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Understanding the main difference in utility of a boost converter (step-up) versus a buck converter (step-down). May I know how if there is an intuitive explanation as to why the capacitors we choose for boost converters are usually of higher capacitance as compared to buck converters?

Is it because a higher voltage output (due to voltage gain >= 1) is expected, hence the corresponding voltage ripple needs to be further filtered and minimized with larger capacitors. Or are there more accurate explanations to this practice?

Thank you.

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  • \$\begingroup\$ I'm not an expert but it would seem to me that it has to do with output ripple and the difference in discontinuous mode short term flyback time for the boost which is often short versus the buck. \$\endgroup\$
    – jonk
    Commented Mar 4, 2018 at 12:18
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    \$\begingroup\$ Shouldn't you change this question to improve the scenarios i.e. you want 10 volts out and the options are 20 volts in (buck) or 5 volts in (boost)? \$\endgroup\$
    – Andy aka
    Commented Mar 4, 2018 at 15:02

2 Answers 2

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In boost convertors, for part of the cycle, all the load current comes from the capacitor.

Buck converters can deliver current continuously through the inductor to the load (alternately from the voltage source and from ground), so only a fraction of the load current ever comes from the capacitor (in continuous current mode).

Therefore, for the same ripple voltage, the capacitor can be that fraction of the size of that in the boost convertor.

(This is not the whole answer; a more accurate answer would take the different duty cycles for different voltage conversion ratios into account too)

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    \$\begingroup\$ What? When the buck switch is off, all the current comes from the output inductor/capacitor and nothing from the input. \$\endgroup\$
    – winny
    Commented Mar 4, 2018 at 12:51
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    \$\begingroup\$ @winny When the buck switch is off, the diode conducts, supplying part of the load current from the inductor, and part from the capacitor. Doesn't change the basic point, but clarified, thanks. \$\endgroup\$
    – user16324
    Commented Mar 4, 2018 at 13:41
  • \$\begingroup\$ Much better. I see buck as the inverse of boost. Where buck is “kind” to the output side due to inductor (in CCM at least), the boost is to the input side and vice versa. \$\endgroup\$
    – winny
    Commented Mar 4, 2018 at 17:34
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Boost converters also transform to have lower input load impedance than buck converter input , Zin(f) so it is improved with lower Xc(f) with low ESR and bigger C.

This serves to improve output regulation by improving input load regulation error on source which is the inverse of impedance ratio load to source.

Think of the SMPS boost regulator like a stepup transformer in (Vout/Vin)^2 for impedance ratio Zout/Zin. (over simplified)

However, if a Buck supply is driving a boost regulator, an ultra low ESR boost input cap is essential to isolate the two regulator switching dynamic loads. Usually buck is PWM and Boost is PFM and the two combined can create a condition called "Chaos" in control systems, which is very loud random noise ( in piezo ceramic ferrite). I once heard my chip inductor singing like a babbling brook in a noisy lab due to piezo effects from chaos instability. Decoupling on Boost input removed the problem, which was only driving Analog bias levels for an AMLCD chip..

The surge current and steady state current is affected by step loads. ESR of the converter plays a big role and energy storage by choosing C value may hurt startup current ( unless soft start used) but improves output transient error.

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