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I'm trying to power this LED filament bulb with a 12V - 2A DC wall adapter.

LED specs:

  • rated at 4W
  • 12 to 24V input

I'm pretty new to this, but if I get it right:

Calculus

  • the LED bulb will draw E/V = 4/12 = 0.33A if power source is at 12V.
  • by measuring voltage drop across LED bulb, I get a reading of 3V.
  • to control current flowing in my circuit, I'm using 1W-rated resistors. I know voltage across resistors will be 12-3 = 9V.
  • So I can deduct that resistors should generate a total resistance of V/I = 9/0.33 = 27 ohms.
  • to do this, I wired 3 groups of resistors in parallel, each group being made of 4 20-ohm resistors in series. That way, I should get a 80-ohm resistance per group, and 80/3 = 26.6 ohms of total resistance. And this setup allows me to have I/V = 0.33/3 * 9/4 = 0.25W crossing each resistor, largely below their 1W rating.

Results enter image description here When I power this circuit, the LED turns on and current crossing it is measured at 0.32V, but after a few seconds:
- the voltage keeps dropping across LED - voltage increases across resistors - current increases across resistors - resistors get warm - LED slowly dims down

I assume that the generated heat from the resistors decreases their resistance, hence an increase in current and voltage crossing the resistors, leaving less and less current to LED. But I'm not sure, and if I'm right, I have no idea how to prevent that. I tried different resistor combinations, always trying to stay way below the resistors' 1W max rating, but it always ended up with voltage slowly decreasing through LED.

Any idea on how to fix that? Thanks in advance!

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  • \$\begingroup\$ what is the purpose of the resistors? \$\endgroup\$
    – jsotola
    Commented Sep 5, 2019 at 22:24
  • \$\begingroup\$ So you're taking an electronic device that's designed to be fed straight from 12V, and you're giving it a voltage significantly less than 12V? Why? Why not just hook the 12V-rated bulb straight to the 12V-rated supply? \$\endgroup\$
    – TimWescott
    Commented Sep 5, 2019 at 22:31
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    \$\begingroup\$ @TimWescott, that was my thinking also ... i just wanted the OP to think about it ... i was hoping that the OP would say "the resistors are for dropping the 12 V to .... wait a minute ... what am i doing?" ... and the proverbial light would come on \$\endgroup\$
    – jsotola
    Commented Sep 5, 2019 at 22:36
  • \$\begingroup\$ thanks guys, I got caught up in some crazy reasoning indeed. My bad! \$\endgroup\$
    – garys
    Commented Sep 5, 2019 at 22:46
  • \$\begingroup\$ "Working with Low voltage 12-24V, AC and DC compatible... NO dimmable, it cann't be working with dimmer" \$\endgroup\$
    – Bruce Abbott
    Commented Sep 5, 2019 at 23:04

2 Answers 2

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Looks like that bulb has an internal circuit that provides constant 3.5W output to the LED filaments, so that it will draw I = 3.5/V where V is the applied voltage (for some voltage in the 12-24V range).

If you apply less than 12V it may no longer be able to provide full power, but it will likely do so for voltages somewhat below 12V.

When you add a resistor (and assuming the bulb can still provide full power) you get I = 3.5/(12V-I*R) which, if you solve for I, is a quadratic equation, and for R > 144/14 ~=10.3 ohms there is no real solution.

So the bulb will probably hover on the edge of not working and the results may not be consistent.

If you want it to operate from 12V, you just apply 12V. Any added resistance just wastes power. The bulb is not intended to be dimmed, as per the Amazon listing.

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  • \$\begingroup\$ I was assuming that the LED bulb needed some resistance added to it, I really got it all mixed up. 12V to 24V it is, then. To be fully transparent re. the actual circuit, I'm powering the bulb with a 5000mAh 5V (2A max output) battery, that's wired to a step-up buck converter. But for some reason, the bulb just flashes and turns off. I think that it's because the voltage output is under 12V for a few ms, causing the LED to fail. I just don't know how to immediately output 12V. Any idea? \$\endgroup\$
    – garys
    Commented Sep 5, 2019 at 22:42
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    \$\begingroup\$ That's a tough one- the circuits often draw a lot of current when starting up, so sometimes the DC-DC might need more power capability or the battery voltage may be dropping. Have you tried turning the DC-DC on first and then the bulb? \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Sep 5, 2019 at 22:45
  • \$\begingroup\$ Just tried it, still same issue. The LED bulb keeps flashing. I also tried changing the potentiometer resistance on the buck converter to increase voltage output (I went up to 16V) but that didn't fix it. \$\endgroup\$
    – garys
    Commented Sep 5, 2019 at 22:55
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    \$\begingroup\$ You could try a capacitor on the DC-DC output plus switching it on first. Something more than 1000uF might work, but too much capacitance and the DC-DC might not start. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Sep 5, 2019 at 22:56
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    \$\begingroup\$ Great advice, thanks, I’ll get a capacitor and try that. \$\endgroup\$
    – garys
    Commented Sep 5, 2019 at 23:02
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enter image description here

Figure 1. According to My 2 μF there's more than just LEDs in there.

The linked article is for mains powered bulbs but you may find something similar hidden in the ES (Edison screw) cap of your low voltage lamps.

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  • \$\begingroup\$ Thanks a lot, there's indeed a lot more that just a few LEDs in the ES, I hadn't realize that. \$\endgroup\$
    – garys
    Commented Sep 5, 2019 at 22:43

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