CoB LED filament analysis

What use would diodes be in series with the assumed end resistors?

Draw a circuit please.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

If the ceramic slab has an upper positive rail and a lower neg rail, there could be a bridge with half the diodes on each end.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Nice to know someone else could see them too!

You ask, you get ...

piglet

Yes, that's probably it.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Congratulations, you did it again! Your adroit analysis is anticlimactic. Because it eliminates my expectations of exotic electronic elements such as miniaturized MOSFETs. Yet, in retrospect, it's simple to see how power elements demand a relatively large footprint in order to handle a high voltage. It's only simple now because you saw it first. Anyhow, your radiographic interpretation eloquently puts everything into perspective. Is it possible for you to give me permission to include it on the webpage with a credit to you? You also confirm my suspicion of a half-wave rectifier at each end. One half-wave connected to Line and another attached to Neutral. This gives me enough insight to include an illustrative kicad schematic on the page. Then the only things left to do are a detailed write-up and a pesky V-I curve. Even if the curve only shows a resistive load it still gives me an excuse to teach myself how to handle a Fluke current probe.

Thank you. Please feel free to use my sketchy schematic as you wanted. The LEDs inside a bridge rectifier is hardly original. Because the resistors drop most of the voltage even cheap 1N914 type diodes would suffice in the bridge as the peak reverse voltage they see is just the Vf of the seven LEDs.

Alright you guys, a couple of current curve images are now available on the webpage. For the present case, in regards to the probe's "P2 < P1" orientation decal, P1 designates Line while P2 indicates Neutral. The probe provides a sharper, cleaner curve compared to the resistor. Both curves illustrate how the silicon filament primarily presents itself as a capacitive load. The filament's high voltage combined with its low amperage degrades resistor curve quality. It took a 10 M ohm resistive current sensor for the half-wave rectification effect to become visible. The probe curve's mostly blue trace shows the peak-to-peak Line voltage. It's used to properly trigger the scope. (Perhaps the scope's line trigger accomplishes the same result with less effort?) The probe curve's mostly green trace shows the current, with peaks and valleys created by half-wave rectification at each end of the filament. The current curve is set to 2mV per division. Does anyone know how to transpose it to mA?

Danke,

It can't be capacitive. It has no mechanism to store energy.

Unless one of the thingies on the end is a cap. A DC curve would resolve that.

A thermal image would be interesting too.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

No, it lights up with DC, so there's no series cap.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

ELI the ICE man. The trace of the current curve clearly leads the voltage trace, so the filament must present a capacitive load. A silicon device that presents a capactive 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.

Danke,

Seems unlikely. It's more likely bad triggering from a not-all-that-closely related waveform.

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.

Since the current probe waveform is so different from the one that uses a resistor, one must be wrong, or likely both.

Stray capacitances are at least six orders of magnitude from causing the phase shift that you are seeing.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

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.

Danke,

Power factor has a diffferent meaning for nonlinear loads; it doesn't necessarily mean phase shift.

I have one official document that says PF is undefined for non-sinusoidal loads.

One common practice is to say that PF = real watts / (RMS volts * RMS amps)

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

For what it's worth, my own expectations were subverted when the current curve indicated a non-resistive load. So, this thread tries to resolve the discrepancy. In my experience, the 320 Vp-p 1 mA combo crushes the current sensing resistor idea. It took a 10 M-ohm resistor in series with the CoB LED to see anything other than a ~320 Vp-p waveform across the so- called sensing resistor. Connecting a current sensing probe to an oscilloscope is even easier than inserting a resistor in series with a CoB LED. Yet given the unknown meaning of the "P2 > P1" decal on the probe, it's plausible for the probe to be connected backwards. This oscillogram depicts a probe reversal:

Danke,

A reasonable current sense resistor might be 10 or 100 ohms.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

You're faffing about. Make a cup of tea, sit down and think for a minute.

What does a forking fantasy about faffing have to do with your bright idea to use a series resistor to create a current curve? 1, 10, 100,

Danke,