I've just bought 10m of 220VAC LED Strip yesterday just to found out that it flicker so bad on mains. My country is using the 220V 50Hz as their mains. Anyone can help me on a way to reduce the flicker? What I'm thinking is adding a 0.47uF 400V caps before the LED.. Is it a good idea? Or are there any way to lessen the effect of flicker? I don't need it to be completely gone (as it is pretty much impossible)

I don't want to throw this strip and buy 12v or 24v ones because I've already bought 10m of em and the store doesn't have a return policy.

EDIT: Since many people asked for a schematics.. I don't have any, but I forgot to say that it is indeed an SMD 5050 LED, will try to find any schematics for that. And here's a pic to it http://imgur.com/16Yg96a For anyone who's wondering what is written on the R, it is 391. Sorry for my camera.

  • 2
    \$\begingroup\$ Does it have a data sheet? If not then welcome to your first important lesson in electronics and probably THE most valuable one you'll ever get. \$\endgroup\$
    – Andy aka
    Nov 17 '16 at 12:00
  • 1
    \$\begingroup\$ Can you add a photo of the components on the strip? \$\endgroup\$
    – Pelle
    Nov 17 '16 at 14:10
  • \$\begingroup\$ Does your strip include a power module that plugs into the mains? If not, I think you'll have to accept the 50 Hz. flicker. \$\endgroup\$
    – glen_geek
    Nov 17 '16 at 15:07
  • 1
    \$\begingroup\$ If your module takes AC in, you cannot use a cap. \$\endgroup\$
    – DoxyLover
    Nov 18 '16 at 1:43
  • \$\begingroup\$ @Andyaka No, it doesn't have any datasheet, well 99% electronics stuff being sold at my country doesn't have it except from those well-known manufacturers. \$\endgroup\$ Nov 20 '16 at 3:33

Rev x

**From reviewing your Photo, I misread 60-D-16R as 16Ω but in fact 391 is correct and \$16E\$ is upside down was my misteak

enter image description here


Plan C

Assuming 33 mA, 390Ω, Vtot = 2.8V*4 + 390Ω*0.033A = 24.07V ...*0.033A= =0.8W

With 60mm pitch 10m yields ~165 in parallel or 165*0.8W/string = 132W max. or 13W/m or 24V@5.5A using a Laptop universal charger ( if you can find... usually all are 19.5 V now)

Old info from insufficient data and poor description for historical ruminance

Plan B ( based on poor specs) - Measuring the Photo the 4LED + 16R , pitch is 60mm and total power @14V= 2.367W/m - Thus 24W per 10 m minimum. cooler.
- Suggested Supply Tolerance 14V+/-0.2V @ 2A min pref 3A - 12V will work but dimmer.

In order to smoothen the flicker let's do some calculations.

  • First you need a bridge rectifier to make rectified AC into DC.
  • Second , the decay time between 10ms pulses needs to attenuate 100Hz by at least 20dB or 10:1 in current to be significantly attenuate flicker.

    • if they did something like use 2 diodes to run 1 string in + and another on - then the flicker rate would be really bad at 50Hz.

So what value C is needed and at what cost?

  • I know that 10% ripple uses RC=8t and R=10~15 Ohms (approx) per 65mW~100mW white LED.
    • For an RMS voltage of 220V , let's assume there are 330 LED's split into 5 strings of 66 in series or 5P66S with a Vf of 66*3V=198Vdc
    • the stripleds resistance when above 198V becomes 166*15R/5 ~500 Ohms
    • thus C=80ms/500 Ohms=160uF !! at >400V rating with a series drop resistor of 20V/(20mA*5P)=200 Ohms @ 2W

You can substitute your values and see if you want to retire it or rewire it with a bridge , **cap, power resistor . have fun and get extra R in case it is too bright or too hot. The resistor must be flame proof and rated for line voltage.

