I'm designing a 10V to 3.3V step-down converter. Looking at the LT8610, the applications example shows two similar circuits that have different switching frequencies.

The Efficiency vs Frequency plot shows that a lower switching frequency is slightly more efficient. Why is this so?

Alternatively, what are the advantages of a higher switching frequency?

LT8610 3.3V Step-Down Converter

Efficiency vs Frequency plot


2 Answers 2


There are turn-on and turn-off losses with every switching cycle, both in driving the switching elements themselves (gate drive loss if we're talking MOSFETs) and in the power train if you're considering a hard-switching topology like the step-down converters pictured in your question.

Reducing the operating frequency reduces the number of these events per unit time - all of which are lossy. Voila, you're saving some power now.

However, the benefits of lower frequency switching aren't free. The result of a lower switching frequency is higher peak current per switching cycle.

There's generally a balancing point between switching / gate losses and conduction losses due to the current. Finding the balance is part of the 'magic' in power supply design.

Higher frequency operation reduces the peak current (which means smaller magnetics) but increases gate and switching losses. Again, it's all about balance.

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    \$\begingroup\$ could you explain what you meant by peak current ? \$\endgroup\$ Commented Aug 17, 2012 at 19:38
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    \$\begingroup\$ Switching regulators chop DC into high frequency AC, then rectify it and make DC again. The high-frequency chopping involves inductors and/or transformers. During the on-time there is a current ramp in the magnetic component - the longer the on-time, the higher the peak will get for a given inductance value. \$\endgroup\$ Commented Aug 17, 2012 at 19:47

MOSFETs can be fairly good switches: they may have a low leakage current when off and a low on-resistance, so in either situation they're very little power dissipation; either the current is low, or the voltage. But to switch the FET on and off it has to go through its active region, and there neither voltage nor current are negligible, and their product is dissipated power. The higher the frequency the more times per second you have these switching losses, so expect 5 times more switching losses at 2 MHz than at 400 kHz.

Higher frequency is useful because the inductor has to store less energy, and can be made smaller. (Energy is Power \$\times\$ Time, and at higher frequency the switching period is shorter.)

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    \$\begingroup\$ good explanation. \$\endgroup\$ Commented Aug 17, 2012 at 19:41

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