Repurposing LED TV backlights from thrown out kerbside TV's

Considering repurposing backlight LED's from discarded broken TV's into a large LED lamp built into a picture frame & run directly from 240 v AC.

Roughly if each LED need 3v to power on , I'd need 80 LED's linked in series to total up to 240 v.

This will eliminate the need for separate power circuitry ( which would bulk up the frame), thus have single power chord out of frame into wall socket.

Will this kill all the LED's with voltage spikes?

Will the final lamp flicker at supply rate of 50hz?

Will it potentially be fire hazard from overheating & shorting?

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Reply to
Taupe
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no.

that depends on the power supply circuitry you use

Unless you use a full wave rectifier it will. (else 100Hz)

yes.

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Reply to
Jasen Betts

You're not going to have control of how many LEDs are in the backlight assembly in the first place anyway. LED forward voltages also vary quite a bit, 3V is at the high end.

You could quite easily have the power circuitry in a separate box and connect that to the panel via another cord.

Without any current limiting, it will kill the LEDs even without a voltage spike.

Yes, though personally I wouldn't be concerned about that.

Yes, but the LEDs will blow up and break the circuit before you get the chance to leave it unattended anyway.

If you're looking for the simple way to do this, what you want is a series resistance which will drop the extra voltage according to Ohms law. R = 80V / I, with I being the current, and best determined by how bright you want the light to be (the maximum rating won't be known, but presumably the backlight will be too bright for its purpose before you get to that). For safety, and efficiency, I'd recommend that you do use a transformer or switch-mode power supply as well.

Note that this won't be as efficient as a constant current LED driver, and the efficency will be worse the greater the gap between the LED backlight's voltage drop and 240VAC (unless a lower voltage power supply is used). You'll also need to use high wattage wire-wound resistor/s in order to handle the wasted power, with their own cooling considerations.

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Reply to
Computer Nerd Kev

R = 80V / I? That's not the Ohms law that I was taught at school, where does the 80 come from?

Simple answer to all this do not stick it across the main with just a dropper resistor, it's a really bad idea.

Reply to
keithr0

can't be done? what if 240v rectified to DC with a current limiting resistor?

thats the point, not to use seperate box power supply

I'll put a fuse in circuit?

Reply to
Taupe

I'll alow myself some buffer & drop voltage to say 2v & increase series to

120 LED's

this is what i'm trying to avoid.

I'll include 3kohm series resitor.

already have existing power supply, thought I'd get rid of it by running directly of 240v AC

LED's (in strips of 8-10 ) were stuck to metal body of TV to cool will do same with metal back for frame.

Reply to
Taupe

assume LED require 2v & 15mA to operate.

current Resistor = v240- 100 x 2v/ 15ma

40 /.015 a =2666 ohm resistor say 3kohm

i'll probably put in 120 LED's to start off with, then work back taking out LED's until we get best brightness.

Reply to
Taupe

** The AC supply peaks at 340V and swings positive and negative compared to earth. Typical white LEDs are bright at 3.3 to 3.5 V.

So you need about 100 LEDs in series.

Some flicker may be noticeable under certain conditions, but 50Hz operation is common for dial lights and works OK.

If you add a bridge rectifier, then the flicker frequency moves to 100Hz and should be invisible.

.... Phil

Reply to
Phil Allison

that would be poswersupply circuitry.... (perhaps I misunderstood the original question)

That should be enough, brightness may change significantly with AC voltage variations.

You probably want to do 70 LEDs and 30V drop in the resistor depending on how bright you want this that resistor could get hot.

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Reply to
Jasen Betts

That won't give predictable results, mains voltage is not constant, and varies slowly minute to minute and also has breif dips and peaks.

eg: you may find that the lamp flickers every time the vacuum-cleaner or refrigerator starts up. This is why people use LED drivers.

Are you going to put the whole backlight lens assembly in the frame or just the LED strip?

it seems that it would be difficult to change the number of LEDs in the backlight assembly.

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  When I tried casting out nines I made a hash of it.
Reply to
Jasen Betts

I got it into my head at one point that he'd said the voltage drop of the backlight was 80V, so I started doing ((240VAC x 1.414) - 80V) / I Then realised that he hadn't mentioned the Vf at all and edited everything to remove all the specifics, but I accidentally left the

80V in there instead of just the V.