  • \$\begingroup\$ From what I get after reading your answer is that I need to rectify the mains to DC and then add a Cap and a R? There is no way to rewire the inside since it is rated for IP65, the most I can think is to make the wire route to all the "Rectifier, Cap and R". Assuming all the calculations you have done, I'd need 160uF Cap in series with a 200 Ohms @ 2W R? --Anyway, I have added a Pic of the strip in my first post. \$\endgroup\$ Nov 20 '16 at 3:40
  • \$\begingroup\$ I'm not going to run the whole 10m in 1 supply then. Since all I got is 14v 1.7A DC supply I think it's best to split the strip into two 5m's? I'll get back ASAP later. \$\endgroup\$ Nov 21 '16 at 17:08
  • \$\begingroup\$ I'm pretty sure the R is rated at 390 Ohms, I'm quite unsure where you got the 16R from. \$\endgroup\$ Nov 21 '16 at 17:16
  • \$\begingroup\$ Is your photo correct? 16 shown on part and 60D-16R shown on FPC \$\endgroup\$ Nov 21 '16 at 17:21
  • \$\begingroup\$ OK it looks I read it upside down 16E=391 yet even 50mA which for half power on 5050 results in \$0.05A^2*390=1W\$ ( does it get burning hot? >100'C) \$\endgroup\$ Nov 21 '16 at 17:54

From the picture that you've posted it looks a like a regular 5050 led strip that is powered from a 12V supply. It also looks like parallel connections of 3 leds in series. Each of these leds take about 30mA. If you have a full 5m strip on a reel which has 300 leds in it, that's 100 parallel strips each with 3 leds in series. Which means you'll need a supply that can supply 100 x 30mA which is 3A.

Note that you won't be able to power them directly from the mains supply. You will need to step it down to 12V. Look for simple rectifier bridge followed by a regulator circuits online. There are plenty of them.

In case you don't want I build one on your own there are plenty of 12V, 3A supplies available online. These supplies usually operate from 100-240V mains AC.

  • \$\begingroup\$ I was suspecting that it could be run from a 12v supply. I haven't got the time, so I'll try it later. They are rated at 220VAC though (Idk, maybe I was fooled or the guy who sell this is a professor). And from what I can see I think it's a series of 4 LED? \$\endgroup\$ Nov 21 '16 at 17:11
  • \$\begingroup\$ Yep it does look like they are 4 leds in series. This actually means you'll need a slightly lower current power supply. (300/4) x 30mA = 2.25A. \$\endgroup\$ Nov 21 '16 at 18:24
  • \$\begingroup\$ I'm quite surprised that 12VDC 1.7A supply doesn't go very well with it. \$\endgroup\$ Nov 22 '16 at 17:50
  • \$\begingroup\$ at 12V the Zener like behavior means it drops from 3V to 2.8V asymptotic curve / 390 Ohms + ~45 ohms ESR ( ~0.2V/435 = 0.5mA ) or just barely ON from my estimate. You can read voltage across 390R to verify It may be (3-2.6)/435=1mA and thus show 390mV . ok? \$\endgroup\$ Nov 22 '16 at 18:00

A detailed circuit diagram is needed for a more definitive answer, but here goes my best guess.
LEDs require "pure" DC to not flicker. Most likely there is a diode rectifying the mains voltage. This only provides pulsed DC, thus causing the flicker.
What you need is to filter the pulsed DC and get it closer to pure DC.
The size of the cap will depend on the amount of current needed for the LEDs. However, the voltage should be at least 600 AC and you need to connect it to the node formed by the diode and the start of the LEDs string. You should also add a 1 megohm across the cap to bleed it off, when not in use.

Based on the additional information, it appears that you have one resistor limiting the current to four LEDs. Each set of resistor and 4 LEDs, drop approximately 12v, so there are probably 20 "sets" (80 LEDs) connected in series (20 x 12v = 240v). If there is no rectifier, it means the LEDs are the rectifier.

If all this is correct, you can reduce the flicker by using a full wave rectifier, a 200uf cap, and a 10 megohm resistor, to rectify and smooth out the DC. If you need more detailed information to build it your self, let me know, but if this sounds unfamiliar to you, I recommend you get a 220v DC, 100ma supply.

  • \$\begingroup\$ Thank you for your interest in helping, I'm sorry that I forgot to add the pic. You can see the strip now (I've added that to the first post) but well You can't see the routes anyway. But I'm sure this Strip doesn't have any diodes in there. All I can see is just LED's and R's. So you mean I need to add a Cap and R in series in between the mains and the strip right? \$\endgroup\$ Nov 20 '16 at 3:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.