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Popular buck converter modules specify a minimum Vin/Vout difference, something like "Vin must be at least 1.5V higher than Vout". But I do not see such requirements for boost converter modules, other than a minimum Vin. Do they have no such requirements and can they convert any Vin, within the range of minimum Vin <= Vin <= Vout, to Vout, like converting 4.9V to 5V?

enter image description here

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    \$\begingroup\$ Not wishing to be rude but, you appear to have accepted an answer that does not address your question. If it's beyond you to get beyond a few lines in an answer that's OK, but, be aware that when asking a searching question you do have somewhat of a responsibility to properly consider answers rather than picking one that might cause your eyes to glaze over the least. \$\endgroup\$
    – Andy aka
    Nov 25 at 6:12
  • \$\begingroup\$ @Andyaka Since I do not know much about electronics, I just picked the one with the most upvotes. When I picked it, it had the most upvotes, and it seems that another answer, since then, has received more upvotes. I will change the answer. \$\endgroup\$ Nov 25 at 12:01
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If you look at the basic circuit of a boost converter, you'll see that the output connects to the incoming voltage via a diode so, if the MOSFET switch did nothing, the output would be roughly tied to the input voltage but about 0.7 volts lower. That sets the baseline for basic boost converters. If you need a slightly higher output voltage than input minus 0.7 volts then the MOSFET is called into action to store energy in the inductor that is cyclically transferred to the output. So, theoretically, with a very fine pulse on the MOSFET , you could start to raise the output to voltages slightly higher than input voltage minus 0.7 volts. Theoretically you could arrange for the output to equal the input voltage and be reasonably regulated but, it will be susceptible to input line noise. If you can live with that then, that's the lower limit.

Of course, the devil will be in the detail in many commercial offerings and some will just about regulate to equal input and output voltage and, some won't.

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  • \$\begingroup\$ A boost converter with a synchronous rectifier would not have that voltage drop, FWIW. Synchronous rectifiers are much more common with buck converters, I will admit. \$\endgroup\$
    – nsayer
    Nov 25 at 3:23
  • \$\begingroup\$ You didn't read what I wrote. I'm talking about a basic circuit AND I'm talking about a circuit where the MOSFET is "at rest". If this were extended to a sync booster and the output MOSFET was also "at rest", it's still just a diode drop to the output. \$\endgroup\$
    – Andy aka
    Nov 25 at 6:06
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It depends on the chip and circuit, but at least the XL6009 chip has no such limit, as it can be used to build a buck-boost regulator, so output voltage can be lower than input voltage.

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  • \$\begingroup\$ For example, can I use those common booster modules that use XL6009 to keep the output voltage constant 5V at the end of a long cable to whose other end a 5V adapter is connected? Since the cable is long, the voltage can drop like 4V~5V. The current would be up to 200 mA. \$\endgroup\$ Nov 25 at 1:56
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Yes, a boost converter has limitations.

What happens in a boost converter is that an inductor is charged (on-State). And that stored energy is then tranferred to the output (Off-State).

enter image description here

Source of image: Wikipedia

If the input voltage is small, the voltage across the inductor will also be small. That means that the inductor will be charged more slowly. So it takes longer to charge the inductor with the same amount of energy.

We could "solve" that by using a smaller inductor value but that increases the current (it has to, how else can we get more energy from the same small input voltage?).

As the current increases, the resistance of the switch will get more important, if you double the current, the switch will have a doubled voltage drop.

Also, there will be a maximum current, see the "Input Current: 4A (max)" in the datasheet of that XL6009. That means that you must use an inductor that has a large enough value to ensure that this 4 A limit is not reached.

These factors limit the amount of power we can convert to a higher output voltage.

If the output is not loaded or loaded with a small current, the amount of energy is small and the things I describe above are not an issue. The converter will work as expected.

However, if we load the converter and it has to deliver more energy to the output, the issues will become important. At some point, the inductor cannot transfer enough energy from input to output and then we will hit the limit of what the converter can do.

If we hit that limit we can "back away" from that by:

  • increasing the input voltage, then the inductor will charge faster, more energy can be delivered to the output.

  • decreasing the load on the output so less current flows and less energy is needed.

  • decreasing the voltage at the output so less energy is needed. This is similar to decreasing the load at the output.

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    \$\begingroup\$ This does not answer the OP's question, as far as I can see. \$\endgroup\$
    – mkeith
    Nov 24 at 22:20
  • \$\begingroup\$ @mkeith Feel free to write a better answer :-) Realize that OP doesn't appear to realize that how the limitations work are fundamentally different and more complex in a boost converter compared to a buck converter or a linear regulator. So that's why I address these limitations. \$\endgroup\$ Nov 25 at 9:37
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These boost specs imply that Vout>Vin +2V just as for buck your comment says,"Vin must be at least 1.5V higher than Vout".

For Buck-Boost regulators there is only a range limitation, however efficiency drops up to 10% when Vout-Vin <1 V for some designs. I verified this using my lab supply which displays V,I,&W for a 20W 12V LED for automotive spot light (very bright) and computing input power to LED around 12V while brightness remained constant but power increased during crossover, then remained constant over a wide automotive voltage range.

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