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Repurposing LED TV backlights from thrown out kerbside TV's

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Taupe
Guest

Fri Feb 08, 2019 12:42 am   



Jasen Betts wrote:
Quote:
On 2019-02-07, Taupe <jonahreal_at_yopmail.com> wrote:
Jasen Betts wrote:
On 2019-02-06, Taupe <jonahreal_at_yopmail.com> wrote:
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.

no.
can't be done?
what if 240v rectified to DC with a current limiting resistor?

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


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

Quote:

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

that depends on the power supply circuitry you use

thats the point, not to use seperate box power supply


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

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

Will it potentially be fire hazard from overheating & shorting?

yes.

I'll put a fuse in circuit?

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.


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

Taupe
Guest

Fri Feb 08, 2019 12:45 am   



Jasen Betts wrote:
Quote:
On 2019-02-07, Taupe <jonahreal_at_yopmail.com> wrote:
keithr0 wrote:
On 2/7/2019 7:36 AM, Computer Nerd Kev wrote:
Taupe <jonahreal_at_yopmail.com> wrote:
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.

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.

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.

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

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

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

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

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

Will it potentially be fire hazard from overheating & shorting?

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.

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

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.


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?


will put back perspex lense to disperse the light.

Quote:

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


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.

Taupe
Guest

Fri Feb 08, 2019 12:45 am   



Computer Nerd Kev wrote:
Quote:
Taupe <jonahreal_at_yopmail.com> wrote:
assume LED require 2v & 15mA to operate.

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.


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

Quote:

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

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

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.


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.

Computer Nerd Kev
Guest

Fri Feb 08, 2019 7:45 am   



Taupe <jonahreal_at_yopmail.com> wrote:
Quote:
Computer Nerd Kev wrote:
Taupe <jonahreal_at_yopmail.com> wrote:
assume LED require 2v & 15mA to operate.

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.

apparently voltage & amperage varies with manufacurer


They sure do.

Quote:
, so I chose figures
within range to get rough idea.

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.

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.


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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