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From: "Don" <g@crcomp.net>
Newsgroups: sci.electronics.design
Subject: Re: CoB LED filament analysis
Date: Tue, 21 Apr 2026 21:05:40 -0000 (UTC)
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Bill Sloman <bill.sloman@ieee.org> wrote:
> On 22/04/2026 2:55 am, Don wrote:
>> john larkin wrote:
>>> john larkin wrote:
>>>> Don wrote:
>>>>> Don wrote:
>
> <snip>
>
>>> No, it lights up with DC, so there's no series cap.
>>
>> ELI the ICE man. The trace of the current curve clearly leads the
>> voltage trace, so the filament must present a capacitive load.
>
> Seems unlikely. It's more likely bad triggering from a
> not-all-that-closely related waveform.
>
>>      A silicon device that presents a capacitive load is different from
>> a series capacitor:
>>
>>      COB LEDs present capacitive loads through parasitic capacitances
>>      inherent to their densely packed chip-on-board structure. These
>>      capacitances arise between closely spaced LED dies, bonding
>>      wires, and the substrate, affecting driver circuits during
>>      switching.
>
> The internal capacitance won't be anything like big enough to show up at
> mains frequency
>
> The likeliest explanation is a bunch of LEDs connected in series with a
> current limiting resistor between each LED. You'd put in four mains
> rated diodes to rectify the current going through the LED/resistor
> string as the applied voltage switched direction.
>
> Close to zero and 180 degrees the mains voltage isn't high enough to
> push any current through the LEDs, but they will start turning on at
> about 20 degrees, and progressively more of the voltage will get soaked
> up by the resistors until the voltage peaks at 90 degrees/270 degrees.
>
> The LEDs and the resistors will heat up a bit during each half cycle -
> more in the middle of the string than at the ends, and that will make
> the current lumpier than you'd expect from a pure sine wave.
>
> It won't make much difference to the resistors but LED forward voltage
> drop is a function of both current and the die temperature, and as the
> junction gets hotter the forward voltage drops by about 2mV/C
>
> The LED/resistor load  also going to make current less sinusoidal than
> the mains supplier would like.

The power factor of cheap COB LED filament can drop to as low as 0.55
leading. [1] A power factor of 0.55 corresponds to a phase angle of
about 56.6 degrees, since PF=cos⁡(θ) and θ=cos⁡−1(0.55).

Looking at the current curves from a different perspective, perhaps
peak polarities are opposite. And a narrow negative peak current only
occurs near the tops of positive peak voltage. This roughly corresponds
to Oscillogram 2. [2]

Note.

[1] <https://blog.1000bulbs.com/home/power-factor-explained>
[2] <https://shuttlelighting.com/the-electrical-and-dimming-behaviour-of-led-filament-lamps/>

Danke,

--
73, Don, WD7Q                                             veritas    _|_
                                                          liberabit   |
https://www.qsl.net/wd7q                                  vos         |