Has anyone here played with by-passing the polarity protection diode on the Nanjg 105c? Or any amc715 driver for that matter.
The below link say that silicon diodes consume about .6v. So eliminating it would increase regulated run time. It mentions that it can cause instability issues sometimes, but doesn’t provide any details.:
I bypassed it on one of my drivers. Can't remember why, but it works just fine.
interesting…
The diode is only in the MCU current path. the LEDs are directly connected to Battery +. Their current does not pass through the diode, so there is no loss there.
There is no benefit to bypassing the diode.
Also, you will mess up you low battery warning, making it kick in at a lower voltage.
It doesn’t drop the voltage to the LEDs because it is not in the led circuit path but only in the path through the IC and Vdd pins of the 7135 chips. Led + bypasses the diode. It might allow the IC and chips to stay fully on a bit longer at the expense of safety margin. You won’t know your cells are dead as soon. This is not the end of the world but does require even more awareness than usual to stop using the cell immediately when it dims.
Thank you relic38 and Rufusbduck. I’ve been trying to find a circuit diagram to figure out what the diode was doing. You two sure helped.
Rufusbducky, I was just reading this thread you started:
That was some amazing work you did there. It’s still over my head at this point, but I will study and try to learn from it.
I have done this when using NiMH batteries, to get that little bit more out of them. I think it was originally started because of NiMH use (3xNiMH).
Circuit diagram has been delivered 
Here
Talk about a clear, easy to follow diagram. A picture is worth a thousand words. Thank you very much relic38.
Thanks, I seem to have a penchant for getting in way too deep but I’m pretty sure that whole thread was trivial to an EE.
If you start with the assumption that it isn’t really complicated but more like tinkertoys it might come easier. I don’t know squat about circuit design but this is more like plugging stereo components together but simpler. Just look at the pictures and try to get familiar with the names of each part and don’t worry about how the chips do their thing only what they do.
When the 7135 is turned on, 350mA flows through it from led- to ground.
The Attiny 13A IC (8 legged chip) turns the 7135 on and off rapidly(this is called PWM) and the percentage of “on” time determines whether we see 100, or 5, or 30%, or the different blinky modes.
The Vdd pin on the 7135 is the switch that the IC turns on and off.
Removing/Bypassing the diode is completely pointless, unfortunately. It eliminates the protection for the MCU and doesn’t do anything productive. The ATTINY13A can operate down to 1.8V, so even with a 0.6V diode drop, it can operate down to 2.4V (and actually, a bit lower - 1.8 is the guaranteed minimum). If the cells are below 2.4V, they won’t be lighting up the LED (white) anyway.
PPtk
In O-L’s case, it would give more run time because it lowers the effective low battery cutoff point. I do not know how much light you can get below 2.8V (normal cutoff) but if you want to do that bypassing the diode will do it.
A safer way to eliminate low battery protection would be to remove the 4.7k Ohm resistor. It will never register low battery after that. I would not recommend this for anything other than non-Li-Ion use.
CRAP! And I just soldered in the 105c for an alkaline-only D-cell Mag, just my luck. I don't have one in front of me to check, which one is the 4.7K? I know the two resistors are right beside each other, is it the inboard or outboard one?
edit: Duh, the one with '4701' on it, of course. Right?
Wish I knew the answer to your question. My knowledge in the area of electronics is pretty weak. There is a clear pic of the resistors you are asking about on Post #45 of the following page:
Does anyone know which of the resisters on pads R1 and R3 is the 4.7k Ohm resistor?
I don't have one currently not installed to look at but had this pic I took the other day. Resistor markings match what's in relic38's napkin diagram, so it's R1 that I need to get rid of.
The one that says 4701, I think.
What about the control pin of the 7135’s? Will they turn on completely at less than 2.8V? That would be the only reason for doing this.
IIRC, you need to get rid of the 4701 resistor. If you get rid of the wrong resistor, you will get constant low battery.
So, when you take it off, keep it handy just in case you need to put it back on
Edit: yup, just looked at that bare board again, it’s the 4701 that goes away to disable low batt detection.