Question about the NANJG driver and the MT-G2

Flashlight Modding and DIY Parts
1

I can understand, (almost), the concept of modding a NANJG driver with a zenier diode and resistor, to bleed excess voltage off the NANJG driver, when using 2xLi-ion batteries with an MT-G2 led.

My question is.... If you add chips to the driver, to raise the amperage, dosen't that mean that the voltage will have to increase? If you went to 6 amps versus 3 amps, (double the 8 chips to 16), wouldn't you have to have closer to 7 volts instead of 6 volts, to get the full 6 amps?. Or if you went up to 24 chips, (9 amps), wouldn't you be closer to 8V or at least 7.5V to get the desired 9 amps? If that be the case, then would the components take that voltage without killing them and wouldn't in mean a different diode and resistor?

Also, it would not make any difference if all the chips were stacked on one board, or if three boards, (master/slave), were to be used, correct? No matter what, even with master slave, the voltage would have to go up, to get the higher amps, so all the boards would still be seeing the same higher voltage?

Yes, No??

2

I would say YES. Voltage will increase to supply higher amperage to an MT-G2. In my own testing with a bench power supply I was seeing 7 volts ±.10 volts when driving an MT-G2 at 5 amps.

3

Hi O-L. That mod you speak of only drops voltage to the brain (MCU). The 7135 chips are still feed the full voltage of the cells.

4

I’m sure comfychair or someone else will be here with a better explanation soon but I find it easiest to look at the setup as two seperate circuits.

One circuit providing power to the MCU which needs to be kept below 6V by the zener and a second basically independent circuit with the 7135s and the LEDs.
The only thing connecting the two is the PWM signal coming from the MCU and connecting to the Vdd pin on the 7135s.

I look at 7135s are basically just smart variable resistors with an on-off switch, the MCU sends that on/off signal to them but otherwise they are entirely self contained and doing their thing.

Here’s a great answer by comfy that covers basically all of this.

5

Ok, I understand, but how many volts will the 7135 chips take before they fry and do you think it possible to reach 9 amps with either a NANJG driver with added chips, or with the three drivers, master/slave? I would think master/slave would be better, just because the heat from the chips would be displaced better on three boards.

EDIT: Then I would have to look at more than two Li-ions to get the voltage?, because two would only get me to about 8 volts max, more or less.

6

This is exactly the same thing I asked in this thread

And got this answer from Comfy.

This is really crucial to understanding how the 7135s work. They don’t care about the total voltage that is being applied across the LED to achieve a certain current. All they care about is dropping that voltage enough to maintain their maximum 350mah drive current. They do this by increasing their resistance, just like a normal variable resistor can be adjusted to drop the voltage on a circuit by a certain amount. If they are asked to drop or burn off too much voltage, say if you were driving a single XML with 2s input then they would simply burn up.

I’m not sure I’m using the exact right technical terms here, totally not an expert on any of this but that’s how I try to understand it myself.
Check out that thread, I’m sure it will help.

7

Yes the next problem that has very little to do with the 7135s is if the input voltage is actually enough to drive the LED at that current. The 7135s do have some inherent resistance even when they are in direct drive mode, I believe they will drop around 0.12v or something like that just by having them in the circuit so that needs to be taken into account.

My 7135 boards are stacked with 16x chips and deliver a solid 6Amps to an MT-G2 from an 8.4v source. I haven’t tried stacking any more but you will eventually end up just getting into the limits of direct driving. Going for 9A, voltage sag from the batteries is going to be your main issue I would think.

With regard to going for more cells, that’s not going to be possible with 7135s. Since they have to burn off the excess voltage to maintain regulation they work best when Input voltage and desired vF led voltage are as close as possible. Remember the bigger that voltage gap between maximum cell voltage and desired led voltage the more heat the little things need to dissipate. If it’s more than a couple of volts they will burn unless extra heatsinking is factored in.

I was looking at going for 3s LifePO4 cells that would get a nominal voltage of 9.9, might be something to consider if driving an MT-G2 at >9A. Then the difference between the vF and input would only be around 1.9v
Ultimately for my 6A driver current that option was still too much of a voltage gap for the 7135s to handle comfortably.

8

Yeah, I don't think you can get 9 amps with 7135's, because you can't get 9 amps with 2S cells and 3S cells is too much voltage differential for the 7135.

EDIT: If you could get 2S to deliver 9amps, then yes, you could use 7135's to regulate it. Provided you adequately heat sink them.

9

Oh it’s should certainly be possible to hit 9A if that MT-G2 graph is correct (hits 9A -@ 7.4vF ) only depends on how beefy your battery is to minimize sagging and for how long you can maintain that input voltage.
@~1v to burn off (max 8.4v input) the 7135s should be relatively happy too :slight_smile:

I wouldn’t think any round cells could handle that drive current while maintaining that voltage range for long though. Certainly not in a simple 2s configuration. But maybe 2s-2p 26650s? I haven’t played enough with good quality 26650s to know really. I’m spoiled with my Lipo battery pack though, those things down even sweat or sag at 20A.

10

The max current you can get with 7135s will depend on what a particular cell or set of cells will do when powering a particular LED, minus a little bit for losses through the 7135s. If 2 cells direct drive into a MTG2 will only do 5.5 amps, and then you test it with two drivers, one with 15 350mA 7135s, and another otherwise identical driver but with 18 7135s, the 18x one will not give any higher current than the one with 15x. 15x is enough to handle more than what the cells are capable of supplying, so adding more won't get you anything. You'd need stronger cells to see the gains of adding more chips.

