Resistor mod on LD-29 driver

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dct73
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Resistor mod on LD-29 driver

I’m sure it must have been discussed here but I’ve never seen it. Can the Ld-29 driver current be increased by a resistor mod? If so does anyone have a pic of which resistor?

Rod911
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Bumping up this thread because I want to know the answer too.

Considering getting a stand alone high amp (ie. 4-5A) is starting to get difficult to find, maybe it’s possible to mod existing 3A drivers to the 4-5A range?

If it is possible for higher amperage, would this driver fit into a Convoy L2 host?

HKJ
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Look near the black wire, there is a R025 resistor. Reducing that a bit will probably increase the current.

For detailed pictures of the pcb, check my review and click on the photos.

My website with reviews of many chargers and batteries (More than 1000): https://lygte-info.dk/

Rod911
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HKJ wrote:

Look near the black wire, there is a R025 resistor. Reducing that a bit will probably increase the current.

For detailed pictures of the pcb, check my review and click on the photos.

So, if I go buy a whole bunch of resistors , and, say, solder in a 100Ω resistor, how can I measure the difference it made (ie. what is the actual output of the driver to the LED)? Which resistor should I start using to get the desired 4-5A?

I have searched around the forum, but there doesn’t seem to be a definitive “how-to” guide on resistor modding here.

HKJ
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Rod911 wrote:
HKJ wrote:

Look near the black wire, there is a R025 resistor. Reducing that a bit will probably increase the current.

For detailed pictures of the pcb, check my review and click on the photos.

So, if I go buy a whole "bunch of resistors":http://www.fasttech.com/products/0/10000032/1007904-metal-film-resistor-... , and, say, solder in a 100Ω resistor, how can I measure the difference it made (ie. what is the actual output of the driver to the LED)? Which resistor should I start using to get the desired 4-5A? I have searched around the forum, but there doesn't seem to be a definitive "how-to" guide on resistor modding here.

The resistor is 0.025 ohm, this means a 100 ohm resistor will not do anything.

Placing another 0.025 ohm on top of the first would theoretically double the current, in praxis it would be less due to resistance in the solder and the pcb traces.

0.025 ohm is a very low resistance, 10 cm 24 AWG wire has about 0.009 ohm (Do not make a coil of thin wire, it will not work).

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Rod911
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HKJ wrote:

The resistor is 0.025 ohm, this means a 100 ohm resistor will not do anything.

Placing another 0.025 ohm on top of the first would theoretically double the current, in praxis it would be less due to resistance in the solder and the pcb traces.

0.025 ohm is a very low resistance, 10 cm 24 AWG wire has about 0.009 ohm (Do not make a coil of thin wire, it will not work).

So, doing something like this from this thread is how to do a resistor mod? Looks ok enough. Just don’t know the maths behind it though. I do have some 22 AWG wire from Int’l Outdoor coming my way, so I should be able to use around 5cm for each leg of the resistor to total 10cm, and, from the losses incurred by resistances throughout the light, I could potentially be seeing around 5A to the LED after all?

HKJ
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Rod911 wrote:
HKJ wrote:

The resistor is 0.025 ohm, this means a 100 ohm resistor will not do anything.

Placing another 0.025 ohm on top of the first would theoretically double the current, in praxis it would be less due to resistance in the solder and the pcb traces.

0.025 ohm is a very low resistance, 10 cm 24 AWG wire has about 0.009 ohm (Do not make a coil of thin wire, it will not work).

So, doing something like this from "this thread":http://budgetlightforum.com/node/18176 is how to do a resistor mod? Looks ok enough. Just don't know the maths behind it though. I do have some 22 AWG wire from Int'l Outdoor coming my way, so I should be able to use around 5cm for each leg of the resistor to total 10cm, and, from the losses incurred by resistances throughout the light, I could potentially be seeing around 5A to the LED after all? !http://64.19.142.12/i26.photobucket.com/albums/c135/BimmerR/driver3_zpsd...

That is not a very good mod, the legs on the resistor is way to long. It is best to solder another SMD resistor on top of the first one (But they can be hard to find in the small values required).

You will want to use resistors somewhere between 0.025 and 0.1 ohm. Calculations will probably not be correct, due to resistance in the solder and PCB.

 

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Rod911
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HKJ wrote:

That is not a very good mod, the legs on the resistor is way to long. It is best to solder another SMD resistor on top of the first one (But they can be hard to find in the small values required).

You will want to use resistors somewhere between 0.025 and 0.1 ohm. Calculations will probably not be correct, due to resistance in the solder and PCB.

 


For the SMD resistor, FT sells them here , however, no 0.025 ohm but 0.027 ohm is in that value pack. Am I correct in saying that, the closer to 0.1 ohm I get to (referring to the resistor used), the safer it is (ie. I won’t suddenly see > 5.5A going into the LED, but, rather, 3.1A from the default 3.0A, using general numbers)?
HKJ
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Rod911 wrote:
For the SMD resistor, FT sells them "here":http://www.fasttech.com/products/0/10000028/1005300-0805-62k-910k-smd-re... , however, no 0.025 ohm but 0.027 ohm is in that value pack.

The problem is not getting smd resistors, but getting values below 0.1 ohm or even below 1 ohm.

Rod911 wrote:
Am I correct in saying that, the closer to 0.1 ohm I get to (referring to the resistor used), the safer it is (ie. I won't suddenly see > 5.5A going into the LED, but, rather, 3.1A from the default 3.0A, using general numbers)?

Yes.

