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I have a 3 high powered LEDs hooked up into a strip to for the load. For simplicity, I am going with a constant voltage driver. The load can be hooked up in 2 configurations:

Config 1:
3 Green LEDs (3.3Vf, 4A) in parallel => 12A and 3.3V load

Config 2:
3 Green LEDs (3.3Vf, 4A) in series => 4A, 13.2V load

I'm using a 7.4 50C (2S) battery, with whatever capacity required. So the two driving approaches would be:

  1. 2S wired to Buck converter for Config 1
  2. 2S wired to Boost converter for Config 2

The load wattage is the same in either case, but the difference is in input-output voltage and the current out of the battery (these high discharge lipo's can handle those currents) is different.

I used the power stage designer tool from TI. See below for the Boost driving approach:

enter image description here

See below for the Buck driving approach:

enter image description here

Honestly, both seem like usable approaches, there's also a HUGE number of dc/dc IC's out there - synchronous and non-synchronous. What is the criteria to consider an approach to go with when I'm looking at simplicity and easier heat management? Also my target run time is 1 min.

I understand a few things:

  • a buck converter is more efficient
  • 12As will require more copper and thicker traces for heat management, thereby potentially requiring more board estate
  • the parallel load circuit is safer for working even with a short

EDIT: Added buck with synchronous switch below: enter image description here

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  • \$\begingroup\$ If you're worried about efficiency, why are you using a diode instead of a synchronous switch? || A buck converter is not inherently more efficient than a boost converter. There are different considerations to make in your choices of components, but you can achieve better than 90% efficiency with either. \$\endgroup\$
    – Hearth
    Commented Mar 28 at 21:04
  • \$\begingroup\$ @Hearth That was my bad. I understand the case about efficiency. Now how about with dealing with high current vs high voltage load - in terms of PCB design what are some things to look at? Also what are the different considerations that you are referring to? \$\endgroup\$
    – roaibrain
    Commented Mar 28 at 21:12
  • \$\begingroup\$ @Hearth a question not specified, how do I calculate the battery capacity required for a 1min runtime for either of the configs? \$\endgroup\$
    – roaibrain
    Commented Mar 28 at 21:20
  • \$\begingroup\$ A SEPIC converter can operate as a buck-boost converter, seamlessly transitioning between the two modes. Constant current drive is preferable when driving LEDs. Constant voltage requires ballast resistors which wastes power. \$\endgroup\$
    – qrk
    Commented Mar 28 at 23:31

3 Answers 3

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I would go with the boost current source:

  • Better efficiency (no current-limiting resistor, because the convertor is the resistor)
  • better stability performance (current stays constant even when LED voltage drop changes with age and temperature)
  • Safer (the convertor is inherently current-limited by design even into a short)
  • Solution footprint should be smaller (I'm not taking this claim to the grave).
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First of all, you should drive the LEDs with constant current, not constant voltage.

Simplicity/complexity: both options are equally complex, except the sync rectifier is slightly more.

Efficiency: boost config has higher input current. This will put higher "stress" on the batteries and may therefore reduce their lifetime. If you don't go for a synchronous boost then the efficiency will be even lower.

Safety: this wasn't in your question but worth considering. Take the possible failures such as LEDs (fail open or fail short) and MOSFETs, necessary protections (especially for the batteries) and consequences (cost, time etc) into account.

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  • \$\begingroup\$ Yea, I know constant current is the go to. But that additional complexity doesn't seem to be a requirement for me. As I don't care about uniform luminosity. I understand that I need to have some fail safe in place as the LEDs will start to draw increasing current as it heats up. There will be a lot of heat management in place. In such a scenario is constant current really needed? Am I right that CC is more complex than CV? Also, thanks for the safety tips, those are excellent considerations to look into. \$\endgroup\$
    – roaibrain
    Commented Mar 28 at 21:43
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    \$\begingroup\$ @roaibrain A CV-driven LED requires a current limiting resistor to set the drive current. This will bring an extra loss. A bigger problem is, as the LED heats up its voltage drop will decrease, therefore the voltage drop across the limiting resistor will increase which results in higher drive current, therefore further temperature rise of the LED. Still worth considering despite the heat management. With CC there's no such problem. As for complexity, CV and CC are equally complex but CC is more efficient since the required current set/sense resistance will be much lower. \$\endgroup\$ Commented Mar 28 at 21:51
  • \$\begingroup\$ @roaibrain Good rule of thumb is that for constant voltage you deliver 1/4 of your power to the resistors, 3/4 to the LEDs. Are you ok with wasting that much power? Complexity is similar either way. \$\endgroup\$ Commented Mar 28 at 22:29
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It's simpler to connect the LEDs in series that way you only need a single boost converter, insteasd of needing three buck converters. If you have not done it now you should read the datasheet for you LEDs, especially the part about connecting high power LEDs in parallel.

Synchronous converters are more efficient and at 4A it should be easy to find a boost chip that includes this feature built in.

LEDs are more efficient at constant current so don't skimp on output filter capacitors.

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  • \$\begingroup\$ What would go wrong with one buck connected to all three? It should work with the LEDs drawing whatever they need in current right? \$\endgroup\$
    – roaibrain
    Commented Mar 29 at 8:07
  • \$\begingroup\$ Why would an LED do that? \$\endgroup\$ Commented Mar 31 at 12:34

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