Not knowing too much about LCD backlights, if they designed them with a voltage drop close enough to the peak mains voltage that the series resistor could be a reasonable wattage, there could be a case for running from mains. If, of course, the wiring and case are suitably insulated or grounded.

I did a bit of research and this doesn't seem to be the case though, this page says that they range from 24V - 66V:

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I also stumbled upon this, which appeals to me for some reason:

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Reply to
Computer Nerd Kev

If you're able to mess about with the individual LEDs, I suggest measuring one to find its actual forward voltage rather than guessing. This is easily done by connecting it via, say 1K, to a 5V power supply and measuring over the LED leads with a multimeter.

I guess you meant (240V - 200V) / 0.015A. If the panel uses LEDs in parallel, the current may need to be higher than that for just one.

Also, 240VAC has a peak voltage of 240 x 1.414 = 339V. So for calculating for maximum current you should use the peak voltage, eg. (339V - 200V) / 0.015A = 9267R = ~10Kohm. Or you could add LEDs in series so that they drop 300V instead of 200V, which also prevents wasting up to (339V - 200V) x 0.015A = 2.1W of power (actually that's not too bad, but more likely figures would be 0.15A and 21W or greater) at the very peak of the mains voltage cycle.

However that way the LEDs are only lit for a breif part of the waveform and the appearance would be more sensitive to variations in the mains than if you had the wider 140V gap above the LED's turn-on voltage. The LEDs would also appear brighter at lower current because they are on for more time, which might allow a greater resistance to be used in series and thereby offset some of the efficiency loss mentioned in the last paragraph.

I have a side-lit LED backlight pulled from a digital photo frame LCD, which I use as a slide viewer. To work out how to power that I simply connected it via a 1K resistor and increased the voltage from my bench supply until I noticed a light. I then tried a lower value series resistor and raised the voltage above the turn-on point to see the stage at which increasing the voltage stopped producing a significant increase in brightness, the current thereby being approximately the maximum. Then I found a brightness that suited the purpose. Finally I ended up using a 100R wire-wound resistor in series with a 18VDC plugpack, and it works well without the resistor getting burning hot, which is all that I was after.

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Reply to
Computer Nerd Kev

The older ones used fluorescent tubes and a high voltage power supply to drive them. The more modern LED ones are more likely to use a constant current supply. The voltage across the panel and the current going into it would depend on how the LEDs are connected, series, parallel, or series parallel, and how many LEDs are involved.

Gawd, how crude! Obviously this was a fluorescent lit display.

I've got a project to take the panel out of an old display that I got from a local op-shop, put it in an old radar set in lieu of the CRT and use a Raspberry pi to simulate a radar picture.

Reply to
keithr0

Yes, the constant current supplies in LED light bulbs are sometimes linear, with the series LED forward voltage high enough that they don't waste too much power. The same could presumably be applied to LCD backlights, but with a SMPS in there for the logic voltages anyway, I doubt that there'd be much gained by doing so.

Easy to spot that one.

Neat. Bonus points if it displays data from flightradar24. :)

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Reply to
Computer Nerd Kev

good point, should allow for 340v peak

Reply to
Taupe

handful components, diodes resistors etc. fixed to frame is OK compared to proper power ciruitry that would bulk up frame

I'd rather drop voltage over more LED's than resitors that lose energy via heat.

Reply to
Taupe

will put back perspex lense to disperse the light.

very fiddly trying to unsolder the tiny LED's from strip,

I'm Just going to Use as they are , problem is working out circuit tracks under white paint.

Reply to
Taupe

apparently voltage & amperage varies with manufacurer , so I chose figures within range to get rough idea.

18VDC seems high, plugpacks usually unreliable from what written on outside to what it actually produces . If it works on 12vDC maybe run off that, & save electricity heating up resistor.
Reply to
Taupe

They sure do.

I can't remember what voltage it began to work at. If you drop the voltage though, you then need to lower the resistance to keep the current through the LEDs the same. If I wanted efficiency I would have used a constant current driver, which effectively sets the optimum values automatically. But wasting a Watt or two for half an hour once every few weeks when I'm sorting out slides isn't something I'm too worried about. Especially given that it's often next to a slide projector which burns a 500W bulb!

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Reply to
Computer Nerd Kev

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