0
\$\begingroup\$

Is it more efficient to drive LEDs at a high current and low voltage, than a lower current and high voltage?

I'm in a situation where I have to design an LED driver for a custom CoB emitter, so can choose the voltage and current as long as it's 22W overall. My input voltage is likely to range from 7-24V. I'll also need the LEDs to be dimmable in some way or another, likely PWM.

I'm not sure whether I should ask for the emitter panel to be made a lower voltage with higher current, or the other way around. It seems after a lot of looking that a reasonably priced simple boost LED driver is hard to come by, but I'm worried that using high current will cause power losses to be greater in the PCB traces and other components with significant ESRs.

So which way is best and why? And while I'm at it, any part recommendations for a cheap-ish driver?

\$\endgroup\$
2
  • \$\begingroup\$ Do you have a particular LED in mind? If so, can you post the part/datasheet so we can have more information? \$\endgroup\$ Commented Aug 12, 2015 at 16:15
  • \$\begingroup\$ Like I said, it's a custom made CoB from a chinese manufacturer. I don't have a datasheet for the particular LEDs on the CoB just yet, but might be able to get one. \$\endgroup\$
    – Sensors
    Commented Aug 12, 2015 at 19:46

1 Answer 1

2
\$\begingroup\$

If your input voltage is going to vary that much, you're going to have to start with a power conversion stage to generate a local stable power source. You have three choices - step down to less than 7 V (let's say 5 V), step up to more than 24 V (let's say 26 V) or a buck-boost type converter which aims for somewhere in the middle (let's say 12 V).

Given the extra complexity of the buck-boost, I would rule it out personally. You then have two different operational currents which you can calculate - 22 W / 5 V = 4.4A or 22 W / 26 V = 850 mA. Neither of these are horribly large given decent amounts of copper trace on a 2oz / sq. ft. board. However, your ohmic power losses in traces / PWM device channel will be just over 27 x higher with the low voltage version - but they may still be insignificant as a percentage of the total output power.

As you can see here, boost converters tend to be slightly less efficient than buck converters, so you're trading losses in the converter stage against losses in the traces / PWM control. Personally I would opt for the high voltage option because it places lower demands on my PWM control device, but overall efficiency is hard to call between the two, and will likely be down to the quality of design execution over the choice of path. If you have the chance, mock up both approaches on a board using a dummy load in place of the LEDs and work out which matches your needs best.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks for the answer, good to know my thought process is at least in the right direction. After much deliberation I've settled on this as my boost converter: farnell.com/datasheets/1882496.pdf So far I can't see any problem with that part that I had found with others. \$\endgroup\$
    – Sensors
    Commented Aug 13, 2015 at 9:51
  • \$\begingroup\$ Looks good to me, and not too costly either. \$\endgroup\$
    – stefandz
    Commented Aug 13, 2015 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.