I am wondering how to match a driver to an led, how is that determined?
The LED is a red XR-C, it has a rated max Vf of 2.2 and a max forward current of 750ma. I tried to swap it into a jacob a60 and it way overpowers the red led and dims it hardcore. Since it would be one 18650 wit would be 4.2v max input and whatever current output that is stock. Is it the voltage that is too high or the current?
The output is 4.2, if input is 4.2. How does this work……? I know, simple stuff but I am confused a bit.
If your driver is current regulated you don’t need to worry about battery voltage and led Vf just the output current of the driver. In your case a 2x7135 single mode driver (input voltage 2.7-6V) would suit.
Be careful with Linear drivers (like the AMC7135 based ones) with an LED that has a Vf that low. The driver chips will get tremendously hot, and likely fail.
If, for instance, the battery voltage sags to 4.0v under that 700mA load, then the driver will have to dissipate 1.8V. At 350mA per chip, that’s 630mW per chip.
When mounted on a two layer FR4 PCB, the temperature rise of a typical SOT-89 package is about 175C per watt. That means, 0.630 * 175 = 110C rise. The AMC7135 Chips are going to rise about 110 degrees C above ambient temperature (likely around 20). They’ll be running at 130C. That’s hot. Not quite at the maximum temperature for an AMC7135, but way too close for comfort.
Using 2 chips the current will always be 700mA so how about adding a 1W, .7ohm resistor to take some of the load from the chips? At 4.2V the resistor would absorb 500mV(700mA x .7ohm) of the excess voltage and the chips would take the rest and at 3.2V the voltage drop would still be above Vf of the led and Vdd of the chips.
interesting tip Rufus! I’m about to retrofit an parallel triple red XP-E to a light that currently has a single white XP-G (behind a red filter), so this is really neat. I thinking that it basically makes the LED appear as a 2.7V device, right? The energy is going to get burned off anyway, so this is just spreading it around?
If my light pulses between 350mA and 1.4A, I need to work out the resistor value at 1.4A I think, which would be 0.5V/1.4A=0.35ohms and 0.7W (I think). That would give a drop of ~1.1V, so <400mW per chip.
At .35A the drop would only be 0.12V and the drop would be higher, at ~1.7V, so ~600mW for that one chip. Hmm.
Doubling the resistance would drop that to ~200mW and ~550mW respectively, but I’d need to step up a power grade for the resistor (or put 2 1W resistors in parallel). What’s best?
awesome, thanks PilotPTK! The amount I don’t know about electronics never ceases to amaze me Interesting things those whatchamacallits, I’m off to do some more reading.
At least I now know what those odd looking things between B+ and the Vdd trace on simple single mode AMC7135 drivers are now Makes sense what they are too - they help drop the battery voltage to the Vdd voltage…
huh. So is there anywhere to add a diode on the board?
I have some diodes at work but dont know specs. They are plain jane black and silver banded diodes, so they all drop .7v? I will test that tomarrow when I have them in hand.
Of course, any voltage drop will hurt efficiency correct? SOOOO the white led, given all the same factors, would be more efficient than the red counterpart? ….
Most single and multi mode drivers already have the diode but it only drops the input voltage to the IC Or the 7135 Vdd pin and is not in the led-7135 circuit path. PilotPtk may have another suggestion but I would get one with axial leads and use it to replace one of the led wires and slip some insulation over the bare leads. It would need to be rated for 1W(2x actual load).
Any passive you add will prevent you from gettng full capacity from the battey. Th best option is a buck driver. There is a nice one at Fasttech tha taks 3 –8.4V in, 700mA out. Cant type let alone get a link on this thing. Tomorrow.
While essentially true, in this case the passive won’t change efficiency when compared to the raw AMC7135 driver by itself. During the discharge cycle, you’re just moving where the heat gets generated from the AMC7135s to the Diode and the 0.7V Drop won’t inhibit you from draining the batteries to their minimum safe voltage (2.75ish) by more that a tiny bit - There is so little energy left in a LiIon when it falls below 2.9V that it’s barely worth mentioning.
Most standard Si diodes drop about .65~~.75V. It’s inherent in their design. Schottky diodes can get much lower (.2~~.5V), which is good in a lot of circumstances, but you actually want the cheap crappy Si in this case so that it drops more voltage.
Because of the low Vf of the red led the battery will be pretty much depleted by the time the led dims from low input voltage even with the added resistance. What size is the driver in the A60?
Any voltage drop will affect run time but not necessarily efficiency as the drop needs to occur somewhere and the resistor merely outsources some of it. A buck driver may or may not be more efficient, it depends on the quality of the driver and some of them are crap and even the better ones are most efficient when Vb is closer to Vf. Do the one that looks easiest or do both and see which gives the longest run time. It’s up to you.
Sorry Matthemuppet, I slid right by your post. It’s much too late for me to check your math and I’d likely goof anyway but I think you have the idea. 7135 chips seem Taylor made for this because as the battery voltage dimishes the apparent resistance of the chips drops to maintain current output so they get more efficient as the battery depletes. For the longest run time we would want to minimize the value of any extra resistance and balance that against the need to protect the chips. The .7V diode drop seems a nice fit for this application. I did some playing around last summer with chips on a heat sink separate from the led heat sink and found the performance was much better since as hot as they get the chips don’t generate nearly as much heat as the led they supply current to. As with LEDs, the better the heat sink provided for the chips, the harder they can be driven. Better than just potting is some cu or Alu bonded to the chips directly. Better still is reflowing the ground tab to a lot more cu than is found on the typical FR4 driver board(think”XML on copper star” but for the chips instead).
If anything I say seems to conflict with PilotPtk, trust him over me. He knows this stuff, I just yak about it.
Your driver should be either voltage or current limited or both.
Usually driver like Nanjg 105C is current limited and of course it has some voltage operating range, but what it sets is the output current which is fixed.
That is a good point, I was getting my wires crossed. I was thinking the LED would begin to roll off at Vf=3.0V, not 2,2V’ish. In this case a standard 1N4007 diode will work well to offload the pain from the 7135’s.
I still think a buck regulator has a chance at being more efficient. It might be worth a try, so here’s the link to it here
Note: I just looked at the spec they list on the right; apparently it’s much more than a 0.7A driver. I doubt it’s suitable here. I have one and if I get a chance I’ll check it out. Reviews seem to confirm the results too.
the irony of being wrong in a post explaining how little I know is not lost on me
Just ordered a pack of those 1N5819 diodes, should have plenty of space, fingers crossed. Here’s the light that’s getting modded. Hoping to get quite a bit more light to improve my visibility in grotty conditions.
thanks to both of you for the help, I’m happy to be learning something new!
I’ve finally found the time to start (but not finish :laughing: my triple parallel red XP-E upgrade and the diodes I ordered work a treat. Without them the Vf of the LED is 1.8-2V (the driver oscillates between 350 and 1400mA), so the driver would be burning off ~1.7V at 1.4A. One diode dropped 0.4 to 0.45V, so I wired 2 in series to give a 0.8 to 0.9V drop (is this ok?), which takes the effective Vf of the LED to 2.6 - 2.9V, about the same as the XP-G R5 CW I took out. The diodes are JB welded to the LED/ driver holder, so they should be able to dissipate any heat generated fairly well.
As soon as I find some time to put it all back together, I’ll post up some pics. Thanks again for all the advice!
Interesting things those whatchamacallits, I’m off to do some more reading.
Makes sense what they are too - they help drop the battery voltage to the Vdd voltage…