Power 5630 LEDs without resistors using AMC7135 LED driver

Hi,

Have a question about using the AMC7135 LED driver and 5630 SMD LEDs.
I have a project were I want to power 18x 5630 LEDs (in parallel) using six AMC7135 LED drivers (also in parallel). This will give me 6x350mA = 2100mA from the AMC7135 drivers. The maximum current for the LEDs is 18x150mA = 2700mA.
I see in the datasheet for the AMC7135 that no resistor is added in series with the LED.

The circuit input voltage is 2,7v to 4,2v from two 18650 batteries (in parallel). From the 5630 LED datasheet, the 5630 LED is rated for 3,4v (max).

Can I safely power the LEDs witout using resistors, only using the AMC7135 drivers for limiting the current? Will limiting the current drop the voltage throught the LEDs?

(If I were to use resistors I would have to add 6.9ohm/0.4W resistors in series with each LED. These SMD resistors would take up a lot of space on the PCB design. Thats why I dont want them)

Thanks for all your help!

Best regards,
Hans

If one of the LEDs has a lower vF it will take more current than the others, causing it to fail prematurely if the difference is too much. That’s not a big issue for a triple or quad if the LED isn’t fragile but for 18 it could easily become one. Total current won’t exceed 2100mA but it doesn’t mean it will distributed evenly, voltage will be whatever is needed to get 2100mA in total (about 120mA per LED). That’s the theory.

I see. I can redesign the circuit so that 1x AMC7135 powers 3x 5630 LEDs, then to this six times so that I get a total of 18x 5630 LEDs.

What do you think about using AMC7135 drivers insted of resistors?

7135 is the same resistor (only smarter) current stable (always 350mA) and can be turned on and off. All your leds will share current, if some leds will fail other will share higher current .
with simple resistors you will not get stabilized output with voltage drop light will dimm

Okey, thanks.
So using the AMC7135 in series with the LEDs, So I dont need the resistor in series with the LED then? Im used to always have resistors in series with LEDs.
I will redesign the circuit so that I only have 3x LEDs for each AMC7135. That way I will lower the risk of LED failure

3x LEDs for each AMC7135… it depends on what is max current for 1 led. Dont group by 3, if 1 fail other 2 leds will share 350mA :laughing:, just leave it as is

One LED is 150mA max. 3x LEDs = 450mA max. Using 1x AMC7135 will safely power 3x LEDs with 350mA (lower than 450mA max).

The big question is if I need to put a resistor between the AMC7135 and LED (s)?

just connect bank of 6 or 7 amc7135s to bank of 18 parallel leds

I have no experience at all designing circuits, take what I say with a grain of salt. If you put a resistor between the 7135 and the LEDs it only needs to be big enough to help stabilize the current so if a LED wants to take 300mA instead of 120 at the average vF it will reduce it’s voltage a bit, let’s say 0.1v, so it won’t take that much current. You don’t need it to drop 4.2V to 3.4V, that’s the 7135’s job. How much resistance is needed would depend on the variation on the LED’s vF curves, sometimes just the circuit resistance is enough but I think the 5630 may be more fragile than the high power LEDs we are used to.

If only 3 LEDs are sharing each 7135 I think the problem should be reduced, beyond limiting the max current to 350mA, if a LED is ‘greedy’ it should take less current when the total is 350mA than if all the LEDs share the full current.

Edit: but take into account what Quadrupel said, if a LED fails for whatever reason the other LEDs will share the current.

Maybe just a small resistor per LED after the 7135? You don’t want your lamp to fail inside a cave after all.

Perfect!

So the AMC7135 can do the voltagedrop-job, nice! Thanks!

And thank you everyone for highlighting that a broke LED will result in higher current through the remaining LEDs

Take a look at Quadrupel’s edit, doesn’t recommend grouping by 3. Quadrapel has experience unlike me.

Good point.
And thanks @Quadrapel

If you put a resistor between the 7135 and the LEDs use the smallest resistance needed to make it reliable as it will just waste power when the battery is low.