KD linear 1400mA regulator does anyone have experience with those?

The driver:

http://www.kaidomain.com/ProductImages.aspx?ProductId=6653

What the heck do they want to say with that picture?

Also is it possible just by removing 1 amc to have 1050mA? Or there is some obsure bridging to do after a single amc removal?

Q1, Q2, etc are used to denote integrated circuits, you will see they are labelled on the board, so by connecting two points you enable the output of another regulator. So you don't even need to remove the chip, just don't enable one of the regulators.

look at the datasheet for the chip, it is basically telling you if you disconnect the power rail from the chip it won't function...

So if i leave it as is i get the 1400mA? If i fiddle with the various bridges i sistematically disable one or more AMC chips...

And the question that was mind boggling for quite some time. Since the common leds Vf is somewhere around 3,2-3,7V does that driver also buck the voltage to suitable level? I really doubt it forwards direct 4,2V from a 18650 minus expected losses?

Or better: What voltage does those drivers output along with the presumed 1400mA if Voltage in is greater by 0.1V versus the Vf of the led?

I'm not very into electronics design and such... have some basic understanding but thats all.

Looking at the pictures you may need to bridge those connections yourself, or if you order the 1400mA one they may do it for you.

LEDs are current mode devices, the voltage doesn't matter if you are using a constant current supply (as long as it is high enough).

I think it regulates the current only. My inexpert understanding is:

As the voltage supplied to the LED rises, the LED will take higher and higher current untill it burns out. On the cree datasheets this happens over quite a small voltage increase.

This driver stops the current rising above a certain level (one third of an amp per 7135-chip). A freshly charged 18650 may supply a forward voltage of ~3.9V, and if there was no driver a xm-l would draw over 3amps from this. Using this driver though it would be regulated to only drawing 1.4amps.

The loss with this type of driver is that any excess voltage supplied to the driver is converted into heat, but I think its still very efficient with li-ions which soon sag below the LED Vf and are effectively direct-driving the LED but not burning it out when freshly charged.

These are available as 1amp drivers already assembled. There are nanjg-brand varieties with multi-modes and single-mode multipacks that are less than $2 per driver.

But a 2.8amp one I bought from KD didn't work properly, and as I hadn't logged into buy it, returning it seems very difficult.

using a linear design in a flashlight does seem silly, I would have thought a buck regulator would have been better.

If you want constant brigthes almost to the end of cell thats the way. Buck/boost are not found anywhere.

Im aware there are also 1050mA drivers available. I need some 1400mA and want to be able to convert it to 1050 when i need one. Not gonna stockup with a zillion of drivers. Still have 2 Ak-47 which are useless if i want a single mode flashlight out of it, otherwise it's great!

Not buck/boost, just buck, this looks right http://www.kaidomain.com/ProductDetails.aspx?ProductId=9908 . Not 100% sure though, I know a lot less about flashlights than I know about electronics.

So essentially those linear regulators stress the common cree leds by means of voltage fed to them?

No, the forward voltage doesn't matter on a diode.

Linear regulators often beat buck regulators in terms of efficacy when driving white power LEDs of off one lithium ion cell. The common AMC7135 based linear regulators used in many flashlights needs a relatively small voltage overhead in order to stay in regulation (~0.13V), making them run in regulation longer than many step down converters on a single li-ion.

Budgeteer, you can customize the current the LED sees by adding or removing AMC chips. In your case you would need to remove one AMC IC from the board, it is fairly easy to do since AMCs are not all that small. Just heat the heat sink tab (on the opposite side to its three "legs") until the solder melts and pull it off.

Thanks to all. I got all the info i needed.

Could anyone point me to a link where you can buy AMC7135's in small quantities at sensible prices? UK/EU would be good :)

8 pc for $ 4.70 here.

40 pc. for $ 18.94 here.

If you plan on buying less than a thousand pieces, the cheapest sources are actually those driver boards. I tried to source some loose 7135s and the best quote I could get for 50 - 100 chips was around $0.60 a piece, plus shipping. It is really much cheaper to order some of the simple 7135-based boards from KD and harvest the chips.

Yup, best ordering boards from China. Farnell doesn't even sell the chips.

Ah, okay, didn't realise that.

Nor do RS or Maplin or Tait Components (Where I get my network hardware supplies)

Farnell and RS were the first places I looked, after not finding them I just sort of assumed they must be there but under a different name. I didn't even try Maplin on the amazingly unlikely chance that they may have had them, I'm sure they would have we about £12 each :P

It does seem a little odd that you can't buy a component that must be produced in vast numbers, I'll have to go the cannibalistic route then.