Led4power.com : LD-4 CC linear drivers, ILC-0/1 illuminated tailcaps, optics, MOSX, copper DTP MCPCBs...

on the LD-x4 drivers what happens if you set the current higher than what the cell or LED will support? Does it just act like a normal FET? If so, is there some extra resistance somewhere making this a bad idea?

It behaves like DD driver on that mode (so called dropout mode of linear power supply). Nothing bad will happen, this is normal condition when battery is more or less empty and (LED Vf + parasitic voltage drops) is higher than battery voltage.

Neven, will we see E-switch support on your drivers? If so, when? If not, why?

Also, why no reverse polarity protection on LD-BVHV?

I'm working on new driver generation, it will take few months,so there will be e-sw support.

Regarding LD-B4HV, there was no space on pcb left for RVP.

Ok, thanks.

To fit a triple MCPCB with Luxeon V under a Carclo optic, does it have to be very modded (as in, as much as with XP-L HD leds)?
Thanks in advance!

I did not mod anything with Carclo optics. They fitted well, although it was a tight fit. The Luxeon V is very flood under Carclo optics. Not my cup of tea. I reverted back to Nichia’s in my Emisar D4. Will be trying out Luminus SST-20 soon… From what I have read they perform well under optics.

Thanks for your input on this! I’m willing to try a Luxeon V on 10511 optics and see what I get, as it has a narrower beam.
I’m not familiarized with the Luminus, but as I’ll order from L4P, maybe I’ll try to get one to see how it looks like!

I have tried all the possible Carclo optic configurations. You just won’t get throw out of this emitter in those specific optics. The physical emitting surface is just too big! It floods beyond the TIR part. You’re better off trying the Luxeon V2, that has a smaller physical emitting surface and should yield much better optical performance.

The Luxeon V performs well in a reflector. I have a Convoy S3 single and I put swapped them in on my BLF Q8. Now it really performs!

General question about the Luxeon MZ:

Looking to do a hotrod build around the LD-B4/MZ DTP using an S2+ host, would the MZ be alright driven up to 12A CC using the 8-12A driver?

I had planned on doing a floody MZ triple, but I’ve read the MZ will handle up to 15A before output drops so now the plan is to use a 45/60deg pebbled TIR and a single MZ emitter driven with the 12A B4, simplifying the build without much loss in output. Is this a sane plan? I imagine a triple would be marginally more efficient at higher output than the single emitter if it were also driven up to 12A.

I did here:

45 degree should look good w/ pebbled tir. It will pull 15Amps in direct drive and 12A will only be a problem for your hands, not the LED :slight_smile:

Oh perfect, that quashes any reservations I had about the emitter handling that current in a small host!

I guess I should also warn you. If you do not use a high current spring or bypass it with a wire you will likely have problems with it softening up to the point it loses connection/melts.

Yeah that’s what I had figured, cheers.

The above statement confuses me. I'm about to order an KW CSLPM1.TG on ∅20mm 3030 DTP MCPCB to be fed with a heavily modified LD-29 driver, does this apply to me?

Thanks.

Cheers ^:)

That note was about LEDs which I have, but not at all possible pcbs (mostly 3x,4x,6x mosX PCBs). For example I still don't have listed SST-20 and Luxeon V2 soldered on 32mm and 34mm mosX boards for ANGIE,CUTE and ANNA optics, but I can do that on request.

So this does not apply to your case.

Hello Neven, you once wrote that can be reduced by replacing resistors of the moon mode. Can you tell me what they are and what value they must have? I have a LD-A4. Many Thanks

FET-switches and 30A drivers…

Resistor in red circle is 600Ohm (R1), resistor in green circle is 300Ohm(R2). When you select 2.5mA moonlight current in UI only 600Ohm resistor is active, when you select 5mA only 300Ohm resistor is active, and when you select 7.5mA both resistors are active.

Moonlight current can be calculated as Im1= (Vbatt-Vled)/R1, Im2=(Vbatt-Vled)/R2, Im3=Im1+Im2.

many thanks. I first order some resistors.