Help with Current Sense Resistor Modding

wight wrote:

. . . I don’t think you have the ability to flash MCUs, but if you did you could use ToyKeeper’s STAR mod which features support for status LEDs. You’d just wire up the 105c to the status LEDs & their limiting resistors on the stock PCB.

Do you know which modified version of STAR has this support? I searched for discussion on the topic, but didn't find any.

Yes. :wink: It’s the version discussed in comfychair’s Roche F6 Hacking thread.

She may have implemented the status LEDs in one or more of her other firmwares, but I know that they are present in this one for Helios-’s Ferrero Rocher DD driver for the F6: http://bazaar.launchpad.net/~toykeeper/flashlight-firmware/trunk/files/head:/ToyKeeper/Ferrero_Rocher/

^ Didn't search that thread. Searched the OSH Park and STAR threads. Thank you. :)

Got the new FET from RMM mounted. Now I'm gonna power up and check currents from battery pack and at LED wires.

Cruddy cell phone pic.

-Garry

Ok here are my measurements at 4.2v battery pack (2 cells parallel @ 4.01v)

off: 0.011A

low: 0.98A

Med:1.30A

High: 1.75A

Now meter in line with LED wires:

off: 0.000A

low: 0.81A

med: 1.33A

high: 1.81A

So I guess my LED measurements are higher because my battery pack wiring & connectors (maybe just my connectors I add to get inline measurements) are adding some resistance. Anyway, I'm pretty much back to stock output. What resistors would I stack / replace? Think my goal was about 3A on high (1.5A per emitter). I know I have R120's and think I have R500's. I have others but I'd have to look.

By the way, I'm correct that the 1.75A I measure gets cut in half for "per emitter" current, right?

Thanks,

-Garry

Iout = Vref / Rsense

Vref = Iout * Rsense

Rsense = Vref / Iout

Therefore for 3A output you want something like 0.033 ohms for Rsense. Three R100’s would be good. Four R120’s is 0.03 ohms for ~3.3A output.

By the way, I’m correct that the 1.75A I measure gets cut in half for “per emitter” current, right? - Yes.

Thanks. I'll try the (4) R120's and then measure again.

-Garry

Ok, I pulled the R100 and stacked (3) R120's (left other R120 in place). My resistors are supposed to be 0805 size from FastTech and they are 1/2 the size of the ones on this driver. Does that seem right? Is it ok?

Anyway, current on the LED wire is now:

Low-1.32A

Med-1.80A

High-1.96A

I'm wondering now if the LED leads are too thin. Holding the lead up to 24awg wire I'd say its close to the same size, maybe a size smaller (26Awg?).

-Garry

Or could I have burned up a resistor? Or do the new Low & Med currents show the resistors are all good?

-Garry

It may be that the thing has a maximum duty cycle it will not exceed. The new Med/Low currents do seem to indicate that the resistor mod is having an effect. Just not the end result you want for some reason. :frowning:

0805 sized sense resistors should be OK with 4 in parallel. They are quite small compared to 1206 sized ones, I prefer to use 1206 when I can. Assuming the 0805’s are good for 1/8-watt or greater, there should be no problem with dumping around 0.3w across 4 of them in parallel.

EDIT: no harm in replacing the leads.

So would you suggest adding another R120 just to see if High increases? Or just replace the leads? Do both?

I need to get some 1206 sized resistors! These are too darn small!

-Garry

By the way, I'm done working on it tonight.

-Garry

I recommend replacing the leads first. If that helps, great. If not, I feel that you have run into a limitation of the software that came preloaded on that driver’s MCU.

Copy/paste from my post in another thread: FT carries 3 values of sense resistor for 1206. Use the dropdown menus. http://www.fasttech.com/products/0/10007371/1615805-1206-0-12r-smd-precision-resistors-100-piece

For anything else you’ll want Mouser/Digikey/etc

Thanks!

-Garry

The modes compressing like that indicate that something else is limiting current. You may be able to just short the sensor bank and still get under 2 amps (probably in all modes). We know that FET is capable of very high currents with low resistance.

I'm having trouble reading the traces. It appears that the top left pin of the MCU in the picture below is the PWM output. It feeds into a 27 ohm resistor (may also have a 10k ohm resistor going to ground) on the way to the FET's Gate. You could try to bypass the MCU's PWM by using a paper clip to bridge the Vcc (bottom right Pin in the pic below) to the PWM Pin. This will be equivalent to a PWM of 255. If current increases significantly, then the MCU may be somehow governing the current to just under 2 amps.

Adding more resistors does seem to be bringing up the lower modes more than raising high. The two LEDs will pass whatever current their relative Vf’s allow. If they’re exactly the same then 1/2 total each. The difference in current measurements isn’t loss in the wires but the additional current used by the rest of the circuit. Voltage drops occur all through a circuit but the current in every part of a given path is constant as long as you account for any branch circuits.

I bridged the upper left MCU pin to the lower right MCU pin and current only increased slightly to 2.196A @ LED wire (still stock wiring). So what does this tell us?

Edit - 2 cells at 3.98v now.

Edit#2 - I get 1.88A prior to bridging. wiggled my connector from battery pack and got 2.24A after bridging now.

-Garry

Ledoman suggested I take voltage measurements both at the battery pack and at the PCB (where battery wire connections are soldered). Perhaps he's onto something! Here are my results:

Low @batt: 3.87v @PCB: 3.67v

Med @batt: 3.80v @PCB: 3.48v

High @batt: 3.76v @PCB: 3.37v

I'm now going to unsolder the power cord from the PCB and solder thicker wires in place and remeasure.

-Garry

Ok. If we assume there are also some losses in the circuit you might get voltage to the leds to low. As you probably know leds needs enough voltage to run at certain currents. In your case the voltage at leds might be below Vf (forward voltage). If this is the case there is no way to get desired current.

Also you need to know: more current means more resistance in the wires ie. higher voltage drop unless the wires are thick enough.

At DX forum discussing about some driver I've been advised there can be significant loss in the connector. It is much more powerful driver pulling up to 6A from the battery pack. At that current minimum 20AWG wire is needed. I've been advised to use Deans Ultra connector instead of tipical 5.5x2.1mm which can have resistance of 100 mOhm or more. Unfortunately Deans Ultra connectors are not waterproff.

So every bit can count.