No - Wait - Stop - Don’t Cut. …just melt the solder at the joint and pull it loose from the board with needle nose. (just lurking here, waiting to see what this board is doing…)
LowLumen is right. Cutting is fine, but I was not suggesting cutting, sucking, or wicking. There is plenty of slack available to remove one leg of the inductor at a time. Just heat them up and pull them out, one at a time.
I see nothing to indicate that you’ve got anything wrong unless it’s wire placement, and that sounded fine. You’ve already tried several different LED setups and you’ve tried a range of input voltages. Continuing to plug at it in that way is a waste of time IMO.
I don’t know why I missed this, but I found that I can re-position the toroid well enough to get some pics that might be good enough, so I’ll post those in a bit, then, if it’s still not good enough to help, I’ll do what I need to get the darn thing off.
Some new, hopefully better info:
I used some new heavier leads from the driver to the emitter (an XM-L2 on Noctigon) and re-tested and I think that this is the behavior I am seeing (and I tried it several times):
- As I increased Vbat (Vin to the driver) when it gets to somewhere around 4.5 - 5V, the current to the driver (Iin/Ibat) starts dropping, and basically continued to drop as I increased Vbat/Vin.
- I have my clamp meter on the emitter lead, and watched Iemitter more carefully this time (sorry about earlier), and also measured Vemitter. I did these Iemitter and Vemitter measurements at Vbat/Vin at 5V - 9V at 1V increments. For all of these measurements, Vemitter stayed almost exactly 3.00V, and Iemitter stayed at 0.89 - 0.90 amps.
In other words, it looks like above a certain Vbat/Vin:
- Iemitter stays at 0.89 - 0.90 amps and Vemitter stays at 3.00V, but
- As Vbat/Vin increases, Ibat/Iin decreases.
I’ll post the pics that I can get in a bit…
Some pics from different angles. Some are duplicates, but just in case, and some have components labelled in RED:
EDIT: These pics are from the 2nd copy of the driver, which I haven’t actually used (notice no emitter leads on it yet). This board appears to have a solder bridge (you can see the big blob of solder in the 2nd pic), which I’ll have to remove and then add the emitter leads and then test, so I’ve been focused on using the 1st board.
If I just pulled one leg out, I think that the through hole would probably just fill when I pull the leg out. I’ll do that if needed, but let me know if the pics I posted are not good enough…
Hi,
Can someone sanity check me?
I was just looking at the pic of the driver from the Cofly thread that I mentioned above:
And, I don’t know if I’m imagining it, but it looks like the hole where the toroid leg goes into and the hole where the positive emitter lead goes into on the above image is reversed compared to the boards that I have (well, the toroid leg hole anyway) - some new pics below where I tried to take the pics in similar orientation to the above pic, so you can compare:
In the pic from the Cofly thread, it kind of looks like the toroid leg is going into the left through hole/pad and the red emitter lead is going into the right through hole/pad, whereas in the 2 pics of my board, the toroid leg is going into the right through hole/pad and the red emitter lead (which I added because the toroid was already in the other hole/pad) is going to the left hole/pad?
It’s kind of hard to tell from the 1st pic, so I’m going to post and see if nofear can post a clearer picture.
Looks like that R200 resistor is your sense resistor: so… I = .18 / R or therabouts if your LED current measures are correct.
NVM the above, I think that is a false alarm. It looks like the two pads where the toroid leg and the emitter lead go to are shorted together anyway, i.e., so it doesn’t matter which pad the toroid leg and the emitter lead are soldered onto :(…
nofearek9 has posted info on news test he did with tailcap current on his thread:
The above was with 2 different pairs of batteries. I don’t know what emitter voltage he was seeing, but let’s just say it was also about 3V, so then his input power would have been:
1 amp x (2 x ~4V) = ~8 watts
If his Vemitter was ~3V, then his Iemitter would be (assuming no power loss between input vs. output):
8 / 3 = 2.67 amps. Good.
Whereas, in my latest measurements, I was seeing about .9 amps Iemitter at 3V Vemitter, or about 2.7 watts at the emitter end.
