Direct mains power LED - Page 1
Forum section: electronics / projects

author: tdarlic
author_rank: Contributor
timestamp: October 05, 2016, 11:45:23 pm
content: Hi guys. I am trying to figure this project that I have tried to put together today. I am trying to power a series of LEDs directly from the mains 240V without transformer. I designed a indication light with a few years ago with a similar circuit and it still works. The one I put together burned in a sec.For C2 I usedhttp://export.farnell.com/webapp/wcs/stores/servlet/ProductDisplay?catalogId=15001&langId=71&urlRequestType=Base&partNumber=2495928&storeId=10152By my calculations this should give about 25 mA per LED line with the voltage on the VCC at about 30V. Forward voltage on the LEDs is 3.3VI tried to put it in the LTSpice and the simulation below is what I got. v_led (green) is VCC on the schematic, V(n002) is voltage between the D2 and D3, V(n048) if you can see it is voltage between D9 and R2. Finally I(D5) is current trough D5. The bridge rectifier on my test jig is soldered fine and for some reason I get 217V on VCC after plugging it to mains and with enough current to just fry all resistors instantly. What am I missing here?

author: moffy
author_rank: Super Contributor
timestamp: October 06, 2016, 12:17:52 am
content: Without a suitable load the voltage will float quite high. You would normally have a clamping zener to limit that. Max current = 240*(2*PI*50*(C1+C2)) A RMS if you are 240v/50Hz.

author: amyk
author_rank: Super Contributor
timestamp: October 06, 2016, 12:52:10 am
content: With that big of a capacitor the inrush current through each LED is going to be exceedingly high - around 6.3A if you happen to connect it at the peak of the cycle. You need an inrush limiting resistor on the input.

author: StillTrying
author_rank: Super Contributor
timestamp: October 06, 2016, 01:08:59 am
content: Simulating it I get 34V and 45mA through each LED, and 4V across each LED.

author: Cyberdragon
author_rank: Super Contributor
timestamp: October 06, 2016, 03:32:49 am
content: Is that 4.4uF?!That's WAY too big for an LED capacitor dropper! The ones in actual LED light bulbs are usually <1uF.You will always get near full input from a capacitor dropper open-circuit. It's when you load it that it reduces voltage because the reactance reduces the voltage. A zener is not necessary, but without it the filter capacitor will burn out if the LEDs fail.

author: sleemanj
author_rank: Super Contributor
timestamp: October 06, 2016, 04:45:47 am
content: bigclive is the defacto expert on capacitive dropper LED powering...http://www.bigclive.com/ledlmp.htmhttps://www.youtube.com/user/bigclivedotcom/search?query=%22capacitive+dropper%22

author: tdarlic
author_rank: Contributor
timestamp: October 06, 2016, 06:51:36 am
content: Quote from: Cyberdragon on October 06, 2016, 03:32:49 amHi, thanks for answering, actually it's 2.2uF only. I've put two different footprints on the schematic as at the time I developed the circuit I did not have the access to stock and I knew we had 2.2 uF on stock and then i just ballparked two fooprints that I thought would match size.Actually I missed both but managed to squeeze this one in it was ~1mm off.

author: tdarlic
author_rank: Contributor
timestamp: October 06, 2016, 08:29:28 am
content: Quote from: moffy on October 06, 2016, 12:17:52 amThanks for replying.That was my calculation exactly. With the 2.2 uF I got 166 mA of current divided trough 6 lines it's about 28 mA per line. My simulation showed the same more or less as you can see in original post. I was in for quite a surprise as I completely nonchalantly plugged this in the LEDs lit up for a few seconds and they they all failed apparently... I'll play today with few high power resistors today and try to get the gist of what's wrong. I'll place an inrush current limiting resistor instead of fuse and than see what's happening.

author: amyk
author_rank: Super Contributor
timestamp: October 06, 2016, 11:10:21 am
content: Quote from: StillTrying on October 06, 2016, 01:08:59 amYou simulated steady-state conditions, but not the initial turn-on transient. In the latter case the capacitor does nothing and the 340V peak is applied through two bridge rectifier diodes and the 50R resistor directly to the string of LEDs. Assuming ~3V forward voltage for the LEDs, that leaves over 300V across the resistor, meaning the current will be 6A per LED string.

author: Gyro
author_rank: Super Contributor
timestamp: October 06, 2016, 11:30:03 am
content: A cascade failure would probably have taken out any strings that survived the initial surge. As each string fails (assuming open circuit) more and more current would be applied to the remaining strings.Edit: I'm not sure how they handle this in commercial parallel string lights, maybe this is the normal end of life failure mode.

author: tdarlic
author_rank: Contributor
timestamp: October 06, 2016, 12:13:38 pm
content: Quote from: amyk on October 06, 2016, 11:10:21 amI figure that's probably what happened that and the cascade failure as the Gyro says below. I was overconfident as I did this before.I should have documented the first project with capacitor dropper. I just assumed it was a fuse instead of resistor on the input. I'll redo the circuit with the resistor and will come back with the result...

author: StillTrying
author_rank: Super Contributor
timestamp: October 06, 2016, 12:32:07 pm
content: Quote from: amyk on October 06, 2016, 11:10:21 amYou're forgetting that the combination of 2u2 / 330uF divides the initial 340V spike by 150, of course if the 330u had a lot of inductance, that would be a bad thing.I get values very close to the OP now that I've changed to just 1 2u2 cap.A bit of resistance near the 2u2 would be safer, and the 330uF could be a lot smaller.Having a fuse on Neutral doesn't seem very safe.

author: tdarlic
author_rank: Contributor
timestamp: October 06, 2016, 01:22:04 pm
content: Quote from: StillTrying on October 06, 2016, 12:32:07 pmThank you all guys for valuable input...The values are chosen from the stock, hence the 6 strings of 8 diodes and 330uF. This neutral line fuse was squeezed in because of size constrains. My thought is that this can be plugged either way so the fuse would be on neutral side 50% of time statistically speakingI've checked the bigclive page. I will whack in an in rush resistor and the inductor too. We have exactly 470uH one and will have few runs to test it.Jeez this was supposed to be just a few hours of work on a side, these LEDs are UVs and this jig is running just for 5 minutes at a time every few days to expose the UV developed compound on a device... The Murphy has struck again...