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Consider a synchronous buck converter that is off, inductor voltage is 0, capacitor voltage is 0.

The moment the high-side switch is turned on for the first time (no matter for high long). There will be a di/dt = (Vin-0)/L flowing through the inductor and through the high side switch.

Soft-start mechanisms reduce the Ton time of the high side and gradually increase it but I don't understand how that helps reduce inrush current. No matter how long you keep the high side on for, at the first instant of turn on, it will take the full di/dt of Vin/L.

So how does soft-start help anything?

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    \$\begingroup\$ Because, if dt is short, di is small. \$\endgroup\$
    – user16324
    Commented May 15, 2022 at 21:24

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It takes a certain amount of current for a certain amount of time, to charge the output capacitor to target voltage.

If the capacitor is allowed to charge to target voltage more slowly, then less current is needed.

The soft start feature does just that; it limits the rate of how fast the capacitor voltage rises, and slower rate needs less current. Without soft start, the capacitor voltage would be charged to target voltage as quickly as possible, which requires more current.

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  • \$\begingroup\$ Yes but at the instant of first turn on for the first time, you would still get a large inrush current since you essentially have a short (considering inductor is 0V across it and cap is fully discharged) - so wouldn't there be a large spike at the very first turn on instant of the high side? \$\endgroup\$
    – alayoiskgfbfqhxjiw
    Commented May 15, 2022 at 22:56
  • \$\begingroup\$ No, not really, because, soft start limits the current draw and there might be multiple different ways different chips implement the soft start. The high side on period ramps current up, but ramp is based on inductance. Limiting the on time and ramping on time up slowly limits the level how much average current flow in the inductor to charge it to target voltage. Controlling the off time determines how often a pulse of current is drawn from input and how low the inductor current drops. \$\endgroup\$
    – Justme
    Commented May 15, 2022 at 23:10
  • \$\begingroup\$ Ok, it makes sense if a chip has some way of controlling the current directly (for example by changing the resistances of the HS FET with multiple fingers turning on slowly?) because that would be like deliberately adding a resistance to the path to ground. But if a chip implements soft start by just minimizing on-time, then I don't think that would solve the inrush current problem? \$\endgroup\$
    – alayoiskgfbfqhxjiw
    Commented May 15, 2022 at 23:14
  • \$\begingroup\$ The on-time minimisation technique to me seems like all it is doing is preventing the capacitor from overshooting. \$\endgroup\$
    – alayoiskgfbfqhxjiw
    Commented May 15, 2022 at 23:18
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    \$\begingroup\$ The inductor is a device that resists current change. Applying a 10us on pulse to inductor will cause current to rise half as much as applying a 20us on pulse. So repeated short pulses ramp up the current in longer time using more pulses, than repeated long pulses, which ramp up the current in shorter time using less pulses. \$\endgroup\$
    – Justme
    Commented May 15, 2022 at 23:22
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It's not the instantaneous peak of inductor current that is limited, but the average inductor current (over at least one switching cycle.

A major reason for softstart is to limit the current (power) delivered to the output decoupling capacitor. By limiting dV/dt at the output, the DCDC's input current (=CLOADdVOUT/dtVOUT/VIN) will be limited and if necessary can be kept not much more than the static DC load on the output.

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  • \$\begingroup\$ Oh so its really so give time for the supply of the buck (or input cap) to handle this? \$\endgroup\$
    – alayoiskgfbfqhxjiw
    Commented May 16, 2022 at 9:44
  • \$\begingroup\$ Also, I think 'justme''s comment about inductors not changing current instantly invalidates my answer. There can't be a sudden spike in current on the turn-on point. \$\endgroup\$
    – alayoiskgfbfqhxjiw
    Commented May 16, 2022 at 10:14
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Consider that the inductor current is 0 (at least I hope so) before the converter is started, so while you’re correct that the initial rate of change of input current is high (the highest I think) the actual current is zero at the leading edge and no more than t_on*di/dot at the end of the first cycle.

Also there is likely to be some filtering at the input of any practical buck converter to prevent the supply from carrying the full current peaks of the inductor, which would further limit the surge currents seem by the supply, but at a much shorter timescale than I think we’re talking about here.

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Your thinking that 100% of the current flows instantly into the capacitor. Your forgeting about the inductor in series between the output of the switch node and the capacitor. Remeber that the dV/dI requires dT. If your chopping up the power going into a LC filter the switch will not be shorted thru ground via the capacitor until the inductor stops resisting the change of current. Keep in mind depending on your switching frequency, capacitor, and inductor will change the dI at the switch. Soft starting will slow down the rate of the charge in the capacitor.

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There are 2 components to inrush current: 1) the inrush current to charge up the input DC bus capacitor and 2) the inrush commutated through the converter from charging up the output capacitors/load capacitors.

The 1st component is unavoidable but can be controlled via an NTC thermistor, or at higher power, by using series resistors that are switched out after startup.

The 2nd component can be avoided by slowly ramping up the output voltage if the chipset supports it.

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