0
\$\begingroup\$

I've searched the web and didn't find an answer that suitably answers my query. I have a dc-dc buck / step-down converter. Input range: 72-144 V DC. Output: 5 V 60 A. My question: Is there a difference in output wattage when I use, say, 72 V as opposed to 144 V input? Will I get the full 5 V 60 A (or close, considering efficiency %, etc) when I use a lower recommended input voltage instead of a higher voltage (or vice versa)?

During my research I came across various comments/statements that the input voltage must be as close to the intended output voltage as possible, whether one is boosting/stepping up or bucking/stepping down. Maybe I'm totally confused about what they were meaning. Or I'm misinterpreting it.

\$\endgroup\$

2 Answers 2

3
\$\begingroup\$

For standard buck and boost converters, an input voltage close to the output voltage usually results in higher efficiency. That means a low input voltage for a buck converter, and a high one for a boost converter.

In terms of power output, though, it's different. It's not uncommon--though not universal--for low input voltage to result in a reduced maximum power output; this is usually due to restricting the current through the switching device or inductor to a maximum to avoid damage. Any reputable manufacturer will have derating information in the datasheet.

\$\endgroup\$
2
\$\begingroup\$

Unless the manufacturer specified anything else, it would be safe to assume the output specs can be met all over the input range.

If you dig into datasheets of several supplies on the market, you may find some that for example the lower part of the input voltage range comes with caveats such as not up to the full temperature range or less hold-up time.

Here is one example from MeanWell RSP-750 (https://www.meanwell.com/Upload/PDF/RSP-750/RSP-750-SPEC.PDF):

enter image description here

Below 100 V AC, there is derating of the output so you can't load it to 100 % of specified output.

As always, check the datasheet for yours.

\$\endgroup\$

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.