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Hi, see my posts here http://budgetlightforum.com/node/18870

 

HKJ is correct, the sensing resistor is 0.025 Ω. The ouput amps to the LED is 2.8A, so therefor the sensing voltage is V = I x R = 2.8 x 0.025 = 0.07V. The sensing voltage will stay constant, so if you want 5A output to the LED, your sensing resistor should be: R = V/I = 0.07/5 = 0.014 Ω. So your best bet would be to remove the 0.025 Ω and add a 0.014 Ω SMD resistor. Unfortunately, it is not always possible to get the exact resistor in small quantities. Thus we stack resistors to get to the desired overall resistance.

These are the resistors you want: http://www.fasttech.com/products/1234405

And then these are your options:

- 2 x 0.1Ω in parallel = 0.05Ω, put that in parallel to the original 0.025Ω = 0.01666Ω = 4.2A = ±1400 lumens
- 3 x 0.1Ω in parallel = 0.03333Ω, put that in parallel to the original 0.025Ω = 0.01428Ω = 4.9A = ±1600 lumens
- 4 x 0.1Ω in parallel = 0.02500Ω, put that in parallel to the original 0.025Ω = 0.01250Ω = 5.6A = ±1700 lumens

I've tested these results, 2 x 0.025Ω in parallel will give you 5.6A to the LED. How good that is for the driver and/or LED will depend on the heatsinking, potting, AWG wires used, and constant runtime of the light.

Hope all is clear Smile

Current Collection:

BLF: BLF-GT90, BLF-GT70 (CW Sliced), BLF GTmini, BLF-LT1; 

BTU: Shocker (3 x SST-40 @ 8A)

Solarforce: L2P (XM-L2 U3 @ 4A), MPP-1 (XP-L HI @ 6A), MPP-3 (3 x XM-L2 U2 @ 12A), M6 (Nichia 319A @ 6A), M8 (XHP-50.2 @ 9A), 9x (9 x XM-L2 U2 @ 2A)

Coming Soon: Lumintop: BLF-GT4; 

Rod911
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Thanks HKJ and Lothar. One final thing though. As I am a complete electronics noob, when Lothar mentioned to solder the 0.1Ω in parallel to the original 0.25Ω resistor, is this what you mean:

The black thing is the original resistor, and the red things are the 0.1Ω resistors soldered on top of the original resistor (grey blocks are solder). I am also looking at using 4 resistors as I do not expect it to be mathematically perfect at 5.6A, but anything from 4.8A to 5.5A should probably be doable in the Convoy L2 host, Noctigon pad and Arctic Silver 5.

I have ordered the LD-29 drivers from fancyflashlights.com instead of FT as the ones from FT look different to the ones HKJ reviewed and the starting modes are also different.

HKJ
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Yes, that is parallel.

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Lothar
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Rod911 wrote:
Thanks HKJ and Lothar. One final thing though. As I am a complete electronics noob, when Lothar mentioned to solder the 0.1Ω in parallel to the original 0.25Ω resistor, is this what you mean: !http://i41.tinypic.com/29n8qza.jpg! The black thing is the original resistor, and the red things are the 0.1Ω resistors soldered on top of the original resistor (grey blocks are solder).

100% correct. Good luck and let us know how it turns out Wink

Current Collection:

BLF: BLF-GT90, BLF-GT70 (CW Sliced), BLF GTmini, BLF-LT1; 

BTU: Shocker (3 x SST-40 @ 8A)

Solarforce: L2P (XM-L2 U3 @ 4A), MPP-1 (XP-L HI @ 6A), MPP-3 (3 x XM-L2 U2 @ 12A), M6 (Nichia 319A @ 6A), M8 (XHP-50.2 @ 9A), 9x (9 x XM-L2 U2 @ 2A)

Coming Soon: Lumintop: BLF-GT4; 

Rod911
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I intend to do a full write up of when I put the Convoy L2 thrower project together. All this is making me look forward to getting back home, along with messages from my sister that my packages are arriving back in Oz (including the Vinh modded BTU shocker)!!! Big Smile

dct73
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My order with the LD-29 drivers never made it out of Hong Kong. Sad I should have ordered the batteries separately.

Bluecvbc
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I have moded two of the ld-29 first one with 5% 0.02 ohm resistor and measured the current at 3.2Amp to the XML2 instead of theoretical 3.5 amp. The second one received a 1% 0.017 and delivered only 3.7 amp but no calculated 4.1Amp.

dct73
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Is that with one battery or two and what cells?

Bluecvbc
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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.

Bluecvbc
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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.

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

dct73
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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.

dct73
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Bluecvbc wrote:
I have moded two of the ld-29 first one with 5% 0.02 ohm resistor and measured the current at 3.2Amp to the XML2 instead of theoretical 3.5 amp. The second one received a 1% 0.017 and delivered only 3.7 amp but no calculated 4.1Amp.

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.

Bluecvbc
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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?

dct73
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That’s about 80% efficiency. Sounds pretty normal, but time to get the potting compound out.

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Lothar wrote:

HKJ is correct, the sensing resistor is 0.025 Ω. The ouput amps to the LED is 2.8A, so therefor the sensing voltage is V = I x R = 2.8 × 0.025 = 0.07V. The sensing voltage will stay constant, so if you want 5A output to the LED, your sensing resistor should be: R = V/I = 0.07/5 = 0.014 Ω. So your best bet would be to remove the 0.025 Ω and add a 0.014 Ω SMD resistor. Unfortunately, it is not always possible to get the exact resistor in small quantities. Thus we stack resistors to get to the desired overall resistance.


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?

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Bluecvbc wrote:
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?

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

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Barkuti wrote:
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?

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tekwyzrd wrote:
… 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 Cool

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

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

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Barkuti wrote:
tekwyzrd wrote:
… 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 8^)

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?

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