Help Finalizing LED Boost converter

[modern_messiah]

Hi all,

I have designed (tried to design maybe?) a 500mA LED boost driver based around the LM3410X LED driver. The completed schematic for my driver is below. Apologies if my nomenclature is off...

The control circuit half of the driver is based around an ATMEL ATTINY 13V micro-controller and incorporates some external parts for battery voltage detection, reverse polarity protection and thermal protection. The PWM input of the LM3410 chip (SOT-23) is driven by pin 7 of the ATMEL chip. I know the control half of the circuit works as it is currently humming along nicely in a simple linear regulator driver I built recently.

Anyway, for the boost converter half I am relatively confident it's correct as it is more or less based around the circuit depicted in design example 4 of the LM3410 datasheet (page 28). The only difference is I have boosted my output current from 340mA to 500mA.

A few questions and notes:

1) I can't for the life of me figure out why they have R1 and R3 in parallel in design example 4. Why not a single resistor as per pretty much every other design example in the datasheet? I went with a single resistor and set it to provide me with my desired LED string current.

2) I've listed the Schottky diode as needing to be greater than 500mA and 10V - this is because I have not started looking for a suitable component yet.

3) I need help picking a suitable inductor. The maths isn't beyond me but my fundamental understanding of inductors is lacking, and I want to make sure I pick a suitable part for the job. Footprint size is an issue so if I can get some parameters to look for (and obviously an explanation as to how these values were arrived at) I can go hunting for the part myself.

That's about all of got. If anyone notices anything wrong with the circuit, or even ways to improve it, I would be very appreciative.

Thanks,

- Matt

 

[Harald Kapp]

I'm not an expert at switch mode supplies. Nevertheless I'll give it a try:

1) Probably to arrive at 600mOhm. They might have used a 604mOhm resistor (E96 series), but that may be harder to come by than a 1 Ohm + 1.5Ohm resistor. I can see no technical reason.

2) So what's the question? Depending on the operating volatge of your LED(string) you may have to use a higher breakdown voltage than 10V. When the switching transistor in the LM3410 is closed, the anode of the Schottky diode is at GND potential, the cathode is at the LED's operating volatge, buffered by the output capacitor. This is the minimum reverse voltage for the diode. Add sufficient margin to ensure the diode doesn't break down under different conditions (temperature, load variation etc.)

3) Have you read the chapter "inductor selection" (page 11 ff.)? Since 500mA isn't that far from the designed 340mA, you could start by using the inductor from the design example. Examine the current through the inductor under light load and full load. If you signs of saturation (signal is no longer triangular but capped or inductor becomes hot) use an inductor with the same inductance but higher current rating.

 

[modern_messiah]

I'm not an expert at switch mode supplies. Nevertheless I'll give it a try:

More of an expert than me

1) Probably to arrive at 600mOhm. They might have used a 604mOhm resistor (E96 series), but that may be harder to come by than a 1 Ohm + 1.5Ohm resistor. I can see no technical reason.

Someone else on a forum elsewhere has suggested it's due to power dissipation, but it's possibly a combination of both? 0.6ohm does give you a 320mA current (based on a 190mV feedback voltage). That being said the power dissipated through the resistor is only 0.07W, which a single resistor can easily handle.

2) So what's the question? Depending on the operating volatge of your LED(string) you may have to use a higher breakdown voltage than 10V. When the switching transistor in the LM3410 is closed, the anode of the Schottky diode is at GND potential, the cathode is at the LED's operating volatge, buffered by the output capacitor. This is the minimum reverse voltage for the diode. Add sufficient margin to ensure the diode doesn't break down under different conditions (temperature, load variation etc.)

I intend to use Nichia 219 emitters. At 500mA their forward voltage is a touch over 3.2V. 3 in series means a maximum string voltage of 9.6V (let's say 9.7V to be safe). So long as I rate my diode above 500mA and 10V I should be OK...sorry, that wasn;t really a question at all was it!?

3) Have you read the chapter "inductor selection" (page 11 ff.)? Since 500mA isn't that far from the designed 340mA, you could start by using the inductor from the design example. Examine the current through the inductor under light load and full load. If you signs of saturation (signal is no longer triangular but capped or inductor becomes hot) use an inductor with the same inductance but higher current rating.

I have read it but I need to sit down and try work through the maths. I don't have the equipment to practically test the inductor, so I'm trying to nail down a suitable part straight off the bat. If I just rate the inductor above the one used in the design example I should be OK. A bit dodgy I know...it will probbaly come back to haunt me.

 

[KrisBlueNZ]

All good advice from Harald here.
I would just add that I'd use a Schottky diode rated for at least 20V to be safe, and connect a zener of say 18V across the LED string, because the LED string, being off-board, could get disconnected or broken, and that would expose the diode to a very high voltage.
Ah, I see there's a diagram further down on page 18 of the data sheet that shows exactly how that should be done.

 

[modern_messiah]

Thanks Kris, I'll take that advice I think!

 

[modern_messiah]

Actually I have one more question - and this is more of a thought experiment than anything else....just a "what if".

What if you tied the DMM pin to VIN (so the LM3410 IC is always enabled), and connected a switch with 2 different resistors in place of the resistor on the FB pin (380 mill-ohm)?

Set the 2 resistors so that one provides the desired maximum current of 500mA, and the other provides a lower current, so for example 40mA.

Press the switch to alternate between the 2, the feedback voltage on the FB pin changes, and a different current flows through the LED strong. High and low without the need for a micro-controller.

Would this work? I'm reading through the datasheet now to try and see if it mentions anything like this but so far nothing.

Thanks,

- Matt

 

[KrisBlueNZ]

The boost converter inductor value is chosen based on the minimum and maximum input voltages, output voltage, and output current. If you change the output current significantly, the converter may not be able to regulate with the chosen inductor value. This is why brightness control is done via PWM, not in the way you suggest.

 

[modern_messiah]

OK - so here's an equally stupid final question. Selecting the inductor value based on the larger of the two desired string currents won't necessarily 'cover' lower string current? I'm assuming it doesn't work that way...

 

[KrisBlueNZ]

I'll pass on that question. Can anyone else answer it?