that's why I posted in the first place - everybody can give their opinion....I'm expecting BudM to be around here soon and nag me for getting it wrong I suck at this

[stj]

Re: Capacitive dropper question

i was just thinking that he would appear to post the formula and explain how it goes!