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.
Capacitive dropper question
Collapse
X
-
Re: Capacitive dropper question
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.Wattevah...Comment
-
Re: Capacitive dropper question
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.Wattevah...Comment
-
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.Comment
-
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.Wattevah...Comment
-
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)Comment
-
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)Wattevah...Comment
-
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"Wattevah...Comment
-
Comment
Related Topics
Collapse
-
by Per HanssonI've noticed a trend of capacitive dropper supplies not lasting.
This supply is super simple: you use a capacitor in series with the load to limit the current and voltage the device can consume.
I have fixed several such power supplies now where the only problem is that the safety rated X2 capacitor used for the implementation has dropped in capacitance.
This of course starves the circuit of power:
In many implementations there is a relay being driven, and thus when the load of the relays coil is added the device simply turns off or rather resets.
I have a... -
by Wolf1977hi to all
in an inverter for motor
on input output modul has power supply 24 volt dc is coming from a electrolytic capacitor of main power supply of inverter modul
parallel has an capacitive smd capacitor that is shunt
i want to ask
first what is the play of this capacitor
and what value and voltage do you think must i use for that -
by magnehHello all.
Not very familiar with transformers and did not find the datasheet for this one.
Would these resistance values from Pimary and secondary normal? (check photo in red)
Maybe someone with more experience can help me here
Appreciate2 Photos -
by JFAI have a Honda generator with a faulty inverter. The inverter is encased in black resin with just the tops of a dozen small electrolytic caps and four 230v 1900uf electrolytic caps poking out of the top of the resin. The plastic cover of one of the 230v caps was burned off but there was no sign of bulging and the capacitor can was in good condition. The heat source looks likely to be from the outside rather than from the component itself but since the inverter is basically unrepairable, I chiseled away and removed the capacitor. I had to destroy the capacitor so had no chance to test it but it...
-
This specification for the Lenovo ThinkPad T440s Grade Barga1n+ Notebook can be useful for upgrading or repairing a laptop that is not working. As a community we are working through our specifications to add valuable data like the T440s Grade Barga1n+ boardview and T440s Grade Barga1n+ schematic. Our users have donated over 1 million documents which are being added to the site. This page will be updated soon with additional information. Alternatively you can request additional help from our users directly on the relevant badcaps forum. Please note that we offer no warranties that any specification,...09-06-2024, 08:47 AM
- Loading...
- No more items.
Comment