The zener/resistor mod isn't affected by any of that. With a single MTG2 you can only use a max of two 4.2v cells because of the input-voltage-vs-forward-voltage balance that has to be kept in that 'safe' range of 2-4 volts. The MCU only has to be above a minimum of ~2.8v, which it always will be - you'd have to drain the cells to 1.4v each for it to be affected, and that would be A Bad Thing To Do. The zener/resistor only limits the voltage at the upper end, as input voltage falls below the zener's rating the zener no longer has any excess voltage to bleed off, so it's effectively taken out of the circuit. But draining the cells that low would be bad too, two cells at 4.6v is 2.3v each, also dangerously low.

But, the low voltage protection no longer functioning isn't really an issue, the light will dim dramatically to the point it's impossible not to notice it before the cells are down into the danger zone.

11

There’s two ways to blow up a 7135, overvoltage and overtemperature.

Overvoltage…

The AMC7135 is rated to 7V absolute max on both Vdd and OUT pins. Vdd gets fed with the Vcc voltage of the ATTiny, so provided the zener mod is properly done to keep the AVR’s Vcc at 5V or below, you’re good. You’ll probably blow up the AVR before you blow up the 7135.

Assuming everything else in the flashlight is zero resistance and the AMC7135 is off, the voltage on OUT is going to be the battery voltage minus the voltage drop of the LED at zero current. According to the MT-G2 datasheet (pages 7/8) the forward voltage drops at 0mA are ~5V and ~7.5V respectively.

Say you’ve got two cells (8.4V charged) and a 6V MT-G2, there’ll be 8.4-6 = 2.4V across the 7135 with the flashlight off. You’re OK. And if you’ve got three cells (12.6V) and a 9V MT-G2, there’ll be 12.6-7.5 = 5.1V across the 7135. You’re still OK.

But if you use 12.6V of batteries and a 6V MT-G2, there’ll be 12.6-5 = 7.6V across the 7135. Now what will 7.6V do? Probably nothing, but as you keep increasing the voltage on the OUT pin of the device the pin will start sinking current, increasing with increasing voltage across the 7135. This voltage/current combo will start to heat up the device.

If the 7135 has a FET output, this would be fine as long as the device doesn’t end up overheating. But I’ve got my suspicions the 7135 has a NPN transistor output, which means avalanche breakdown can happen, and that may not require much current (potentially just a few mA). Paralleling 7135s won’t save you here as different parts can have different breakdown voltages meaning breakdown current won’t be shared.

So 12.6V of batteries and a 6V MT-G2 might work - really it depends on the point when the 7135 breaks down. Personally I wouldn’t do it.

Overtemperature…

When the 7135s are on, they’ll dissipate heat. Again assuming zero resistance for the rest of the flashlight, the power dissipated in the 7135s will be LED current * (battery voltage - LED forward voltage at current).

With a 8.4V battery voltage and a 6V MT-G2 at 3A, the MT-G2 forward voltage curve says 6.2V. This gives (8.4-6.2)*3 = 6.6 watts to be dissipated in the 7135s. With a 12.8V battery voltage and a 9V MT-G2 at 2A (forward voltage 9.3V) you’re dissipating (12.8-9.3)*2 = 7 watts. But with a 12.8V battery voltage and a 6V MT-G2 at 3A (6.2V), you’re dissipating (12.8-6.2) = 19.8 watts!

How much power the driver can dissipate is proportional to the # of 7135s and how well the driver is heatsinked, but dissipating 19.8 watts is going to be a heck of a lot harder than 6.6 watts.

Overall…

If your driver fails, chances are the 7135s will fail short-circuit, turning your driver into a direct-drive driver. This will cause a very bright LED if you’re lucky, an open-circuit blown LED if you’re half lucky, or a shorted LED if you’re unlucky. The last situation shorts out your cells, in which case I hope you use protected cells.

12

2 Samsung 20R’s will supply more than 9A. I have 2 in my Solarforce M8 with an MT-G2 running direct drive and it pulls 11.88A from the 2 Samsungs.

Knock yourself out Justin. Richard (RMM) has these pretty cheap and you can have em in a couple of days. I paid $20 for 3 of em shipped.

13

Well, I guess I will guinea pig it then. I have an MT-G2 on a noctigon and a 3' piece of 1-3/4" Aluminum round bar stock, with plenty of 14 ga wire. The 20R batteries might come by the week-end, (wishful thinking, since it's now Thursday), so when they get here, I will see what I can do with the 3 drivers in master/slave. I should have over 9 amps from them.

The light I want to do this to is this one, which should get here in a week or so... https://budgetlightforum.com/t/-/20312 I want to do reflector tests on the light, to see if I need to drop it down to die height or below die height, to get the tightest spot from the led.

I was figuring on 2s/2p with it and the Panasonics, but I will have four of the 20Rs, so there should be no excuse about the batteries not hitting 9 amps.

14

Looking forward to it :slight_smile:

15

20Rs will barely even chuckle at a 4.5A load. If you haven't used those cells before, you're in for a shock (hopefully not literally)!