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Update: I couldn't desolder any LEDs from the donor board without destroying them. They simply refuse to come off. This raises the question for me:

Are they even soldered there or manufactured as they are together, and cannot be desoldered?

My flashlight's LED is broken, the tip of the LED's housing is exposed and it does not light up.

As the title says, will it be feasible using a single LED from a 220V LED light bulb to repair my 1.5V flashlight?

How do I know the voltage rating (or more precisely I need the turn on voltage below 1.5V(?) (maybe not? --> see Edit2 below) ) of each LED inside my sub 1$ LED lamp (9W in my case)?

Since it is extremely cheap this bulb does not fully have a switching regulator. It is just a "regulator" that probably has huge ripples, which we don't see because it sneakily blanks at 60Hz.

enter image description here

Here is the picture of the schematic:

Bulb schematic

There are almost a dozen LEDs in there, even if I take one out and replace it with a short, the lamp will be overdriven a bit but surely work (with acceptable loss in its runtime, it was cheap anyway.)

Edit: This method won't work on a 1.5V AA powered LED flashlight because the LED lamp has LEDs which only turn on barely at 4.5V. I tested with my 12V computer power supply rail and only 2 of the series LEDs turn on. I couldn't go up to 3. Threshold voltages are very high on these types of cheap lighting LEDs for some reason. Why?

Edit2: My flashlight has a simple circuit doing something to the raw 1.5V before feeding it to the LED. Does anyone have any idea what this does? I wonder how many volts my LED had as threshold before it broke down.

1.5V AA flashlight LED driver PCB

Side view of 1.5V AA flashlight LED driver PCB

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    \$\begingroup\$ You can estimate the forward voltage of the LEDs in your light by adding up the number of diodes and dividing by 220*sqrt(2). This will be a relatively high voltage, maybe 15 or 21v per diode and so probably not usable in a flashlight. \$\endgroup\$ Commented Dec 15, 2021 at 2:30
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    \$\begingroup\$ Are you sure your flashlight is 1.5 V and not something higher? The minimum forward voltage to turn on an LED depends on the color, with red being the visible color with the lowest forward voltage. The nominal forward voltage of a red LED is above 1.5 V. One chart lists the nominal forward voltage for a red LED as 1.9 volts. It may be possible in a very dark room to barely make out that such an LED is "on" even at 1.5 volts, but it will definitely not be at it's brightest. Perhaps your flashlight has multiple cells? Or a Li-Ion cell? Or there is a DC to DC converter in it? \$\endgroup\$ Commented Dec 15, 2021 at 3:29
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    \$\begingroup\$ if it is the same size and luminous power rating and colour , sure why not, but do you know which LED in the string is bad? \$\endgroup\$ Commented Dec 15, 2021 at 4:52
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    \$\begingroup\$ Don't overthink it. Try it. If it works, great. If it doesn't, no big loss. \$\endgroup\$
    – Kyle B
    Commented Dec 15, 2021 at 5:42
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    \$\begingroup\$ @NerdyNerdie it is a boost circuit. There is no white LED with a threshold voltage low enough to run on 1.5 volts. They are all above 2.8 volts. So they need some form of boost converter to get from 1.5 to 2.8+ volts. It probably won't go to the range of 4-6 V though. \$\endgroup\$
    – Arsenal
    Commented Dec 15, 2021 at 12:29

5 Answers 5

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With demand for more TV backlight LEDs, production of < 3V white LEDs now offers 6V, 9V. These are chips with cascaded single diode multiples of 3V and look identical on 2835 SMD, unless you look closely.

https://lumileds.com/products/mid-power-leds/luxeon-2835-architectural/

Your Dad's flashlight will have an SMD inductor that has a darker colour than caps or resistors that with the cap, diode and a 2 transistor oscillator arranged as a voltage boost regulator.

I have hundreds of bags full of ultrabright white 5mm LEDs for someone with a good application.

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This method won't work on a 1.5V AA powered Led flashlight because the LED lamp has LED's which only turn on barely at 4.5V. I tested with my 12V computer power supply rail and only 2 of the series LEDs turn on. I couldn't go up to 3. Threshold voltages are very high on these types of cheap lighting LEDs for some reason. Why?

Cheap light bulbs run off of rectified line voltage using a linear regulator as the driver. Since a linear regulator dissipates heat in proportion to the voltage dropped across it, most of the voltage must be dropped by the LEDs. Since you have 220v RMS line voltage and 20 LEDs, each one must drop about 15v at full brightness.

If they used normal 3v LEDs, about 100 would be required.

Edit for 2023 question:

Update: I couldn't desolder any LEDs from the donor board without destroying them. They simply refuse to come off... This raises the question for me: Are they even soldered there or manufactured as they are together, and cannot be desoldered?

They're soldered like any other circuit component, but that is a metal core PCB so you will need to heat up the entire PCB to remove the diodes. If you have a preheater, set to 150C and then use hot air at 350C or so. If you don't have a preheater you can still do it with hot air, but it will take a long time. In either case remove the capacitor first to avoid popping it (which could be an eye hazard).

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  • \$\begingroup\$ I've never seen a cheap LED bulb that actually bothered to have a linear regulator--they usually have little more than a capacitive dropper. \$\endgroup\$
    – Hearth
    Commented Dec 15, 2021 at 16:33
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Short answer: no, the LED won't work in your flashlight. As others have stated, LED chips in household LED lamps have a higher voltage. I believe internally they have multiple LEDs on the die to boost the operating voltage.

Content creator BigClive on Youtube has several fascinating teardowns on LED bulbs. Very educational.

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Your led bulb has multiple chips, but each of these chips has multiple led junctions on it. Essentially multiple leds in a single package. As there are multiple series leds, the average forward voltage is going to be much higher.

Your led flashlight is using a standard boost circuit or IC to boost the 1.5V to 3.x volts needed to light the single led. It is like a joule thief.

You would not be able to properly use the chips from the bulb in the flashlight because the target voltages are different. You can try, the circuit in the flashlight may be able to boost high enough, but it may not work without modifying the circuit component values.

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The good news is that all white LEDs run at the same voltage, around 3.0 to 3.2V. However the maximum current rating will vary between LED chips.

If you have the soldering skills it may be possible to transfer an LED from the lamp.

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    \$\begingroup\$ Virtually all light bulbs use 9, 15 or 21 v LEDs composed of multiple diodes per die wired in series. A 3v LED wouldn't be practical in a cheap mains powered device since it would require a transformer or Buck converter. \$\endgroup\$ Commented Dec 15, 2021 at 20:02
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    \$\begingroup\$ I have taken many of these light bulbs apart, and I've never seen 3V LED's in one of these, just 6V, 9V, and 12V. I've never seen 21V, as user1850479 has, but any multiple of 3V is possible. What looks like a single "LED" is really multiple royal-blue or UV LED dies wired in series and then covered with the "yellow" fluorescent coating. So your answer is half correct -- all LED dies run at 3v. But these "LED's" are really internally strings of 3V LED dies wired in series, two for 6V, I believe, in this case. \$\endgroup\$ Commented Apr 14, 2022 at 23:22

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