I just added 3 7135 chips to my 105C on my Convoy C8. It should have been a 2.8A driver and I was hoping to push it closer to 4. I had taken a reading at the tailcap before modding and could only muster up 1.9A. After adding 3 chips I was able to pull 2.1.
I was reading through the forums and found that replacing the leads on my DMM should give a more accurate reading. So, I used a couple of short pieces of copper wire used for a subwoofer ground instead. Now I was able to pull 2.6A. Still low though, and I honestly didn’t think it looked any brighter after I added the chips.
I have some braided copper wire coming in the next week so that I can mod the springs. Could this be the cause of the low reading? That would mean the spring on the driver is the problem. Seems odd that they would make a driver advertised at such-and-such amps, but then limit this with the spring. And, yes, I do know this stuff comes from China, so… :bigsmile:
What kind of cell do you have? Some cheaper cells just can't push that much current. I had an Ultrafire that wasn't able to get 2A through a direct drive XM-L.
I might add that you mention that it didn’t look ant brighter either. It may not. Brightness is hard to discern unless there is a side by side comparison. That is unless of course it’s 10 times brighter.
BTW, this might seem too obvious, but if you want to know what twice as bright looks like, just shine 2 lights on the same spot. I know it sounds silly, but it is an easy way to know
Thanks for the links and suggestions. At this point I won’t sweat it and just wait until my braided wire gets delivered. I was testing using Panasonic and LG cells…both high draining.
I don’t think anyone addressed this specifically: The driver and spring are certainly 100% capable of 2.8 amps. The problem is that your LED has a high Vf needed for that current (out of the driver’s control!) and your battery is not providing a voltage significantly higher than the Vf needed under load. This is a common problem and has nothing to do with the driver. Resistance mods reduce losses to try and make up for the problem. Reducing losses is always a good thing, but you could also attack the problem by choosing an emitter with a lower Vf (this will be difficult for you to do) or a battery with a higher voltage under load (this is easy to do!).
Yeah... I have a light with a first-gen XM-L emitter and it has never been able to pull the full current. I've never verified, but I think it's just an unusually high Vf holding me back. That light has a bad reflector though so I've never addressed it any further.
I understand that they now have some pills that can help with electrile disfunction. Side effects seem to include a compulsion to throw footballs through tire swings and a compulsion to take baths in public places.
I need to revisit this thread again. I’ve read the posts and recommended links as well. I decided to test a few different batteries. I tried a Panasonic NCR, LG ICR, a green Sony laptop pull, and an UltraFire cell that came with a light a couple of year ago and I have never used it.
The LG cell is brand new and I got it specifically for this, but it actually performed towards the bottom. I can’t believe this, but the Ultracrap cell registered better than the Panasonic by about .2A.
I have 2 Panasonic NCR protected cells behind it and can only muster 1.7A at the tail. So, I would think there is enough voltage in this case to push the full 3A. Also, to fit the protected cells I’m not running a spring on the driver, so shouldn’t be any resistance there. Only thing I can think of at this point would be the stock wires on the driver going to the emitter. Thanks.
That is 1.7A at 7.2v, test the Amps at the led and you will find that it is getting around 3A.
With 2 cells in series double the tail Amps and take a bit off for the driver, 3 cells or more are the same, multiply by the number and allow for the driver is as close as you will get without measuring at the led.
1.7 amps at the tail cap running 2 cells is 3+ amps to the emmiter.Your running twice the voltage, therefore you only need half the current to get the wattage output to the diode.
