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?
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).
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+.
I probably was not 100% clear, I shouldn’t have posed that the way I did.
I suspect that all 3 of those leads are connected regardless of the blob. Do you have a continuity test (beep) mode on your DMM? You could check them that way on the blob-less board. You can also just use a resistance mode. Taking a close look at the blob-less driver PCB will likely show that traces connect all 3 leads.
So what I’m asking is this: have you verified beyond a shadow of a doubt that they are not connected via PCB traces on the blob-less driver?
Correction to what I said above: On the one board that I have leads soldered onto, the leg of the toroid is soldered to one end of the 151 component, but they are not solder bridged to the larger brown component.
On the other board, the one I haven’t tested yet, all 3 components (one leg of the toroid, one end of the 151 component, and one end of the brown component) are all bridged.
To your last paragraph, no, not yet, but I will do that.
Here I go… dumpster diving time :)!
R200 is a 0.2 ohm resistor.
Sounds good, thank you for checking.
GL! Here’s your chance to prove why you need to horde that stuff
R200 is 0.2 Ohm: .9 = .18 / .2
Oops. Thus proving… wow, good thing we don’t depend on me around here! I got moving too fast. Let me go back and edit (cross out) a few things… OK, done. I didn’t cross out where I was doing the datasheet math, but that was wrong too since I used incorrect units. The datasheet’s math would be more like 0.25/0.2 = 1.25A
Thanks for correcting me.
The buck drivers I have worked all use .25 , but (more efficient) drivers from sandwich shoppe have used .05 sense… so I was using .18 based on the current measures here.
This will all get sorted out sooner or later here, thanks.