If I was to use a buck converter to step-down my input voltage would I have a greater efficiency if my voltages are close together e.g. 7.2v to 5v compared to 12v to 5v?

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    \$\begingroup\$ the best bet is to look at the efficiency curves in the datasheet. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 23, 2016 at 19:57
  • \$\begingroup\$ How much current? \$\endgroup\$ Commented Feb 23, 2016 at 20:08
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    \$\begingroup\$ Nick has it. Without knowing anything else, at least the current, or better the current wrt the max current for the regulator, must be known. \$\endgroup\$ Commented Feb 23, 2016 at 21:27

4 Answers 4


No, not really, within limits. All your voltages are low enough so that special "high voltage" techniques and parts don't need to be used. With only 7.2 V in, you have to think carefully about making sure the switch can be driven to fully on. That means getting a FET with low gate voltage requirements, or making a higher voltage.

For a buck converter, the best is probably when the output voltage is 1/2 the input voltage. That minimizes short pulses on either side since the duty cycle will be 50%, at least in continuous mode.

Another part that affects efficiency is the output voltage by itself. At 5 V out, even the drop across a Schottky diode is significant. Synchronous rectification is therefore important if you are trying to push efficiency.

All in all, I'd rather have 12 V than 7.2 V as input for a buck switcher that has to make 5 V out. However, both can be used quite efficiently. There is no strong preference, especially if you are using a off the shelf buck converter chip. In that case, just get a chip intended for the voltage range, with the efficiency you want.

As Ali80 points out in his answer, this is assuming synchronous rectification, which many off the shelf chips now include. If not, the diode losses dominate and the lower the input voltage the better. See Ali's answer for details.

With synchronous rectification, there will still be some preference to higher or lower input voltage. Usually this doesn't matter much, which is why I said "not really" in the first sentence. If this level of detail matters to you, then you need to carefully read the datasheet for whatever part you are using, and compare datasheets carefully to decide what part to use in the first place.

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    \$\begingroup\$ Whoever downvoted this, please explain what you think is incorrect, misleading, or badly written. I have looked it over, and it still seems right to me. \$\endgroup\$ Commented Feb 23, 2016 at 21:13
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    \$\begingroup\$ Haters gonna hate. Have my +1 and please keep up the great work. \$\endgroup\$ Commented Feb 23, 2016 at 21:26
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    \$\begingroup\$ I didnt downvote but the mosfet is way more efficient than the schotky so I prefer the fet to conduct way more hence the input and output voltages be as near as practical :) so in my opinion you answer is incorrect \$\endgroup\$
    – Ali80
    Commented Feb 23, 2016 at 22:01
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    \$\begingroup\$ Im surprised you didn't tell op "Read Datasheet, #%&@ You, Close Vote". It's clearly answerable in the sheet. \$\endgroup\$
    – Passerby
    Commented Feb 23, 2016 at 22:57
  • \$\begingroup\$ That'd be real helpful for people looking for an answer like this in future if the answer was "Read Datasheet, #%&@ You, Close Vote". \$\endgroup\$ Commented Feb 24, 2016 at 9:59

You have three major sources of power loss in buck topology:

  1. inductor's conduction loss and switching loss
  2. mosfet switching loss and conduction loss
  3. diode switching loss and conduction loss

inductor's resistive loss is pretty much constant but inductor's core loss increase if we increase the input voltage(although negligible compared to other losses).

If we increase the input voltage, mosfet swithing times increase(rise time and fall time) and hence switching losses increase in the mosfet, although it doesn't affect diode switching loss if we assume continous conduction mode.

so what we have left are mosfet and diode conduction losses, mosfets are generally better at conducting and have much lower losses, (that's why in some high efficiency low ouput voltage buck convertes we use mosfet instead of the output diode for rectification.) so if we have to choose between mosfet and diode which to conduct more, mosfet is obviously a better choise, by increasing the input voltage we generall decrease mosfets conduction time and increase diodes conduction time and that's not good for efficiency either

So the lower input voltage is generally better for efficiency. Here are efficiency curves for TI's TPS54340: TPS540340 buck converter efficiency curves


When the voltages are closer together the Buck switch has a greater duty cycle and hence more conduction loss simply because it spends more time conducting.When you have a big input voltage while keeping V Out the same the switching losses increase . At your low input voltages the mosfet on resistance will be low and the prospective switching losses will also be low so you wont notice a big efficiency difference.I have done a lot of wide range stuff which means a big buck ratio at the top end of V in .I implemented a switching loss reduction scheme to preserve the efficiency at high V in .Remember that switching losses are a function of peak voltage and peak current at turn on and turn off .Also as V in increases the coil works harder because the current and hence flux ramps up faster .On your application the coil should be fine .


The voltage drops across the components that matter (switch and diode) are more or less constant, therefore there are fewer losses proportionally speaking (which is directly linked with efficiency) when the input voltage is higher. Therefore no, efficiency drops when a lower input voltage is used.

  • \$\begingroup\$ There is a limit though. If input voltage is too high, then duty cycle gets very small (for a fixed-frequency regulator) and switching transient losses start to dominate. \$\endgroup\$
    – The Photon
    Commented Feb 23, 2016 at 19:52

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