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+.
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.
Not a bad idea of course, when I was using the wrong units it was just really confusing to me where 0.18 came from. Now that I realize my mistake your logic is clear!
I fished around on the internet a little and it seems that the “LEDA” chip is definitely the QX9922.
http://club.dx.com/forums/forums.dx/threadid.1205677
[quote=desolder]
It’s the QX9920:
http://www.szroya.com/uploadFile/download/200886204048427.PDF
The IC is rated for 24V, but the input capacitor is only rated for 16V. So it appears the “5V-12V” description is the correct one.
If anyone is curious, here’s the thread claiming that it’s the QX9920, apparently from a representative of the company:
http://translate.google.com/translate?hl=en&sl=zh-CN&u=http://bbs.dianyuan.com/topic/581194&prev=/search%3Fq%3DLEDA%2BSOT-23-6%26hl%3Den%26client%3Dfirefox-a%26hs%3DuZE%26rls%3Dorg.mozilla:en-US:official%26prmd%3Dimvns&sa=X&ei=S0cUUPWkOcbsqAGBl4CYCw&ved=0CFoQ7gEwAAp
[/quote]
desolder is misleading when he says that the chip is rated for 24v, the chip can run on 2-6.5v. The diagram in the datasheet shows a 2.5v - 5.5v Zener Diode in use along with the appropriate resistor in order to keep the chip inside that range.
No doubt that's what some of the extra stuff on the PCB in this thread is. I don't see the advantage of QX9920 over QX5241.
Hi,
Sorry, too many posts have gone by, so I’ll answer the ones I know about here so we can kind of start fresh:
- [wight, I think]: That was a good call on the solder blob, i.e., you were right, that all 3 components (one end of the toroid, one end of the 151 component, and one end of the brown component) are shorted together, both the blob-less and the blob-ful board.
- [wight, again, I think]: I didn’t find an R200, but I did find an R300, so I figured I’d try that, parallel. From the post above, R300 would be .3 ohms?
So, I have R200 and R300 in parallel, i.e, so equivalent of 0.12 ohms.
Is that correct?
Anyway, I tested the driver after that, and BINGO, Iemitter on high went to ~1.62 amps (from previous 0.90 amps)!
So:
0.2 ohms => 0.9 amps
0.12 ohms => 1.62 amps
I took the Vbat/Vin quite high, > 10V, I think, but I got excited, and didn’t really watch the correlation of the Ibat/Iin vs. the Iemitter, but I did see Ibat/Iin start dropping at some point as I raised Vbat/Vin.
I’ll do that later.
So, given the above, what resistor value do you all recommend to replace the R200 “safely”?
Thanks,
Jim
EDIT: Can we get this driver to 3 amps at the emitter, or maybe even higher?
EDIT 2: And yes, I still do have all 5 modes…
It’s possible, based on your results, that the sense voltage listed in the datasheet is no longer accurate. Maybe it’s using the same sense voltage used by QX5241, approximately 205mV. Or it could be 180-290mV, who knows. Really it’s just as likely that somewhere in this mess of a driver there’s something messing up the normal sense voltage setup.
I’d dig up an R100 and put that in parallel with the R200 (for 0.0666666666666 Ohms). That should give you approximately the current you want. (maybe 2.7 - 2.9A based on what we’ve seen so far, maybe 3.75A based on what the datasheet claims (250mV sense)) If current is too high, discontinue use until you’ve increased resistance again ;-).
Pushing this current too high will generate problems related to the SS34 diode, the inductor, the no-doubt-crappy FET, etc.
EDIT: yes, you are correct about R300 and about R200 and R300 in parallel.
I just wanted to confirm: I need to look for a 0.1 ohm, 1206 resistor, right?