It definitely seems like the driver I have is underperforming compared to the one in nofearek9’s Cofly?
Just bump up that R200 sense resistor to the formula: I = .18 / R
Do you think I should try bridging that resistor with something in parallel? If so, what value would you suggest? I have some resistors, but I don’t remember what values, but maybe I can pull something from one of the drivers in my junk box?
Also, earlier it was not recommended to short across the current sense resistor, but what about jumpering just for a short test? Would that cause a problem (like “poof”?)?
Just me, but I would never short a sense resistor. Also, the SS34 diode is rated to 3 Amps.
First thing; sanity check: the orientation of the red wire and the leg from the inductor do not matter. They don’t even need to connect to the PCB, just each other. The location on the PCB is provided for tidiness and physical strength.
The circuit does not look familiar to me.
The markings on [what I assume is] the buck controller look exactly the way QX9920 is marked. I don’t know what the markings mean, but the numbers change - maybe it’s batch numbers? The pinout may not match, it’s very difficult to tell because many traces go under other components.
The unmarked chip is most likely just some crap “flashlight modes” chip. It’s not very important, assuming I am correct. If it’s a familiar model you might be able to short pins on it in order to change mode groups.
Do you have a multimeter? If so it would be a good thing to do to check voltage at all pins on the “LEDA” chip in both “High” and “Medium” modes. Check them against ground.
What if I find another R200 resistor? Would soldering that in parallel be ok?
What exactly is the value (ohms) of the R200 resistor?
Thanks,
Jim
I agree about bumping R200, but I’m not quite sure where that formula is coming from? It doesn’t match the current results, so I think we should not use it.
It seems that the formula in a QX9920 datasheet is reasonably close. 250mV / Rcs = Ileds (250/200 = 1.25)
ohaya, don’t you have two of these? You can try to rob an R200 from one and stack it on the other. (250/100 = 2.5A)
200 Ohms. [EDIT: wrong. Seeing a pattern here?] Putting two in parallel is good, that’s what I suggest doing next.
I think that where LowLumen was coming from was .18/200 => .9 amps (at the emitter), which I was seeing in the last set of tests (which are hopefully correct now :().
Yes, I have 2, but in the end, ideally, I’d like to end up with 2 working drivers (optimally), so if I can rob an R200 from another board, I’d rather do that. I’ll do some digging around my junk box(es).
Thanks,
Jim
Also, I don’t see any solder blobs that are obviously a problem. The short leg of the inductor, one leg of the SS34 diode, and one leg of the 151 power resistor are all probably supposed to be soldered together anyway, or that’s how it looks from here…
Are they not all connected on the other driver?
Re. your last paragraph/sentence: You are correct, those pins/leads are NOT bridged/connected on the one driver that I added the emitter leads to yet.
Those pins/leads are only bridged/connected via the blob on the one driver that I haven’t added the emitter leads to yet (and haven’t tested yet).
I was going to post about this: I chose to add the leads to the board that I did add the leads to because that blob didn’t look “right”. At various times, I’ve been thinking maybe the blob SHOULD be there, and maybe the board that doesn’t have the blob is the one that’s “wrong”, and maybe I should test the blob-ful board also and maybe that one will give me 3 amps, but then I decided to focus on the blob-less board first.
OK, but that math is not correct. 0.18/200 = 0.0009 [EDIT: wrong]
If you can’t find any R200 in the junk box, do not despair. I’d just grab one from the other driver and keep rolling, replacing it will not be a big deal. “R200 is cheap 50pcs/$1 shipped~~“:http://www.ebay.com/itm/50pcs-SMD-1206-Resistor-200-Ohm-200R-/350972597504 [EDIT: wrong, I linked to 200R~~ the digikey/mouser info should be right] I’m assuming that it’s 1206 sized, but I could be wrong. Anyway you probably won’t end up wanting R200 since that’s apparently a much too high number! 1206 resistors of a more known quality are about $0.10-$0.25 each in QTY/10 on Mouser or Digikey, but you’ve really got to get your order together for it to make sense because shipping is $6+.