Resistor mod on LD-29 driver

That was with two 18650. Current draw from the batteries is about 2.2. Didn’t calculate efficiency coz I didn’t measure the wilt age across the led or the batteries. I can do it if you need it.

Ok did the measurements. Ld-29 modded with 1% 0.017 ohm resistor delivers 3.67 amps to an XPG2 at 3.63 volts that is mounted on a 20mm sinkpad for total of 13.36 watts while drawing 2.25amps from 2s 18650 with voltage under that load of 7.3 volts for total power draw of 16.425 watts. So the efficacy at that drive current is about 81% and loss about 3 watts in the driver. Not bad but could be better.

Thanks for trying. Thats a nice little bump in output, especially for those that already own this driver and are able to mod.

Thanks Blue. If my drivers ever come I’m going to try the same. They should have been here but I had batteries on the same order and it got sent back.

Just eyeballing that it looks like each one is 10% less than expected. Maybe I’ll calculate the resistor needed for 10% more current than I want and hope for the best.

Just tested the driver with two 0.025 resistors stacked for 0.0125 ohms and got about 4.8amps and slowly dropping down to 4.55 amps over about a minute. Driver got extremely hot, the inductor the most.

Edit: did some more measurements and calculations. Total power lost in driver at the 4.6 amp current is almost 4.5watts!!!
Draw from two 18650’s is 3Amps at 7.15 volts for total of 21.35 watts.
Led is 4.6Amps at 3,67 volts for 16.88 watts. Total loss in driver is 4.47 watts. Can you say hot driver?

That’s about 80% efficiency. Sounds pretty normal, but time to get the potting compound out.

I want to get 3.8A on LED.
As I understood, I should remove native resistor 0.025 and install 0.018.
(R = V/I = 0.07/3.8=0.018)

Is my calculations correct?

For buck and boost drivers, efficiency, usually, is more or less inversely proportional to input/output voltage delta (difference), so you can manage more power in the driver if you can keep this voltage delta minimized.
I’m currently planning on soldering 0’05Ω in parallel with the 0’025Ω nominal one, for a theoretical current delivery of 4’2A if we do not take into account battery contacts, cables, springs and other resistances into the equation.
Should deliver around 3’7A at 6’25V to an XHP50.

Cheers

I thought the ld-29 was intended for 3v emitters. How do you plan to get 6.25v?

No. The LD-29 is a buck driver, which means it is a current controlled switching power supply. It uses a logic IC which monitors voltage drop across a shunt resistor (0’07V in this case). That means:

  • If the measured Vdrop is above reference (0’07V), too much current is flowing; output voltage is adjusted proportionally downwards (theoretically multiplied by the quotient between reference voltage and the measured one at the shunt’s terminals).
  • If the measured Vdrop is below reference (0’07V), too little current flows; output voltage is adjusted proportionally upwards (again, theoretically multiplied by the quotient between reference voltage and the measured one at the shunt’s terminals).

Quintessentially: the ratio between the reference voltage and the measured one at the shunt’s terminals determines the figure you need to multiply the voltage output with, to reach the desired current flow ratio.

With 2 li-ions in series, you’re guaranteed to have at least equal or above led voltage output at the input, even for a 6ish voltage drop led.
Hope this helps.

Cheers :sunglasses:

Thanks Barkuti, that might be the best explanation Ive read describing the operation of a buck driver.

@Barkuti Thanks. This gives me another option for my Convoy L5.

But not all buck controllers will do higher than ~4v output, no matter the input voltage. The LD-29 used to have that limitation. Have recent versions been changed?

I suspect this limitation would be present in the older version used in the L5. Can anyone confirm?

Buck drivers also have overhead, meaning that a certain voltage above emitter Vf is needed to remain in regulation. Often around a volt so that 6V emitter might need 7V from the cells to maintain full output.

Mmm…

Just saw this:

1st version LD-29 (?) review at lygte-info.dk

Which differs from: LD-29 at FT

Newer LD-29 review at lygte-info.dk

Some of you may have one of those at hand; could you measure the no-load output voltage of this newer version while powering it up with 2S li-ion voltage range? Well, just feed it with 8 or 9V for the sake of simplicity. I'd like to make sure it can drive 2S2P XHPs before breaking the bank…

Cheers ^:)

@Barkuti 404 error for the lygte-info pages. Tried to do the links myself and got 404 again. For some reason extra text is being inserted breaking the link. Here’s the page with the “list of driver reviews”:Index of led driver tests

Seems the Simple Post Editor chokes with those %n code character substitutions. Re-edited the thing on the advanced one (pain in the arse as it is now, everything has to be reworked). :confounded:

Cheers

From what I see on the LD-29’s photographs, the shunt resistor size seems to be “1206” imperial units, or about 3’2×1’6mm according to the size tables I’ve found (should be 3’048×1’524mm).
May someone confirm?

If that is the case, I see it as a somewhat poor choice from a design standpoint because rated power for that footprint is ¼W (doesn’t it?) and that thing has to dissipate around 0’196W in a usually hot environment: no thermal derating headroom as I previously said.

Cheers fellas :crown: