Recent thread topic transitioned to a tentative opinion on a work-in- progress webpage:
Radiography imparts interesting insight. Second opinions welcome! The current waveform flopped. Any advice on how to use a Fluke
No idea. Can you just put a resistor in series and do a A-B on the scope?
It looks like there are just a couple of resistors in series with the LED's. If the current waveform follows the voltage supply after the LEDs start conducting that points to a resistor to provide current limiting. If it was something active it would reach the current limit and stay there. Check if the current is supplied in pulses, indicating no supply smoothing. If these are high intensity LEDs they may only need a mA or two to provide adequate illumination. Most of the silicon drivers are simply a LDO regulator with a depletion MOSFET wrapped around it, with an enable input on the regulator so that the intensity can be controlled by a PWM signal.
There's a fellow on Youtube (HairyDave) who tears down a lot of COB displays, it might be worth checking if he's had a lokk at one like your's.
Sorry - should be bigclivedotcom. (Although he is hairy as well as big...)
Very cool. Thanks.
I suspect that the things on the end are resistors. If so, the current waveform will be low duty-cycle rounded bumps. Might be half-wave.
Run it between AC hot and ground. Put a resistor in the low side and scope the voltage across it. Try 10 or 100 ohms maybe. That's perfectly safe.
The current will be low enough that that beast of a 1000 amp current probe won't even notice it.
A thermal image would be interesting. If I had one, I could thermal image it, measure the current waveform, and do a high-res Xray with our big Nikon.
John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics
Why not just slap it across the AC line?
John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics
Looks like radiograph shows resistors at each end. Each resistor connects to two wire bonded diode chips then a series string of seven led chips. So basically seven LEDs inside a bridge rectifier with series resistors to limit current.
Likely no rectifier and more than seven LEDs. The current waveform would reveal much. Or a couple of DC V-I curves.
John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics
The diodes are visible next the resistors even the bond wires are faintly visible in the photo.
You are probably correct, but easy enough for Don to check by using a DC supply to check conduction with both polarities.
Yes. Graph and don't guess.
John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics
OK you guys, lots of excellent ideas! The DC idea's easiest, so it's first. Big Clive's youtube link now appears as a Footnote on the pertinent page:
The webpage also includes a couple of new images where 68 VDC is applied to barely illuminate the filament's CoBs, first in one direction, and then the opposite. How do you graph such empirical data? Anyhow, as expected, seven filament CoBs illuminate regardless of DC polarity. Perhaps each of the seven CoBs contains a couple of LEDs, cross connected to conduct current through the anode of one LED, while the cathode of its coupled LED blocks current? JM, elsewhere you mention fourteen illuminated CoBs in Big Clive's video. It seems significant that fourteen CoBs are also illuminated in my filament (seven for each DC polarity). The filament also contains a small hole on one lead to denote its anode. But it's pointless because the filament conducts in either direction. Does one factory in China create CoB filaments for both AC and DC applications? So users end up with filaments filled with fourteen CoBs and an anode hole in one lead?
No, you only have seven LEDs - they always see the same polarity due to the diode bridge (which you have proved to exist). The lamp in the hairyones video is a chinese knock off of the Osram GW T3LSF2.DM
Your device has a bridge and current limiting resistors inbuilt, something like the sketch at 2.36m in the youtube video.
The xray sure doesn't look like there is a bridge rectifier.
If 68v is the turnon threshold, and a blue LED needs, say 2.8 to light up a bit, there are roughly 24 LEDs in series.
What someone should do is graph current vs voltage, both polarities.
John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics
The four rectifiers creating the bridge rectifier may not be mounted together as a discrete bridge rectifier, but that doesn't stop them working together to create the same effect.
That would mean twice as many LEDs, which would be a much more expensive solution.
There might be, if anybody could see them on the X-rays. Google says that blue LED drop between 3.0V and 3.4V. Seven of the them would drop between 21 and 24 V and enough resistance to soak up the remaining 44V would make the current tolerably stable against self-heating in the LEDs (whose forward voltage drops as the junction temperature increases).
What John Larkin needs to do is to think a bit harder.
There's no obvious bridge in the xray, but resolution isn't very good.
There are ways to decap a plastic-potted IC.
Look at the V-I curve on that part. It conducts 1 mA at about 2.8 volts, and would be visible at way lower current. 1 uA might be visible in dim room light.
Well, I do have a day job.
John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics
Can you see the four diodes now ...?
piglet
All he needs to do is measure the current at a couple of dc voltages (both enough to illuminate the LEDs) and calculate the series R from delta V/ delta I. What else is there to know?
You beat me to it.
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