1
\$\begingroup\$

Should I use an inverting buck-boost or a positive buck-boost topology?

Input voltage range = 19-22VDC, Output voltage range = 10-56VDC, Application = LED driver

The app not I have says the inverting topology fits applications with an input voltage range that overlaps the output voltage, and that the positive buck-boost supports applications where the input voltage range crosses the output voltage.

I feel like with those descriptions either topology would work. Although, the input voltage doesn't exceed the output voltage. Which one would be better?

Thanks!

\$\endgroup\$
  • \$\begingroup\$ If it's a proper buck-boost, it should be able to handle those ranges fine. What are your most common usage scenarios? Is the most likely/common input going to be 20V and the most common output going to be higher? The most common use scenarios are important to design for, for best efficiency. Even though the device CAN handle variations on the input/output, think about the most common requirements first. If the device spends 90% of it's time in a certain configuration, design for best efficiency in that mode. \$\endgroup\$ – KyranF Nov 21 '14 at 20:23
0
\$\begingroup\$

The appnote you are citing sounds as if it uses slightly different terms to describe the same working conditions for both inverting and positive buck-boost topologies.

A classic buck-boost (simple flyback toplogy) that uses a single winding inductor always inverts its voltage. To make it non-inverting, a second take-off winding is typically added, which can be wired the other way around to make the voltage come out the same as the input. With a two winding buck-boost (flyback) you can also completely isolate the secondary. You can also change the turns ratio to optimize for different input/output voltage ratios.

In your case, if you only want to drive LEDs, and they don't have to be referenced to GND in any particular way, there's no reason you can't use an inverting buck-boost, which, by only requiring a single-winding inductor, might be simpler.

However, as another answerer has pointed out, the ranges you give for your input and output voltages are quite wide. Without knowing more about the predominant operating conditions, it's impossible to give advice to help optimize efficiency.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the answer (it won't let me vote yet). I believe the inverting buck-boost is the way to go. The FAE I talked to also suggested the inverting, if not only because of it being simpler/fewer parts. The voltage ranges aren't really out of the ordinary if you look at LED drivers. In order to not have a lot of skus the wide range is needed to accommodate different LED loads. More than likely this will be used somewhere in the middle. That said I agree with the optimal efficiency usage remark. Thanks! \$\endgroup\$ – DigitalNinja Nov 24 '14 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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