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What's inside an E10 LED bulb (a 3V flashlight bulb) that allows the LED to light without worrying about current? Is there a tiny LED driver inside the bulb?

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    \$\begingroup\$ Not all E10 bulbs are the same, some are low voltage, others are not. Which bulb are you asking about, add a link / datasheet. Probably there are only one or more resistors and maybe also a few diodes if the bulb is suited for AC. \$\endgroup\$ Commented Jan 11, 2022 at 21:58
  • \$\begingroup\$ They don't worry about it. It's a flashlight. When the LED melts down, you buy another flashlight. The fact you feed it only 3V is how they avoid instantaneous failure. \$\endgroup\$
    – Kyle B
    Commented Jan 11, 2022 at 23:32
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    \$\begingroup\$ @KyleB very funny :) but not very technical \$\endgroup\$ Commented Jan 12, 2022 at 0:43

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The ESR or Series bulk resistance needs to be sufficient to limit current at a constant fresh battery voltage less than the rated current. Chips of this size tend to be in the range of 65 mW and the 10% current threshold is often around 2.8V while rated current such as 20 mA is around 3.05 to 3.1V The tolerance on this value depends on the process controls of the epiwafers purchased by the supplier to make the lamps.

I'll let you in on a little secret I learnt while distributing over 1million LEDs after my freedom 55.

  • The semiconductor diode bulk resistance Rs = k/Pd rated for k= 1 reducing to 0.25 +/- x% for high power LEDs. Thus for Pd rated at 65 mW the resistance is 1/0.065= 15 ohms so for 20mA and Vt=2.8V Vf= Vt+IR= 2.8 + 0.02A * 15 ohms = 3.1

There is an NTC thermal effect of the white LED diode, just like all diodes, but the value is about -4% of Vt=2.8 or -112 mV for a Tj rise of 100'C with Rs=15 ohms 0.112V/15 ohms= 7.5 mA current change.

This is why the Alkaline 2x 1.5V cells will not burn out the LEDs but they will gradually decay in brightness with the batteries and reduce current at the same time. So the design is inherently stable by the choice of small LEDs.

If it were a 5W LED then the ESR would be between 100 and 200 mohms and might have a slightly lower Vt from higher efficacy, this would NOT be the same case as high power LEDs operate at a lower Vf due to a lower Rs. This is the key factor inversely related to current rating and power.

To operate several LEDs in parallel at the same voltage, you would have to know the tolerance of Rs for the LEDs being used and add a Rs equal to that tolerance to stabilize the currents and prevent thermal runaway. But that's a another question for another day.

I recently repaired an expensive kitchen pendant lamp which lasted all of 2 yrs until at least one LED in the centre with 20x 6V SMT 3580 LEDS burnt out. I replaced it with a 10W LED and operated it at 5W using 9V instead o f 12V. Using a digital controlled lab supply I used the current at 9V CV and adjusted it for CC and then it displayed the drop in voltage due to thermal rise. with it mounted to the internal heatsink with thermal paste and with 4 screws. The drop was 170 mV which translates to a Tjcn rise of 45 'C. on a constant current of 0.549A. I intend to operate it at CV of 9V and since I measured to confirm Rs and know the increase in power raises Tj, I can not predict and verify the new junction temperature and know it will be stable with adequate margin for at least a 10 yr lifespan.

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The E10 designation is just a socket size,

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so your question ends up being very general, as every manufacturer puts something different into their product. If you look at this Google search for "E10 LED Bulb" images, you will see that there are many incarnations, and the only thing they have in common is the size of the screw socket. They may have a simple resistor inside, or a "317 through hole adjustable positive linear voltage regulator" like an LM317 or a switching regulator like a Joule Thief, which is a self-oscillating voltage-boost circuit, but they may also have a buck circuit, or any other kind of power converter circuit imaginable, and may also be targeted for any voltage imaginable, so your question is really almost too general for the answer to be useful. Hope this helps.

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