I’ve modded an Solarforce K3 with MT-G2 by changing the driver to an 105C style with stacked chips. Taking the 3.04A driver to 4.40A by adding 4 7135 chips. I’m now making changes to the heat sinking ability of this light and am wondering just how far this can be taken? If I stack 4 chips on the other side of the board for a total of 16 chips, is that detrimental? What should I know about stacking chips?
Of course the idea of an MT-G2 running at ~6A in the K3 is intriquing, but is it valid?
Please help me understand the process, the pro’s and con’s of stacking these 7135 regulator chips.
More has a plus side in this case as the 105c driver has been modified to take 8.4V from 2 cells. By adding more regulator chips, there’s more to share the burden and they’re scrubbing less per chip, so perhaps less heat is being generated which would mean less power being scavenged by the chips themselves….if I’m thinking correctly.
Also, the Vf of the emitter goes up as it’s power consumption rises, so again there’d be less for the chips to scrub.
But then again, the MT-G2 at 6A will be making enough heat to go around, and then some!
I know by going direct drive, you can get more amps out of the same cell than with 4.5A worth of 7135's - each one has a voltage loss, so it's been said here. Maybe that's a good thing for driving a MT-G2 on 2 cells, I'm assuming...
Like any other setup the current it'll flow in direct drive depends on the cells and how much the voltage sags. With my plain 18650s that's only a little over 5 amps.
Try your cells in DD, see what it does. That will let you know at what point adding more chips won't gain anything extra. You won't hurt the LED, you'll melt the wires off before you kill an MTG2 I think. I ran one at 9 amps and it just laughed at me.
Each chip drops the full excess voltage so as the number of chips rises the Vf does as well reducing the excess voltage. The math doesn’t really work out in favor of the light unless it can handle a HUGE amount of heat. Try two examples : 12 chips at 4.5A and 16 chips at 6A. Using Djozz’s Vf data for the MTG-2 at 4.5A Vf is 5.2V and at 6A it’s 6.2V. If the cells can sustain 7.8V at 4.5A then there will be 2.6V excess or 1W per chip in wasted heat and 12W total in the driver. If the cells drop to 7.4V at 6A then there’s 1.2V excess or.5W per chip or 8W total. Using Djozz’s excellent work and my purely hypothetical bs and completely ignoring the fact that the led will generate much more heat at higher currents one might assume its better to overdrive the led than under drive. I can’t say how useless this excercise was but it sure was fun. A real test will need to include Vb, Ve, and Vd to see where the wattage goes.
Will the emitter take direct drive from 2 cells at 8.4V? I have cells that will drive 20A, don’t think we want those in direct drive, do we?
Edit: I put a Samsung INR18650 20R in my Maglight with springs bypassed and running an Elektrolumens Triple XM-L2 U2 in DD and got a very quick 10.1A before I pulled the plug to protect my 10A DMM.
The current capacity of the cell has nothing to do with the current it will run in direct drive, the voltage determines how much current will flow. More voltage = more current. More voltage sag under load = less current.
If you connect it to a DC power supply the size of the sun and capable of thirty trillion amps, but at only 6.3 volts, it will not shove any more than 3.3 amps through the emitter.
+1, with all of the information and misinformation, it can get confusing. I do not try to get into it much, as I am totally misinformed and have decided I like it that way. At least I have an excuse since I know nothing....
The answer is to have someone develop a "cheap" driver for the dammed MT-G2. Either that or just foot the bucks for a TaskLed h6flex and put 6600mA to it done right.
Any buck driver given an input voltage higher than the LED's voltage is 'a driver for the MTG2'. This is only being done because such things are rare as rocking horse sh1t in the 17mm size, if you're not limited to 17mm there are tons and tons of decent choices out there.
How many is too many has been answered - too many is when adding more doesn't increase the current over what it runs at in direct drive. And in direct drive from those cells it runs at...? Did I miss that part?
But you did redirect the question without answering it.
I do see the logic in what you present. But am confused as to why it’s ok to hook a 6V emitter up to 8.4V of power supply. You’re making me think, and I’m remembering that hooking a 3V emitter to a 4.2V power supply is almost exactly the same deal, and what happens is that the 3V emitter is overdriven, sometimes to the point of blink and we get to start over. It would seem then, that for me, the issue is that I do not understand the relationship between power supply (V) and current (A) as it pertains to the finite (Vf). In my own humble experience something in there usually produces smoke that is unpleasant to the nose and makes my wife fuss at me. Which is why I ask questions in area’s that I know I’m not well versed.
Looks like I might have to seek my answer elsewhere.
